Section 10.2 Calculus with Parametric Curves

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Section 10.2 Calculus with Parametric Curves

880 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.2 Calculus with Parametric Curves

1.  = 

1 + ,  =√

1 +  ⇒ 

 = 1

2(1 + )^−12 = 1 2√

1 + , 

 =(1 + )(1) − (1) (1 + )² = 1

(1 + )², and



 = 

 =1(2√ 1 +  )

1(1 + )² = (1 + )² 2√

1 + = 1

2(1 + )^32.

2.  = ^,  =  + sin  ⇒ 

 = 1 + cos , 

 = ^+ ^= ^( + 1), and 

 = 

 = 1 + cos 

^( + 1).

3.  = ³+ 1,  = ⁴+ ;  = −1. 

 = 4³+ 1, 

 = 3², and 

 = 

 = 4³+ 1

3² . When  = −1, ( ) = (0 0) and  = −33 = −1, so an equation of the tangent to the curve at the point corresponding to  = −1 is

 − 0 = −1( − 0), or  = −.

4.  =√,  = ²− 2;  = 4.

 = 2 − 2,

 = 1 2√

, and

 =

 = (2 − 2)2√

 = 4( − 1)√. When  = 4, ( ) = (2 8)and  = 4(3)(2) = 24, so an equation of the tangent to the curve at the point corresponding to  = 4 is

 − 8 = 24( − 2), or  = 24 − 40.

5.  = ⁴+ 1,  = ³+ ;  = −1 

 = 3²+ 1, 

 = 4³, and 

 = 

 =3²+ 1

4³ . When  = −1,

( ) = (2 −2) and  = ₋₄⁴ = −1, so an equation of the tangent to the curve at the point corresponding to  = −1 is  − (−2) = (−1)( − 2), or  = −.

6.  = ^sin ,  = ^2;  = 0. 

 = 2^2, 

 = ^( cos ) + (sin )^= ^( cos  + sin ), and



 = 

 = 2^2

^( cos  + sin ) = 2^

 cos  + sin . When  = 0, ( ) = (0 1) and  = 2, so an equation of the tangent to the curve at the point corresponding to  = 0 is  − 1 =²( − 0), or  = ² + 1.

7. (a)  = 1 + ln ,  = ²+ 2; (1 3). 

 = 2 

 = 1

and

 = 

 = 2

1 = 2². At (1 3),

 = 1 + ln  = 1 ⇒ ln  = 0 ⇒  = 1 and 

= 2, so an equation of the tangent is  − 3 = 2( − 1), or  = 2 + 1.

(b)  = 1 + ln  ⇒ ln  =  − 1 ⇒  = ^⁻¹, so  = ²+ 2 = (^⁻¹)²+ 2 = ^2⁻²+ 2, and ⁰= ^2⁻²· 2.

At (1 3), ⁰= ²⁽¹⁾⁻²· 2 = 2, so an equation of the tangent is  − 3 = 2( − 1), or  = 2 + 1.

8. (a)  = 1 +√,  = ^²; (2 ). 

 = ^²· 2, 

 = 1 2√

, and 

 =

 = 2^² 1

2√

 = 4^32^². At (2 ),

 = 1 +√

 = 2 ⇒ √

 = 1 ⇒  = 1 and

 = 4, so an equation of the tangent is  −  = 4( − 2), or  = 4 − 7.

10.2 Calculus with Parametric Curves

1.  = 

1 + ,  =√

1 +  ⇒ 

 = 1

2(1 + )^−12 = 1 2√

1 + , 

 =(1 + )(1) − (1) (1 + )² = 1

(1 + )², and



 = 

 =1(2√ 1 +  )

1(1 + )² = (1 + )² 2√

1 + = 1

2(1 + )^32.

2.  = ^,  =  + sin  ⇒ 

 = 1 + cos , 

 = ^+ ^= ^( + 1), and 

 = 

 = 1 + cos 

^( + 1).

3.  = ³+ 1,  = ⁴+ ;  = −1. 

 = 4³+ 1, 

 = 3², and 

 = 

 = 4³+ 1

3² . When  = −1, ( ) = (0 0) and  = −33 = −1, so an equation of the tangent to the curve at the point corresponding to  = −1 is

 − 0 = −1( − 0), or  = −.

4.  =√,  = ²− 2;  = 4.

