Section 10.2 Calculus with Parametric Curves 18. Find

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Section 10.2 Calculus with Parametric Curves

18. Find dy/dx and d²y/dx². For which values of t is the curve concave upward?

x = t²+ 1, y = e^t− 1 Solution:

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 957 16.  = ³+ 1,  = ²−  ⇒ 

 = 

 = 2 − 1 3² = 2

3 − 1 3² ⇒

²

² =











 = − 2 3² + 2

3³

3² =

2 − 2

3³

3² = 2(1 − )

9⁵ . The curve is CU when ²

²  0, that is, when 0    1.

17.  = ^,  = ^− ⇒ 

 = 

 = −^−+ ^−

^ = ^−(1 − )

^ = ^−2(1 − ) ⇒

²

² =











 = ^−2(−1) + (1 − )(−2^−2)

^ = ^−2(−1 − 2 + 2)

^ = ^−3(2 − 3). The curve is CU when

²

²  0, that is, when   ³₂.

18.  = ²+ 1,  = ^− 1 ⇒ 

 = 

= ^

2 ⇒ ²

² =











 =

2^− ^· 2 (2)²

2 = 2^( − 1)

(2)³ = ^( − 1) 4³ . The curve is CU when ²

²  0, that is, when   0 or   1.

19.  =  − ln ,  =  + ln  [note that   0] ⇒ 

 = 

 = 1 + 1

1 − 1 =  + 1

 − 1 ⇒

²

² =











 =

( − 1)(1) − ( + 1)(1) ( − 1)²

( − 1) = −2

( − 1)³. The curve is CU when ²

²  0, that is, when 0    1.

20.  = cos ,  = sin 2, 0     ⇒ 

 = 

= 2 cos 2

− sin  ⇒

²

² =











 =

(− sin )(−4 sin 2) − (2 cos 2)(− cos ) (− sin )²

− sin  = (sin )(8 sin  cos ) + [2(1 − 2 sin²)](cos ) (− sin ) sin²

= (cos )(8 sin² + 2 − 4 sin²)

(− sin ) sin² = −cos 

sin  · 4 sin² + 2

sin² [ (− cot ) · positive expression]

The curve is CU when ²

²  0, that is, when − cot   0 ⇔ cot   0 ⇔ ^2    .

21.  = ³− 3,  = ²− 3. 

 = 2, so

 = 0 ⇔  = 0 ⇔ ( ) = (0 −3). 

 = 3²− 3 = 3( + 1)( − 1), so 

 = 0 ⇔

 = −1 or 1 ⇔ ( ) = (2 −2) or (−2 −2). The curve has a horizontal tangent at (0 −3) and vertical tangents at (2 −2) and (−2 −2).

31. (a) Find the slope of the tangent line to the trochoid x = rθ − d sin θ, y = r − d cos θ in terms of θ. (See Exercise 10.1.49.)

(b) Show that if d < r, then the trochoid does not have a vertical tangent.

Solution:

884 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

 = 2(2²− 3 − 2) = 0 ⇔ 2(2 + 1)( − 2) = 0 ⇔  = 0, −¹2 or 2. It seems that we have missed one vertical tangent, and indeed if we plot the curve on the -interval [−12 22] we see that there is another vertical tangent at (−8 6).

24. We graph the curve  = ⁴+ 4³− 8²,  = 2²−  in the viewing rectangle [−37 02] by [−02 14]. It appears that there is a horizontal tangent at about (−04 −01), and vertical tangents at about (−3 1) and (0 0).

We calculate

 = 

 = 4 − 1

4³+ 12²− 16, so there is a horizontal tangent where  = 4 − 1 = 0 ⇔  = ¹4. This point (the lowest point) is shown in the ﬁrst graph. There are vertical tangents where  = 4³+ 12²− 16 = 0 ⇔ 4(²+ 3 − 4) = 0 ⇔ 4( + 4)( − 1) = 0. We have missed one vertical tangent corresponding to  = −4, and if we plot the graph for  ∈ [−5 3], we see that the curve has another vertical tangent line at approximately (−128 36).

25.  = cos ,  = sin  cos .  = − sin ,

 = − sin² + cos² = cos 2. ( ) = (0 0) ⇔ cos  = 0 ⇔  is an odd multiple of^₂. When  = ^₂,  = −1 and  = −1, so  = 1.

