1 Parametric Equations and Polar Coordinates

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-parametric-equations-and-polar-coordinates

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Introduction

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-introduction

Chapter Outline

1.1 Parametric Equations
1.2 Calculus of Parametric Curves
1.3 Polar Coordinates
1.4 Area and Arc Length in Polar Coordinates
1.5 Conic Sections

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1.1 Parametric Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-1-parametric-equations

1.1 Parametric Equations

1.1.1 Plot a curve described by parametric equations.
1.1.2 Convert the parametric equations of a curve into the form $y = f(x).$
1.1.3 Recognize the parametric equations of basic curves, such as a line and a circle.
1.1.4 Recognize the parametric equations of a cycloid.

In this section we examine parametric equations and their graphs. In the two-dimensional coordinate system, parametric equations are useful for describing curves that are not necessarily functions. The parameter is an independent variable that both *x* and *y* depend on, and as the parameter increases, the values of *x* and *y* trace out a path along a plane curve. For example, if the parameter is *t* (a common choice), then *t* might represent time. Then *x* and *y* are defined as functions of time, and $\left( {x(t),y(t)} \right)$ can describe the position in the plane of a given object as it moves along a curved path.

Parametric Equations and Their Graphs

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra $\frac{1}{4}$ day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

Figure 1.2 depicts Earth’s orbit around the Sun during one year. The point labeled $F_{2}$ is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure 1.3). Then each *x* value on the graph is a value of position as a function of time, and each *y* value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

We can determine the functions for $x(t)$ and $y(t),$ thereby parameterizing the orbit of Earth around the Sun. The variable $t$ is called an independent parameter and, in this context, represents time relative to the beginning of each year.

A curve in the $\left( {x,y} \right)$ plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

If *x* and *y* are continuous functions of *t* on an interval *I*, then the equations

$$x = x(t)\ \text{and}\ y = y(t)$$

are called parametric equations and *t* is called the parameter. The set of points $\left( {x,y} \right)$ obtained as *t* varies over the interval *I* is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or *plane curve*, and is denoted by *C*.

Notice in this definition that *x* and *y* are used in two ways. The first is as functions of the independent variable *t.* As *t* varies over the interval *I*, the functions $x(t)$ and $y(t)$ generate a set of ordered pairs $\left( {x,y} \right).$ This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, *x* and *y* are variables. It is important to distinguish the variables *x* and *y* from the functions $x(t)$ and $y(t).$

Graphing a Parametrically Defined Curve

Sketch the curves described by the following parametric equations:

1. $x(t) = t - 1,\quad y(t) = 2t + 4,\quad-3 \leq t \leq 2$

2. $x(t) = t^{2} - 3,\quad y(t) = 2t + 1,\quad-2 \leq t \leq 3$

3. $x(t) = 4\ \text{cos}\ t,\quad y(t) = 4\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution

1. To create a graph of this curve, first set up a table of values. Since the independent variable in both $x(t)$ and $y(t)$ is *t*, let *t* appear in the first column. Then $x(t)$ and $y(t)$ will appear in the second and third columns of the table.

| *t* | $x(t)$ | $y(t)$ |

|-----|--------|--------|

| −3 | −4 | −2 |

| −2 | −3 | 0 |

| −1 | −2 | 2 |

| 0 | −1 | 4 |

| 1 | 0 | 6 |

| 2 | 1 | 8 |

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure 1.4. The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as *t* varies from −3 to 2.

2. To create a graph of this curve, again set up a table of values.

| *t* | $x(t)$ | $y(t)$ |

|-----|--------|--------|

| −2 | 1 | −3 |

| −1 | −2 | −1 |

| 0 | −3 | 1 |

| 1 | −2 | 3 |

| 2 | 1 | 5 |

| 3 | 6 | 7 |

The second and third columns in this table give a set of points to be plotted (Figure 1.5). The first point on the graph (corresponding to $t = -2)$ has coordinates $\left( {1,-3} \right),$ and the last point (corresponding to $t = 3)$ has coordinates $\left( {6,7} \right).$ As *t* progresses from −2 to 3, the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.

3. In this case, use multiples of $\pi\text{/}6$ for *t* and create another table of values:

| *t* | $x(t)$ | $y(t)$ | | *t* | $x(t)$ | $y(t)$ |

|------------------|---------------------------|-------------------------|-----|-------------------|---------------------------|---------------------------|

| 0 | 4 | 0 | | $\frac{7\pi}{6}$ | $-2\sqrt{3} \approx -3.5$ | 2 |

| $\frac{\pi}{6}$ | $2\sqrt{3} \approx 3.5$ | $2$ | | $\frac{4\pi}{3}$ | −2 | $-2\sqrt{3} \approx -3.5$ |

| $\frac{\pi}{3}$ | $2$ | $2\sqrt{3} \approx 3.5$ | | $\frac{3\pi}{2}$ | 0 | −4 |

| $\frac{\pi}{2}$ | 0 | 4 | | $\frac{5\pi}{3}$ | 2 | $-2\sqrt{3} \approx -3.5$ |

| $\frac{2\pi}{3}$ | −2 | $2\sqrt{3} \approx 3.5$ | | $\frac{11\pi}{6}$ | $2\sqrt{3} \approx 3.5$ | 2 |

| $\frac{5\pi}{6}$ | $-2\sqrt{3} \approx -3.5$ | 2 | | $2\pi$ | 4 | 0 |

| $\pi$ | −4 | 0 | | | | |

The graph of this plane curve appears in the following graph.

This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates $\left( {4,0} \right).$

Sketch the curve described by the parametric equations

$$x(t) = 3t + 2,\quad y(t) = t^{2} - 1,\quad-3 \leq t \leq 2.$$

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables *x* and *y.* Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in Example 1.1b. are

$$x(t) = t^{2} - 3,\quad y(t) = 2t + 1,\quad-2 \leq t \leq 3.$$

Solving the second equation for *t* gives

$$t = \frac{y - 1}{2}.$$

This can be substituted into the first equation:

$$x = \left( \frac{y - 1}{2} \right)^{2} - 3 = \frac{y^{2} - 2y + 1}{4} - 3 = \frac{y^{2} - 2y - 11}{4}.$$

This equation describes *x* as a function of *y.* These steps give an example of *eliminating the parameter*. The graph of this function is a parabola opening to the right. Recall that the plane curve started at $\left( {1,-3} \right)$ and ended at $\left( {6,7} \right).$ These terminations were due to the restriction on the parameter *t.*

Eliminating the Parameter

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.

1. $x(t) = \sqrt{2t + 4},\quad y(t) = 2t + 1,\quad-2 \leq t \leq 6$

2. $x(t) = 4\ \text{cos}\ t,\quad y(t) = 3\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution

1. To eliminate the parameter, we can solve either of the equations for *t.* For example, solving the first equation for *t* gives

$$\begin{array}{rll}

x & = & \sqrt{2t + 4} \\

x^{2} & = & {2t + 4} \\

{x^{2} - 4} & = & {2t} \\

t & = & {\frac{x^{2} - 4}{2}.}

\end{array}$$

Note that when we square both sides it is important to observe that $x \geq 0.$ Substituting $t = \frac{x^{2} - 4}{2}$ this into $y(t)$ yields

$$\begin{array}{rll}

{y(t)} & = & {2t + 1} \\

y & = & {2\left( \frac{x^{2} - 4}{2} \right) + 1} \\

y & = & {x^{2} - 4 + 1} \\

y & = & {x^{2} - 3.}

\end{array}$$

This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter *t*. When $t = -2,$ $x = \sqrt{2(-2) + 4} = 0,$ and when $t = 6,$ $x = \sqrt{2(6) + 4} = 4.$ The graph of this plane curve follows.

2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are

$$x(t) = 4\ \text{cos}\ t\ \text{and}\ y(t) = 3\ \text{sin}\ t.$$

Solving either equation for *t* directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the *t*) gives us

$$\text{cos}\ t = \frac{x}{4}\ \text{and}\ \text{sin}\ t = \frac{y}{3}.$$

Now use the Pythagorean identity $\text{cos}^{2}t + \text{sin}^{2}t = 1$ and replace the expressions for $\text{sin}\ t$ and $\text{cos}\ t$ with the equivalent expressions in terms of *x* and *y*. This gives

$$\begin{array}{rll}

{\left( \frac{x}{4} \right)^{2} + \left( \frac{y}{3} \right)^{2}} & = & 1 \\

{\frac{x^{2}}{16} + \frac{y^{2}}{9}} & = & 1.

\end{array}$$

This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.

As *t* progresses from $0$ to $2\pi,$ a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.

$$x(t) = 2 + \frac{3}{t},\quad y(t) = t - 1,\quad 2 \leq t \leq 6$$

So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.

Parameterizing a Curve

Find two different pairs of parametric equations to represent the graph of $y = 2x^{2} - 3.$

Solution

First, it is always possible to parameterize a curve by defining $x(t) = t,$ then replacing *x* with *t* in the equation for $y(t).$ This gives the parameterization

$$x(t) = t,\quad y(t) = 2t^{2} - 3.$$

Since there is no restriction on the domain in the original graph, there is no restriction on the values of *t.*

We have complete freedom in the choice for the second parameterization. For example, we can choose $x(t) = 3t - 2.$ The only thing we need to check is that there are no restrictions imposed on *x*; that is, the range of $x(t)$ is all real numbers. This is the case for $x(t) = 3t - 2.$ Now since $y = 2x^{2} - 3,$ we can substitute $x(t) = 3t - 2$ for *x.* This gives

$$\begin{array}{cl}

{y(t)} & {= 2\left( {3t - 2} \right)^{2} - 3} \\

& {= 2\left( {9t^{2} - 12t + 4} \right) - 3} \\

& {= 18t^{2} - 24t + 8 - 3} \\

& {= 18t^{2} - 24t + 5.}

\end{array}$$

Therefore, a second parameterization of the curve can be written as

$$x(t) = 3t - 2\ \text{and}\ y(t) = 18t^{2} - 24t + 5.$$

Find two different sets of parametric equations to represent the graph of $y = x^{2} + 2x.$

Cycloids and Other Parametric Curves

Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure 1.9). A cycloid generated by a circle (or bicycle wheel) of radius *a* is given by the parametric equations

$$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$

To see why this is true, consider the path that the center of the wheel takes. The center moves along the *x*-axis at a constant height equal to the radius of the wheel. If the radius is *a*, then the coordinates of the center can be given by the equations

$$x(t) = at,\quad y(t) = a$$

for any value of $t.$ Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by

$$x(t) = \text{−}a\ \text{sin}\ t,\quad y(t) = \text{−}a\ \text{cos}\ t.$$

(The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.

$$x(t) = a\left( {t - \text{sin}\ t} \right),\quad y(t) = a\left( {1 - \text{cos}\ t} \right).$$

Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in Figure 1.10. In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point on the edge of the green circle traces out the red graph, which is called a hypocycloid.

The general parametric equations for a hypocycloid are

$$\begin{array}{l}

{x(t) = \left( {a - b} \right)\ \text{cos}\ t + b\ \text{cos}\left( \frac{a - b}{b} \right)\ t} \\

{y(t) = \left( {a - b} \right)\ \text{sin}\ t - b\ \text{sin}\left( \frac{a - b}{b} \right)\ t.}

\end{array}$$

These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this case we assume the radius of the larger circle is *a* and the radius of the smaller circle is *b.* Then the center of the wheel travels along a circle of radius $a - b.$ This fact explains the first term in each equation above. The period of the second trigonometric function in both $x(t)$ and $y(t)$ is equal to $\frac{2\pi b}{a - b}.$

The ratio $\frac{a}{b}$ is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated in Figure 1.11. This ratio can lead to some very interesting graphs, depending on whether or not the ratio is rational. Figure 1.10 corresponds to $a = 4$ and $b = 1.$ The result is a hypocycloid with four cusps. Figure 1.11 shows some other possibilities. The last two hypocycloids have irrational values for $\frac{a}{b}.$ In these cases the hypocycloids have an infinite number of cusps, so they never return to their starting point. These are examples of what are known as space-filling curves.

The Witch of Agnesi

Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartes or the spiral of Archimedes. However, perhaps the strangest name for a curve is the witch of Agnesi. Why a witch?

Maria Gaetana Agnesi (1718–1799) was one of the few recognized women mathematicians of eighteenth-century Italy. She wrote a popular book on analytic geometry, published in 1748, which included an interesting curve that had been studied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which he later called the “versoria,” a Latin term for a rope used in sailing. Agnesi used the Italian term for this rope, “versiera,” but in Latin, this same word means a “female goblin.” When Agnesi’s book was translated into English in 1801, the translator used the term “witch” for the curve, instead of rope. The name “witch of Agnesi” has stuck ever since.

The witch of Agnesi is a curve defined as follows: Start with a circle of radius *a* so that the points $(0,0)$ and $(0,2a)$ are points on the circle (Figure 1.12). Let *O* denote the origin. Choose any other point *A* on the circle, and draw the secant line *OA*. Let *B* denote the point at which the line *OA* intersects the horizontal line through $(0,2a).$ The vertical line through *B* intersects the horizontal line through *A* at the point *P*. As the point *A* varies, the path that the point *P* travels is the witch of Agnesi curve for the given circle.

Witch of Agnesi curves have applications in physics, including modeling water waves and distributions of spectral lines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In this project you will parameterize these curves.

1. On the figure, label the following points, lengths, and angle:

1. *C* is the point on the *x*-axis with the same *x*-coordinate as *A*.

2. *x* is the *x*-coordinate of *P*, and *y* is the *y*-coordinate of *P*.

3. *E* is the point $(0,a).$

4. *F* is the point on the line segment *OA* such that the line segment *EF* is perpendicular to the line segment *OA*.

5. *b* is the distance from *O* to *F*.

6. *c* is the distance from *F* to *A*.

7. *d* is the distance from *O* to *B*.

8. $\theta$ is the measure of angle $\text{∠}COA.$

The goal of this project is to parameterize the witch using $\theta$ as a parameter. To do this, write equations for *x* and *y* in terms of only $\theta.$

2. Show that $d = \frac{2a}{\text{sin}\ \theta}.$

3. Note that $x = d\ \text{cos}\ \theta.$ Show that $x = 2a\ \text{cot}\ \theta.$ When you do this, you will have parameterized the *x*-coordinate of the curve with respect to $\theta.$ If you can get a similar equation for *y*, you will have parameterized the curve.

4. In terms of $\theta,$ what is the angle $\text{∠}EOA?$

5. Show that $b + c = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right).$

6. Show that $y = 2a\ \text{cos}\left( {\frac{\pi}{2} - \theta} \right)\ \text{sin}\ \theta.$

7. Show that $y = 2a\ \text{sin}^{2}\theta.$ You have now parameterized the *y*-coordinate of the curve with respect to $\theta.$

8. Conclude that a parameterization of the given witch curve is

$$x = 2a\ \text{cot}\ \theta,y = 2a\ \text{sin}^{2}\theta, - \infty < \theta < \infty.$$

9. Use your parameterization to show that the given witch curve is the graph of the function $f(x) = \frac{8a^{3}}{x^{2} + 4a^{2}}.$

Travels with My Ant: The Curtate and Prolate Cycloids

Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a wheel traces as the wheel rolls along a straight path. In this project we look at two different variations of the cycloid, called the curtate and prolate cycloids.

First, let’s revisit the derivation of the parametric equations for a cycloid. Recall that we considered a tenacious ant trying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the very edge, where the tire touches the ground. As the wheel rolls, the ant moves with the edge of the tire (Figure 1.13).

As we have discussed, we have a lot of flexibility when parameterizing a curve. In this case we let our parameter *t* represent the angle the tire has rotated through. Looking at Figure 1.13, we see that after the tire has rotated through an angle of *t*, the position of the center of the wheel, $C = \left( {x_{C},y_{C}} \right),$ is given by

$$x_{C} = at\ \text{and}\ y_{C} = a.$$

Furthermore, letting $A = \left( {x_{A},y_{A}} \right)$ denote the position of the ant, we note that

$$x_{C} - x_{A} = a\ \text{sin}\ t\ \text{and}\ y_{C} - y_{A} = a\ \text{cos}\ t.$$

Then

$$\begin{array}{l}

{x_{A} = x_{C} - a\ \text{sin}\ t = at - a\ \text{sin}\ t = a(t - \text{sin}\ t)} \\

{y_{A} = y_{C} - a\ \text{cos}\ t = a - a\ \text{cos}\ t = a(1 - \text{cos}\ t).}

\end{array}$$

Note that these are the same parametric representations we had before, but we have now assigned a physical meaning to the parametric variable *t*.

After a while the ant is getting dizzy from going round and round on the edge of the tire. So he climbs up one of the spokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path of motion. The new path has less up-and-down motion and is called a curtate cycloid (Figure 1.14). As shown in the figure, we let *b* denote the distance along the spoke from the center of the wheel to the ant. As before, we let *t* represent the angle the tire has rotated through. Additionally, we let $C = \left( {x_{C},y_{C}} \right)$ represent the position of the center of the wheel and $A = \left( {x_{A},y_{A}} \right)$ represent the position of the ant.

1. What is the position of the center of the wheel after the tire has rotated through an angle of *t*?

2. Use geometry to find expressions for $x_{C} - x_{A}$ and for $y_{C} - y_{A}.$

3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtate cycloid?

Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from his home. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headed back in the right direction. So the ant heads over to the train tracks to wait. After a while, a train goes by, heading in the right direction, and he manages to jump up and just catch the edge of the train wheel (without getting squished!).

The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so he decides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keep the wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, the distance from the center of the wheel to the ant is actually greater than the radius of the wheel (Figure 1.15).

The setup here is essentially the same as when the ant climbed up the spoke on the bicycle wheel. We let *b* denote the distance from the center of the wheel to the ant, and we let *t* represent the angle the tire has rotated through. Additionally, we let $C = \left( {x_{C},y_{C}} \right)$ represent the position of the center of the wheel and $A = \left( {x_{A},y_{A}} \right)$ represent the position of the ant (Figure 1.15).

When the distance from the center of the wheel to the ant is greater than the radius of the wheel, his path of motion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.

4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of the ant.

5. What do you notice about your answer to part 3 and your answer to part 4?

Notice that the ant is actually traveling backward at times (the “loops” in the graph), even though the train continues to move forward. He is probably going to be *really* dizzy by the time he gets home!

Section 1.1 Exercises

For the following exercises, sketch the curves below by eliminating the parameter *t*. Give the orientation of the curve.

$x = t^{2} + 2t,$ $y = t + 1$

2\.

$x = \text{cos}(t),y = \text{sin}(t),\left( {0,2\pi} \right\rbrack$

$x = 2t + 4,y = t - 1$

4\.

$x = 3 - t,y = 2t - 3,1.5 \leq t \leq 3$

For the following exercises, eliminate the parameter and sketch the graphs.

$x = 2t^{2},\quad y = t^{4} + 1$

For the following exercises, use technology (CAS or calculator) to sketch the parametric equations.

6\.

\[T\] $\begin{array}{ll}

{x = t^{2} + t,} & {y = t^{2} - 1}

\end{array}$

\[T\] $\begin{array}{ll}

{x = e^{\text{−}t},} & {y = e^{2t} - 1}

\end{array}$

8\.

\[T\] $\begin{array}{ll}

{x = 3\ \text{cos}\ t,} & {y = 4\ \text{sin}\ t}

\end{array}$

\[T\] $\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{cos}\ t}

\end{array}$

For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph.

10\.

$x = e^{t},\quad y = e^{2t} + 1$

11.