 = 2 − 2,

 = 1 2√

, and

 =

 = (2 − 2)2√

 = 4( − 1)√. When  = 4, ( ) = (2 8)and  = 4(3)(2) = 24, so an equation of the tangent to the curve at the point corresponding to  = 4 is

 − 8 = 24( − 2), or  = 24 − 40.

5.  = ⁴+ 1,  = ³+ ;  = −1 

 = 3²+ 1, 

 = 4³, and 

 = 

 =3²+ 1

4³ . When  = −1,

( ) = (2 −2) and  = ₋₄⁴ = −1, so an equation of the tangent to the curve at the point corresponding to  = −1 is  − (−2) = (−1)( − 2), or  = −.

6.  = ^sin ,  = ^2;  = 0. 

 = 2^2, 

 = ^( cos ) + (sin )^= ^( cos  + sin ), and



 = 

 = 2^2

^( cos  + sin ) = 2^

 cos  + sin . When  = 0, ( ) = (0 1) and  = 2, so an equation of the tangent to the curve at the point corresponding to  = 0 is  − 1 =²( − 0), or  = ² + 1.

7. (a)  = 1 + ln ,  = ²+ 2; (1 3). 

 = 2 

 = 1

and

 = 

 = 2

1 = 2². At (1 3),

 = 1 + ln  = 1 ⇒ ln  = 0 ⇒  = 1 and 

= 2, so an equation of the tangent is  − 3 = 2( − 1), or  = 2 + 1.

(b)  = 1 + ln  ⇒ ln  =  − 1 ⇒  = ^⁻¹, so  = ²+ 2 = (^⁻¹)²+ 2 = ^2⁻²+ 2, and ⁰= ^2⁻²· 2.

At (1 3), ⁰= ²⁽¹⁾⁻²· 2 = 2, so an equation of the tangent is  − 3 = 2( − 1), or  = 2 + 1.

8. (a)  = 1 +√

,  = ^²; (2 ). 

 = ^²· 2, 

 = 1 2√

, and 

 =

 = 2^² 1

2√

 = 4^32^². At (2 ),

 = 1 +√

 = 2 ⇒ √

 = 1 ⇒  = 1 and

 = 4, so an equation of the tangent is  −  = 4( − 2), or  = 4 − 7.

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 881 (b)  = 1 +√

 ⇒ √

 =  − 1 ⇒  = ( − 1)², so  = ^²= ^(⁻¹⁾⁴, and ⁰= ^(⁻¹⁾⁴· 4( − 1)³. At (2 ), ⁰=  · 4 = 4, so an equation of the tangent is  −  = 4( − 2), or  = 4 − 7.

9. = ²− ,  = ²+  + 1; (0 3). 

 =

 =2 + 1

2 − 1. To ﬁnd the value of  corresponding to the point (0 3), solve  = 0 ⇒

²−  = 0 ⇒ ( − 1) = 0 ⇒  = 0 or  = 1. Only  = 1 gives

 = 3. With  = 1,  = 3, and an equation of the tangent is

 − 3 = 3( − 0), or  = 3 + 3.

10. = sin ,  = ²+ ; (0 2). 

= 

 = 2 + 1

 cos . To ﬁnd the value of  corresponding to the point (0 2), solve  = 2 ⇒

²+  − 2 = 0 ⇒ ( + 2)( − 1) = 0 ⇒  = −2 or  = 1.

Either value gives  = −3, so an equation of the tangent is

 − 2 = −³( − 0), or  = −³ + 2.

11.  = ²+ 1,  = ²+  ⇒ 

 =

 =2 + 1

2 = 1 + 1

2 ⇒ ²

² =











 = −1(2²) 2 = − 1

4³. The curve is CU when²

²  0, that is, when   0.

12. = ³− 12,  = ²− 1 ⇒ 

 = 

 = 2

3²− 12 ⇒

²

² =











 =

(3²− 12) · 2 − 2(6) (3²− 12)²

3²− 12 = −6²− 24

(3²− 12)³ = −6(²+ 4)

3³(²− 4)³ = −2(²+ 4) 9(²− 4)³ . Thus, the curve is CU when ²− 4  0 ⇒ ||  2 ⇒ −2    2.

13. = 2 sin ,  = 3 cos , 0    2.



 =

 = −3 sin  2 cos  = −3

2tan , so²

² =











 =−³2sec² 2 cos  = −3

4sec³.