When  =^3₂ ,  = 1 and  = −1. So  = −1. Thus,  =  and

 = − are both tangent to the curve at (0 0).

26.  = −2 cos ,  = sin  + sin 2. From the graph, it appears that the curve crosses itself at the point (1 0). If this is true, then  = 1 ⇔

−2 cos  = 1 ⇔ cos  = −¹2 ⇔  = ^23 or ^4₃ for 0 ≤  ≤ 2.

Substituting either value of  into  gives  = 0, conﬁrming that (1 0) is the point where the curve crosses itself. 

 = 

= cos  + 2 cos 2

2 sin  . When  = 2

3 ,

 = −12 + 2(−12) 2(√

32) = −32

√3 = −

√3

2 , so an equation of the tangent line is  − 0 = −

√3

2 ( − 1), or  = −

√3 2  +

√3

2 . Similarly, when  = 4

3 , an equation of the tangent line is  =

√3 2  −

√3 2 . 27.  =  −  sin ,  =  −  cos .

(a) 

 =  −  cos ,

 =  sin , so 

 =  sin 

 −  cos .

(b) If 0    , then | cos | ≤   , so  −  cos  ≥  −   0. This shows that  never vanishes, so the trochoid can have no vertical tangent if   .

38. Find the area enclosed by the given parametric curve and the y-axis.

the x-axis.

35-36 Find the area enclosed by the given parametric curve and

35. x = t³+ 1, y = 2t - t²

y

。

36. x = sin t, y = sin t cos t, 0 三 t這 π/2

y

。 ^x

the y-axis.

37-38 Find the area enclosed by the given parametric curve and

37. x = sin弓，

y = cos t y

。

38. x = t²- 2t y =

Ji

y

X

。

39. Use the parametric equations of an ellipse, x = αcos

e

^,

y ⁼ b sin

e

^,0 '"三。三三 27T, to find the area that it encloses.

40. Find the area of the region enclosed by the loop of the curve x = 1 - t², Y = t - t³

y

1 x

SECTION 10.2 仁alculuswith Parametric Curves 681

41. Find the area under one arch of the trochoid of Exer- cise 10.1.49 for the case d < r.

42. Let q]t be the region enclosed by the loop of the curve in Example 1.

(a) Find the area of q]t.

(b) If q]t is rotated about the x-axis, find the volume of the resulting solid.

(c) Find the centroid of q]t.

固的-46 Set up an integral 伽trepresents the length of the part of the parametric curve shown in the graph. Then use a calculator (or computer) to find the length co叮ectto four decimal places.

43. x = 3t² 戶 y = t²- 2t

2

。

44. x = t + e一九 y = t²+ t

y 6

x

2

-A

AU

x

45. x = t - 2 sin t, y = 1 - 2 cos t, 0 '"三 I三 47T

y

x Solution:

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 961

³− 3 + 2 = 0 ⇔ ( − 1)²( + 2) = 0 ⇔  = 1 or −2. Hence, the desired equations are  − 3 =  − 4, or

 =  − 1, tangent to the curve at (4 3), and  − (−15) = −2( − 13), or  = −2 + 11, tangent to the curve at (13 −15).

35. The curve  = ³+ 1,  = 2 − ²= (2 − ) intersects the axis when  = 0, that is, when  = 0 and  = 2. The corresponding values of  are 1 and 9. The shaded area is given by

 =9

=1

(− ^)  =

 =2

=0 [() − 0] ⁰()  =

 2

0 (2 − ²)(3²) 

= 32

0(2³− ⁴)  = 3

1

2⁴− ¹5⁵2 0= 3

8 −³²5

= ²⁴₅

36. The curve  = sin ,  = sin  cos , 0 ≤  ≤ 2 intersects the axis when  = 0, that is, when  = 0 and  = 2.

The corresponding values of  are 0 and 1, so the area enclosed by the curve and the axis is given by

 =1

=0

  =

 =2

=0

() ⁰()  =

 2 0

sin  cos  (cos ) = −^c

 0 1

² =₁

3³1 0= ¹₃

37. The curve  = sin²,  = cos  intersects the axis when  = 0, that is, when  = 0 and  = . (Any integer multiple of  will result in  = 0, though we choose two values of  over which the curve is traced out once.) The corresponding values of  are 1 and −1, so the area enclosed by the curve and the axis is given by

