11.2   Tangent Lines and Arc Length for Parametric and Polar Curves

In this section we will derive the formulas required to find slopes, tangent lines, and arc lengths of parametric and polar curves.

Tangent Lines to Parametric Curves

We will be concerned in this section with curves that are given by parametric equations

in which and have continuous first derivatives with respect to t. It can be proved that if , then y is a differentiable function of x, in which case the chain rule implies that

(1)  

This formula makes it possible to find directly from the parametric equations without eliminating the parameter.


 Example 1  Find the slope of the tangent line to the unit circle
at the point where (Figure 11.2.1).

Figure 11.2.1    

Solution.  From 1, the slope at a general point on the circle is

(2)  

Thus, the slope at is

 

Note that Formula 2 makes sense geometrically because the radius from the origin to the point has slope . Thus the tangent line at P, being perpendicular to the radius, has slope

(Figure 11.2.2).

Figure 11.2.2    

 

It follows from Formula 1 that the tangent line to a parametric curve will be horizontal at those points where and , since at such points. Two different situations occur when . At points where and , the right side of 1 has a nonzero numerator and a zero denominator; we will agree that the curve has infinite slope and a vertical tangent line at such points. At points where and are both zero, the right side of 1 becomes an indeterminate form; we call such points singular points. No general statement can be made about the behavior of parametric curves at singular points; they must be analyzed case by case.


 Example 2  In a disastrous first flight, an experimental paper airplane follows the trajectory
but crashes into a wall at time (Figure 11.2.3).

Figure 11.2.3    

(a)  

At what times was the airplane flying horizontally?

(b)  

At what times was it flying vertically?

Solution(a).  The airplane was flying horizontally at those times when and . From the given trajectory we have

(3)  

Setting yields the equation , or, more simply, . This equation has four solutions in the time interval :
Since for these values of t (verify), the airplane was flying horizontally at times
which is consistent with Figure 11.2.3.

Solution (b).  The airplane was flying vertically at those times when and . Setting in 3 yields the equation

This equation has three solutions in the time interval (Figure 11.2.4):
Since is not zero at these points (why?), it follows that the airplane was flying vertically at times
which again is consistent with Figure 11.2.3.

Figure 11.2.4    


 Example 3  The curve represented by the parametric equations
is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows that . The graph of this equation, shown in Figure 11.2.5, consists of two branches: an upper branch obtained by graphing and a lower branch obtained by graphing . The two branches meet at the origin, which corresponds to in the parametric equations. This is a singular point because the derivatives and are both zero there.

Figure 11.2.5    


 Example 4  Without eliminating the parameter, find and at (1, 1) and on the semicubical parabola given by the parametric equations in Example 3.

Solution.  From 1 we have

(4)  

and from 1 applied to we have

(5)  

Since the point (1, 1) on the curve corresponds to in the parametric equations, it follows from 4 and 5 that
Similarly, the point corresponds to in the parametric equations, so applying 4 and 5 again yields

Note that the values we obtained for the first and second derivatives are consistent with the graph in Figure 11.2.5, since at (1, 1) on the upper branch the tangent line has positive slope and the curve is concave up, and at on the lower branch the tangent line has negative slope and the curve is concave down.

Finally, observe that we were able to apply Formulas 4 and 5 for both and , even though the points (1, 1) and lie on different branches. In contrast, had we chosen to perform the same computations by eliminating the parameter, we would have had to obtain separate derivative formulas for and .

Tangent Lines to Polar Curves

Our next objective is to find a method for obtaining slopes of tangent lines to polar curves of the form in which r is a differentiable function of θ. We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter θ by substituting for r in the equations and . This yields

from which we obtain

(6)  

Thus, if and are continuous and if , then y is a differentiable function of x, and Formula 1 with θ in place of t yields

(7)  


 Example 5  Find the slope of the tangent line to the circle at the point where .