The curve is CU when sec³  0 ⇒ sec   0 ⇒ cos   0 ⇒ ^2    ^3₂ . 14. = cos 2,  = cos , 0    .



 =

 = −sin 

−2 sin 2 = sin 

2 · 2 sin  cos  = 1 4 cos  =1

4sec , so²

² =











 =

1

4sec  tan 

−4 sin  cos = − 1 16sec³

The curve is CU when sec³  0 ⇒ sec   0 ⇒ cos   0 ⇒ ^2    .

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 881 (b)  = 1 +√

 ⇒ √

 =  − 1 ⇒  = ( − 1)², so  = ^²= ^(⁻¹⁾⁴, and ⁰= ^(⁻¹⁾⁴· 4( − 1)³. At (2 ), ⁰=  · 4 = 4, so an equation of the tangent is  −  = 4( − 2), or  = 4 − 7.

9.  = ²− ,  = ²+  + 1; (0 3). 

 = 

= 2 + 1

2 − 1. To ﬁnd the value of  corresponding to the point (0 3), solve  = 0 ⇒

²−  = 0 ⇒ ( − 1) = 0 ⇒  = 0 or  = 1. Only  = 1 gives

 = 3. With  = 1,  = 3, and an equation of the tangent is

 − 3 = 3( − 0), or  = 3 + 3.

10.  = sin ,  = ²+ ; (0 2). 

 = 

 = 2 + 1

 cos . To ﬁnd the value of  corresponding to the point (0 2), solve  = 2 ⇒

²+  − 2 = 0 ⇒ ( + 2)( − 1) = 0 ⇒  = −2 or  = 1.

Either value gives  = −3, so an equation of the tangent is

 − 2 = −³( − 0), or  = −³ + 2.

11.  = ²+ 1,  = ²+  ⇒ 

 = 

 =2 + 1

2 = 1 + 1

2 ⇒ ²

² =











 = −1(2²) 2 = − 1

4³. The curve is CU when²

²  0, that is, when   0.

12.  = ³− 12,  = ²− 1 ⇒ 

 = 

 = 2

3²− 12 ⇒

²

² =











 =

(3²− 12) · 2 − 2(6) (3²− 12)²

3²− 12 = −6²− 24

(3²− 12)³ = −6(²+ 4)

3³(²− 4)³ = −2(²+ 4) 9(²− 4)³. Thus, the curve is CU when ²− 4  0 ⇒ ||  2 ⇒ −2    2.

13.  = 2 sin ,  = 3 cos , 0    2.



 = 

 = −3 sin  2 cos  = −3

2tan , so²

² =











 = −³2sec² 2 cos  = −3

4sec³.

The curve is CU when sec³  0 ⇒ sec   0 ⇒ cos   0 ⇒ ^2    ^3₂. 14.  = cos 2,  = cos , 0    .



 = 

 = −sin 

−2 sin 2 = sin 

2 · 2 sin  cos = 1 4 cos = 1

4sec , so²

² =











 =

1

4sec  tan 

−4 sin  cos  = −1 16sec³

The curve is CU when sec³  0 ⇒ sec   0 ⇒ cos   0 ⇒ ^2    .

882 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 15.  =  − 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.

16.  = 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    .

17.  = 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).

18.  = ³− 3,  = ³− 3². 

 = 3²− 6 = 3( − 2), so

 = 0 ⇔

 = 0or 2 ⇔ ( ) = (0 0) or (2 −4). 

 = 3²− 3 = 3( + 1)( − 1), so

 = 0 ⇔  = −1 or 1 ⇔ ( ) = (2 −4) or (−2 −2). The curve has horizontal tangents at (0 0) and (2 −4), and vertical tangents at (2 −4) and (−2 −2).

19.  = cos ,  = cos 3. The whole curve is traced out for 0 ≤  ≤ .



 = −3 sin 3, so

= 0 ⇔ sin 3 = 0 ⇔ 3 = 0, , 2, or 3 ⇔

 = 0, ^₃, ^2₃ , or  ⇔ ( ) = (1 1),1 2 −1,

−¹2 1

, or (−1 −1).



 = − sin , so

 = 0 ⇔ sin  = 0 ⇔  = 0 or  ⇔ ( ) = (1 1)or (−1 −1). Both

and 

 equal 0 when  = 0 and .