$x = 6\ \text{sin}(2\theta),y = 4\ \text{cos}(2\theta)$

12\.

$\begin{array}{ll}

{x = \text{cos}\ \theta,} & {y = 2\ \text{sin}(2\theta)}

\end{array}$

13.

$\begin{array}{ll}

{x = 3 - 2\ \text{cos}\ \theta,} & {y = -5 + 3\ \text{sin}\ \theta}

\end{array}$

14\.

$\begin{array}{ll}

{x = 4 + 2\ \text{cos}\ \theta,} & {y = -1 + \text{sin}\ \theta}

\end{array}$

15.

$\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{tan}\ t}

\end{array}$

16\.

$\begin{array}{ll}

{x = \text{ln}(2t),} & {y = t^{2}}

\end{array}$

17.

$\begin{array}{ll}

{x = e^{t},} & {y = e^{2t}}

\end{array}$

18\.

$\begin{array}{ll}

{x = e^{-2t},} & {y = e^{3t}}

\end{array}$

19.

$\begin{array}{ll}

{x = t^{3},} & {y = 3\ \text{ln}\ t}

\end{array}$

20\.

$\begin{array}{ll}

{x = 4\ \text{sec}\ \theta,} & {y = 3\ \text{tan}\ \theta}

\end{array}$

For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form.

21.

$\begin{array}{ll}

{x = t^{2} - 1,} & {y = \frac{t}{2}}

\end{array}$

22\.

$\begin{array}{ll}

{x = \frac{1}{\sqrt{t + 1}},} & {y = \frac{t}{1 + t},t > -1}

\end{array}$

23.

$x = 4\ \text{cos}\ \theta,y = 3\ \text{sin}\ \theta,\theta \in \left( {0,2\pi} \right\rbrack$

24\.

$\begin{array}{ll}

{x = \text{cosh}\ t,} & {y = \text{sinh}\ t}

\end{array}$

25.

$\begin{array}{ll}

{x = 2t - 3,} & {y = 6t - 7}

\end{array}$

26\.

$\begin{array}{ll}

{x = t^{2},} & {y = t^{3}}

\end{array}$

27.

$\begin{array}{ll}

{x = 1 + \text{cos}\ t,} & {y = 3 - \text{sin}\ t}

\end{array}$

28\.

$\begin{array}{ll}

{x = \sqrt{t},} & {y = 2t + 4}

\end{array}$

29.

$\begin{array}{ll}

{x = \text{sec}\ t,} & {y = \text{tan}\ t,\pi \leq t < \frac{3\pi}{2}}

\end{array}$

30\.

$\begin{array}{ll}

{x = 2\ \text{cosh}\ t,} & {y = 4\ \text{sinh}\ t}

\end{array}$

31.

$\begin{array}{ll}

{x = \text{cos}(2t),} & {y = \text{sin}\ t}

\end{array}$

32\.

$x = 4t + 3,y = 16t^{2} - 9$

33.

$\begin{array}{ll}

{x = t^{2},} & {y = 2\ \text{ln}\ t,t \geq 1}

\end{array}$

34\.

$\begin{array}{ll}

{x = t^{3},} & {y = 3\ \text{ln}\ t,t \geq 1}

\end{array}$

35.

$\begin{array}{ll}

{x = t^{n},} & {y = n\ \text{ln}\ t,t \geq 1,}

\end{array}$ where *n* is a natural number

36\.

$\begin{array}{l}

{x = \text{ln}(5t)} \\

{y = \text{ln}(t^{2})}

\end{array}$ where $1 \leq t \leq e$

37.

$\begin{array}{l}

{x = 2\ \text{sin}(8t)} \\

{y = 2\ \text{cos}(8t)}

\end{array}$

38\.

$\begin{array}{l}

{x = \text{tan}\ t} \\

{y = \text{sec}^{2}t - 1}

\end{array}$

For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of equations represents.

39.

$\begin{array}{l}

{x = 3t + 4} \\

{y = 5t - 2}

\end{array}$

40\.

$\begin{array}{l}

{x - 4 = 5t} \\

{y + 2 = t}

\end{array}$

41.

$\begin{array}{l}

{x = 2t + 1} \\

{y = t^{2} - 3}

\end{array}$

42\.

$\begin{array}{l}

{x = 3\ \text{cos}\ t} \\

{y = 3\ \text{sin}\ t}

\end{array}$

43.

$\begin{array}{l}

{x = 2\ \text{cos}(3t)} \\

{y = 2\ \text{sin}(3t)}

\end{array}$

44\.

$\begin{array}{l}

{x = \text{cosh}\ t} \\

{y = \text{sinh}\ t}

\end{array}$

45.

$\begin{array}{l}

{x = 3\ \text{cos}\ t} \\

{y = 4\ \text{sin}\ t}

\end{array}$

46\.

$\begin{array}{l}

{x = 2\ \text{cos}(3t)} \\

{y = 5\ \text{sin}(3t)}

\end{array}$

47.

$\begin{array}{l}

{x = 3\ \text{cosh}(4t)} \\

{y = 4\ \text{sinh}(4t)}

\end{array}$

48\.

$\begin{array}{l}

{x = 2\ \text{cosh}\ t} \\

{y = 2\ \text{sinh}\ t}

\end{array}$

49\.

Show that $\begin{array}{l}

{x = h + r\ \text{cos}\ \theta} \\

{y = k + r\ \text{sin}\ \theta}

\end{array}$ represents the equation of a circle.

50\.

Use the equations in the preceding problem to find a set of parametric equations for a circle whose radius is 5 and whose center is $\left( {-2,\ 3} \right).$

For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation.

51.

\[T\] $\begin{array}{l}

{x = \theta + \text{sin}\ \theta} \\

{y = 1 - \text{cos}\ \theta}

\end{array}$

52\.

\[T\] $\begin{array}{l}

{x = 2t - 2\ \text{sin}\ t} \\

{y = 2 - 2\ \text{cos}\ t}

\end{array}$

53.

\[T\] $\begin{array}{l}

{x = t - 0.5\ \text{sin}\ t} \\

{y = 1 - 1.5\ \text{cos}\ t}

\end{array}$

54\.

An airplane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by $x = 100t,y = -4.9t^{2} + 4000,t \geq 0$ where the origin is the point on the ground directly beneath the plane at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target?

55.

The trajectory of a bullet is given by $x = v_{0}\left( {\text{cos}\ \alpha} \right)\ t,y = v_{0}\left( {\text{sin}\ \alpha} \right)\ t - \frac{1}{2}gt^{2}$ where $v_{0} = 500\ \text{m/s,}$ $g = 9.8 = 9.8{\ \text{m/s}}^{2},$ and $\alpha = 30\ \text{degrees}.$ When will the bullet hit the ground? How far from the gun will the bullet hit the ground?

56\.

\[T\] Use technology to sketch the curve represented by $x = \text{sin}(4t),y = \text{sin}(3t),0 \leq t \leq 2\pi.$

57.

\[T\] Use technology to sketch $x = 2\ \text{tan}(t),y = 3\ \text{sec}(t),\text{−}\pi < t < \pi.$

58\.

Sketch the curve known as an *epitrochoid*, which gives the path of a point on a circle of radius *b* as it rolls on the outside of a circle of radius *a*. The equations are

$\begin{array}{l}

{x = (a + b)\text{cos}\ t - c \cdot \text{cos}\left\lbrack \frac{(a + b)t}{b} \right\rbrack} \\

{y = (a + b)\text{sin}\ t - c \cdot \text{sin}\left\lbrack \frac{(a + b)t}{b} \right\rbrack.}

\end{array}$

Let $a = 1,b = 2,c = 1.$

59.

\[T\] Use technology to sketch the spiral curve given by $x = t\ \text{cos}(t),y = t\ \text{sin}(t)$ from $-2\pi \leq t \leq 2\pi.$

60\.

\[T\] Use technology to graph the curve given by the parametric equations $x = 2\ \text{cot}(t),y = 1 - \text{cos}(2t),\text{−}\pi\text{/}2 \leq t \leq \pi\text{/}2.$ This curve is known as the witch of Agnesi.

61.

\[T\] Sketch the curve given by parametric equations $\begin{array}{l}

{x = \text{cosh}(t)} \\

{y = \text{sinh}(t),}

\end{array}$ where $-2 \leq t \leq 2.$

---

1.2 Calculus of Parametric Curves

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-2-calculus-of-parametric-curves

1.2 Calculus of Parametric Curves

1.2.1 Determine derivatives and equations of tangents for parametric curves.
1.2.2 Find the area under a parametric curve.
1.2.3 Use the equation for arc length of a parametric curve.
1.2.4 Apply the formula for surface area to a volume generated by a parametric curve.

Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve $\left( {x(t),y(t)} \right),$ then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.

Derivatives of Parametric Equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

$$x(t) = 2t + 3,\quad y(t) = 3t - 4,\quad-2 \leq t \leq 3.$$

The graph of this curve appears in Figure 1.16. It is a line segment starting at $\left( {-1,-10} \right)$ and ending at $(9,5).$

We can eliminate the parameter by first solving the equation $x(t) = 2t + 3$ for *t*:

$$\begin{array}{rll}

{x(t)} & = & {2t + 3} \\

{x - 3} & = & {2t} \\

t & = & {\frac{x - 3}{2}.}

\end{array}$$

Substituting this into $y(t),$ we obtain

$$\begin{array}{rll}

{y(t)} & = & {3t - 4} \\

y & = & {3\left( \frac{x - 3}{2} \right) - 4} \\

y & = & {\frac{3x}{2} - \frac{9}{2} - 4} \\

y & = & {\frac{3x}{2} - \frac{17}{2}.}

\end{array}$$

The slope of this line is given by $\frac{dy}{dx} = \frac{3}{2}.$ Next we calculate $x^{\prime}(t)$ and $y^{\prime}(t).$ This gives $x^{\prime}(t) = 2$ and $y^{\prime}(t) = 3.$ Notice that $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{3}{2}.$ This is no coincidence, as outlined in the following theorem.

Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations $x = x(t)$ and $y = y(t).$ Suppose that $x^{\prime}(t)$ and $y^{\prime}(t)$ exist, and assume that $x^{\prime}(t) \neq 0.$ Then the derivative $\frac{dy}{dx}$ is given by

$$\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$$ (1.1)

Proof

This theorem can be proven using the Chain Rule. In particular, assume that the parameter *t* can be eliminated, yielding a differentiable function $y = F(x).$ Then $y(t) = F\left( {x(t)} \right).$ Differentiating both sides of this equation using the Chain Rule yields

$$y^{\prime}(t) = F^{\prime}\left( {x(t)} \right)x^{\prime}(t),$$

$$F^{\prime}\left( {x(t)} \right) = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$$

But $F^{\prime}\left( {x(t)} \right) = \frac{dy}{dx},$ which proves the theorem.

□

Equation 1.1 can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function $y = f(x)$ is any point $x = x_{0}$ such that either $f^{\prime}\left( x_{0} \right) = 0$ or $f^{\prime}\left( x_{0} \right)$ does not exist. Equation 1.1 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function $y = f(x)$ or not.

Finding the Derivative of a Parametric Curve

Calculate the derivative $\frac{dy}{dx}$ for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.

1. $x(t) = t^{2} - 3,\quad y(t) = 2t - 1,\quad-3 \leq t \leq 4$

2. $x(t) = 2t + 1,\quad y(t) = t^{3} - 3t + 4,\quad-2 \leq t \leq 2$

3. $x(t) = 5\ \text{cos}\ t,\quad y(t) = 5\ \text{sin}\ t,\quad 0 \leq t \leq 2\pi$

Solution

1. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

$$\begin{array}{l}

{x^{\prime}(t) = 2t} \\

{y^{\prime}(t) = 2.}

\end{array}$$

Next substitute these into the equation:

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{2}{2t}} \\

{\frac{dy}{dx} = \frac{1}{t}.}

\end{array}$$

This derivative is undefined when $t = 0.$ Calculating $x(0)$ and $y(0)$ gives $x(0) = (0)^{2} - 3 = -3$ and $y(0) = 2(0) - 1 = -1,$ which corresponds to the point $\left( {-3,-1} \right)$ on the graph. The graph of this curve is a parabola opening to the right, and the point $\left( {-3,-1} \right)$ is its vertex as shown.

2. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

$$\begin{array}{l}

{x^{\prime}(t) = 2} \\

{y^{\prime}(t) = 3t^{2} - 3.}

\end{array}$$

Next substitute these into the equation:

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{3t^{2} - 3}{2}.}

\end{array}$$

This derivative is zero when $t = \pm 1.$ When $t = -1$ we have

$$x(-1) = 2(-1) + 1 = -1\ \text{and}\ y(-1) = (-1)^{3} - 3(-1) + 4 = -1 + 3 + 4 = 6,$$

which corresponds to the point $\left( {-1,6} \right)$ on the graph. When $t = 1$ we have

$$x(1) = 2(1) + 1 = 3\ \text{and}\ y(1) = (1)^{3} - 3(1) + 4 = 1 - 3 + 4 = 2,$$

which corresponds to the point $\left( {3,2} \right)$ on the graph. The point $\left( {3,2} \right)$ is a relative minimum and the point $\left( {-1,6} \right)$ is a relative maximum, as seen in the following graph.

3. To apply Equation 1.1, first calculate $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

$$\begin{array}{l}

{x^{\prime}(t) = -5\ \text{sin}\ t} \\

{y^{\prime}(t) = 5\ \text{cos}\ t.}

\end{array}$$

Next substitute these into the equation:

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{5\ \text{cos}\ t}{-5\ \text{sin}\ t}} \\

{\frac{dy}{dx} = \text{−}\text{cot}\ t.}

\end{array}$$

This derivative is zero when $\text{cos}\ t = 0$ and is undefined when $\text{sin}\ t = 0.$ This gives $t = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},\text{and}\ 2\pi$ as critical points for *t.* Substituting each of these into $x(t)$ and $y(t),$ we obtain

| $t$ | $x(t)$ | $y(t)$ |

|------------------|--------|--------|

| 0 | 5 | 0 |

| $\frac{\pi}{2}$ | 0 | 5 |

| $\pi$ | −5 | 0 |

| $\frac{3\pi}{2}$ | 0 | −5 |

| $2\pi$ | 5 | 0 |

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 1.19). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

Calculate the derivative ${dy}\text{/}{dx}$ for the plane curve defined by the equations

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 3$$

and locate any critical points on its graph.

Finding a Tangent Line

Find an equation of the tangent line to the curve defined by the equations

$$x(t) = t^{2} - 3,\quad y(t) = 2t - 1,\quad-3 \leq t \leq 4\ \text{when}\ t = 2.$$

Solution

First find the slope of the tangent line using Equation 1.1, which means calculating $x^{\prime}(t)$ and $y^{\prime}(t)\text{:}$

$$\begin{array}{l}

{x^{\prime}(t) = 2t} \\

{y^{\prime}(t) = 2.}

\end{array}$$

Next substitute these into the equation:

$$\begin{array}{l}

{\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}}} \\

{\frac{dy}{dx} = \frac{2}{2t}} \\

{\frac{dy}{dx} = \frac{1}{t}.}

\end{array}$$

When $t = 2,$ $\frac{dy}{dx} = \frac{1}{2},$ so this is the slope of the tangent line. Calculating $x(2)$ and $y(2)$ gives

$$x(2) = (2)^{2} - 3 = 1\ \text{and}\ y(2) = 2(2) - 1 = 3,$$

which corresponds to the point $\left( {1,3} \right)$ on the graph (Figure 1.20). Now use the point-slope form of the equation of a line to find an equation of the tangent line:

$$\begin{array}{rll}

{y - y_{0}} & = & {m\left( {x - x_{0}} \right)} \\

{y - 3} & = & {\frac{1}{2}\left( {x - 1} \right)} \\

{y - 3} & = & {\frac{1}{2}x - \frac{1}{2}} \\

y & = & {\frac{1}{2}x + \frac{5}{2}.}

\end{array}$$

Find an equation of the tangent line to the curve defined by the equations

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 10\ \text{when}\ t = 5.$$

Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y = f(x)$ is defined to be the derivative of the first derivative; that is,

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left\lbrack \frac{dy}{dx} \right\rbrack.$$

Since $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}},$ we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}.$ This gives us

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}}.$$ (1.2)

If we know ${dy}\text{/}{dx}$ as a function of *t,* then this formula is straightforward to apply.

Finding a Second Derivative

Calculate the second derivative ${d^{2}y}\text{/}{dx^{2}}$ for the plane curve defined by the parametric equations $x(t) = t^{2} - 3,y(t) = 2t - 1,-3 \leq t \leq 4.$

Solution

From Example 1.4 we know that $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}.$ Using Equation 1.2, we obtain

$$\frac{d^{2}y}{dx^{2}} = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}} = \frac{\left( {d\text{/}{dt}} \right)\left( {1\text{/}t} \right)}{2t} = \frac{\text{−}t^{-2}}{2t} = - \frac{1}{2t^{3}}.$$

Calculate the second derivative ${d^{2}y}\text{/}{dx^{2}}$ for the plane curve defined by the equations

$$x(t) = t^{2} - 4t,\quad y(t) = 2t^{3} - 6t,\quad-2 \leq t \leq 3$$

and locate any critical points on its graph.

Integrals Involving Parametric Equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x(t) = t - \text{sin}\ t,\quad y(t) = 1 - \text{cos}\ t.$ Suppose we want to find the area of the shaded region in the following graph.

To derive a formula for the area under the curve defined by the functions

$$x = x(t),\quad y = y(t),\quad a \leq t \leq b,$$

we assume that $x(t)$ is increasing on the interval $t~ \in ~\lbrack a,~b\rbrack$ and $x(t)$ is differentiable and start with an equal partition of the interval $a \leq t \leq b.$ Suppose $t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b$ and consider the following graph.

We use rectangles to approximate the area under the curve. The height of the $i$th rectangle is $y\left( t_{i–1} \right)$, so an approximation to the area is

$$\begin{array}{l}

\begin{matrix}

& {\sum\limits_{i = 1}^{n}y\left( t_{i - 1} \right)\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)} \\

= & {\sum\limits_{i = 1}^{n}y\left( t_{i - 1} \right)\frac{\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)}{\left( t_{i} - t_{i - 1} \right)}\left( t_{i} - t_{i - 1} \right)} \\

& \left. \rightarrow\int_{a}^{b}y(t)x'(t)dt~\text{as~max}\left\{ \left( t_{i} - t_{i - 1} \right) \right\}\rightarrow 0 \right.

\end{matrix}

\end{array}$$

This follows from results obtained in Calculus 1 for the function $y\left( t_{i - 1} \right)\frac{\left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right)}{\left( t_{i} - t_{i - 1} \right)}.$

Then a Riemann sum for the area is

$$A_{n} = \sum\limits_{i = 1}^{n}y\left( \overset{—}{t_{i}} \right)\ \left( x\left( t_{i} \right) - x\left( t_{i - 1} \right) \right).$$

Multiplying and dividing each area by $t_{i} - t_{i - 1}$ gives

$$A_{n} = {\sum\limits_{i = 1}^{n}{y\left( {x\left( {\overset{–}{t}}_{i} \right)} \right)\ \left( \frac{x\left( t_{i} \right) - x\left( t_{i - 1} \right)}{t_{i} - t_{i - 1}} \right)}}\left( {t_{i} - t_{i - 1}} \right) = {\sum\limits_{i = 1}^{n}{y\left( {x\left( {\overset{–}{t}}_{i} \right)} \right)\ \left( \frac{x\left( t_{i} \right) - x\left( t_{i - 1} \right)}{\text{Δ}t} \right)}}\text{Δ}t.$$

Taking the limit as $n$ approaches infinity gives

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}.$$

If $x$ is a decreasing function for $a \leq t \leq b$, a similar derivation will show that the area is given by $\begin{matrix}

{- \int_{a}^{b}y(t)x'(t)dt} & =

\end{matrix}\int_{a}^{b}y(t)x'(t)dt$

This leads to the following theorem.