Solution.  From 7 with we obtain (verify)

Thus, at the point where the slope of the tangent line is
which implies that the circle has a horizontal tangent line at the point where (Figure 11.2.6).

Figure 11.2.6    


 Example 6  Find the points on the cardioid at which there is a horizontal tangent line, a vertical tangent line, or a singular point.

Solution.  A horizontal tangent line will occur where and , a vertical tangent line where and , and a singular point where and . We could find these derivatives from the formulas in 6. However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting in the conversion formulas and . This yields

Differentiating these equations with respect to θ and then simplifying yields (verify)
Thus, if or , and if or . We leave it for you to solve these equations and show that the solutions of on the interval are
and the solutions of on the interval are
Thus, horizontal tangent lines occur at and ; vertical tangent lines occur at , π, and ; and singular points occur at and (Figure 11.2.7). Note, however, that at both singular points, so there is really only one singular point on the cardioid—the pole.

Figure 11.2.7    

Tangent Lines to Polar Curves at the Origin

Formula 7 reveals some useful information about the behavior of a polar curve that passes through the origin. If we assume that and when , then it follows from Formula 7 that the slope of the tangent line to the curve at is

(Figure 11.2.8). However, tan is also the slope of the line , so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result.

Figure 11.2.8    

11.2.1 Theorem.  If the polar curve passes through the origin at , and if at , then the line is tangent to the curve at the origin.

This theorem tells us that equations of the tangent lines at the origin to the curve can be obtained by solving the equation . It is important to keep in mind, however, that may be zero for more than one value of θ, so there may be more than one tangent line at the origin. This is illustrated in the next example.


 Example 7  The three-petal rose in Figure 11.2.9 has three tangent lines at the origin, which can be found by solving the equation
It was shown in Exercise 81 of Section 11.1 that the complete rose is traced once as θ varies over the interval , so we need only look for solutions in this interval. We leave it for you to confirm that these solutions are
Since for these values of θ, these three lines are tangent to the rose at the origin, which is consistent with the figure.

Figure 11.2.9    

ARC Length of a Polar Curve

A formula for the arc length of a polar curve can be derived by expressing the curve in parametric form and applying Formula (6) of Section 7.4 for the arc length of a parametric curve. We leave it as an exercise to show the following.

11.2.2 ARC Length Formula for Polar Curves.  If no segment of the polar curve is traced more than once as θ increases from α to β, and if is continuous for , then the arc length L from to is

(8)  


 Example 8  Find the arc length of the spiral in Figure 11.2.10 between and .

Figure 11.2.10    

Solution.  


 Example 9  Find the total arc length of the cardioid .

Solution.  The cardioid is traced out once as θ varies from to . Thus,

Since changes sign at π, we must split the last integral into the sum of two integrals: the integral from 0 to π plus the integral from π to 2π. However, the integral from π to 2π is equal to the integral from 0 to π, since the cardioid is symmetric about the polar axis (Figure 11.2.11). Thus,

Figure 11.2.11    

Quick Check Exercises 11.2

1.  

(a)  

To find directly from the parametric equations
we can use the formula = _________ .

(b)  

Find and directly from the parametric equations , .

2.  

(a)  

To obtain directly from the polar equation , we can use the formula
= _________

(b)  

Use the formula in part (a) to find directly from the polar equation .

3.  

(a)  

What conditions on and guarantee that the line is tangent to the polar curve at the origin?

(b)  

What are the values of in [0, 2π] at which the lines are tangent at the origin to the four-petal rose ?

4.  

(a)  

To find the arc length L of the polar curve , we can use the formula L = _________ .

(b)  

The polar curve has arc length L = _________ .

Exercise Set  11.2                  Graphing Utility           CAS

Focus on Concepts
 

1.  

(a)  

Find the slope of the tangent line to the parametric curve , at and at without eliminating the parameter.

(b)  

Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

2.  

(a)  

Find the slope of the tangent line to the parametric curve , at and at without eliminating the parameter.

(b)  

Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

3.  