To ﬁnd the slope when  = 0, we ﬁnd lim

→0



 = lim

→0

−3 sin 3

− sin 

= limH

→0

−9 cos 3

− cos  = 9, which is the same slope when  = .

Thus, the curve has horizontal tangents at1

2 −1and

−¹2 1, and there are no vertical tangents.

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 885

28.  =  cos³,  =  sin³.

(a) 

 = −3 cos² sin , 

 = 3 sin² cos , so 

 = −sin 

cos  = − tan .

(b) The tangent is horizontal ⇔  = 0 ⇔ tan  = 0 ⇔  =  ⇔ ( ) = (± 0).

The tangent is vertical ⇔ cos  = 0 ⇔  is an odd multiple of^2 ⇔ ( ) = (0 ±)  (c)  = ±1 ⇔ tan  = ±1 ⇔  is an odd multiple of ^4 ⇔ ( ) =

±^√4² ±^√4² [All sign choices are valid.]

29.  = 3²+ 1,  = ³− 1 ⇒ 

 = 

= 3² 6 = 

2. The tangent line has slope1 2when 

2 =1

2 ⇔  = 1, so the point is (4 0).

30.  = 3²+ 1,  = 2³+ 1, 

 = 6,

 = 6², so

 =6²

6 =  [even where  = 0].

So at the point corresponding to parameter value , an equation of the tangent line is  − (2³+ 1) = [ − (3²+ 1)].

If this line is to pass through (4 3), we must have 3 − (2³+ 1) = [4 − (3²+ 1)] ⇔ 2³− 2 = 3³− 3 ⇔

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).

31. 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

= 

32. 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

33. 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

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 887 40.  =  +√

,  =  −√

, 0 ≤  ≤ 1. 

 = 1 + 1 2√

 and 

 = 1 − 1 2√

, so





2

+





2

=

 1 + 1

2√



2

+

 1 − 1

2√



2

= 1 + 1

√+ 1

4+ 1 − 1

√+ 1

4= 2 + 1 2. Thus,  =

 



()²+ ()² =

 1 0

 2 + 1

2 = lim

→0⁺

 1



 2 + 1

2 ≈ 20915.

41.  = 1 + 3²,  = 4 + 2³, 0 ≤  ≤ 1.  = 6 and  = 6², so ()²+ ()²= 36²+ 36⁴.

Thus,  =

 1 0

36²+ 36⁴ =

 1 0

6

1 + ² = 6

 2 1

√1

2 [ = 1 + ²,  = 2 ]

= 3

2 3^322

1= 2(2^32− 1) = 2 2√

2 − 1

42.  = ^− ,  = 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.

43.  =  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. 44.  = ^+ ^−,  = 5 − 2, 0 ≤  ≤ 3.  = ^− ^−and  = −2, so

()²+ ()²= ^2− 2 + ^−2+ 4 = ^2+ 2 + ^−2= (^+ ^−)². Thus,  =3

0(^+ ^−)  =

^− ^−3

0= ³− ⁻³− (1 − 1) = ³− ⁻³.

45.  = ^cos ,  = ^sin , 0 ≤  ≤ .

_



2

+



2

= [^(cos  − sin )]²+ [^(sin  + cos )]²

= (^)²(cos² − 2 cos  sin  + sin²) + (^)²(sin² + 2 sin  cos  + cos²

= ^2(2 cos² + 2 sin²) = 2^2

Thus,  = 0

√2^2 = 0

√2 ^ =√ 2

^

0 =√

2 (^− 1).

46.  = cos  + ln(tan¹₂),  = sin , 4 ≤  ≤ 34.



 = − sin  +

1

2sec²(2)

tan(2) = − sin  + 1

2 sin(2) cos(2) = − sin  + 1

sin  and

 = cos , so





2

+





2

= sin² − 2 + 1

sin²+ cos² = 1 − 2 + csc² = cot². Thus,

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 887 40.  =  +√

,  =  −√

, 0 ≤  ≤ 1. 

 = 1 + 1 2√

 and 

 = 1 − 1 2√

, so





2

+





2

=

 1 + 1

2√



2

+

 1 − 1

2√



2

= 1 + 1

√+ 1

4+ 1 − 1

√+ 1

4= 2 + 1 2. Thus,  = 



()²+ ()² =

 1 0

 2 + 1

2 = lim

→0⁺

 1



 2 + 1

2 ≈ 20915.