Area under a Parametric Curve

Consider the non-self-intersecting plane curve defined by the parametric equations

$$x = x(t),\quad y = y(t),\quad a \leq t \leq b$$

and assume that $x(t)$ is differentiable. The area under this curve is given by

$$A = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}.$$ (1.3)

Finding the Area under a Parametric Curve

Find the area under the curve of the cycloid defined by the equations

$$x(t) = t - \text{sin}\ t,\quad y(t) = 1 - \text{cos}\ t,\quad 0 \leq t \leq 2\pi.$$

Solution

Using Equation 1.3, we have

$$\begin{array}{cl}

A & {= {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {1 - \text{cos}\ t} \right)\left( {1 - \text{cos}\ t} \right)\ dt}}} \\

& {= {\int_{0}^{2\pi}{(1 - 2\ \text{cos}\ t + \text{cos}^{2}t)dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {1 - 2\ \text{cos}\ t + \frac{1 + \text{cos}\ 2t}{2}} \right)\ dt}}} \\

& {= {\int_{0}^{2\pi}{\left( {\frac{3}{2} - 2\ \text{cos}\ t + \frac{\text{cos}\ 2t}{2}} \right)\ dt}}} \\

& {= \left. {\frac{3t}{2} - 2\ \text{sin}\ t + \frac{\text{sin}\ 2t}{4}} \right|_{0}^{2\pi}} \\

& {= 3\pi.}

\end{array}$$

Find the area under the curve of the hypocycloid defined by the equations

$$x(t) = 3\ \text{cos}\ t + \text{cos}\ 3t,\quad y(t) = 3\ \text{sin}\ t - \text{sin}\ 3t,\quad 0 \leq t \leq \pi.$$

Arc Length of a Parametric Curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point *A* to point *B* along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Given a plane curve defined by the functions $x = x(t),y = y(t),a \leq t \leq b,$ we start by partitioning the interval $\lbrack a,b\rbrack$ into *n* equal subintervals: $t_{0} = a < t_{1} < t_{2} < \text{⋯} < t_{n} = b.$ The width of each subinterval is given by $\text{Δ}t = {{(b - a)}\text{/}n}.$ We can calculate the length of each line segment:

$$\begin{array}{l}

{d_{1} = \sqrt{\left( {x\left( t_{1} \right) - x\left( t_{0} \right)} \right)^{2} + \left( {y\left( t_{1} \right) - y\left( t_{0} \right)} \right)^{2}}} \\

{d_{2} = \sqrt{\left( {x\left( t_{2} \right) - x\left( t_{1} \right)} \right)^{2} + \left( {y\left( t_{2} \right) - y\left( t_{1} \right)} \right)^{2}}\ \text{etc}.}

\end{array}$$

Then add these up. We let *s* denote the exact arc length and $s_{n}$ denote the approximation by *n* line segments:

$$s \approx {\sum\limits_{k = 1}^{n}s_{k}} = {\sum\limits_{k = 1}^{n}\sqrt{\left( {x\left( t_{k} \right) - x\left( t_{k - 1} \right)} \right)^{2} + \left( {y\left( t_{k} \right) - y\left( t_{k - 1} \right)} \right)^{2}}}.$$ (1.4)

If we assume that $x(t)$ and $y(t)$ are differentiable functions of *t,* then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval $\lbrack t_{k - 1},t_{k}\rbrack$ there exist ${\hat{t}}_{k}$ and ${\widetilde{t}}_{k}$ such that

$$\begin{array}{l}

{x\left( t_{k} \right) - x\left( t_{k - 1} \right) = x^{\prime}\left( {\hat{t}}_{k} \right)\left( {t_{k} - t_{k - 1}} \right) = x^{\prime}\left( {\hat{t}}_{k} \right)\text{Δ}t} \\

{y\left( t_{k} \right) - y\left( t_{k - 1} \right) = y^{\prime}\left( {\widetilde{t}}_{k} \right)\left( {t_{k} - t_{k - 1}} \right) = y^{\prime}\left( {\widetilde{t}}_{k} \right)\text{Δ}t.}

\end{array}$$

Therefore Equation 1.4 becomes

$$\begin{array}{cl}

s & {\approx {\sum\limits_{k = 1}^{n}s_{k}}} \\

& {= {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)\text{Δ}t} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)\text{Δ}t} \right)^{2}}}} \\

& {= {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2}\left( {\text{Δ}t} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}\left( {\text{Δ}t} \right)^{2}}}} \\

& {= \left( {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}}} \right)\text{Δ}t.}

\end{array}$$

This is a Riemann sum that approximates the arc length over a partition of the interval $\lbrack a,b\rbrack.$ If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

$$\begin{array}{cl}

s & {= \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{k = 1}^{n}s_{k}}} \\

& {= \underset{n\rightarrow\infty}{\text{lim}}\left( {\sum\limits_{k = 1}^{n}\sqrt{\left( {x^{\prime}\left( {\hat{t}}_{k} \right)} \right)^{2} + \left( {y^{\prime}\left( {\widetilde{t}}_{k} \right)} \right)^{2}}} \right)\text{Δ}t} \\

& {= {\int_{a}^{b}{\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.}

\end{array}$$

When taking the limit, the values of ${\hat{t}}_{k}$ and ${\widetilde{t}}_{k}$ are both contained within the same ever-shrinking interval of width $\text{Δ}t,$ so they must converge to the same value.

We can summarize this method in the following theorem.

Arc Length of a Parametric Curve

Consider the plane curve defined by the parametric equations

$$x = x(t),\quad y = y(t),\quad t_{1} \leq t \leq t_{2}$$

and assume that $x(t)$ and $y(t)$ are differentiable functions of *t.* Then the arc length of this curve is given by

$$s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$ (1.5)

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function $y = F(x).$ Then $y(t) = F\left( {x(t)} \right)$ and the Chain Rule gives $y^{\prime}(t) = F^{\prime}\left( {x(t)} \right)x^{\prime}(t).$ Substituting this into Equation 1.5 gives

$$\begin{array}{cl}

s & {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( {F^{\prime}(x)\frac{dx}{dt}} \right)^{2}}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2}\left( {1 + \left( {F^{\prime}(x)} \right)^{2}} \right)}dt}}} \\

& {= {\int_{t_{1}}^{t_{2}}{x^{\prime}(t)\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dt}}.}

\end{array}$$

Here we have assumed that $x^{\prime}(t) > 0,$ which is a reasonable assumption. The Chain Rule gives $dx = x^{\prime}(t)\ dt,$ and letting $a = x\left( t_{1} \right)$ and $b = x\left( t_{2} \right)$ we obtain the formula

$$s = {\int_{a}^{b}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}},$$

which is the formula for arc length obtained in the Introduction to the Applications of Integration.

Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations

$$x(t) = 3\ \text{cos}\ t,\quad y(t) = 3\ \text{sin}\ t,\quad 0 \leq t \leq \pi.$$

Solution

The values $t = 0$ to $t = \pi$ trace out the red curve in Figure 1.23. To determine its length, use Equation 1.5:

$$\begin{array}{cl}

s & {= {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{\left( {-3\ \text{sin}\ t} \right)^{2} + \left( {3\ \text{cos}\ t} \right)^{2}}dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{9\ \text{sin}^{2}t + 9\ \text{cos}^{2}t}\ dt}}} \\

& {= {\int_{0}^{\pi}{\sqrt{9\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}dt}}} \\

& {= {\int_{0}^{\pi}{3dt}} = \left. {3t} \right|_{0}^{\pi} = 3\pi.}

\end{array}$$

Note that the formula for the arc length of a semicircle is $\pi r$ and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.

Find the arc length of the curve defined by the equations

$$x(t) = 3t^{2},\quad y(t) = 2t^{3},\quad 1 \leq t \leq 3.$$

We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive *x*-axis, the parametric equations for this curve can be written as

$$x(t) = 140t,\quad y(t) = -16t^{2} + 2t$$

where *t* represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions $x(t)$ and $y(t)$ using *v* as an independent variable, so as to eliminate any confusion with the parameter *t*:

$$x(v) = 140v,\quad y(v) = -16v^{2} + 2v.$$

Then we write the arc length formula as follows:

$$\begin{array}{cl}

{s(t)} & {= {\int_{0}^{t}{\sqrt{\left( \frac{dx}{dv} \right)^{2} + \left( \frac{dy}{dv} \right)^{2}}dv}}} \\

& {= {\int_{0}^{t}{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}}.}

\end{array}$$

The variable *v* acts as a dummy variable that disappears after integration, leaving the arc length as a function of time *t.* To integrate this expression we can use a formula from Appendix A,

$${\int{\sqrt{a^{2} + u^{2}}du}} = \frac{u}{2}\sqrt{a^{2} + u^{2}} + \frac{a^{2}}{2}\text{ln}\left| {u + \sqrt{a^{2} + u^{2}}} \right| + C.$$

We set $a = 140$ and $u = -32v + 2.$ This gives $du = -32dv,$ so $dv = - \frac{1}{32}du.$ Therefore

$$\begin{array}{cl}

{\int{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}} & {= - \frac{1}{32}{\int{\sqrt{a^{2} + u^{2}}du}}} \\

& {= - \frac{1}{32}\left\lbrack \begin{array}{l}

{\frac{\left( {-32v + 2} \right)}{2}\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}} \\

{+ \frac{140^{2}}{2}\text{ln}\left| {\left( {-32v + 2} \right) + \sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}} \right|}

\end{array} \right\rbrack + C}

\end{array}$$

and

$$\begin{array}{cl}

{s(t)} & {= - \frac{1}{32}\left\lbrack {\frac{\left( {-32t + 2} \right)}{2}\sqrt{140^{2} + \left( {-32t + 2} \right)^{2}} + \frac{140^{2}}{2}\text{ln}\left| {\left( {-32t + 2} \right) + \sqrt{140^{2} + \left( {-32t + 2} \right)^{2}}} \right|} \right\rbrack} \\

& {\mspace{11mu} + \frac{1}{32}\left\lbrack {\sqrt{140^{2} + 2^{2}} + \frac{140^{2}}{2}\text{ln}\left| {2 + \sqrt{140^{2} + 2^{2}}} \right|} \right\rbrack} \\

& {= \left( {\frac{t}{2} - \frac{1}{32}} \right)\sqrt{1024t^{2} - 128t + 19604} - \frac{1225}{4}\text{ln}\left| {\left( {-32t + 2} \right) + \sqrt{1024t^{2} - 128t + 19604}} \right|} \\

& {\mspace{11mu} + \frac{\sqrt{19604}}{32} + \frac{1225}{4}\text{ln}\left( {2 + \sqrt{19604}} \right).}

\end{array}$$

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to *t.* While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

$$\frac{d}{dx}{\int_{a}^{x}{f(u)\ du}} = f(x).$$

Therefore

$$\begin{array}{cl}

{s^{\prime}(t)} & {= \frac{d}{dt}\left\lbrack {s(t)} \right\rbrack} \\

& {= \frac{d}{dt}\left\lbrack {\int_{0}^{t}{\sqrt{140^{2} + \left( {-32v + 2} \right)^{2}}dv}} \right\rbrack} \\

& {= \sqrt{140^{2} + \left( {-32t + 2} \right)^{2}}} \\

& {= \sqrt{1024t^{2} - 128t + 19604}} \\

& {= 2\sqrt{256t^{2} - 32t + 4901}.}

\end{array}$$

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

$$\begin{array}{cl}

{s\left( \frac{1}{3} \right)} & {= \left( {\frac{1\text{/}3}{2} - \frac{1}{32}} \right)\sqrt{1024\left( \frac{1}{3} \right)^{2} - 128\left( \frac{1}{3} \right) + 19604}} \\

& {\mspace{11mu} - \frac{1225}{4}\text{ln}\left| {\left( {-32\left( \frac{1}{3} \right) + 2} \right) + \sqrt{1024\left( \frac{1}{3} \right)^{2} - 128\left( \frac{1}{3} \right) + 19604}} \right|} \\

& {\mspace{11mu} + \frac{\sqrt{19604}}{32} + \frac{1225}{4}\text{ln}\left( {2 + \sqrt{19604}} \right)} \\

& {\approx 46.69\ \text{feet}.}

\end{array}$$

This value is just over three quarters of the way to home plate. The speed of the ball is

$$s^{\prime}\left( \frac{1}{3} \right) = 2\sqrt{256\left( \frac{1}{3} \right)^{2} - 16\left( \frac{1}{3} \right) + 4901} \approx 140.34\ \text{ft/s}.$$

This speed translates to approximately 95 mph—a major-league fastball.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function $y = f(x)$ from $x = a$ to $x = b,$ revolved around the *x*-axis:

$$S = 2\pi{\int_{a}^{b}{f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}.$$

We now consider a volume of revolution generated by revolving a parametrically defined curve $x = x(t),y = y(t),a \leq t \leq b$ around the *x*-axis as shown in the following figure.

The analogous formula for a parametrically defined curve is

$$S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$$ (1.6)

provided that $y(t)$ is not negative on $\lbrack a,b\rbrack.$

Finding Surface Area

Find the surface area of a sphere of radius *r* centered at the origin.

Solution

We start with the curve defined by the equations

$$x(t) = r\ \text{cos}\ t,\quad y(t) = r\ \text{sin}\ t,\quad 0 \leq t \leq \pi.$$

This generates an upper semicircle of radius *r* centered at the origin as shown in the following graph.

When this curve is revolved around the *x*-axis, it generates a sphere of radius *r*. To calculate the surface area of the sphere, we use Equation 1.6:

$$\begin{array}{cl}

S & {= 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}} \\

& {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{\left( {\text{−}r\ \text{sin}\ t} \right)^{2} + \left( {r\ \text{cos}\ t} \right)^{2}}dt}}} \\

& {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{r^{2}\text{sin}^{2}t + r^{2}\text{cos}^{2}t}\ dt}}} \\

& {= 2\pi{\int_{0}^{\pi}{r\ \text{sin}\ t\sqrt{r^{2}\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}dt}}} \\

& {= 2\pi{\int_{0}^{\pi}{r^{2}\text{sin}\ t\ dt}}} \\

& {= 2\pi r^{2}(\left. {\text{−}\text{cos}\ t} \right|_{0}^{\pi})} \\

& {= 2\pi r^{2}\left( {\text{−}\text{cos}\ \pi + \text{cos}\ 0} \right)} \\

& {= 4\pi r^{2}.}

\end{array}$$

This is, in fact, the formula for the surface area of a sphere.

Find the surface area generated when the plane curve defined by the equations

$$x(t) = t^{3},\quad y(t) = t^{2},\quad 0 \leq t \leq 1$$

is revolved around the *x*-axis.

Section 1.2 Exercises

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

62\.

$\begin{array}{ll}

{x = 3 + t,} & {y = 1 - t}

\end{array}$

63.

$\begin{array}{ll}

{x = 8 + 2t,} & {y = 1}

\end{array}$

64\.

$\begin{array}{ll}

{x = 4 - 3t,} & {y = -2 + 6t}

\end{array}$

65.

$\begin{array}{ll}

{x = -5t + 7,} & {y = 3t - 1}

\end{array}$

For the following exercises, determine the slope of the tangent line, then find an equation of the tangent line at the given value of the parameter.

66\.

$\begin{array}{ll}

{x = 3\ \text{sin}\ t,} & {y = 3\ \text{cos}\ t,\quad t = \frac{\pi}{4}}

\end{array}$

67.

$\begin{array}{ll}

{x = \text{cos}\ t,} & {y = 8\ \text{sin}\ t,}

\end{array}t = \frac{\pi}{2}$

68\.

$\begin{array}{ll}

{x = 2t,} & {y = t^{3},\quad t = -1}

\end{array}$

69.

$\begin{array}{ll}

{x = t + \frac{1}{t},} & {y = t - \frac{1}{t},\quad t = 1}

\end{array}$

70\.

$\begin{array}{ll}

{x = \sqrt{t},} & {y = 2t,\quad t = 4}

\end{array}$

For the following exercises, find all points on the curve that have the given slope.

71.

$\begin{array}{ll}

{x = 4\ \text{cos}\ t,} & {y = 4\ \text{sin}\ t,}

\end{array}$ slope = 0.5

72\.

$\begin{array}{ll}

{x = 2\ \text{cos}\ t,} & {y = 8\ \text{sin}\ t,\ \text{slope} = -1}

\end{array}$

73.

$\begin{array}{ll}

{x = t + \frac{1}{t},} & {y = t - \frac{1}{t},\ \text{slope} = 1}

\end{array}$

74\.

$\begin{array}{ll}

{x = 2 + \sqrt{t},} & {y = 2 - 4t,\ \text{slope} = 0}

\end{array}$

For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter *t*.

75.

$\begin{array}{ll}

{x = e^{\sqrt{t}},} & {y = 1 - \text{ln}\ t^{2},\quad t = 1}

\end{array}$

76\.

$\begin{array}{ll}

{x = t\ \text{ln}\ t,} & {y = \text{sin}^{2}t,}

\end{array}t = \frac{\pi}{4}$

77.

$\begin{array}{ll}

{x = e^{t},} & {y = {(t - 1)}^{2},\quad\text{at}(1,1)}

\end{array}$

78\.

For $x = \text{sin}(2t),y = 2\ \text{sin}\ t$ where $0 \leq t < 2\pi.$ Find all values of *t* at which a horizontal tangent line exists.

79.

For $x = \text{sin}(2t),y = 2\ \text{sin}\ t$ where $0 \leq t < 2\pi.$ Find all values of *t* at which a vertical tangent line exists.

80\.

Find all points on the curve $x = 4\ \text{sin}(t),y = 4\ \text{cos}(t)$ that have the slope of $0.5$

81.

Find $\frac{dy}{dx}$ for $x = \text{sin}(t),y = \text{cos}(t).$

82\.

Find an equation of the tangent line to $x = \text{sin}(t),y = \text{cos}(t)$ at $t = \frac{\pi}{4}.$

83.

For the curve $x = 4t,y = 3t - 2,$ find the slope and concavity of the curve at $t = 3.$

84\.

For the parametric curve whose equation is $x = 4\ \text{cos}\ \theta,y = 4\ \text{sin}\ \theta,$ find the slope and concavity of the curve at $\theta = \frac{\pi}{4}.$

85.

Find the slope and concavity for the curve whose equation is $x = 2 + \text{sec}\ \theta,y = 1 + 2\ \text{tan}\ \theta$ at $\theta = \frac{\pi}{6}.$

86\.

Find all points on the curve $x = t + 4,y = t^{3} - 3t$ at which there are vertical and horizontal tangents.

87.

Find all points on the curve $x = \text{sec}\ \theta,y = \text{tan}\ \theta$ at which horizontal and vertical tangents exist.

For the following exercises, find ${d^{2}y}\text{/}{dx^{2}.}$

88\.

$\begin{array}{ll}

{x = t^{4} - 1,} & {y = t - t^{2}}

\end{array}$

89.

$\begin{array}{ll}

{x = \text{sin}\left( {\pi t} \right),} & {y = \text{cos}\left( {\pi t} \right)}

\end{array}$

90\.

