t 000sin2 2 1 2cos2 - media.pearsoncmg.com · 972 chapter 13 vector-valued functions and motion in...

Copyright 2018 Pearson Education, Inc. 959

CHAPTER 13 VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE

13.1 CURVES IN SPACE AND THEIR TANGENTS

1. 5 52 22 3 4 2 3 4lim sin cos tan sin cos tant

tt

i j k i j k

12 i j k

2. 332 21

lim sin ln 2 1 sin ln1t

t t t

i j k i j k

i j

3. 1

2 21

1 111 2ln 1 11 1 1

lim tan lim lim tan 1t

t

tt tt tt t t

t

i j k i j k 1

2 42 i j k

4. 2 3 2sin tan 8 cos 2 tan secsin 2 2 1 2cos 20 0 0

lim lim lim ( 4)t t t t t tt t t tt t t

t

i j k i j k

4 i k

5. 1x t and 2 2 21 ( 1) 1 2 ;y t y x x x 2 2 2d ddt dtt r vv i j a j v i j and 2a j at

1t

6. 1t

tx and 1

11 1 1

111;y

yt y xy x y

2 2 3 31 1 2 2

( 1) ( 1)d ddt dtt t t t

r vv i j a i j

4 4 v i j and 16 16 a i j at 12t

7. tx e and 2 22 29 9 ;ty e y x 2 284

9 9 3 4d t t t tdt e e e e rv i j a i j v i j and 3 8 a i j at

ln 3t

8. 2 219cos 2 and 3sin 2 1;x t y t x y 2sin 2 6cos 2d

dt t t rv i j

4cos 2 12sin 2 6ddt t t va i j v j and 4 a i at 0t

9. (cos ) (sin )ddt t t rv i j and (sin ) (cos )d

dt t t va i j

2 24 4 2 2for ,t v i j and 2 2

4 2 2 ; a i j

for 2 2 2, andt v j a i

10. 2 22sin 2cosd t tdt rv i j and 2 2cos sind t t

dt va i j

for , ( ) 2t v i and ( ) ; a j

for 3 32 2, 2 2t v i j and 2 23

2 2 2 a i j

960 Chapter 13 Vector-Valued Functions and Motion in Space

Copyright 2018 Pearson Education, Inc.

11. (1 cos ) (sin )ddt t t rv i j and (sin ) (cos )d

dt t t va i j

for , ( ) 2t v i and ( ) ; a j

for 3 32 2,t v i j and 3

2 a i

12. 2ddt t rv i j and 2d

dt va j for 1,t , ( 1) 2 v i j and

( 1) 2 ; a j for 0, (0)t v i and (0) 2 ;a j for 1, (1) 2t v i j and (1) 2a j

13. 2

22( 1) 1 2 2 2 2 ;d d

dt dtt t t t r rr i j k v i j k a j Speed: 22 2| (1) | 1 2(1) 2 3; v

Direction: (1) 2(1) 2 1 2 2 1 2 2| (1)| 3 3 3 3 3 3 3(1) 3 v i j kv i j k v i j k

14. 2 3 2

222 2

32 2 2(1 ) 2 ;dt t t d

dt dtt t t r rr i j k v i j k a j k

Speed: 2 22(1)2 22

| (1)| 1 1 2; v

Direction:

2(1) 22

1(1) 1 1 1 1 1 1| (1)| 2 2 2 2 22 2

(1) 2

i j kv

v i j k v i j k

15. 2

2(2 cos ) (3 sin ) 4 ( 2 sin ) (3 cos ) 4 ( 2 cos ) (3 sin ) ;d ddt dt

t t t t t t t r rr i j k v i j k a i j

Speed: 2 2 22 2 22 sin 3 cos 4 2 5; v

Direction: 2

2

32 4 1 2 1 22 2 22 5 2 5 2 5 5 5 5 5

sin cos 2 5

v

vi j k i k v i k

16. 24 43 3(sec ) (tan ) sec tan secd

dtt t t t t t rr i j k v i j k

2

22 3 2sec tan sec 2 sec tan ;d

dtt t t t t ra i j

Speed: 22 22 46 6 6 6 3sec tan sec 2; v

Direction:

2 4

6 6 6 36

6

sec tan sec 1 2 2 1 2 22 3 3 3 6 3 3 32

i j kv

vi j k v i j k

17. 2 2

2 22 2 2

2 1 ( 1)2 ln( 1) 2 2 ;dt d

dt t dt tt t t t

r rr i j k v i j k a i j k

Speed: 2 2 221 1(1) 2(1) 1 6; v

Direction: 21 1 2(1) (1)(1) 1 2 1 1 2 1

| (1)| 6 6 6 6 6 6 6(1) 6

i j kv

v i j k v i j k

13.1 Curves in Space and Their Tangents 961


18. 12 cos 3 2 sin 3 6 sin 3 6 cos 3t ddte t t e t t rr i j k v i j k

2

2 18 cos 3 18 sin 3 ;tddt

e t t ra i j k

Speed: 2 2 20(0) 6 sin 3(0) 6 cos 3(0) 37;e v

Direction:

0 6 sin 3(0) 6 cos 3(0)(0) 6 61 1(0) 37 37 37 37 37

(0) 37e

i j kv

v i k v i k

19. 3 3 2t v i j k and 2 (0) 3 3 a k v i j and 22 2(0) 2 | (0)| 3 3 0 12 a k v and

22| (0)| 2 2; (0) (0) 0 cos 0 a v a

20. 2 22 2 32t v i j and 2 2

2 232 (0) a j v i j and 2 22 2

2 2(0) 32 | (0)| 1 a j v and

2 2 16 2 2 32 1(32) 2 4| (0)| ( 32) 32; (0) (0) ( 32) 16 2 cos a v a

21. 2 2

1/222 11 1

1tt t

t t

v i j k and

2

2 2 3/22 2 22 2 2 1

1 1 1(0)t t

t t t

a i j k v j and

(0) 2 (0) 1 a i k v and 2 22(0) 2 1 5; (0) (0) 0 cos 0 a v a

22. 1/2 1/22 2 13 3 3(1 ) (1 )t t v i j k and 1/2 1/21 1 2 2 1

3 3 3 3 3(1 ) (1 ) (0)a t t i j v i j k and

2 2 21 1 2 2 13 3 3 3 3(0) (0) 1 a i j v and 2 2 21 1 2 2

3 3 3 9 9(0) ; (0) (0) 0 a v a

2cos 0

23. 2( ) (sin ) cos ( ) (cos ) (2 sin ) ;t tt t t t e t t t t e r i j k v i j k 0 00 ( )t t v i k and

0 0 0, 1,1 0 , 1,t P x t t y r and 1z t are parametric equations of the tangent line

24. 2 3 2( ) (2 1) ( ) 2 2 3 ;t t t t t t t r i j k v i j k 0 2 (2) 4 2 12t v i j k and 0 0( ) (4, 3, 8)t P r 4 4 , 3 2 ,x t y t and 8 12z t are parametric equations of the tangent line

25. 21 31 1

02 3( 2)( ) (ln ) ( ln ) ( ) (ln 1) ; 1 (1)t

t t tt t t t t t t

r i j k v i j k v i j k and

1 10 0 3 3(0,0,0) 0 , 0 ,t P x t t y t t r and 0z t t are parametric equations of the tangent

line

26. 0 02( ) (cos ) (sin ) (sin 2 ) ( ) ( sin ) (cos ) (2 cos 2 ) ; ( ) 2t t t t t t t t t t r i j k v i j k v i k and

0 0( ) (0, 1, 0) 0 , 1,t P x t t y r and 0 2 2z t t are parametric equations of the tangent line