For the parametric curve in Exercise 1, make a conjecture about the sign of at and at , and confirm your conjecture without eliminating the parameter.

4.  

For the parametric curve in Exercise 2, make a conjecture about the sign of at and at , and confirm your conjecture without eliminating the parameter.

5-10  Find and at the given point without eliminating the parameter.

5.  

, ;

6.  

, ;

7.  

, ;

8.  

, ;

9.  

, ;

10.  

, ;

11.  

(a)  

Find the equation of the tangent line to the curve
at without eliminating the parameter.

(b)  

Find the equation of the tangent line in part (a) by eliminating the parameter.

12.  

(a)  

Find the equation of the tangent line to the curve
at without eliminating the parameter.

(b)  

Find the equation of the tangent line in part (a) by eliminating the parameter.

13-14  Find all values of t at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line.

13.  

,

14.  

,

Focus on Concepts
 

15.  

The Lissajous curve

crosses itself at the origin. (See the accompanying figure.) Find equations for the two tangent lines at the origin.

Figure Ex-15    

16.  

The prolate cycloid

crosses itself at a point on the x-axis. (See the accompanying figure.) Find equations for the two tangent lines at that point.

Figure Ex-16    

17.  

Show that the curve , intersects itself at the point (4, 0), and find equations for the two tangent lines to the curve at the point of intersection.

18.  

Show that the curve with parametric equations

intersects itself at the point (3, 1), and find equations for the two tangent lines to the curve at the point of intersection.

19.  

(a)  

Use a graphing utility to generate the graph of the parametric curve
and make a conjecture about the values of t at which singular points occur.

(b)  

Confirm your conjecture in part (a) by calculating appropriate derivatives.

20.  

(a)  

At what values of θ would you expect the cycloid in Figure 1.8.14 to have singular points?

(b)  

Confirm your answer in part (a) by calculating appropriate derivatives.

21-26  Find the slope of the tangent line to the polar curve for the given value of θ.

21.  

;

22.  

;

23.  

;

24.  

;

25.  

;

26.  

;

27-28  Calculate the slopes of the tangent lines indicated in the accompanying figures.

27.  

Figure Ex-27    

28.  

Figure Ex-28    

29-30  Find polar coordinates of all points at which the polar curve has a horizontal or a vertical tangent line.

29.  

30.  

31-32  Use a graphing utility to make a conjecture about the number of points on the polar curve at which there is a horizontal tangent line, and confirm your conjecture by finding appropriate derivatives.

31.  

32.  

33-38  Sketch the polar curve and find polar equations of the tangent lines to the curve at the pole.

33.  

34.  

35.  

36.  

37.  

38.  

39-44  Use Formula 8 to calculate the arc length of the polar curve.

39.  

The entire circle

40.  

The entire circle

41.  

The entire cardioid

42.  

from to

43.  

from to

44.  

from to

45.  

(a)  

What is the slope of the tangent line at time t to the trajectory of the paper airplane in Example 2?

(b)  

What was the airplane's approximate angle of inclination when it crashed into the wall?

46.  

Suppose that a bee follows the trajectory

(a)  

At what times was the bee flying horizontally?

(b)  

At what times was the bee flying vertically?

47.  

(a)  

Show that the arc length of one petal of the rose is given by

(b)  

Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the four-petal rose .

(c)  

Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the n-petal rose for ; then make a conjecture about the limit of these arc lengths as .

48.  

(a)  

Sketch the spiral .

(b)  

Find an improper integral for the total arc length of the spiral.

(c)  

Show that the integral converges and find the total arc length of the spiral.

49-54  If and are continuous functions, and if no segment of the curve
is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the x-axis is
and the area of the surface generated by revolving the curve about the y-axis is
[The derivations are similar to those used to obtain Formulas (4) and (5) in Section 7.5.] Use the formulas above in these exercises.

49.  

Find the area of the surface generated by revolving , about the x-axis.

50.  

Find the area of the surface generated by revolving the curve , about the x-axis.