$\begin{array}{ll}

{x = e^{\text{−}t},} & {y = t}

\end{array}e^{2t}$

For the following exercises, find points on the curve at which tangent line is horizontal or vertical.

91.

$\begin{array}{ll}

{x = t(t^{2} - 3),} & {y = 3(t^{2} - 3)}

\end{array}$

92\.

$\begin{array}{ll}

{x = \frac{3t}{1 + t^{3}},} & {y = \frac{3t^{2}}{1 + t^{3}}}

\end{array}$

For the following exercises, find ${dy}\text{/}{dx}$ at the value of the parameter.

93.

$\begin{array}{ll}

{x = \text{cos}\ t,} & {y = \text{sin}\ t,\quad t = \frac{3\pi}{4}}

\end{array}$

94\.

$\begin{array}{ll}

{x = \sqrt{t},} & {y = 2t + 4,\quad t = 9}

\end{array}$

95.

$\begin{array}{ll}

{x = 4\ \text{cos}\left( {2\pi s} \right),} & {y = 3\ \text{sin}}

\end{array}\left( {2\pi s} \right),\quad s = - \frac{1}{4}$

For the following exercises, find ${d^{2}y}\text{/}{dx^{2}}$ at the given point without eliminating the parameter.

96\.

$\begin{array}{lll}

{x = \frac{1}{2}t^{2},} & {y = \frac{1}{3}t^{3},} & {t = 2}

\end{array}$

97.

$x = \sqrt{t},\quad y = 2t + 4,\quad t = 1$

98\.

Find *t* intervals on which the curve $x = 3t^{2},y = t^{3} - t$ is concave up as well as concave down.

99.

Determine the concavity of the curve $x = 2t + \text{ln}\ t,y = 2t - \text{ln}\ t.$

100\.

Sketch and find the area under one arch of the cycloid $x = r\left( {\theta - \text{sin}\ \theta} \right),y = r\left( {1 - \text{cos}\ \theta} \right).$

101.

Find the area bounded by the curve $x = \text{cos}\ t,y = e^{t},0 \leq t \leq \frac{\pi}{2}$ and the lines $y = 1$ and $x = 0.$

102\.

Find the area enclosed by the ellipse $x = a\ \text{cos}\ \theta,y = b\ \text{sin}\ \theta,0 \leq \theta < 2\pi.$

103.

Find the area of the region bounded by $x = 2\ \text{sin}^{2}\theta,y = 2\ \text{sin}^{2}\theta\ \text{tan}\ \theta,$ for $0 \leq \theta \leq \frac{\pi}{2}.$

For the following exercises, find the area of the regions bounded by the parametric curves and the indicated values of the parameter.

104\.

$x = 2\ \text{cot}\ \theta,y = 2\ \text{sin}^{2}\theta,0 \leq \theta \leq \pi$

105.

\[T\] $x = 2a\ \text{cos}\ t - a\ \text{cos}(2t),y = 2a\ \text{sin}\ t - a\ \text{sin}(2t),0 \leq t < 2\pi$

106\.

\[T\] $x = a\ \text{sin}(2t),y = b\ \text{sin}(t),0 \leq t < 2\pi$ (the “hourglass”)

107.

\[T\] $x = 2a\ \text{cos}\ t - a\ \text{sin}(2t),y = b\ \text{sin}\ t,0 \leq t < 2\pi$ (the “teardrop”)

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.

108\.

$x = 4t + 3,\quad y = 3t - 2,\quad 0 \leq t \leq 2$

109.

$\begin{array}{lll}

{x = \frac{1}{3}t^{3},} & {y = \frac{1}{2}t^{2},} & {0 \leq t \leq 1}

\end{array}$

110\.

$\begin{array}{lll}

{x = \text{cos}(2t),} & {y = \text{sin}(2t),} & {0 \leq t \leq \frac{\pi}{2}}

\end{array}$

111.

$\begin{array}{lll}

{x = 1 + t^{2},} & {y = \left( {1 + t} \right)^{3},} & {0 \leq t \leq 1}

\end{array}$

112\.

$\begin{array}{lll}

{x = e^{t}\text{cos}\ t,} & {y = e^{t}\text{sin}\ t,} & {0 \leq t \leq \frac{\pi}{2}}

\end{array}$ (Use a CAS for this and express the answer as a decimal rounded to three places.)

113.

$x = a\ \text{cos}^{3}\theta,y = a\ \text{sin}^{3}\theta$ on the interval $\lbrack 0,2\pi)$ (the hypocycloid)

114\.

Find the length of one arch of the cycloid $x = 4\left( {t - \text{sin}\ t} \right),y = 4\left( {1 - \text{cos}\ t} \right).$

115.

Find the distance traveled by a particle with position $\left( {x,y} \right)$ as *t* varies in the given time interval: $\begin{array}{lll}

{x = \text{sin}^{2}t,} & {y = \text{cos}^{2}t,} & {0 \leq t \leq 3\pi}

\end{array}.$

116\.

Find the length of one arch of the cycloid $x = \theta - \text{sin}\ \theta,y = 1 - \text{cos}\ \theta.$

117\.

Show that the total length of the ellipse $x = 4\ \text{sin}\ \theta,y = 3\ \text{cos}\ \theta$ is $L = 16{\int_{0}^{\pi\text{/}2}{\sqrt{1 - e^{2}\text{sin}^{2}\theta}\ d\theta,}}$ where $e = \frac{c}{a}$ and $c = \sqrt{a^{2} - b^{2}}.$

118\.

Find the length of the curve $x = e^{t} - t,y = 4e^{t\text{/}2},-8 \leq t \leq 3.$

For the following exercises, find the area of the surface obtained by rotating the given curve about the *x*-axis.

119.

$\begin{array}{lll}

{x = t^{3},} & {y = t^{2},} & {0 \leq t \leq 1}

\end{array}$

120\.

$\begin{array}{lll}

{x = a\ \text{cos}^{3}\theta,} & {y = a\ \text{sin}^{3}\theta,} & {0 \leq \theta \leq}

\end{array}\frac{\pi}{2}$

121.

\[T\] Use a CAS to find the area of the surface generated by rotating $x = t + t^{3},y = t - \frac{1}{t^{2}},1 \leq t \leq 2$ about the *x*-axis. (Answer to three decimal places.)

122\.

Find the surface area obtained by rotating $x = 3t^{2},y = 2t^{3},0 \leq t \leq 5$ about the *y*-axis.

123.

Find the area of the surface generated by revolving $x = t^{2},y = 2t,0 \leq t \leq 4$ about the *x*-axis.

124\.

Find the surface area generated by revolving $x = t^{2},y = 2t^{2},0 \leq t \leq 1$ about the *y*-axis.

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1.3 Polar Coordinates

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-3-polar-coordinates

1.3 Polar Coordinates

1.3.1 Locate points in a plane by using polar coordinates.
1.3.2 Convert points between rectangular and polar coordinates.
1.3.3 Sketch polar curves from given equations.
1.3.4 Convert equations between rectangular and polar coordinates.
1.3.5 Identify symmetry in polar curves and equations.

The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a *one-to-one mapping* from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.

Defining Polar Coordinates

To find the coordinates of a point in the polar coordinate system, consider Figure 1.27. The point $P$ has Cartesian coordinates $\left( {x,y} \right).$ The line segment connecting the origin to the point $P$ measures the distance from the origin to $P$ and has length $r.$ The angle between the positive $x$-axis and the line segment has measure $\theta.$ This observation suggests a natural correspondence between the coordinate pair $\left( {x,y} \right)$ and the values $r$ and $\theta.$ This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane has two values (hence the term *ordered pair*) associated with it. In the polar coordinate system, each point also has two values associated with it: $r$ and $\theta.$

Using right-triangle trigonometry, the following equations are true for the point $P\text{:}$

$$\text{cos}\ \theta = \frac{x}{r}\ \text{so}\ x = r\ \text{cos}\ \theta$$ $$\text{sin}\ \theta = \frac{y}{r}\ \text{so}\ y = r\ \text{sin}\ \theta.$$

Furthermore,

$$r^{2} = x^{2} + y^{2}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

Each point $\left( {x,y} \right)$ in the Cartesian coordinate system can therefore be represented as an ordered pair $\left( {r,\theta} \right)$ in the polar coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. Every point in the plane can be represented in this form.

Note that the equation $\text{tan}\ \theta = {y\text{/}x}$ has an infinite number of solutions for any ordered pair $\left( {x,y} \right).$ However, if we restrict the solutions to values between $0$ and $2\pi$ then we can assign a unique solution to the quadrant in which the original point $\left( {x,y} \right)$ is located. Then the corresponding value of *r* is positive, so $r^{2} = x^{2} + y^{2}.$

Converting Points between Coordinate Systems

Given a point $P$ in the plane with Cartesian coordinates $\left( {x,y} \right)$ and polar coordinates $\left( {r,\theta} \right),$ the following conversion formulas hold true:

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta,$$ (1.7) $$r^{2} = x^{2} + y^{2}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$ (1.8)

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

Converting between Rectangular and Polar Coordinates

Convert each of the following points into polar coordinates.

1. $(1,1)$

2. $(-3,4)$

3. $\left( {0,3} \right)$

4. $(5\sqrt{3},-5)$

Convert each of the following points into rectangular coordinates.

5. $(3,{\pi\text{/}3})$

6. $(2,{{3\pi}\text{/}2})$

7. $(6,{{-5\pi}\text{/}6})$

Solution

1. Use $x = 1$ and $y = 1$ in Equation 1.8:

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {1^{2} + 1^{2}} \\

r & = & \sqrt{2}

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{1}{1} = 1} \\

\theta & = & {\frac{\pi}{4}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {\sqrt{2},\frac{\pi}{4}} \right)$ in polar coordinates.

2. Use $x = -3$ and $y = 4$ in Equation 1.8:

$$\begin{matrix}

\begin{matrix}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {(-3)^{2} + (4)^{2}} \\

r & = & 5

\end{matrix} & & & \text{and} & & & \begin{matrix}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {- \frac{4}{3}} \\

\theta & = & {\pi - \arctan\left( \frac{4}{3} \right)} \\

& \approx & {2.21.}

\end{matrix}

\end{matrix}$$

The point $( - 3,~4)$ lies in Quadrant $\text{II}$. Subtract the value of the reference angle, $\text{arctan}\left( \frac{4}{3} \right)$, from $\pi$ to find the radian measure of $\theta$.

Therefore this point can be represented as $\left( {5,2.21} \right)$ in polar coordinates.

3. Use $x = 0$ and $y = 3$ in Equation 1.8:

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {(3)^{2} + (0)^{2}} \\

& = & {9 + 0} \\

r & = & 3

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{3}{0}.}

\end{array}

\end{array}$$

Direct application of the second equation leads to division by zero. Graphing the point $\left( {0,3} \right)$ on the rectangular coordinate system reveals that the point is located on the positive *y*-axis. The angle between the positive *x*-axis and the positive *y*-axis is $\frac{\pi}{2}.$ Therefore this point can be represented as $\left( {3,\frac{\pi}{2}} \right)$ in polar coordinates.

4. Use $x = 5\sqrt{3}$ and $y = -5$ in Equation 1.8:

$$\begin{array}{lcclccl}

\begin{array}{cll}

r^{2} & = & {x^{2} + y^{2}} \\

& = & {\left( {5\sqrt{3}} \right)^{2} + (-5)^{2}} \\

& = & {75 + 25} \\

r & = & 10

\end{array} & & & \text{and} & & & \begin{array}{cll}

{\text{tan}\ \theta} & = & \frac{y}{x} \\

& = & {\frac{-5}{5\sqrt{3}} = - \frac{\sqrt{3}}{3}} \\

\theta & = & {- \frac{\pi}{6}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {10, - \frac{\pi}{6}} \right)$ in polar coordinates.

5. Use $r = 3$ and $\theta = \frac{\pi}{3}$ in Equation 1.7:

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {3\ \text{cos}\left( \frac{\pi}{3} \right)} \\

& = & {3\left( \frac{1}{2} \right) = \frac{3}{2}}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {3\ \text{sin}\left( \frac{\pi}{3} \right)} \\

& = & {3\left( \frac{\sqrt{3}}{2} \right) = \frac{3\sqrt{3}}{2}.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {\frac{3}{2},\ \frac{3\sqrt{3}}{2}} \right)$ in rectangular coordinates.

6. Use $r = 2$ and $\theta = \frac{3\pi}{2}$ in Equation 1.7:

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {2\ \text{cos}\left( \frac{3\pi}{2} \right)} \\

& = & {2(0) = 0}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {2\ \text{sin}\left( \frac{3\pi}{2} \right)} \\

& = & {2(-1) = -2.}

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {0,-2} \right)$ in rectangular coordinates.

7. Use $r = 6$ and $\theta = - \frac{5\pi}{6}$ in Equation 1.7:

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {6\ \text{cos}\left( {- \frac{5\pi}{6}} \right)} \\

& = & {6\left( {- \frac{\sqrt{3}}{2}} \right)} \\

& = & {-3\sqrt{3}}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {6\ \text{sin}\left( {- \frac{5\pi}{6}} \right)} \\

& = & {6\left( {- \frac{1}{2}} \right)} \\

& = & -3.

\end{array}

\end{array}$$

Therefore this point can be represented as $\left( {-3\sqrt{3},-3} \right)$ in rectangular coordinates.

Convert $\left( {-8,-8} \right)$ into polar coordinates and $\left( {4,\frac{2\pi}{3}} \right)$ into rectangular coordinates.

The polar representation of a point is not unique. For example, the polar coordinates $\left( {2,\frac{\pi}{3}} \right)$ and $\left( {2,\frac{7\pi}{3}} \right)$ both represent the point $\left( {1,\sqrt{3}} \right)$ in the rectangular system. Also, the value of $r$ can be negative. Therefore, the point with polar coordinates $\left( {-2,\frac{4\pi}{3}} \right)$ also represents the point $\left( {1,\sqrt{3}} \right)$ in the rectangular system, as we can see by using Equation 1.8:

$$\begin{array}{lcclccl}

\begin{array}{cll}

x & = & {r\ \text{cos}\ \theta} \\

& = & {-2\ \text{cos}\left( \frac{4\pi}{3} \right)} \\

& = & {-2\left( {- \frac{1}{2}} \right) = 1}

\end{array} & & & \text{and} & & & \begin{array}{cll}

y & = & {r\ \text{sin}\ \theta} \\

& = & {-2\ \text{sin}\left( \frac{4\pi}{3} \right)} \\

& = & {-2\left( {- \frac{\sqrt{3}}{2}} \right) = \sqrt{3}.}

\end{array}

\end{array}$$

Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.

Note that the polar representation of a point in the plane also has a visual interpretation. In particular, $r$ is the directed distance that the point lies from the origin, and $\theta$ measures the angle that the line segment from the origin to the point makes with the positive $x$-axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.

The line segment starting from the center of the graph going to the right (called the positive *x*-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the coordinate system, and corresponds to $r = 0.$ The innermost circle shown in Figure 1.28 contains all points a distance of 1 unit from the pole, and is represented by the equation $r = 1.$ Then $r = 2$ is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of $r$ is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.

Plotting Points in the Polar Plane

Plot each of the following points on the polar plane.

1. $\left( {2,\frac{\pi}{4}} \right)$

2. $\left( {-3,\frac{2\pi}{3}} \right)$

3. $\left( {4,\frac{5\pi}{4}} \right)$

Solution

The three points are plotted in the following figure.

Plot $\left( {4,\frac{5\pi}{3}} \right)$ and $\left( {-3, - \frac{7\pi}{2}} \right)$ on the polar plane.

Polar Curves

Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function $y = f(x)$ and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function $r = f(\theta).$

The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable $(\theta$ in this case) and calculate the corresponding values of the dependent variable $r.$ This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.

Plotting a Curve in Polar Coordinates

1. Create a table with two columns. The first column is for $\theta,$ and the second column is for $r.$

2. Create a list of values for $\theta.$

3. Calculate the corresponding $r$ values for each $\theta.$

4. Plot each ordered pair $\left( {r,\theta} \right)$ on the coordinate axes.

5. Connect the points and look for a pattern.

Watch this video for more information on sketching polar curves.

Graphing a Function in Polar Coordinates

Graph the curve defined by the function $r = 4\ \text{sin}\ \theta.$ Identify the curve and rewrite the equation in rectangular coordinates.

Solution

Because the function is a multiple of a sine function, it is periodic with period $2\pi,$ so use values for $\theta$ between 0 and $2\pi.$ The result of steps 1–3 appear in the following table. Figure 1.30 shows the graph based on this table.

| $\theta$ | $r = 4\ \text{sin}\ \theta$ | | $\theta$ | $r = 4\ \text{sin}\ \theta$ |

|------------------|-----------------------------|-----|-------------------|-----------------------------|

| 0 | 0 | | $\pi$ | 0 |

| $\frac{\pi}{6}$ | $2$ | | $\frac{7\pi}{6}$ | $-2$ |

| $\frac{\pi}{4}$ | $2\sqrt{2} \approx 2.8$ | | $\frac{5\pi}{4}$ | $-2\sqrt{2} \approx -2.8$ |

| $\frac{\pi}{3}$ | $2\sqrt{3} \approx 3.4$ | | $\frac{4\pi}{3}$ | $-2\sqrt{3} \approx -3.4$ |

| $\frac{\pi}{2}$ | $4$ | | $\frac{3\pi}{2}$ | $-4$ |

| $\frac{2\pi}{3}$ | $2\sqrt{3} \approx 3.4$ | | $\frac{5\pi}{3}$ | $-2\sqrt{3} \approx -3.4$ |

| $\frac{3\pi}{4}$ | $2\sqrt{2} \approx 2.8$ | | $\frac{7\pi}{4}$ | $-2\sqrt{2} \approx -2.8$ |

| $\frac{5\pi}{6}$ | $2$ | | $\frac{11\pi}{6}$ | $-2$ |

| | | | $2\pi$ | 0 |

This is the graph of a circle. The equation $r = 4\ \text{sin}\ \theta$ can be converted into rectangular coordinates by first multiplying both sides by $r.$ This gives the equation $r^{2} = 4r\ \text{sin}\ \theta.$ Next use the facts that $r^{2} = x^{2} + y^{2}$ and $y = r\ \text{sin}\ \theta.$ This gives $x^{2} + y^{2} = 4y.$ To put this equation into standard form, subtract $4y$ from both sides of the equation and complete the square:

$$\begin{array}{rll}

{x^{2} + y^{2} - 4y} & = & 0 \\

{x^{2} + \left( {y^{2} - 4y} \right)} & = & 0 \\

{x^{2} + \left( {y^{2} - 4y + 4} \right)} & = & {0 + 4} \\

{x^{2} + \left( {y - 2} \right)^{2}} & = & 4.

\end{array}$$

This is the equation of a circle with radius 2 and center $\left( {0,2} \right)$ in the rectangular coordinate system.

Create a graph of the curve defined by the function $r = 4 + 4\ \text{cos}\ \theta.$

The graph in Example 1.12 was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in Equation 1.8. Example 1.14 gives some more examples of functions for transforming from polar to rectangular coordinates.

Transforming Polar Equations to Rectangular Coordinates

Rewrite each of the following equations in rectangular coordinates and identify the graph.