27. 2( ) ( ) (1 ) (2 3)t t t t r i j k ( ) (2 ) 2 ;t t r i j k

let t a be the point of tangency 2( ) ( ) (1 ) (2 3)a a a a r i j k and a direction vector for the tangent line is ( ) (2 ) 2a a r i j k

parametric equations for the tangent line are 2( ) (2 ) ,x a a s (1 ) ,y a s (2 3) 2 ;z a s the point is

( 8, 2, 1) 2 2 8,a as 1 2,a s 2 3 2 1a s 1s a 2 2 (1 ) 8a a a ( 4)( 2) 0a a 4a or 2a the values of t are 4t and 2.t

28. 3 223( ) ( ) (3)t t t r i j k

( ) ;t t r i k let t a be the point of tangency

3 223( ) ( ) (3)a a a r i j k

and a direction vector for the tangent line is ( )a a r i k parametric

equations for the tangent line are ( ) (1) ,x a s (3) (0) ,y s 3 223 ;z a a s the point is 8

30, 3,

0,a s 3 2 823 3a as s a 3 2 3 2 82 1

3 3 3( )a a a a 4a the value of t is 4.t

29. 2 2( ) (2 ) ( ) ( )t t t t r i j k ( ) (2) (2 ) (2 ) ;t t t r i j k

let t a be the point of tangency 2 2( ) (2 ) ( ) ( )a a a a r i j k

and a direction vector for the tangent line is ( ) (2) (2 ) (2 )a a a r i j k

parametric equations for the tangent line are (2 ) (2) ,x a s 2( ) (2 ) ,y a a s 2( ) ( 2 ) ;z a a s the point is

0, 4, 4 2 2 0,a s 2 2 4,a as 2 2 4a as s a 2 2 ( ) 4a a a 2 4a 2a or 2a the values of t are 2t and 2.t

30. 2( ) ( ) ( ) (ln )t t t t r i j k 1( ) ( 1) (2 ) ;tt t r i j k

let t a be the point of tangency 2( ) ( ) ( ) (ln )a a a a r i j k

and a direction vector for the tangent line is 1( ) ( 1) (2 ) aa a r i j k

parametric equations for the tangent line are ( ) ( 1) ,x a s 2( ) (2 ) ,y a a s 1(ln ) ;az a s the point is

2, 5, 3 2,a s 2 2 5,a as ln 3saa 2s a 2 2 ( 2) 5a a a

( 5)( 1) 0a a 5a or 1a the value of t is 1.t

31. E 32. B 33. D 34. F 35. C 36. A

37. (a) ( ) (sin ) (cos ) ( ) (cos ) (sin ) ;t t t t t t v i j a i j

(i) 2 2( ) ( sin ) (cos ) 1t t t v constant speed; (ii) (sin )(cos ) (cos )(sin ) 0t t t t v a yes, orthogonal; (iii) counterclockwise movement; (iv) yes, (0) 0 r i j (b) ( ) (2 sin 2 ) (2 cos 2 ) ( ) (4 cos 2 ) (4 sin 2 ) ;t t t t t t v i j a i j

(i) 2 2( ) 4 sin 2 4 cos 2 2t t t v constant speed; (ii) 8 sin 2 cos 2 8 cos 2 sin 2 0t t t t v a yes, orthogonal; (iii) counterclockwise movement; (iv) yes, (0) 0 r i j

(c) 2 2 2 2( ) sin cos ( ) cos sin ;t t t t t t v i j a i j

(i) 2 22 2( ) sin cos 1t t t v constant speed;



(ii) 2 2 2 2sin cos cos sin 0t t t t v a yes, orthogonal;

(iii) counterclockwise movement; (iv) no, (0) 0 r i j instead of 0i j (d) ( ) (sin ) (cos ) ( ) (cos ) (sin ) ;t t t t t t v i j a i j