 =1

=−1

  =

 =0

=

() ⁰()  =

 0



sin² (−sin )  =

  0

sin³ =⁶⁷



−1

3(2 + sin²) cos 



0

= ²₃−

−²3

= ⁴₃

38. The curve  = ²− 2 = ( − 2),  =√

intersects the axis when  = 0, that is, when

 = 0and  = 2. The corresponding values of  are 0 and√2. The shaded area is given by

 =√ 2

=0

(− ^)  =

 =2

=0 [0 − ()] ⁰()  = −

 2 0

(²− 2)

 1 2√





= −2 0

₁

2^32− ^12

 = −₁

5^52−²3^322 0

= −

1

5· 2^52−²3 · 2^32

= −2^124 5− ⁴3



= −√ 2

−15⁸

= ₁₅⁸ √ 2

39.  =  cos ,  =  sin , 0 ≤  ≤ 2. By symmetry of the ellipse about the  and axes,

 = 4

 =

=0

  = 4

 =0

=2 sin  (− sin )  = 42

0 sin²  = 42 0

1

2(1 − cos 2) 

= 2

 −¹2sin 22

0 = 2_

2

= 

40. By symmetry, the area of the shaded region is twice the area of the shaded portion above the axis. The top half of the loop is described by  = 1 − ²,  =  − ³= (1 − )(1 + ), 0 ≤  ≤ 1 with intercepts 0 and 1 corresponding to  = 1 and

 = 0, respectively. Thus, the area of the shaded region is 21

0   = 20

1 () ⁰()  = 20

1( − ³)(−2)  = 41

0(²− ⁴)  = 41

3³−¹5⁵1 0 = 41

3 −¹5

= ₁₅⁸.

1

(2)

50. Find the exact length of the curve.

x = 3 cos t − cos 3t, y = 3 sin t − sin 3t, 0 ≤ t ≤ π Solution:

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 963

45.  =  − 2 sin ,  = 1 − 2 cos , 0 ≤  ≤ 4.  = 1 − 2 cos  and  = 2 sin , so

()²+ ()²= (1 − 2 cos )²+ (2 sin )²= 1 − 4 cos  + 4 cos² + 4 sin² = 5 − 4 cos . Thus,

 =



()²+ ()² =4

0

√5 − 4 cos    267298.

46.  =  cos ,  =  − 5 sin .  = cos  −  sin  and  = 1 − 5 cos , so

()²+ ()²= (cos  −  sin )²+ (1 − 5 cos )². Observe that when  = −, ( ) = ( −) and when  = , ( ) = (− ). Thus,  =



()²+ ()² =

−

(cos  −  sin )²+ (1 − 5 cos )²  228546.

47.  = ²₃³,  = ²− 2, 0 ≤  ≤ 3.  = 2²and  = 2, so ()²+ ()² = 4⁴+ 4² = 4²(²+ 1).

Thus,

 =

 



()²+ ()² =

 3 0

4²(²+ 1)  =

 3 0

2

²+ 1 

=

 10 1

√  [ = ²+ 1  = 2 ] =₂

3^3210

1 = ²₃(10^32− 1) = ²3

10√ 10 − 1

48.  = ^− ,  = 4^2, 0 ≤  ≤ 2.  = ^− 1 and  = 2^2, so

()²+ ()²= (^− 1)²+ (2^2)²= ^2− 2^+ 1 + 4^= ^2+ 2^+ 1 = (^+ 1)². Thus,

 =

 2 0

(^+ 1)² =

 2 0

^+ 1  = 2 0

(^+ 1)  =

^+ 2

0= (²+ 2) − (1 + 0) = ²+ 1.

49.  =  sin ,  =  cos , 0 ≤  ≤ 1. 

 =  cos  + sin and

 = − sin  + cos , so





2

+





2

= ²cos² + 2 sin  cos  + sin² + ²sin² − 2 sin  cos  + cos²

= ²(cos² + sin²) + sin² + cos² = ²+ 1.

Thus,  =1 0

√²+ 1 =²¹1 2√

²+ 1 +¹₂ln

 +√

²+ 1¹

0= ¹₂√

2 +¹₂ln 1 +√

2.

50.  = 3 cos  − cos 3,  = 3 sin  − sin 3, 0 ≤  ≤ . 