1. $\theta = \frac{\pi}{3}$

2. $r = 3$

3. $r = 6\ \text{cos}\ \theta - 8\ \text{sin}\ \theta$

Solution

1. Take the tangent of both sides. This gives $\text{tan}\ \theta = \text{tan}({\pi\text{/}3}) = \sqrt{3}.$ Since $\text{tan}\ \theta = {y\text{/}x}$ we can replace the left-hand side of this equation by ${y\text{/}x}.$ This gives ${y\text{/}x} = \sqrt{3},$ which can be rewritten as $y = x\sqrt{3}.$ This is the equation of a straight line passing through the origin with slope $\sqrt{3}.$ In general, any polar equation of the form $\theta = K$ represents a straight line through the pole with slope equal to $\text{tan}\ K.$

2. First, square both sides of the equation. This gives $r^{2} = 9.$ Next replace $r^{2}$ with $x^{2} + y^{2}.$ This gives the equation $x^{2} + y^{2} = 9,$ which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form $r = k$ where *k* is a positive constant represents a circle of radius *k* centered at the origin. (*Note*: when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, $\left( {-3,\frac{\pi}{3}} \right)$ is the same point as $\left( {3,\frac{4\pi}{3}} \right).)$

3. Multiply both sides of the equation by $r.$ This leads to $r^{2} = 6r\ \text{cos}\ \theta - 8r\ \text{sin}\ \theta.$ Next use the formulas

$$r^{2} = x^{2} + y^{2},\quad x = r\ \text{cos}\ \theta,\quad y = r\ \text{sin}\ \theta.$$

This gives

$$\begin{array}{rll}

r^{2} & = & {6\left( {r\ \text{cos}\ \theta} \right) - 8\left( {r\ \text{sin}\ \theta} \right)} \\

{x^{2} + y^{2}} & = & {6x - 8y.}

\end{array}$$

To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.

$$\begin{array}{rll}

{x^{2} + y^{2}} & = & {6x - 8y} \\

{x^{2} - 6x + y^{2} + 8y} & = & 0 \\

{\left( {x^{2} - 6x} \right) + \left( {y^{2} + 8y} \right)} & = & 0 \\

{\left( {x^{2} - 6x + 9} \right) + \left( {y^{2} + 8y + 16} \right)} & = & {9 + 16} \\

{\left( {x - 3} \right)^{2} + \left( {y + 4} \right)^{2}} & = & 25.

\end{array}$$

This is the equation of a circle with center at $\left( {3,-4} \right)$ and radius 5. Notice that the circle passes through the origin since the center is 5 units away.

Rewrite the equation $r = \text{sec}\ \theta\ \text{tan}\ \theta$ in rectangular coordinates and identify its graph.

We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, *a* and *b* are arbitrary constants.

A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which $a = b$ or $a = \text{−}b.$ The rose is a very interesting curve. Notice that the graph of $r = 3\ \text{sin}\ 2\theta$ has four petals. However, the graph of $r = 3\ \text{sin}\ 3\theta$ has three petals as shown.

If the coefficient of $\theta$ is even, the graph has twice as many petals as the coefficient. If the coefficient of $\theta$ is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of $\theta$ is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started (Figure 1.34(a)). However, if the coefficient is irrational, then the curve never closes (Figure 1.34(b)). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive *r* axis are slightly thicker. This is because the petal does not quite match up with the starting point.

Chapter Opener: Describing a Spiral

Recall the chambered nautilus introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular coordinates. Figure 1.35 shows a spiral in rectangular coordinates. How can we describe this curve mathematically?

Solution

As the point *P* travels around the spiral in a counterclockwise direction, its distance *d* from the origin increases. Assume that the distance *d* is a constant multiple *k* of the angle $\theta$ that the line segment *OP* makes with the positive *r*-axis. Therefore $d\left( {P,O} \right) = k\theta,$ where $O$ is the origin. Now use the distance formula and some trigonometry:

$$\begin{array}{rll}

{d\left( {P,O} \right)} & = & {k\theta} \\

\sqrt{\left( {x - 0} \right)^{2} + \left( {y - 0} \right)^{2}} & = & {k\ \text{arctan}\left( \frac{y}{x} \right)} \\

\sqrt{x^{2} + y^{2}} & = & {k\ \text{arctan}\left( \frac{y}{x} \right)} \\

{\text{arctan}\left( \frac{y}{x} \right)} & = & \frac{\sqrt{x^{2} + y^{2}}}{k} \\

y & = & {x\ \text{tan}\left( \frac{\sqrt{x^{2} + y^{2}}}{k} \right).}

\end{array}$$

Although this equation describes the spiral, it is not possible to solve it directly for either *x* or *y*. However, if we use polar coordinates, the equation becomes much simpler. In particular, $d\left( {P,O} \right) = r,$ and $\theta$ is the second coordinate. Therefore the equation for the spiral becomes $r = k\theta.$ Note that when $\theta = 0$ we also have $r = 0,$ so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes $r = a + k\theta$ for arbitrary constants $a$ and $k.$ This is referred to as an Archimedean spiral, after the Greek mathematician Archimedes.

Another type of spiral is the logarithmic spiral, described by the function $r = a \cdot b^{\theta}.$ A graph of the function $r = 1.2\left( 1.25^{\theta} \right)$ is given in Figure 1.36. This spiral describes the shell shape of the chambered nautilus.

Suppose a curve is described in the polar coordinate system via the function $r = f(\theta).$ Since we have conversion formulas from polar to rectangular coordinates given by

$$\begin{array}{l}

{x = r\ \text{cos}\ \theta} \\

{y = r\ \text{sin}\ \theta,}

\end{array}$$

it is possible to rewrite these formulas using the function

$$\begin{array}{l}

{x = f(\theta)\ \text{cos}\ \theta} \\

{y = f(\theta)\ \text{sin}\ \theta.}

\end{array}$$

This step gives a parameterization of the curve in rectangular coordinates using $\theta$ as the parameter. For example, the spiral formula $r = a + b\theta$ from Figure 1.31 becomes

$$\begin{array}{l}

{x = \left( {a + b\theta} \right)\ \text{cos}\ \theta} \\

{y = \left( {a + b\theta} \right)\ \text{sin}\ \theta.}

\end{array}$$

Letting $\theta$ range from $\text{−}\infty$ to $\infty$ generates the entire spiral.

Symmetry in Polar Coordinates

When studying symmetry of functions in rectangular coordinates (i.e., in the form $y = f(x)),$ we talk about symmetry with respect to the *y*-axis and symmetry with respect to the origin. In particular, if $f\left( {\text{−}x} \right) = f(x)$ for all $x$ in the domain of $f,$ then $f$ is an even function and its graph is symmetric with respect to the *y*-axis. If $f\left( {\text{−}x} \right) = \text{−}f(x)$ for all $x$ in the domain of $f,$ then $f$ is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.

Symmetry in Polar Curves and Equations

Consider a curve generated by the function $r = f(\theta)$ in polar coordinates.

1. The curve is symmetric about the polar axis if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\text{−}\theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged by replacing $\theta$ with $\text{−}\theta.$

2. The curve is symmetric about the pole if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\pi + \theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged when replacing $r$ with $\text{−}r,$ or $\theta$ with $\pi + \theta.$

3. The curve is symmetric about the vertical line $\theta = \frac{\pi}{2}$ if for every point $\left( {r,\theta} \right)$ on the graph, the point $\left( {r,\pi - \theta} \right)$ is also on the graph. Similarly, the equation $r = f(\theta)$ is unchanged when $\theta$ is replaced by $\pi - \theta.$

The following table shows examples of each type of symmetry.

Using Symmetry to Graph a Polar Equation

Find the symmetry of the rose defined by the equation $r = 3\ \text{sin}\left( {2\theta} \right)$ and create a graph.

Solution

Suppose the point $\left( {r,\theta} \right)$ is on the graph of $r = 3\ \text{sin}\left( {2\theta} \right).$

1. To test for symmetry about the polar axis, first try replacing $\theta$ with $\text{−}\theta.$ This gives $r = 3\ \text{sin}\left( {2\left( {\text{−}\theta} \right)} \right) = -3\ \text{sin}\left( {2\theta} \right).$ Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing $r$ with $\text{−}r$ and $\theta$ with $\pi - \theta$ yields

$$\begin{array}{l}

{\text{−}r = 3\ \text{sin}\left( {2\left( {\pi - \theta} \right)} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {2\pi - 2\theta} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {-2\theta} \right)} \\

{\text{−}r = -3\ \text{sin}\ 2\theta.}

\end{array}$$

Multiplying both sides of this equation by $-1$ gives $r = 3\ \text{sin}\ 2\theta,$ which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.

2. To test for symmetry with respect to the pole, first replace $r$ with $\text{−}r,$ which yields $\text{−}r = 3\ \text{sin}\left( {2\theta} \right).$ Multiplying both sides by −1 gives $r = -3\ \text{sin}\left( {2\theta} \right),$ which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing $\theta$ with $\theta + \pi$ gives

$$\begin{array}{cl}

r & {= 3\ \text{sin}\left( {2\left( {\theta + \pi} \right)} \right)} \\

& {= 3\ \text{sin}\left( {2\theta + 2\pi} \right)} \\

& {= 3\left( {\text{sin}\ 2\theta\ \text{cos}\ 2\pi + \text{cos}\ 2\theta\ \text{sin}\ 2\pi} \right)} \\

& {= 3\ \text{sin}\ 2\theta.}

\end{array}$$

Since this agrees with the original equation, the graph is symmetric about the pole.

3. To test for symmetry with respect to the vertical line $\theta = \frac{\pi}{2},$ first replace both $r$ with $\text{−}r$ and $\theta$ with $\text{−}\theta.$

$$\begin{array}{l}

{\text{−}r = 3\ \text{sin}\left( {2\left( {\text{−}\theta} \right)} \right)} \\

{\text{−}r = 3\ \text{sin}\left( {-2\theta} \right)} \\

{\text{−}r = -3\ \text{sin}\ 2\theta.}

\end{array}$$

Multiplying both sides of this equation by $-1$ gives $r = 3\ \text{sin}\ 2\theta,$ which is the original equation. Therefore the graph is symmetric about the vertical line $\theta = \frac{\pi}{2}.$

This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of $\theta$ between 0 and $\pi\text{/}2$ and then reflect the resulting graph.

| $\theta$ | $r$ |

|-----------------|-----------------------------------|

| $0$ | $0$ |

| $\frac{\pi}{6}$ | $\frac{3\sqrt{3}}{2} \approx 2.6$ |

| $\frac{\pi}{4}$ | $3$ |

| $\frac{\pi}{3}$ | $\frac{3\sqrt{3}}{2} \approx 2.6$ |

| $\frac{\pi}{2}$ | $0$ |

This gives one petal of the rose, as shown in the following graph.

Reflecting this image into the other three quadrants gives the entire graph as shown.

Determine the symmetry of the graph determined by the equation $r = 2\ \text{cos}\left( {3\theta} \right)$ and create a graph.

Section 1.3 Exercises

In the following exercises, plot the point whose polar coordinates are given by first constructing the angle $\theta$ and then marking off the distance *r* along the ray.

125.

$\left( {3,\frac{\pi}{6}} \right)$

126\.

$\left( {-2,\frac{5\pi}{3}} \right)$

127.

$\left( {0,\frac{7\pi}{6}} \right)$

128\.

$\left( {-4,\frac{3\pi}{4}} \right)$

129.

$\left( {1,\frac{\pi}{4}} \right)$

130\.

$\left( {2,\frac{5\pi}{6}} \right)$

131.

$\left( {1,\frac{\pi}{2}} \right)$

For the following exercises, consider the polar graph below. Give two sets of polar coordinates for each point.

132\.

Coordinates of point *A*.

133.

Coordinates of point *B*.

134\.

Coordinates of point *C*.

135.

Coordinates of point *D*.

For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the point in $\left( {0,2\pi} \right\rbrack.$ Round to three decimal places.

136\.

$\left( {2,\ 2} \right)$

137.

$\left( {3,-4} \right)$

138\.

$\left( {8,\ 15} \right)$

139.

$\left( {-6,\ 8} \right)$

140\.

$\left( {4,\ 3} \right)$

141.

$\left( {3,\text{−}\sqrt{3}} \right)$

For the following exercises, find rectangular coordinates for the given point in polar coordinates.

142\.

$\left( {2,\frac{5\pi}{4}} \right)$

143.

$\left( {-2,\frac{\pi}{6}} \right)$

144\.

$\left( {5,\frac{\pi}{3}} \right)$

145.

$\left( {1,\frac{7\pi}{6}} \right)$

146\.

$\left( {-3,\frac{3\pi}{4}} \right)$

147.

$\left( {0,\frac{\pi}{2}} \right)$

148\.

$\left( {-4.5,6.5} \right)$

For the following exercises, determine whether the graphs of the polar equation are symmetric with respect to the $x$-axis, the $y$-axis, or the origin.

149.

$r = 3\ \text{sin}(2\theta)$

150\.

$r^{2} = 9\ \text{cos}\ \theta$

151.

$r = \text{cos}\left( \frac{\theta}{5} \right)$

152\.

$r = 2\ \text{sec}\ \theta$

153.

$r = 1 + \text{cos}\ \theta$

For the following exercises, describe the graph of each polar equation. Confirm each description by converting into a rectangular equation.

154\.

$r = 3$

155.

$\theta = \frac{\pi}{4}$

156\.

$r = \text{sec}\ \theta$

157.

$r = \text{csc}\ \theta$

For the following exercises, convert the rectangular equation to polar form and sketch its graph.

158\.

$x^{2} + y^{2} = 16$

159.

$x^{2} - y^{2} = 16$

160\.

$x = 8$

For the following exercises, convert the rectangular equation to polar form and sketch its graph.

161.

$3x - y = 2$

162\.

$y^{2} = 4x$

For the following exercises, convert the polar equation to rectangular form and sketch its graph.

163.

$r = 4\ \text{sin}\ \theta$

164\.

$r = 6\ \text{cos}\ \theta$

165.

$r = \theta$

166\.

$r = \text{cot}\ \theta\ \text{csc}\ \theta$

For the following exercises, sketch a graph of the polar equation and identify any symmetry.

167.

$r = 1 + \text{sin}\ \theta$

168\.

$r = 3 - 2\ \text{cos}\ \theta$

169.

$r = 2 - 2\ \text{sin}\ \theta$

170\.

$r = 5 - 4\ \text{sin}\ \theta$

171.

$r = 3\ \text{cos}\left( {2\theta} \right)$

172\.

$r = 3\ \text{sin}\left( {2\theta} \right)$

173.

$r = 2\ \text{cos}\left( {3\theta} \right)$

174\.

$r = 3\ \text{cos}\left( \frac{\theta}{2} \right)$

175.

$r^{2} = 4\ \text{cos}\left( {2\theta} \right)$

176\.

$r^{2} = 4\ \text{sin}\ \theta$

177.

$r = 2\theta$

178\.

\[T\] The graph of $r = 2\ \text{cos}(2\theta)\text{sec}(\theta).$ is called a *strophoid.* Use a graphing utility to sketch the graph, and, from the graph, determine the asymptote.

179.

\[T\] Use a graphing utility and sketch the graph of $r = \frac{6}{2\ \text{sin}\ \theta - 3\ \text{cos}\ \theta}.$

180\.

\[T\] Use a graphing utility to graph $r = \frac{1}{1 - \text{cos}\ \theta}.$

181.

\[T\] Use technology to graph $r = e^{\text{sin}(\theta)} - 2\ \text{cos}\left( {4\theta} \right).$

182\.

\[T\] Use technology to plot $r = \text{sin}\left( \frac{3\theta}{7} \right)$ (use the interval $0 \leq \theta \leq 14\pi).$

183.

Without using technology, sketch the polar curve $\theta = \frac{2\pi}{3}.$

184\.

\[T\] Use a graphing utility to plot $r = \theta\ \text{sin}\ \theta$ for $\text{−}\pi \leq \theta \leq \pi.$

185.

\[T\] Use technology to plot $r = e^{-0.1\theta}$ for $-10 \leq \theta \leq 10.$

186\.

\[T\] There is a curve known as the “*Black Hole*.” Use technology to plot $r = e^{-0.01\theta}$ for $-100 \leq \theta \leq 100.$

187.

\[T\] Use the results of the preceding two problems to explore the graphs of $r = e^{-0.001\theta}$ and $r = e^{-0.0001\theta}$ for $|\theta| > 100.$

---

1.4 Area and Arc Length in Polar Coordinates

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-4-area-and-arc-length-in-polar-coordinates

1.4 Area and Arc Length in Polar Coordinates

1.4.1 Apply the formula for area of a region in polar coordinates.
1.4.2 Determine the arc length of a polar curve.

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function $y = f(x)$ defined from $x = a$ to $x = b$ where $f(x) > 0$ on this interval, the area between the curve and the *x*-axis is given by $A = {\int_{a}^{b}{f(x)\ dx}}.$ This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by $L = {\int_{a}^{b}{\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}dx}}.$ In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ Our first step is to partition the interval $\lbrack\alpha,\beta\rbrack$ into *n* equal-width subintervals. The width of each subinterval is given by the formula $\text{Δ}\theta = {{(\beta - \alpha)}\text{/}n},$ and the *i*th partition point $\theta_{i}$ is given by the formula $\theta_{i} = \alpha + i\text{Δ}\theta.$ Each partition point $\theta = \theta_{i}$ defines a line with slope $\text{tan}\theta_{i}$ passing through the pole as shown in the following graph.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

Recall that the area of a circle is $A = \pi r^{2}.$ When measuring angles in radians, 360 degrees is equal to $2\pi$ radians. Therefore a fraction of a circle can be measured by the central angle $\theta.$ The fraction of the circle is given by $\frac{\theta}{2\pi},$ so the area of the sector is this fraction multiplied by the total area:

$$A = \left( \frac{\theta}{2\pi} \right)\ \pi r^{2} = \frac{1}{2}\theta r^{2}.$$

Since the radius of a typical sector in Figure 1.39 is given by $r_{i} = f\left( \theta_{i} \right),$ the area of the *i*th sector is given by

$$A_{i} = \frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}.$$

Therefore a Riemann sum that approximates the area is given by

$$A_{n} = {\sum\limits_{i = 1}^{n}A_{i}} \approx {\sum\limits_{i = 1}^{n}{\frac{1}{2}\left( {\text{Δ}\theta} \right)\left( {f\left( \theta_{i} \right)} \right)^{2}}}.$$

We take the limit as $n\rightarrow\infty$ to get the exact area:

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = \frac{1}{2}{\int_{\alpha}^{\beta}{\left( {f(\theta)} \right)^{2}d\theta}}.$$

This gives the following theorem.

Area of a Region Bounded by a Polar Curve

Suppose $f$ is continuous and nonnegative on the interval $\alpha \leq \theta \leq \beta$ with $0 < \beta - \alpha \leq 2\pi.$ The area of the region bounded by the graph of $r = f(\theta)$ between the radial lines $\theta = \alpha$ and $\theta = \beta$ is

$$A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}.$$ (1.9)

Finding an Area of a Polar Region

Find the area of one petal of the rose defined by the equation $r = 3\ \text{sin}\left( {2\theta} \right).$

Solution

The graph of $r = 3\ \text{sin}\left( {2\theta} \right)$ follows.