(i) 2 2( ) ( sin ) ( cos ) 1t t t v constant speed; (ii) (sin )(cos ) (cos )(sin ) 0t t t t v a yes, orthogonal; (iii) clockwise movement; (iv) yes, (0) 0 r i j (e) ( ) (2 sin ) (2 cos ) ( ) (2 sin 2 cos ) (2 cos 2 sin ) ;t t t t t t t t t t t t v i j a i j

(i) 2 2 2 2 2( ) (2 sin ) (2 cos ) 4 sin cos 2 | | 2 , 0t t t t t t t t t t t v variable speed;

(ii) 2 2 2 24 sin sin cos 4 cos cos sin 4 0t t t t t t t t t t t v a in general not orthogonal in

general; (iii) counterclockwise movement; (iv) yes, (0) 0 r i j

38. Let 2 2 p i j k denote the position vector of the point (2, 2, 1) and let, 1 12 2

u i j and

1 1 13 3 3

. v i j k Then ( ) (cos ) (sin ) .t t t r p u v Note that (2, 2, 1) is a point on the plane and

2 n i j k is normal to the plane. Moreover, u and v are orthogonal unit vectors with 0 u n v n u and v are parallel to the plane. Therefore, ( )tr identifies a point that lies in the plane for each t. Also, for each

t, (cos ) (sin )t tu v is a unit vector. Starting at the point 1 12 2

2 , 2 , 1 the vector ( )tr traces out a circle

of radius 1 and center (2, 2, 1) in the plane 2 2.x y z

39. The velocity vector is tangent to the graph of 2 2y x at the point (2, 2), has length 5, and a positive i

component. Now, 2 2 12 2 2(2,2)

2 2 2dy dydx dxy x y the tangent vector lies in the direction of the

vector 12 i j the velocity vector is 1 5

4 2

5 51 12 21

2 5 5

v i j i j i j

40. (a)

(b) (1 cos ) (sin )t t v i j and (sin ) (cos ) ;t t a i j 2 22 2(1 cos ) sin 2 2 cost t t v v is at a

max when cos 1 , 3 , 5 , etc.,t t and at these values of t, 2 24 max | | 4 2; v v v is at

a min when cos 1 0, 2 , 4 , etc.,t t and at these values of 2, 0 min 0;t v v 2 2 2sin cos 1t t a for every max min 1 1t a a



41. 2 2 0 0d d dddt dt dt dt r r rr r r r r r r is a constant r r r is constant

42. (a) ( ) ( ) ( )d d d dd ddt dt dt dt dt dt u u v wu v w v w u v w v w u w v

( )d d ddt dt dt u v wv w u w u v

(b) 2 2 2 2 3 3

2 2 2 2 3 3 ,d d d d d d d d d d dddt dt dt dt dt dtdt dt dt dt dt dt

r r r r r r r r r r rr r r r since ( ) 0 A A B

and ( ) 0 A B B for any vectors A and B

43. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) df dgd dhdt dt dt dtf t g t h t c cf t cg t ch t c c c c u i j k u i j k u i j k

ddf dg dhdt dt dt dtc c ui j k

(b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x t y t z t f t f t x t f t y t f t z t u i j k u i j k

( ) ( ) ( ) ( ) ( ) ( ) ( )df df dy dfd dx dzdt dt dt dt dt dt dtf t x t f t y t f t z t f t