41.  = 1 + 3²,  = 4 + 2³, 0 ≤  ≤ 1.  = 6 and  = 6², so ()²+ ()²= 36²+ 36⁴.

Thus,  = 1 0

36²+ 36⁴ =

 1 0

6

1 + ² = 6

 2 1

√₁

2 [ = 1 + ²,  = 2 ]

= 3

2 3^322

1= 2(2^32− 1) = 2 2√

2 − 1

42.  = ^− ,  = 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.

43.  =  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 . 44.  = ^+ ^−,  = 5 − 2, 0 ≤  ≤ 3.  = ^− ^−and  = −2, so

()²+ ()²= ^2− 2 + ^−2+ 4 = ^2+ 2 + ^−2= (^+ ^−)². Thus,  =3

0(^+ ^−)  =

^− ^−3

0= ³− ⁻³− (1 − 1) = ³− ⁻³.

45.  = ^cos ,  = ^sin , 0 ≤  ≤ .





2

+



²

= [^(cos  − sin )]²+ [^(sin  + cos )]²

= (^)²(cos² − 2 cos  sin  + sin²) + (^)²(sin² + 2 sin  cos  + cos²

= ^2(2 cos² + 2 sin²) = 2^2

Thus,  = 0

√2^2 = 0

√2 ^ =√ 2

^

0 =√

2 (^− 1).

46.  = cos  + ln(tan¹₂),  = sin , 4 ≤  ≤ 34.



 = − sin  +

1

2sec²(2)

tan(2) = − sin  + 1

2 sin(2) cos(2) = − sin  + 1

sin  and

 = cos , so





2

+





2

= sin² − 2 + 1

sin²+ cos² = 1 − 2 + csc² = cot². Thus,

1

(2)

Some earlier versions of Maple (as well as Mathematica) cannot do the integral exactly, so we use the command evalf(Int(sqrt(diff(x,t)ˆ2+diff(y,t)ˆ2),t=0..4*Pi)); to estimate the length, and find that the arc length is approximately 29403. Derive’s Para_arc_length function in the utility file Int_apps simplifies the integral to 114

0



−4 cos  cos_11

2

− 4 sin  sin_11

2

+ 5 .

56. (a) It appears that as  → ∞, ( ) →₁

2¹₂

, and as  → −∞, ( ) →

−¹2 −¹2

.

(b) By the Fundamental Theorem of Calculus,  = cos 2²and

 = sin_

2², so by Theorem 5, the length of the curve from the origin to the point with parameter value  is

 = 0





2

+



²

 = 0

 cos²

2²

+ sin² 2²



=

01  =  [or − if   0]

We have used  as the dummy variable so as not to confuse it with the upper limit of integration.

57.  =  sin ,  =  cos , 0 ≤  ≤ 2.  =  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

 =

2  =2

0 2 cos √

²+ 1  ≈ 47394.

58.  = sin ,  = sin 2, 0 ≤  ≤ 2.  = cos  and  = 2 cos 2, so ()²+ ()²= cos² + 4 cos²2.

 =

2  =2

0 2 sin 2√

cos² + 4 cos²2  ≈ 80285.

59.  =  + ^,  = ^−, 0 ≤  ≤ 1.

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

 =

2  =1

0 2^−√

1 + 2^+ ^2+ ^−2 ≈ 106705.

60.  = 1 + ^,  = (²+ 1)^, 0 ≤  ≤ 1.

_



2

+



2

= (^+ ^)²+ [(²+ 1)^+ ^(2)]²= [^( + 1)]²+ [^(²+ 2 + 1)]²

= ^2( + 1)²+ ^2( + 1)⁴= ^2( + 1)²[1 + ( + 1)²], so

 =

2  =1

0 2(²+ 1)^

^2( + 1)²(²+ 2 + 2)  =1

0 2(²+ 1)^2( + 1)√

²+ 2 + 2  ≈ 1035999

61.  = 3 − ³,  = 3², 0 ≤  ≤ 1. _



2

+



2

= (3 − 3²)²+ (6)²= 9(1 + 2²+ ⁴) = [3(1 + ²)]².

 =1

0 2 · 3²· 3(1 + ²)  = 181

0(²+ ⁴)  = 18₁

3³+¹₅⁵1 0 =⁴⁸₅

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()) =  ().