When $\theta = 0$ we have $r = 3\ \text{sin}\left( {2(0)} \right) = 0.$ The next value for which $r = 0$ is $\theta = \pi\text{/}2.$ This can be seen by solving the equation $3\ \text{sin}(2\theta) = 0$ for $\theta.$ Therefore the values $\theta = 0$ to $\theta = \pi\text{/}2$ trace out the first petal of the rose. To find the area inside this petal, use Equation 1.9 with $f(\theta) = 3\ \text{sin}\left( {2\theta} \right),$ $\alpha = 0,$ and $\beta = \pi\text{/}2\text{:}$

$$\begin{array}{cl}

A & {= \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}}} \\

& {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{\left\lbrack {3\ \text{sin}\left( {2\theta} \right)} \right\rbrack^{2}d\theta}}} \\

& {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}.}

\end{array}$$

To evaluate this integral, use the formula $\text{sin}^{2}\alpha = \left( {1 - \text{cos}(2\alpha)} \right)\text{/}2$ with $\alpha = 2\theta\text{:}$

$$\begin{array}{cl}

A & {= \frac{1}{2}{\int_{0}^{\pi\text{/}2}{9\ \text{sin}^{2}\left( {2\theta} \right)\ d\theta}}} \\

& {= \frac{9}{2}{\int_{0}^{\pi\text{/}2}{\frac{\left( {1 - \text{cos}\left( {4\theta} \right)} \right)}{2}d\theta}}} \\

& {= \frac{9}{4}\left( {\int_{0}^{\pi\text{/}2}{1 - \text{cos}\left( {4\theta} \right)\ d\theta}} \right)} \\

& {= \frac{9}{4}\left( \left. {\theta - \frac{\text{sin}\left( {4\theta} \right)}{4}} \right) \right._{0}^{\pi\text{/}2}} \\

& {= \frac{9}{4}\left( {\frac{\pi}{2} - \frac{\text{sin}\ 2\pi}{4}} \right) - \frac{9}{4}\left( {0 - \frac{\text{sin}\ 4(0)}{4}} \right)} \\

& {= \frac{9\pi}{8}.}

\end{array}$$

Find the area inside the cardioid defined by the equation $r = 1 - \text{cos}\ \theta.$

Example 1.16 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Finding the Area between Two Polar Curves

Find the area outside the cardioid $r = 2 + 2\ \text{sin}\ \theta$ and inside the circle $r = 6\ \text{sin}\ \theta.$

Solution

First draw a graph containing both curves as shown.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for $\theta\text{:}$

$$\begin{array}{rll}

{6\ \text{sin}\ \theta} & = & {2 + 2\ \text{sin}\ \theta} \\

{4\ \text{sin}\ \theta} & = & 2 \\

{\text{sin}\ \theta} & = & {\frac{1}{2}.}

\end{array}$$

This gives the solutions $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ which are the limits of integration. The circle $r = 3\ \text{sin}\ \theta$ is the red graph, which is the outer function, and the cardioid $r = 2 + 2\ \text{sin}\ \theta$ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6},$ then subtract the area inside the cardioid between $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}\text{:}$

$$\begin{array}{cl}

A & {= \text{circle} - \text{cardioid}} \\

& {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {6\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack {2 + 2\ \text{sin}\ \theta} \right\rbrack^{2}d\theta}}} \\

& {= \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{36\ \text{sin}^{2}\theta\ d\theta}} - \frac{1}{2}{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 4 + 8\ \text{sin}\ \theta + 4\ \text{sin}^{2}\theta \right\rbrack\ d\theta}}} \\

& {= 18{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\frac{1 - \text{cos}\left( {2\theta} \right)}{2}d\theta}} - 2{\int_{\pi\text{/}6}^{{5\pi}\text{/}6}{\left\lbrack 1 + 2\ \text{sin}\ \theta + \frac{1 - \text{cos}\left( {2\theta} \right)}{2} \right\rbrack d\theta}}} \\

& {= 9\left\lbrack {\theta - \frac{\text{sin}\left( {2\theta} \right)}{2}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6} - 2\left\lbrack {\frac{3\theta}{2} - 2\ \text{cos}\ \theta - \frac{\text{sin}\left( {2\theta} \right)}{4}} \right\rbrack_{\pi\text{/}6}^{5\pi\text{/}6}} \\

& {= 9\left( {\frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) - 9\left( {\frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\

& {\mspace{7mu}\text{−}\left( {3\left( \frac{5\pi}{6} \right) - 4\ \text{cos}\ \frac{5\pi}{6} - \frac{\text{sin}\ 2(5\pi\text{/}6)}{2}} \right) + \left( {3\left( \frac{\pi}{6} \right) - 4\ \text{cos}\ \frac{\pi}{6} - \frac{\text{sin}\ 2(\pi\text{/}6)}{2}} \right)} \\

& {= 4\pi.}

\end{array}$$

Find the area inside the circle $r = 4\ \text{cos}\ \theta$ and outside the circle $r = 2.$

In Example 1.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $\theta$ yielded two solutions: $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}.$ However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of $\theta.$ For example, for the cardioid we get

$$\begin{array}{rll}

{2 + 2\ \text{sin}\ \theta} & = & 0 \\

{\text{sin}\ \theta} & = & {-1,}

\end{array}$$

so the values for $\theta$ that solve this equation are $\theta = \frac{3\pi}{2} + 2n\pi,$ where *n* is any integer. For the circle we get

$$6\ \text{sin}\ \theta = 0.$$

The solutions to this equation are of the form $\theta = n\pi$ for any integer value of *n.* These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc Length in Polar Curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve $\left( {x(t),y(t)} \right)$ for $a \leq t \leq b$ is given by

$$L = {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$$

In polar coordinates we define the curve by the equation $r = f(\theta),$ where $\alpha \leq \theta \leq \beta.$ In order to adapt the arc length formula for a polar curve, we use the equations

$$x = r\ \text{cos}\ \theta = f(\theta)\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta = f(\theta)\ \text{sin}\ \theta,$$

and we replace the parameter *t* by $\theta.$ Then

$$\begin{array}{l}

{\frac{dx}{d\theta} = f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \\

{\frac{dy}{d\theta} = f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta.}

\end{array}$$

We replace $dt$ by $d\theta,$ and the lower and upper limits of integration are $\alpha$ and $\beta,$ respectively. Then the arc length formula becomes

$$\begin{array}{cl}

L & {= {\int_{a}^{b}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}} \\

& {= {\int_{\alpha}^{\beta}{\sqrt{\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}}d\theta}}} \\

& {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)\ \text{cos}\ \theta - f(\theta)\ \text{sin}\ \theta} \right)^{2} + \left( {f^{\prime}(\theta)\ \text{sin}\ \theta + f(\theta)\ \text{cos}\ \theta} \right)^{2}}d\theta}}} \\

& {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) + \left( {f(\theta)} \right)^{2}\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right)}d\theta}}} \\

& {= {\int_{\alpha}^{\beta}{\sqrt{\left( {f^{\prime}(\theta)} \right)^{2} + \left( {f(\theta)} \right)^{2}}d\theta}}} \\

& {= {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}.}

\end{array}$$

This gives us the following theorem.

Arc Length of a Curve Defined by a Polar Function

Let $f$ be a function whose derivative is continuous on an interval $\alpha \leq \theta \leq \beta.$ The length of the graph of $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is

$$L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta.}}}}$$ (1.10)

Finding the Arc Length of a Polar Curve

Find the arc length of the cardioid $r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta.$

Solution

When $\theta = 0,r = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu} 0 = 4.$ Furthermore, as $\theta$ goes from $0$ to $2\textit{π}\text{,}$ the cardioid is traced out exactly once. Therefore these are the limits of integration. Using $f{(\theta)} = 2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta,$ $\alpha = 0,$ and $\beta = 2\textit{π}\text{,}$ Equation 1.10 becomes

$$\begin{array}{cl}

L & {= {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}\ d\theta} \\

& {= {\int_{0}^{2\pi}\sqrt{\left\lbrack {2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right\rbrack^{2} + \left\lbrack {- 2\mspace{2mu}\text{sin}\mspace{2mu}\theta} \right\rbrack^{2}}}\ d\theta} \\

& {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\mspace{2mu}\text{cos}^{2}\mspace{2mu}\theta + 4\mspace{2mu}\text{sin}^{2}\mspace{2mu}\theta}}\ d\theta} \\

& {= {\int_{0}^{2\pi}\sqrt{4 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta + 4\left( {\text{cos}^{2}\mspace{2mu}\theta + \text{sin}^{2}\mspace{2mu}\theta} \right)}}\ d\theta} \\

& {= {\int_{0}^{2\pi}\sqrt{8 + 8\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta} \\

& {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta}}\ d\theta.}

\end{array}$$

Next, using the identity $\text{cos}\left( {2\alpha} \right) = 2\mspace{2mu}\text{cos}^{2}\alpha - 1,$ add 1 to both sides and multiply by 2. This gives $2 + 2\mspace{2mu}\text{cos}\left( {2\alpha} \right) = 4\mspace{2mu}\text{cos}^{2}\alpha.$ Substituting $\alpha = {\theta\text{/}2}$ gives $2 + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta = 4\mspace{2mu}\text{cos}^{2}({\theta\text{/}{2),}}$ so the integral becomes

$$\begin{array}{cl}

L & {= 2{\int_{0}^{2\pi}\sqrt{2 + 2\ \text{cos}\ \theta}}d\theta} \\

& {= 2{\int_{0}^{2\pi}{\sqrt{4\ \text{cos}^{2}\left( \frac{\theta}{2} \right)}d\theta}}} \\

& {= 2{\int_{0}^{2\pi}{\left. 2 \middle| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}.}

\end{array}$$

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from $0$ to $\pi$ and double the answer. This strategy works because cosine is positive between $0$ and $\frac{\pi}{2}.$ Thus,

$$\begin{array}{cl}

L & {= 4{\int_{0}^{2\pi}{\left| {\text{cos}\left( \frac{\theta}{2} \right)} \right|d\theta}}} \\

& {= 8{\int_{0}^{\pi}{\text{cos}\left( \frac{\theta}{2} \right)\ d\theta}}} \\

& {= 8\left. \left( {2\ \text{sin}\left( \frac{\theta}{2} \right)} \right. \right)_{0}^{\pi}} \\

& {= 16.}

\end{array}$$

Find the total arc length of $r = 3\ \text{sin}\ \theta.$

Section 1.4 Exercises

For the following exercises, determine a definite integral that represents the area.

188\.

Region enclosed by $r = 4$

189.

Region enclosed by $r = 3\ \text{sin}\ \theta$

190\.

Region in the first quadrant within the cardioid $r = 1 + \text{sin}\ \theta$

191.

Region enclosed by one petal of $r = 8\ \text{sin}(2\theta)$

192\.

Region enclosed by one petal of $r = \text{cos}(3\theta)$

193.

Region below the polar axis and enclosed by $r = 1 - \text{sin}\ \theta$

194\.

Region in the first quadrant enclosed by $r = 2 - \text{cos}\ \theta$

195.

Region enclosed by the inner loop of $r = 2 - 3\ \text{sin}\ \theta$

196\.

Region enclosed by the inner loop of $r = 3 - 4\ \text{cos}\ \theta$

197.

Region enclosed by $r = 1 - 2\ \text{cos}\ \theta$ and outside the inner loop

198\.

Region common to $r = 3\ \text{sin}\ \theta\ \text{and}\ r = 2 - \text{sin}\ \theta$

199.

Region common to $r = 2\ \text{and}\ r = 4\ \text{cos}\ \theta$

200\.

Region common to $r = 3\ \text{cos}\ \theta\ \text{and}\ r = 3\ \text{sin}\ \theta$

For the following exercises, find the area of the described region.

201.

Enclosed by $r = 6\ \text{sin}\ \theta$

202\.

Above the polar axis enclosed by $r = 2 + \text{sin}\ \theta$

203.

Below the polar axis and enclosed by $r = 2 - \text{cos}\ \theta$

204\.

Enclosed by one petal of $r = 4\ \text{cos}\left( {3\theta} \right)$

205.

Enclosed by one petal of $r = 3\ \text{cos}\left( {2\theta} \right)$

206\.

Enclosed by $r = 1 + \text{sin}\ \theta$

207.

Enclosed by the inner loop of $r = 3 + 6\ \text{cos}\ \theta$

208\.

Enclosed by $r = 2 + 4\ \text{cos}\ \theta$ and outside the inner loop

209.

Common interior of $r = 4\ \text{sin}\left( {2\theta} \right)\ \text{and}\ r = 2$

210\.

Common interior of $r = 3 - 2\ \text{sin}\ \theta\ \text{and}\ r = -3 + 2\ \text{sin}\ \theta$

211.

Common interior of $r = 6\ \text{sin}\ \theta\ \text{and}\ r = 3$

212\.

Inside $r = 1 + \text{cos}\ \theta$ and outside $r = \text{cos}\ \theta$

213.

Common interior of $r = 2 + 2\ \text{cos}\ \theta\ \text{and}\ r = 2\ \text{sin}\ \theta$

For the following exercises, find a definite integral that represents the arc length.

214\.

$r = 4\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

215.

$r = 1 + \text{sin}\ \theta$ on the interval $0 \leq \theta \leq 2\pi$

216\.

$r = 2\ \text{sec}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{3}$

217.

$r = e^{\theta}\text{on the interval}\ 0 \leq \theta \leq 1$

For the following exercises, find the length of the curve over the given interval.

218\.

$r = 6\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

219.

$r = e^{3\theta}\text{on the interval}\ 0 \leq \theta \leq 2$

220\.

$r = 6\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

221.

$r = 8 + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

222\.

$r = 1 - \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq 2\pi$

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

223.

\[T\] $r = 3\theta\ \text{on the interval}\ 0 \leq \theta \leq \frac{\pi}{2}$

224\.

\[T\] $r = \frac{2}{\theta}\ \text{on the interval}\ \pi \leq \theta \leq 2\pi$

225.

\[T\] $r = \text{sin}^{2}\left( \frac{\theta}{2} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

226\.

\[T\] $r = 2\theta^{2}\ \text{on the interval}\ 0 \leq \theta \leq \pi$

227.

\[T\] $r = \text{sin}\left( {3\ \text{cos}\ \theta} \right)\ \text{on the interval}\ 0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

228\.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

229.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

230\.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

231.

$r = 3\ \text{sin}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

232\.

$r = \text{sin}\ \theta + \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

233.

$r = 6\ \text{sin}\ \theta + 8\ \text{cos}\ \theta\ \text{on the interval}\ 0 \leq \theta \leq \pi$

234\.

Verify that if $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta$ then $\frac{dy}{d\theta} = f\prime(\theta)\text{sin}\ \theta + f(\theta)\text{cos}\ \theta.$

For the following exercises, find the slope of a tangent line to a polar curve $r = f(\theta).$ Let $x = r\ \text{cos}\ \theta = f(\theta)\text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta = f(\theta)\text{sin}\ \theta,$ so the polar equation $r = f(\theta)$ is now written in parametric form.

235.

Use the definition of the derivative $\frac{dy}{dx} = \frac{{dy}\text{/}{d\theta}}{{dx}\text{/}{d\theta}}$ and the product rule to derive the derivative of a polar equation.

236\.

$r = 1 - \text{sin}\ \theta;$ $\left( {\frac{1}{2},\frac{\pi}{6}} \right)$

237.

$r = 4\ \text{cos}\ \theta;$ $\left( {2,\frac{\pi}{3}} \right)$

238\.

$r = 8\ \text{sin}\ \theta;$ $\left( {4,\frac{5\pi}{6}} \right)$

239.

$r = 4 + \text{sin}\ \theta;$ $\left( {3,\frac{3\pi}{2}} \right)$

240\.

$r = 6 + 3\ \text{cos}\ \theta;$ $\left( {3,\pi} \right)$

241.

$r = 4\ \text{cos}\left( {2\theta} \right);$ tips of the leaves

242\.

$r = 2\ \text{sin}\left( {3\theta} \right);$ tips of the leaves

243.

$r = 2\theta;$ $\left( {\frac{\pi}{2},\frac{\pi}{4}} \right)$

244\.

Find the points on the interval $\text{−}\pi \leq \theta \leq \pi$ at which the cardioid $r = 1 - \text{cos}\ \theta$ has a vertical or horizontal tangent line.

245.

For the cardioid $r = 1 + \text{sin}\ \theta,$ find the slope of the tangent line when $\theta = \frac{\pi}{3}.$

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of $\theta.$

246\.

$r = 3\ \text{cos}\ \theta,\theta = \frac{\pi}{3}$

247.

$r = \theta,$ $\theta = \frac{\pi}{2}$

248\.

$r = \text{ln}\ \theta,$ $\theta = e$

249.

\[T\] Use technology: $r = 2 + 4\ \text{cos}\ \theta$ at $\theta = \frac{\pi}{6}$

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

250\.

$r = 4\ \text{cos}\ \theta$

251.

$r^{2} = 4\ \text{cos}\left( {2\theta} \right)$

252\.

$r = 2\ \text{sin}(2\theta)$

253.

The cardioid $r = 1 + \text{sin}\ \theta$

254\.

Show that the curve $r = \text{sin}\ \theta\ \text{tan}\ \theta$ (called a *cissoid of Diocles*) has the line $x = 1$ as a vertical asymptote.

---

1.5 Conic Sections

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/1-5-conic-sections

1.5 Conic Sections

1.5.1 Identify the equation of a parabola in standard form with given focus and directrix.
1.5.2 Identify the equation of an ellipse in standard form with given foci.
1.5.3 Identify the equation of a hyperbola in standard form with given foci.
1.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
1.5.5 Write the polar equation of a conic section with eccentricity $e$.
1.5.6 Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the *y*-axis as shown.

Conic sections are generated by the intersection of a plane with a cone (Figure 1.44). If the plane intersects both nappes, then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than $90\text{°}),$ then the conic section is an ellipse.

Parabolas

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

A parabola is the set of all points whose distance from a fixed point, called the focus, is equal to the distance from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.

A graph of a typical parabola appears in Figure 1.45. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point *P* with coordinates $\left( {x_{1},y_{1}} \right)$ and point *Q* with coordinates $\left( {x_{2},{\ \text{y}}_{2}} \right),$ the distance between them is given by the formula

$$d\left( {P,Q} \right) = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2}}.$$

Then from the definition of a parabola and Figure 1.45, we get

$$\begin{array}{rll}

{d\left( {F,P} \right)} & = & {d\left( {P,Q} \right)} \\

\sqrt{\left( {0 - x} \right)^{2} + \left( {p - y} \right)^{2}} & = & {\sqrt{\left( {x - x} \right)^{2} + \left( {\text{−}p - y} \right)^{2}}.}

\end{array}$$

Squaring both sides and simplifying yields

$$\begin{array}{rll}

{x^{2} + \left( {p - y} \right)^{2}} & = & {0^{2} + \left( {\text{−}p - y} \right)^{2}} \\

{x^{2} + p^{2} - 2py + y^{2}} & = & {p^{2} + 2py + y^{2}} \\

{x^{2} - 2py} & = & {2py} \\

x^{2} & = & {4py.}

\end{array}$$

Now suppose we want to relocate the vertex. We use the variables $\left( {h,k} \right)$ to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates $\left( {h,k + p} \right)$ and the directrix has the equation $y = k - p.$ Going through the same derivation yields the formula $\left( {x - h} \right)^{2} = 4p\left( {y - k} \right).$ Solving this equation for *y* leads to the following theorem.

Equations for Parabolas

Given a parabola opening upward with vertex located at $\left( {h,k} \right)$ and focus located at $\left( {h,k + p} \right),$ where *p* is a constant, the equation for the parabola is given by

$$y = \frac{1}{4p}\left( {x - h} \right)^{2} + k.$$ (1.11)

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

In addition, the equation of a parabola can be written in the general form, though in this form the values of *h*, *k*, and *p* are not immediately recognizable. The general form of a parabola is written as

$$ax^{2} + bx + cy + d = 0\quad\text{or}\quad ay^{2} + bx + cy + d = 0.$$

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Converting the Equation of a Parabola from General into Standard Form

Put the equation $x^{2} - 4x - 8y + 12 = 0$ into standard form and graph the resulting parabola.