u i j k

( ) ( ) ( ) ( ) ( ) ddf dy dfdx dzdt dt dt dt dt dtx t y t z t f t f t

ui j k i j k u

44. Let 1 2 3( ) ( ) ( )f t f t f t u i j k and 1 2 3( ) ( ) ( ) .g t g t g t v i j k

Then 1 1 2 2 3 3( ) ( ) ( ) ( ) ( ) ( )f t g t f t g t f t g t u v i j k

1 1 2 2 3 3( ) ( ) ( ) ( ) ( ) ( )d

dt f t g t f t g t f t g t u v i j k

1 2 3 1 2 3( ) ( ) ( ) ( ) ( ) ( ) ;d d

dt dtf t f t f t g t g t g t u vi j k i j k

1 1 2 2 3 3( ) ( ) ( ) ( ) ( ) ( )f t g t f t g t f t g t u v i j k

1 1 2 2 3 3( ) ( ) ( ) ( ) ( ) ( )d

dt f t g t f t g t f t g t u v i j k

1 2 3 1 2 3( ) ( ) ( ) ( ) ( ) ( ) d d

dt dtf t f t f t g t g t g t u vi j k i j k

45. Suppose r is continuous at 0.t t Then 0 0

0lim ( ) ( ) lim ( ) ( ) ( )t t t t

t t f t g t h t

r r i j k

0 0 0

0 0 0 0 0 0lim ( ) , lim ( ) , and lim ( ) , ,t t t t t t

f t g t h t f t f t g t g t h t h t f g

i j k and h are

continuous at 0.t t

46. 0 0 0 0 0 0 0

0 0 0

1 2 1 2 3 1 2 3 1 2

1 2 31 2 3

lim [ ( ) ( )] lim ( ) ( ) ( ) lim ( ) lim ( ) lim ( ) lim ( ) lim ( )( ) ( ) ( )

lim ( ) lim ( ) lim ( )t t t t t t t t t t t t t t

t t t t t t

t t f t f t f t f t f t f t t tg t g t g t

g t g t g t

i j k i j k

r r r r

A B

47. 0r t exists 0 0 0f t g t h t i j k exists 0 0 0, ,f t g t h t all exist , , andf g h are continuous at 0 ( )t t t r is continuous at 0t t



48. a b c u C i j k with a, b, c real constants 0 0 0 0d da db dcdt dt dt dt u i j k i j k

49–52. Example CAS commands:

Maple: > with( plots );

r : t - [sin(t)-t*cos(t),cos(t) t*sin(t),t^2];

t0 : 3*Pi/2; 1o : 0;

hi : 6*Pi;

P1: spacecurve( r(t), t 1o..hi, axes boxed, thickness 3 ):

display( P1, title "#49(a) (Section 13.1)" );

Dr : unapply( diff(r(t),t), t ); # (b)

Dr(t0); # (c)

q1: expand( r(t0) Dr(t0)*(t-t0) );

T : unapply( q1, t );

P2 : spacecurve( T(t), t 1o..hi, axes boxed, thickness 3, color black ):

display( [P1,P2], title "#49(d) (Section 13.1)" );


Maple: a : 'a'; b : 'b';

r : (a,b,t) - [cos(a*t),sin(a*t),b*t];

Dr : unapply( diff(r(a,b,t),t), (a,b,t) );

t0 : 3*Pi/2;

q1: expand( r(a,b,t0) Dr(a,b,t0)*(t-t0) );

T : unapply( q1, (a,b,t) );

1o : 0;

hi : 4*Pi; P : NULL: for a in [ 1, 2, 4, 6 ] do

P1: spacecurve( r(a,1,t), t 1o..hi, thickness 3 ):

P2 : spacecurve( T(a,1,t), t 1o..hi, thickness 3, color black ):

P : P, display( [P1,P2], axes boxed, title sprintf("#53 (Section 13.1)\n a %a",a) );

end do: display( [P], insequence true );




Mathematica: (assigned functions, parameters, and intervals will vary) The x-y-z components for the curve are entered as a list of functions of t. The unit vectors i, j, k are not

inserted. If a graph is too small, highlight it and drag out a corner or side to make it larger. Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem.