51.  

Find the area of the surface generated by revolving the curve , about the y-axis.

52.  

Find the area of the surface generated by revolving , about the y-axis.

53.  

By revolving the semicircle

about the x-axis, show that the surface area of a sphere of radius r is .

54.  

The equations

represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the x-axis is given by .

Focus on Concepts
 

55.  

As illustrated in the accompanying figure, suppose that a rod with one end fixed at the pole of a polar coordinate system rotates counterclockwise at the constant rate of 1 rad/s. At time a bug on the rod is 10 mm from the pole and is moving outward along the rod at the constant speed of 2 mm/s.

(a)  

Find an equation of the form for the path of motion of the bug, assuming that when .

(b)  

Find the distance the bug travels along the path in part (a) during the first 5 s. Round your answer to the nearest tenth of a millimeter.

Figure Ex-55    

56.  

Use Formula (6) of Section 7.4 to derive Formula 8.

57.  

The amusement park rides illustrated in the accompanying figure consist of two connected rotating arms of length 1—an inner arm that rotates counterclockwise at 1 radian per second and an outer arm that can be programmed to rotate either clockwise at 2 radians per second (the Scrambler ride) or counterclockwise at 2 radians per second (the Calypso ride). The center of the rider cage is at the end of the outer arm.

(a)  

Show that in the Scrambler ride the center of the cage has parametric equations

(b)  

Find parametric equations for the center of the cage in the Calypso ride, and use a graphing utility to confirm that the center traces the curve shown in the accompanying figure.

(c)  

Do you think that a rider travels the same distance in one revolution of the Scrambler ride as in one revolution of the Calypso ride? Justify your conclusion.

Figure Ex-57    

58.  

Use a graphing utility to explore the effect of changing the rotation rates and the arm lengths in Exercise 57.

59.  

(a)  

If a thread is unwound from a fixed circle while being held taut (i.e., tangent to the circle), then the end of the thread traces a curve called an involute of a circle. Show that if the circle is centered at the origin, has radius a, and the end of the thread is initially at the point , then the involute can be expressed parametrically as
where is the angle shown in part (a) of the accompanying figure.

(b)  

Assuming that the dog in part (b) of the accompanying figure unwinds its leash while keeping it taut, for what values of in the interval will the dog be walking North? South? East? West?

(c)  

Use a graphing utility to generate the curve traced by the dog, and show that it is consistent with your answer in part (b).

Figure Ex-59    

60.  

Recall from Section 6.9 that the Fresnel sine and cosine functions are defined as

The following parametric curve, which is used to study amplitudes of light waves in optics, is called a clothoid or Cornu spiral in honor of the French scientist Marie Alfred Cornu (1841–1902):

(a)  

Use a CAS to graph the Cornu spiral.

(b)  

Describe the behavior of the spiral as and as .

(c)  

Find the arc length of the spiral for .

61.  

As illustrated in the accompanying figure, let be a point on the polar curve , let ψ be the smallest counterclockwise angle from the extended radius to the tangent line at P, and let be the angle of inclination of the tangent line. Derive the formula

by substituting tan for in Formula 7 and applying the trigonometric identity

Figure Ex-61    

62-63  Use the formula for ψ obtained in Exercise 61.

62.  

(a)  

Use the trigonometric identity
to show that if is a point on the cardioid
then .

(b)  

Sketch the cardioid and show the angle ψ at the points where the cardioid crosses the y-axis.

(c)  

Find the angleψ at the points where the cardioid crosses the y-axis.

63.  

Show that for a logarithmic spiral , the angle from the radial line to the tangent line is constant along the spiral (see the accompanying figure). [Note: For this reason, logarithmic spirals are sometimes called equiangular spirals.]

Figure Ex-63    

Quick Check Answers 11.2

1.  

(a)  


(b)  


2.  

(a)  


(b)  


3.  

(a)  


(b)  


4.  

(a)  


(b)  

1