Solution

Since *y* is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for *y,* which will put the equation into standard form. To do that, first add $8y$ to both sides of the equation:

$$8y = x^{2} - 4x + 12.$$

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

$$8y = \left( {x^{2} - 4x} \right) + 12.$$

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

$$8y = \left( {x^{2} - 4x + 4} \right) + 12 - 4.$$

Now combine like terms and factor the quantity inside the parentheses:

$$8y = \left( {x - 2} \right)^{2} + 8.$$

Finally, divide by 8:

$$y = \frac{1}{8}\left( {x - 2} \right)^{2} + 1.$$

This equation is now in standard form. Comparing this to Equation 1.11 gives $h = 2,$ $k = 1,$ and $p = 2.$ The parabola opens up, with vertex at $\left( {2,1} \right),$ focus at $\left( {2,3} \right),$ and directrix $y = -1.$ The graph of this parabola appears as follows.

Put the equation $2y^{2} - x + 12y + 16 = 0$ into standard form and graph the resulting parabola.

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

Ellipses

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

An *ellipse* is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

A graph of a typical ellipse is shown in Figure 1.48. In this figure the foci are labeled as $F$ and $F^{\prime}.$ Both are the same fixed distance from the origin, and this distance is represented by the variable *c*. Therefore the coordinates of $F$ are $\left( {c,0} \right)$ and the coordinates of $F^{\prime}$ are $\left( {\text{−}c,0} \right).$ The points $P$ and $P^{\prime}$ are located at the ends of the major axis of the ellipse, and have coordinates $\left( {a,0} \right)$ and $\left( {\text{−}a,0} \right),$ respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2*a.* Furthermore, $P$ and $P^{\prime}$ are called the vertices of the ellipse. The points $Q$ and $Q^{\prime}$ are located at the ends of the minor axis of the ellipse, and have coordinates $\left( {0,b} \right)$ and $\left( {0,\text{−}b} \right),$ respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point *P.* Since the coordinates of point *P* are $\left( {a,0} \right),$ the sum of the distances is

$$d\left( {P,F} \right) + d\left( {P,F^{\prime}} \right) = \left( {a - c} \right) + \left( {a + c} \right) = 2a.$$

Therefore the sum of the distances from an arbitrary point *A* with coordinates $\left( {x,y} \right)$ is also equal to 2*a.* Using the distance formula, we get

$$\begin{array}{rll}

{d\left( {A,F} \right) + d\left( {A,F^{\prime}} \right)} & = & {2a} \\

{\sqrt{\left( {x - c} \right)^{2} + y^{2}} + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {2a.}

\end{array}$$

Subtract the second radical from both sides and square both sides:

$$\begin{array}{rll}

\sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a - \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\

{\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\

{x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\

{\text{−}2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.}

\end{array}$$

Now isolate the radical on the right-hand side and square again:

$$\begin{array}{rll}

{- 2cx} & = & {4a^{2} - 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\

{4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {4a^{2} + 4cx} \\

\sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {a + \frac{cx}{a}} \\

{\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\

{x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\

{x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.}

\end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$$\begin{array}{rll}

& & \\

{x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\

{\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.}

\end{array}$$

Divide both sides by $a^{2} - c^{2}.$ This gives the equation

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

If we refer back to Figure 1.48, then the length of each of the two green line segments is equal to *a*. This is true because the sum of the distances from the point *Q* to the foci $F\ \text{and}\ F^{\prime}$ is equal to 2*a*, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length *a* and leg lengths *b* and *c*. From the Pythagorean theorem, $c^{2} = b^{2} = a^{2}$ and $b^{2} + a^{2} = c^{2}.$ Therefore the equation of the ellipse becomes

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the ellipse is moved from the origin to a point $\left( {h,k} \right),$ we have the following standard form of an ellipse.

Equation of an Ellipse in Standard Form

Consider the ellipse with center $\left( {h,k} \right),$ a horizontal major axis with length 2*a*, and a vertical minor axis with length 2*b*. Then the equation of this ellipse in standard form is

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (1.12)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices are $x = h \pm \frac{a^{2}}{c}.$

If the major axis is vertical, then the equation of the ellipse becomes

$$\frac{\left( {x - h} \right)^{2}}{b^{2}} + \frac{\left( {y - k} \right)^{2}}{a^{2}} = 1$$ (1.13)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} - b^{2}.$ The equations of the directrices in this case are $y = k \pm \frac{a^{2}}{c}.$

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of an Ellipse

Put the equation $9x^{2} + 4y^{2} - 36x + 24y + 36 = 0$ into standard form and graph the resulting ellipse.

Solution

First subtract 36 from both sides of the equation:

$$9x^{2} + 4y^{2} - 36x + 24y = -36.$$

Next group the *x* terms together and the *y* terms together, and factor out the common factor:

$$\begin{array}{rll}

{\left( {9x^{2} - 36x} \right) + \left( {4y^{2} + 24y} \right)} & = & -36 \\

{9\left( {x^{2} - 4x} \right) + 4\left( {y^{2} + 6y} \right)} & = & -36.

\end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{-4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{6}{2} \right)^{2} = 9.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

$$\begin{array}{l}

{9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = -36 + 36 + 36} \\

{9\left( {x^{2} - 4x + 4} \right) + 4\left( {y^{2} + 6y + 9} \right) = 36.}

\end{array}$$

Now factor both sets of parentheses and divide by 36:

$$\begin{array}{rll}

& & \\

{9\left( {x - 2} \right)^{2} + 4\left( {y + 3} \right)^{2}} & = & 36 \\

{\frac{9\left( {x - 2} \right)^{2}}{36} + \frac{4\left( {y + 3} \right)^{2}}{36}} & = & 1 \\

{\frac{\left( {x - 2} \right)^{2}}{4} + \frac{\left( {y + 3} \right)^{2}}{9}} & = & 1.

\end{array}$$

The equation is now in standard form. Comparing this to Equation 1.14 gives $h = 2,$ $k = -3,$ $a = 3,$ and $b = 2.$ This is a vertical ellipse with center at $\left( {2,-3} \right),$ major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

Put the equation $9x^{2} + 16y^{2} + 18x - 64y - 71 = 0$ into standard form and graph the resulting ellipse.

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure 1.50(a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 1.50(b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

Hyperbolas

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears as follows.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let *P* be a point on the hyperbola with coordinates $\left( {x,y} \right).$ Then the definition of the hyperbola gives $\left| {d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} \right| = \text{constant}.$ To simplify the derivation, assume that *P* is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $\left( {a,0} \right),$ so

$$d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right) = \left( {c + a} \right) - \left( {c - a} \right) = 2a.$$

This equation is therefore true for any point on the hyperbola. Returning to the coordinates $\left( {x,y} \right)$ for *P*:

$$\begin{array}{rll}

{d\left( {P,F_{1}} \right) - d\left( {P,F_{2}} \right)} & = & {2a} \\

{\sqrt{\left( {x + c} \right)^{2} + y^{2}} - \sqrt{\left( {x - c} \right)^{2} + y^{2}}} & = & {2a.}

\end{array}$$

Add the second radical from both sides and square both sides:

$$\begin{array}{rll}

\sqrt{\left( {x - c} \right)^{2} + y^{2}} & = & {2a + \sqrt{\left( {x + c} \right)^{2} + y^{2}}} \\

{\left( {x - c} \right)^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + \left( {x + c} \right)^{2} + y^{2}} \\

{x^{2} - 2cx + c^{2} + y^{2}} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + x^{2} + 2cx + c^{2} + y^{2}} \\

{\text{−}2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx.}

\end{array}$$

Now isolate the radical on the right-hand side and square again:

$$\begin{array}{rll}

{- 2cx} & = & {4a^{2} + 4a\sqrt{\left( {x + c} \right)^{2} + y^{2}} + 2cx} \\

{4a\sqrt{\left( {x + c} \right)^{2} + y^{2}}} & = & {-4a^{2} - 4cx} \\

\sqrt{\left( {x + c} \right)^{2} + y^{2}} & = & {\text{−}a - \frac{cx}{a}} \\

{\left( {x + c} \right)^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\

{x^{2} + 2cx + c^{2} + y^{2}} & = & {a^{2} + 2cx + \frac{c^{2}x^{2}}{a^{2}}} \\

{x^{2} + c^{2} + y^{2}} & = & {a^{2} + \frac{c^{2}x^{2}}{a^{2}}.}

\end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$$\begin{array}{rll}

& & \\

{x^{2} - \frac{c^{2}x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}} \\

{\frac{\left( {a^{2} - c^{2}} \right)x^{2}}{a^{2}} + y^{2}} & = & {a^{2} - c^{2}.}

\end{array}$$

Finally, divide both sides by $a^{2} - c^{2}.$ This gives the equation

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1.$$

We now define *b* so that $b^{2} = c^{2} - a^{2}.$ This is possible because $c > a.$ Therefore the equation of the hyperbola becomes

$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$

Finally, if the center of the hyperbola is moved from the origin to the point $\left( {h,k} \right),$ we have the following standard form of a hyperbola.

Equation of a Hyperbola in Standard Form

Consider the hyperbola with center $\left( {h,k} \right),$ a horizontal major axis, and a vertical minor axis. Then the equation of this hyperbola is

$$\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$$ (1.14)

and the foci are located at $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{b}{a}\left( {x - h} \right).$ The equations of the directrices are

$$x = h \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = h \pm \frac{a^{2}}{c}.$$

If the major axis is vertical, then the equation of the hyperbola becomes

$$\frac{\left( {y - k} \right)^{2}}{a^{2}} - \frac{\left( {x - h} \right)^{2}}{b^{2}} = 1$$ (1.15)

and the foci are located at $\left( {h,k \pm c} \right),$ where $c^{2} = a^{2} + b^{2}.$ The equations of the asymptotes are given by $y = k \pm \frac{a}{b}\left( {x - h} \right).$ The equations of the directrices are

$$y = k \pm \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = k \pm \frac{a^{2}}{c}.$$

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form $Ax^{2} + By^{2} + Cx + Dy + E = 0,$ where *A* and *B* have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Finding the Standard Form of a Hyperbola

Put the equation $9x^{2} - 16y^{2} + 36x + 32y - 124 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Solution

First add 124 to both sides of the equation:

$$9x^{2} - 16y^{2} + 36x + 32y = 124.$$

Next group the *x* terms together and the *y* terms together, then factor out the common factors:

$$\begin{array}{rll}

{\left( {9x^{2} + 36x} \right) - \left( {16y^{2} - 32y} \right)} & = & 124 \\

{9\left( {x^{2} + 4x} \right) - 16\left( {y^{2} - 2y} \right)} & = & 124.

\end{array}$$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of *x* and square it. This gives $\left( \frac{4}{2} \right)^{2} = 4.$ In the second set of parentheses, take half the coefficient of *y* and square it. This gives $\left( \frac{-2}{2} \right)^{2} = 1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

$$\begin{array}{l}

{9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 124 + 36 - 16} \\

{9\left( {x^{2} + 4x + 4} \right) - 16\left( {y^{2} - 2y + 1} \right) = 144.}

\end{array}$$

Next factor both sets of parentheses and divide by 144:

$$\begin{array}{rll}

& & \\

{9\left( {x + 2} \right)^{2} - 16\left( {y - 1} \right)^{2}} & = & 144 \\

{\frac{9\left( {x + 2} \right)^{2}}{144} - \frac{16\left( {y - 1} \right)^{2}}{144}} & = & 1 \\

{\frac{\left( {x + 2} \right)^{2}}{16} - \frac{\left( {y - 1} \right)^{2}}{9}} & = & 1.

\end{array}$$

The equation is now in standard form. Comparing this to Equation 1.15 gives $h = -2,$ $k = 1,$ $a = 4,$ and $b = 3.$ This is a horizontal hyperbola with center at $\left( {-2,1} \right)$ and asymptotes given by the equations $y = 1 \pm \frac{3}{4}\left( {x + 2} \right).$ The graph of this hyperbola appears in the following figure.

Put the equation $4y^{2} - 9x^{2} + 16y + 18x - 29 = 0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

Eccentricity and Directrix

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

The eccentricity *e* of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

1. If $e = 1,$ the conic is a parabola.

2. If $e < 1,$ it is an ellipse.

3. If $e > 1,$ it is a hyperbola.

The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.

The three conic sections with their directrices appear in the following figure.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $x = \text{±}\frac{a^{2}}{c}.$ The right vertex of the ellipse is located at $\left( {a,0} \right)$ and the right focus is $\left( {c,0} \right).$ Therefore the distance from the vertex to the focus is $a - c$ and the distance from the vertex to the right directrix is $\frac{a^{2}}{c} - a.$ This gives the eccentricity as

$$e = \frac{a - c}{\frac{a^{2}}{c} - a} = \frac{c\left( {a - c} \right)}{a^{2} - ac} = \frac{c\left( {a - c} \right)}{a\left( {a - c} \right)} = \frac{c}{a}.$$

Since $c < a,$ this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at $x = \text{±}\frac{a^{2}}{c},$ and a similar calculation shows that the eccentricity of a hyperbola is also $e = \frac{c}{a}.$ However in this case we have $c > a,$ so the eccentricity of a hyperbola is greater than 1.

Determining Eccentricity of a Conic Section

Determine the eccentricity of the ellipse described by the equation

$$\frac{\left( {x - 3} \right)^{2}}{16} + \frac{\left( {y + 2} \right)^{2}}{25} = 1.$$

Solution

From the equation we see that $a = 5$ and $b = 4.$ The value of *c* can be calculated using the equation $a^{2} = b^{2} + c^{2}$ for an ellipse. Substituting the values of *a* and *b* and solving for *c* gives $c = 3.$ Therefore the eccentricity of the ellipse is $e = \frac{c}{a} = \frac{3}{5} = 0.6.$

Determine the eccentricity of the hyperbola described by the equation

$$\frac{\left( {y - 3} \right)^{2}}{49} - \frac{\left( {x + 2} \right)^{2}}{25} = 1.$$

Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section *p* is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where *a* is the length of the semi-major axis (i.e., half the length of the major axis), *c* is the distance from the origin to the focus, and *e* is the eccentricity. In the case of a parabola, *a* represents the distance from the vertex to the focus.

| Conic | *e* | *p* |

|-----------|-------------|-------------------------------------------------------------------|

| Ellipse | $0 < e < 1$ | $\frac{a^{2} - c^{2}}{c} = \frac{a\left( {1 - e^{2}} \right)}{e}$ |

| Parabola | $e = 1$ | $2a$ |

| Hyperbola | $e > 1$ | $\frac{c^{2} - a^{2}}{c} = \frac{a\left( {e^{2} - 1} \right)}{e}$ |

Table 1.1 Eccentricities and Focal Parameters of the Conic Sections

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Polar Equation of Conic Sections

The polar equation of a conic section with focal parameter *p* is given by

$$r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}\ \text{or}\ r = \frac{ep}{1 \pm e\ \text{sin}\ \theta}.$$

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., $e = 0).$

Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

$$r = \frac{3}{1 + 2\ \text{cos}\ \theta}.$$

Solution

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter *p* can be calculated by using the equation $ep = 3.$ Since $e = 2,$ this gives $p = \frac{3}{2}.$ The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $\theta$ and create a table of values. Then we can graph the hyperbola (Figure 1.55).

| $\theta$ | $r$ | $\theta$ | $r$ |

|------------------|------------------------------------------|------------------|------------------------------------------|

| 0 | 1 | $\pi$ | −3 |

| $\frac{\pi}{4}$ | $\frac{3}{1 + \sqrt{2}} \approx 1.2426$ | $\frac{5\pi}{4}$ | $\frac{3}{1 - \sqrt{2}} \approx -7.2426$ |

| $\frac{\pi}{2}$ | 3 | $\frac{3\pi}{2}$ | 3 |

| $\frac{3\pi}{4}$ | $\frac{3}{1 - \sqrt{2}} \approx -7.2426$ | $\frac{7\pi}{4}$ | $\frac{3}{1 + \sqrt{2}} \approx 1.2426$ |

Identify and create a graph of the conic section described by the equation

$$r = \frac{4}{1 - 0.8\ \text{sin}\ \theta}.$$

General Equations of Degree Two

A general equation of degree two can be written in the form

$$Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0.$$

The graph of an equation of this form is a conic section. If $B \neq 0$ then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section $4AC - B^{2}.$ One of the following cases must be true:

1. $4AC - B^{2} > 0.$ If so, the graph is an ellipse.

2. $4AC - B^{2} = 0.$ If so, the graph is a parabola.

3. $4AC - B^{2} < 0.$ If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is $xy = 1.$ This equation can be solved for *y* to obtain $y = \frac{1}{x}.$ The graph of this function is called a *rectangular hyperbola* as shown.

The asymptotes of this hyperbola are the *x* and *y* coordinate axes. To determine the angle $\theta$ of rotation of the conic section, we use the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$ In this case $A = C = 0$ and $B = 1,$ so $\text{cot}\ 2\theta = {{(0 - 0)}\text{/}{1 = 0}}$ and $\theta = 45\text{°}.$ The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime},$ and are given by the formulas

$$\begin{array}{rll}

A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\

B^{\prime} & = & 0 \\

C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\

D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\

E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\

F^{\prime} & = & {F.}

\end{array}$$

The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant $4AC - B^{2}.$

2. Determine $\theta$ using the formula $\text{cot}\ 2\theta = \frac{A - C}{B}.$

3. Calculate $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

4. Rewrite the original equation using $A^{\prime},B^{\prime},C^{\prime},D^{\prime},E^{\prime},\ \text{and}\ F^{\prime}.$

5. Draw a graph using the rotated equation.

Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$$13x^{2} - 6\sqrt{3}xy + 7y^{2} - 256 = 0.$$

Solution

In this equation, $A = 13,B = -6\sqrt{3},C = 7,D = 0,E = 0,$ and $F = -256.$ The discriminant of this equation is $4AC - B^{2} = 4(13)(7) - \left( {-6\sqrt{3}} \right)^{2} = 364 - 108 = 256.$ Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use $\text{cot}\ 2\theta = \frac{A - C}{B}.$ This gives

$$\begin{array}{cl}

{\text{cot}\ 2\theta} & {= \frac{A - C}{B}} \\

& {= \frac{13 - 7}{-6\sqrt{3}}} \\

& {= - \frac{\sqrt{3}}{3}.}

\end{array}$$

Therefore $2\theta = 120^{\text{o}}$ and $\theta = 60^{\text{o}},$ which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

$$\begin{array}{cll}

A^{\prime} & = & {A\ \text{cos}^{2}\theta + B\ \text{cos}\ \theta\ \text{sin}\ \theta + C\ \text{sin}^{2}\theta} \\

& = & {13\text{cos}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{cos}\ 60\ \text{sin}\ 60 + 7\text{sin}^{2}60} \\

& = & {13\left( \frac{1}{2} \right)^{2} - 6\sqrt{3}\left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right) + 7\left( \frac{\sqrt{3}}{2} \right)^{2}} \\

& = & 4, \\

B^{\prime} & = & {0,} \\

C^{\prime} & = & {A\ \text{sin}^{2}\theta - B\ \text{sin}\ \theta\ \text{cos}\ \theta + C\ \text{cos}^{2}\theta} \\

& = & {13\text{sin}^{2}60 + \left( {-6\sqrt{3}} \right)\ \text{sin}\ 60\ \text{cos}\ 60 = 7\text{cos}^{2}60} \\

& = & {\left( \frac{\sqrt{3}}{2} \right)^{2} + 6\sqrt{3}\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{2} \right) + 7\left( \frac{1}{2} \right)^{2}} \\

& = & 16, \\

D^{\prime} & = & {D\ \text{cos}\ \theta + E\ \text{sin}\ \theta} \\

& = & {(0)\ \text{cos}\ 60 + (0)\ \text{sin}\ 60} \\

& = & 0, \\

E^{\prime} & = & {\text{−}D\ \text{sin}\ \theta + E\ \text{cos}\ \theta} \\

& = & {\text{−}(0)\ \text{sin}\ 60 + (0)\ \text{cos}\ 60} \\

& = & 0, \\

F^{\prime} & = & F \\

& = & -256.