 = −3 sin  + 3 sin 3 and 

 = 3 cos  − 3 cos 3, so





2

+





2

= 9 sin² − 18 sin  sin 3 + 9 sin²3 + 9 cos² − 18 cos  cos 3 + 9 cos²3

= 9(cos² + sin²) − 18(cos  cos 3 + sin  sin 3) + 9(cos²3 + sin²3)

= 9(1) − 18 cos( − 3) + 9(1) = 18 − 18 cos(−2) = 18(1 − cos 2)

= 18[1 − (1 − 2 sin²)] = 36 sin².

Thus,  = 0

√36 sin²  = 6

0 |sin |  = 6

0 sin   = −6 cos 

0 = −6 (−1 − 1) = 12.

73. Find the exact area of the surface obtained by rotating the given curve about the x-axis.

x = a cos³θ, y = a sin³θ, 0 ≤ θ ≤π 2 Solution:

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 891 62.  = 2²+ 1,  = 8√

, 1 ≤  ≤ 3.





2

+





2

=

 4 − 1

²

2

+

 4

√

2

= 16²−8

 + 1

⁴ +16

 = 16²+8

 + 1

⁴ =

 4 + 1

²

2

.

 =

 3 1

2_



² +



²

 =

 3 1

2 8√



4 + 1

²

2

 = 16

 3 1

^12(4 + ⁻²) 

= 16

 3 1

(4^32+ ^−32)  = 16

8

5^52− 2^−123

1= 16

72 5

√3 −²3

√3

− (⁸5 − 2)

= 16₂₀₆

15

√3 +₁₅⁶

= ^32₁₅ 103√

3 + 3 63.  =  cos³,  =  sin³, 0 ≤  ≤ ^2. 



2

+



²

= (−3 cos² sin )²+ (3 sin² cos )²= 9²sin² cos².

 =2

0 2 ·  sin³ · 3 sin  cos   = 6²2

0 sin⁴ cos   = ⁶₅²

sin⁵2

0 = ⁶₅² 64.  = 2 cos  − cos 2,  = 2 sin  − sin 2 ⇒





2

+



²

= (−2 sin  + 2 sin 2)²+ (2 cos  − 2 cos 2)²

= 4[(sin² − 2 sin  sin 2 + sin²2) + (cos² − 2 cos  cos 2 + cos²2)]

= 4[1 + 1 − 2(cos 2 cos  + sin 2 sin )] = 8[1 − cos(2 − )] = 8(1 − cos ) We plot the graph with parameter interval [0 2], and see that we should only integrate

between 0 and . (If the interval [0 2] were taken, the surface of revolution would be generated twice.) Also note that  = 2 sin  − sin 2 = 2 sin (1 − cos ). So

 =

0 2 · 2 sin (1 − cos ) 2√ 2√

1 − cos  

= 8√ 2

0 (1 − cos )^32sin   = 8√ 22

0

√³

  = 1− cos 

 = sin  



= 8√ 2₂

5

^522 0= ¹⁶₅√

2(2^52) = ¹²⁸₅ 

65.  = 3²,  = 2³, 0 ≤  ≤ 5 ⇒ _



2

+



²

= (6)²+ (6²)²= 36²(1 + ²) ⇒

 =5 0 2

()²+ ()² =5

0 2(3²)6√

1 + ² = 185 0 ²√

1 + ²2 

= 1826

1 ( − 1)√ 

 = 1 + ²

 = 2 



= 1826

1 (^32− ^12)  = 18

2

5^52−²3^3226 1

= 182

5· 676√

26 −²3· 26√ 26

−2 5 −²3

= ²⁴₅  949√

26 + 1 66.  = ^− ,  = 4^2, 0 ≤  ≤ 1. 



² +



²

= (^− 1)²+ (2^2)²= ^2+ 2^+ 1 = (^+ 1)².

 =1

0 2(^− )

(^− 1)²+ (2^2)² =1

0 2(^− )(^+ 1)

= 21

2^2+ ^− ( − 1)^−¹2²1

0= (²+ 2 − 6)

67. If ⁰is continuous and ⁰() 6= 0 for  ≤  ≤ , then either ⁰()  0for all  in [ ] or ⁰()  0for all  in [ ]. Thus,  is monotonic (in fact, strictly increasing or strictly decreasing) on [ ]. It follows that  has an inverse. Set  =  ◦ ⁻¹, that is, deﬁne  by  () = (⁻¹()). Then  = () ⇒ ⁻¹() = , so  = () = (⁻¹()) =  ().

2