\end{array}$$

The equation of the conic in the rotated coordinate system becomes

$$\begin{array}{rll}

& & \\

{4\left( x^{\prime} \right)^{2} + 16\left( y^{\prime} \right)^{2}} & = & 256 \\

{\frac{\left( x^{\prime} \right)^{2}}{64} + \frac{\left( y^{\prime} \right)^{2}}{16}} & = & 1.

\end{array}$$

A graph of this conic section appears as follows.

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$$3x^{2} + 5xy - 2y^{2} - 125 = 0.$$

Section 1.5 Exercises

For the following exercises, determine the equation of the parabola using the information given.

255.

Focus $\left( {4,0} \right)$ and directrix $x = -4$

256\.

Focus $\left( {0,-3} \right)$ and directrix $y = 3$

257.

Focus $\left( {0,0.5} \right)$ and directrix $y = -0.5$

258\.

Focus $\left( {2,\ 3} \right)$ and directrix $x = -2$

259.

Focus $\left( {0,2} \right)$ and directrix $y = 4$

260\.

Focus $\left( {-1,4} \right)$ and directrix $x = 5$

261.

Focus $\left( {-3,5} \right)$ and directrix $y = 1$

262\.

Focus $\left( {\frac{5}{2},-4} \right)$ and directrix $x = \frac{7}{2}$

For the following exercises, determine the equation of the ellipse using the information given.

263.

Endpoints of major axis at $\left( {4,0} \right),\left( {-4,0} \right)$ and foci located at $\left( {2,0} \right),\left( {-2,0} \right)$

264\.

Endpoints of major axis at $\left( {0,5} \right),\left( {0,-5} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

265.

Endpoints of minor axis at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {3,0} \right),\left( {-3,0} \right)$

266\.

Endpoints of major axis at $\left( {-3,3} \right),\left( {7,3} \right)$ and foci located at $\left( {-2,3} \right),\left( {6,3} \right)$

267.

Endpoints of major axis at $\left( {-3,5} \right),\left( {-3,-3} \right)$ and foci located at $\left( {-3,3} \right),\left( {-3,-1} \right)$

268\.

Endpoints of minor axis at $\left( {0,0} \right),\left( {0,4} \right)$ and foci located at $\left( {5,2} \right),\left( {-5,2} \right)$

269.

Foci located at $\left( {2,0} \right),\ \left( {-2,0} \right)$ and eccentricity of $\frac{1}{2}$

270\.

Foci located at $\left( {0,-3} \right),\ \left( {0,3} \right)$ and eccentricity of $\frac{3}{4}$

For the following exercises, determine the equation of the hyperbola using the information given.

271.

Vertices located at $\left( {5,0} \right),\left( {-5,0} \right)$ and foci located at $\left( {6,0} \right),\left( {-6,0} \right)$

272\.

Vertices located at $\left( {0,2} \right),\left( {0,-2} \right)$ and foci located at $\left( {0,3} \right),\left( {0,-3} \right)$

273.

Endpoints of the conjugate axis located at $\left( {0,3} \right),\left( {0,-3} \right)$ and foci located $\left( {4,0} \right),\left( {-4,0} \right)$

274\.

Vertices located at $\left( {0,1} \right),\left( {6,1} \right)$ and focus located at $\left( {8,1} \right)$

275.

Vertices located at $\left( {-2,0} \right),\left( {-2,-4} \right)$ and focus located at $\left( {-2,-8} \right)$

276\.

Endpoints of the conjugate axis located at $\left( {3,2} \right),\left( {3,4} \right)$ and focus located at $\left( {3,7} \right)$

277.

Foci located at $( - 6,0),(6,0)$ and eccentricity of 3

278\.

$\left( {0,10} \right),\left( {0,-10} \right)$ and eccentricity of 2.5

For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.

279.

$r = \frac{-1}{1 + \text{cos}\ \theta}$

280\.

$r = \frac{8}{2 - \text{sin}\ \theta}$

281.

$r = \frac{5}{2 + \text{sin}\ \theta}$

282\.

$r = \frac{5}{-1 + 2\ \text{sin}\ \theta}$

283.

$r = \frac{3}{2 - 6\ \text{sin}\ \theta}$

284\.

$r = \frac{3}{-4 + 3\ \text{sin}\ \theta}$

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

285.

$\text{Directrix:}\ x = 4;e = \frac{1}{5}$

286\.

$\text{Directrix:}\ x = -4;e = 5$

287.

$\text{Directrix: y} = 2;e = 2$

288\.

$\text{Directrix: y} = -2;e = \frac{1}{2}$

For the following exercises, sketch the graph of each conic.

289.

$r = \frac{1}{1 + \text{sin}\ \theta}$

290\.

$r = \frac{1}{1 - \text{cos}\ \theta}$

291.

$r = \frac{4}{1 + \text{cos}\ \theta}$

292\.

$r = \frac{10}{5 + 4\ \text{sin}\ \theta}$

293.

$r = \frac{15}{3 - 2\ \text{cos}\ \theta}$

294\.

$r = \frac{32}{3 + 5\ \text{sin}\ \theta}$

295.

$r(2 + \text{sin}\ \theta) = 4$

296\.

$r = \frac{3}{2 + 6\ \text{sin}\ \theta}$

297.

$r = \frac{3}{-4 + 2\ \text{sin}\ \theta}$

298\.

$\begin{array}{l}

{\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1} \\

\end{array}$

299.

$\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1$

300\.

$4x^{2} + 9y^{2} = 36$

301.

$25x^{2} - 4y^{2} = 100$

302\.

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

303.

$x^{2} = 12y$

304\.

$y^{2} = 20x$

305.

$12x = 5y^{2}$

For the following equations, determine which of the conic sections is described.

306\.

$xy = 4$

307.

$x^{2} + 4xy - 2y^{2} - 6 = 0$

308\.

$x^{2} + 2\sqrt{3}xy + 3y^{2} - 6 = 0$

309.

$x^{2} - xy + y^{2} - 2 = 0$

310\.

$34x^{2} - 24xy + 41y^{2} - 25 = 0$

311.

$52x^{2} - 72xy + 73y^{2} + 40x + 30y - 75 = 0$

312\.

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as $x^{2} = 4y.$ At what coordinates should you place the lightbulb?

313.

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

314\.

Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

315.

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

316\.

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.

317.

A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the height at the center of the gallery?

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

318\.

Halley’s Comet: length of major axis = 35.88, eccentricity = 0.967

319.

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

320\.

Mars: length of major axis = 3.049, eccentricity = 0.0934

321.

Jupiter: length of major axis = 10.408, eccentricity = 0.0484

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Chapter Review

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Key Terms

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Key Terms

angular coordinate

$\theta$ the angle formed by a line segment connecting the origin to a point in the polar coordinate system with the positive radial (*x*) axis, measured counterclockwise

cardioid

a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is $r = a\left( {1 + \text{sin}\ \theta} \right)$ or $r = a\left( {1 + \text{cos}\ \theta} \right)$

conic section

a conic section is any curve formed by the intersection of a plane with a cone of two nappes

cusp

a pointed end or part where two curves meet

cycloid

the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage

directrix

a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two

discriminant

the value $4AC - B^{2},$ which is used to identify a conic when the equation contains a term involving $xy,$ is called a discriminant

eccentricity

the eccentricity is defined as the distance from any point on the conic section to its focus divided by the perpendicular distance from that point to the nearest directrix

focal parameter

the focal parameter is the distance from a focus of a conic section to the nearest directrix

focus

a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse and a hyperbola have two

general form

an equation of a conic section written as a general second-degree equation

limaçon

the graph of the equation $r = a + b\ \text{sin}\ \theta$ or $r = a + b\ \text{cos}\ \theta.$ If $a = b$ then the graph is a cardioid

major axis

the major axis of a conic section passes through the vertex in the case of a parabola or through the two vertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverse axis

minor axis

the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or at the vertex in the case of the parabola; also called the conjugate axis

nappe

a nappe is one half of a double cone

orientation

the direction that a point moves on a graph as the parameter increases

parameter

an independent variable that both *x* and *y* depend on in a parametric curve; usually represented by the variable *t*

parameterization of a curve

rewriting the equation of a curve defined by a function $y = f(x)$ as parametric equations

parametric curve

the graph of the parametric equations $x(t)$ and $y(t)$ over an interval $a \leq t \leq b$ combined with the equations

parametric equations

the equations $x = x(t)$ and $y = y(t)$ that define a parametric curve

polar axis

the horizontal axis in the polar coordinate system corresponding to $r \geq 0$

polar coordinate system

a system for locating points in the plane. The coordinates are $r,$ the radial coordinate, and $\theta,$ the angular coordinate

polar equation

an equation or function relating the radial coordinate to the angular coordinate in the polar coordinate system

pole

the central point of the polar coordinate system, equivalent to the origin of a Cartesian system

radial coordinate

$r$ the coordinate in the polar coordinate system that measures the distance from a point in the plane to the pole

rose

graph of the polar equation $r = a\ \text{cos}\ 2\theta$ or $r = a\ \text{sin}\ 2\theta$ for a positive constant *a*

space-filling curve

a curve that completely occupies a two-dimensional subset of the real plane

standard form

an equation of a conic section showing its properties, such as location of the vertex or lengths of major and minor axes

vertex

a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has two vertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch

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Book title: Calculus Volume 3
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Key Equations

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Key Equations

Calculus Volume 3Key Equations

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Key Equations

| | |

|-----------------------------------------------------|---------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Derivative of parametric equations | $\frac{dy}{dx} = \frac{{dy}\text{/}{dt}}{{dx}\text{/}{dt}} = \frac{y^{\prime}(t)}{x^{\prime}(t)}$ |

| Second-order derivative of parametric equations | $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{\left( {d\text{/}{dt}} \right)\left( {{dy}\text{/}{dx}} \right)}{{dx}\text{/}{dt}}$ |

| Area under a parametric curve | $A = {\int_{a}^{b}{y(t)x^{\prime}(t)\ dt}}$ |

| Arc length of a parametric curve | $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}$ |

| Surface area generated by a parametric curve | $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}$ |

| | |

|-----------------------------------------------|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Area of a region bounded by a polar curve | $A = \frac{1}{2}{\int_{\alpha}^{\beta}{\left\lbrack {f(\theta)} \right\rbrack^{2}d\theta}} = \frac{1}{2}{\int_{\alpha}^{\beta}{r^{2}d\theta}}$ |

| Arc length of a polar curve | $L = {\int_{\alpha}^{\beta}\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}$ |

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Authors: Gilbert Strang, Edwin “Jed” Herman
Publisher/website: OpenStax
Book title: Calculus Volume 3
Publication date: Mar 30, 2016
Location: Houston, Texas
Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
Section URL: https://openstax.org/books/calculus-volume-3/pages/1-key-equations

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Key Concepts

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Key Concepts

1.1 Parametric Equations

Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
There is always more than one way to parameterize a curve.
Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

1.2 Calculus of Parametric Curves

The derivative of the parametrically defined curve $x = x(t)$ and $y = y(t)$ can be calculated using the formula $\frac{dy}{dx} = \frac{y^{\prime}(t)}{x^{\prime}(t)}.$ Using the derivative, we can find the equation of a tangent line to a parametric curve.
The area between a parametric curve and the *x*-axis can be determined by using the formula $A = {\int_{t_{1}}^{t_{2}}{y(t)x^{\prime}(t)\ dt}}.$
The arc length of a parametric curve can be calculated by using the formula $s = {\int_{t_{1}}^{t_{2}}{\sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2}}dt}}.$
The surface area of a volume of revolution revolved around the *x*-axis is given by $S = 2\pi{\int_{a}^{b}{y(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$ If the curve is revolved around the *y*-axis, then the formula is $S = 2\pi{\int_{a}^{b}{x(t)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}dt}}.$

1.3 Polar Coordinates

The polar coordinate system provides an alternative way to locate points in the plane.
Convert points between rectangular and polar coordinates using the formulas

$$x = r\ \text{cos}\ \theta\ \text{and}\ y = r\ \text{sin}\ \theta$$

and

$$r = \sqrt{x^{2} + y^{2}}\ \text{and}\ \text{tan}\ \theta = \frac{y}{x}.$$

To sketch a polar curve from a given polar function, make a table of values and take advantage of periodic properties.
Use the conversion formulas to convert equations between rectangular and polar coordinates.
Identify symmetry in polar curves, which can occur through the pole, the horizontal axis, or the vertical axis.

1.4 Area and Arc Length in Polar Coordinates

The area of a region in polar coordinates defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $A = \frac{1}{2}{\int_{\alpha}^{\beta}\left\lbrack {f(\theta)} \right\rbrack}^{2}d\theta.$
To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
The arc length of a polar curve defined by the equation $r = f(\theta)$ with $\alpha \leq \theta \leq \beta$ is given by the integral $L = {\int_{\alpha}^{\beta}{\sqrt{\left\lbrack {f(\theta)} \right\rbrack^{2} + \left\lbrack {f^{\prime}(\theta)} \right\rbrack^{2}}d\theta = {\int_{\alpha}^{\beta}{\sqrt{r^{2} + \left( \frac{dr}{d\theta} \right)^{2}}d\theta}}}}.$

1.5 Conic Sections

The equation of a vertical parabola in standard form with given focus and directrix is $y = \frac{1}{4p}\left( {x - h} \right)^{2} + k$ where *p* is the distance from the vertex to the focus and $\left( {h,k} \right)$ are the coordinates of the vertex.
The equation of a horizontal ellipse in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} + \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the major axis has length 2*a,* the minor axis has length 2*b*, and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} - b^{2}.$
The equation of a horizontal hyperbola in standard form is $\frac{\left( {x - h} \right)^{2}}{a^{2}} - \frac{\left( {y - k} \right)^{2}}{b^{2}} = 1$ where the center has coordinates $\left( {h,k} \right),$ the vertices are located at $\left( {h \pm a,k} \right),$ and the coordinates of the foci are $\left( {h \pm c,k} \right),$ where $c^{2} = a^{2} + b^{2}.$
The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
The polar equation of a conic section with eccentricity *e* is $r = \frac{ep}{1 \pm e\ \text{cos}\ \theta}$ or $r = \frac{ep}{1 \pm e\ \text{sin}\ \theta},$ where *p* represents the focal parameter.
To identify a conic generated by the equation $Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0,$ first calculate the discriminant $D = 4AC - B^{2}.$ If $D > 0$ then the conic is an ellipse, if $D = 0$ then the conic is a parabola, and if $D < 0$ then the conic is a hyperbola.

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Book title: Calculus Volume 3
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Location: Houston, Texas
Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
Section URL: https://openstax.org/books/calculus-volume-3/pages/1-key-concepts

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Review Exercises

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Review Exercises

*True or False?* Justify your answer with a proof or a counterexample.

322\.

The rectangular coordinates of the point $\left( {4,\frac{5\pi}{6}} \right)$ are $\left( {2\sqrt{3},-2} \right).$

323.

The equations $x = \text{cosh}(3t),$ $y = 2\ \text{sinh}(3t)$ represent a hyperbola.

324\.

The arc length of the spiral given by $r = \frac{\theta}{2}$ for $0 \leq \theta \leq 3\pi$ is $\frac{9}{4}\pi^{3}.$

325.

Given $x = f(t)$ and $y = g(t),$ if $\frac{dx}{dy} = \frac{dy}{dx},$ then $f(t) = g(t) + \text{C,}$ where C is a constant.

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.

326\.

$x = 1 + t,$ $y = t^{2} - 1,$ $-1 \leq t \leq 1$

327.

$x = e^{t},$ $y = 1 - e^{3t},$ $0 \leq t \leq 1$

328\.

$x = \text{sin}\ \theta,$ $y = 1 - \text{csc}\ \theta,$ $0 \leq \theta \leq 2\pi$

329.

$x = 4\ \text{cos}\ \phi,$ $y = 1 - \text{sin}\ \phi,$ $0 \leq \phi \leq 2\pi$

For the following exercises, sketch the polar curve and determine what type of symmetry exists, if any.

330\.

$r = 4\ \text{sin}\left( \frac{\theta}{3} \right)$

331.

$r = 5\ \text{cos}\left( {5\theta} \right)$

For the following exercises, find the polar equation for the curve given as a Cartesian equation.

332\.

$x + y = 5$

333.

$y^{2} = 4 + x^{2}$

For the following exercises, find an equation of the tangent line to the given curve. Graph both the function and its tangent line.

334\.

$x = \text{ln}(t),$ $y = t^{2} - 1,$ $t = 1$

335.

$r = 3 + \text{cos}\left( {2\theta} \right),$ $\theta = \frac{3\pi}{4}$

336\.

Find $\frac{dy}{dx},$ $\frac{dx}{dy},$ and $\frac{d^{2}x}{dy^{2}}$ of $y = \left( {2 + e^{\text{−}t}} \right),$ $x = 1 - \text{sin}(t)$

For the following exercises, find the area of the region.

337.

$x = t^{2},$ $y = \text{ln}(t),$ $0 \leq t \leq e$

338\.

$r = 1 - \text{sin}\ \theta$ in the first quadrant

For the following exercises, find the arc length of the curve over the given interval.

339.

$x = 3t + 4,$ $y = 9t - 2,$ $0 \leq t \leq 3$

340\.

$r = 6\ \text{cos}\ \theta,$ $0 \leq \theta \leq 2\pi.$ Check your answer by geometry.

For the following exercises, find the Cartesian equation describing the given shapes.

341.

A parabola with focus $(2,-5)$ and directrix $x = 6$

342\.

An ellipse with a major axis length of 10 and foci at $\left( {-7,2} \right)$ and $\left( {1,2} \right)$

343.

A hyperbola with vertices at $(–2,3)$ and $(-2,-5)$ and foci at $(-2,-6)$ and $(-2,4)$

For the following exercises, determine the eccentricity and identify the conic. Sketch the conic.

344\.

$r = \frac{6}{1 + 3\ \text{cos}(\theta)}$

345.

$r = \frac{4}{3 - 2\ \text{cos}\ \theta}$

346\.

$r = \frac{7}{5 - 5\ \text{cos}\ \theta}$

347.

Determine the Cartesian equation describing the orbit of Pluto, the most eccentric orbit around the Sun. The length of the major axis is 39.26 AU and minor axis is 38.07 AU. What is the eccentricity?

348\.

The C/1980 E1 comet was observed in 1980. Given an eccentricity of 1.057 and a perihelion (point of closest approach to the Sun) of 3.364 AU, find the Cartesian equations describing the comet’s trajectory. Are we guaranteed to see this comet again? (*Hint*: Consider the Sun at point $(0,0).)$

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Authors: Gilbert Strang, Edwin “Jed” Herman
Publisher/website: OpenStax
Book title: Calculus Volume 3
Publication date: Mar 30, 2016
Location: Houston, Texas
Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
Section URL: https://openstax.org/books/calculus-volume-3/pages/1-review-exercises

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2 Vectors in Space →