Clear[r, v, t, x, y, z]

_r[t ] { Sin[t] t Cos[t], Cos[t] t Sin[t], t^2}

t0 3π/2; tmin 0; tmax 6π;

ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel {x, y, z}];

_v[t ] r'[t]

_tanline[t ] v[t0] t r[t0]

ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel {x, y, z}]; For 53 and 54, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t. Clear[r, v, t, x, y, z, a, b] _ _ _r[t ,a ,b ]: {Cos[a t], Sin[a t], b t} t0 3π/2; tmin 0; tmax 4π; _ _ _v[t ,a ,b ] D[r[t, a, b], t] _ _ _tanline[t ,a ,b ] v[t0, a, b] t r[t0, a, b] pa1 ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}]; pa2 ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}]; pa4 ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}]; pa6 ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel {x, y, z}]; Show[GraphicsRow[{pa1, pa2, pa4, pa6}]]

13.2 INTEGRALS OF VECTOR FUNCTIONS; PROJECTILE MOTION

1. 4 21 11 13 3104 2 4 20 0 0

7 ( 1) 7 7t tt t dt t t i j k i j k i j k

2. 2

2 2 22 2 3/2 141 1 1 1

(6 6 ) 3 6 3 2 4 3 4 2 2 2t

t t dt t t t t i j k i j k i j k

3. /4 /4 /4 /42 2 2/4 /4 /4 2/4

(sin ) (1 cos ) sec cos sin tan 2t t t dt t t t t

i j k i j k j k

4. /3 /30 0

(sec tan ) (tan ) (2sin cos ) (sec tan ) (tan ) (sin 2 )t t t t t dt t t t t dt

i j k i j k

/3/3 /3 310 0 2 40

sec ln(cos ) cos 2 (ln 2)t t t i j k i j k

5. 44 4 41 1 1 11 15 2 21 1

ln ln (5 ) ln (ln 4) (ln 4) (ln 2)t t t dt t t t i j k i j k i j k

13.2 Integrals of Vector Functions; Projectile Motion 967


6. 22

1 11 1 13 3240 1 0 01

2sin 3 tantt

dt t t

i k i k i k

7. 2 2 1 11 1 1 11

02 20 00

t t t t e eete e dt e e t i j k i j k i i k

8. ln 3 ln 3ln 3 ln 3

11 1 1ln lnt t t t tte e t dt te e e t t t i j k i j k

3 ln 3 1 3 ln 3 ln(ln 3) 1 1e i j k

9. /2 /22 1 12 20 0

cos sin 2 sin cos sin 2 cos 2t t t dt t t t dt i j k i j k

/2 /2/2 1 1 10 2 2 4 40 0

sin cos sin 2t t t t i j k i j k

10. /4 /42 20 0

sec tan sin sec sec 1 sint t t t dt t t t t dt i j k i j k

/4 /4 /4 10 0 40 4 2 2

ln sec tan tan cos sin ln 1+ 2 1t t t t t t t i j k i j k

11. 2 2 2

2 2 2 ;t t tt t t dt r i j k i j k C (0) 0 0 0 2 3 2 3 r i j k C i j k C i j k

2 2 2

2 2 21 2 3t t t r i j k

12. 2 2 2 3163180 180 16 90 90 ;t t t dt t t t r i j i j C 2 2 316

3(0) 90(0) 90(0) (0) 100 r i j C j

2 2 3163100 90 90 100t t t C j r i j

13. 1/2 3/23 12 1( 1) ( 1) ln( 1) ;t t

tt e dt t e t

r i j k i j k C

3/2 0 3/2(0) (0 1) ln(0 1) ( 1) 1 1 1 ln( 1)te t e t r i j k C k C i j k r i j k

14. 4 2 33 2 2 24 2 34 ) 2 2 ;t t tt t t t dt t r i j k i j k C 34 2 2(0)20 0

4 2 3(0) 2(0) r i j k C i j

4 2 32 24 2 32 1 1t t tt C i j r i j k

15. 1 1 12 2 2( ) tan cos (sec2 ) ln sec 2sin ln sec2 tan 2 ;t t t t dt t t t t r i j k i j k C

12(0) ln sec0 (2sin 0) ln sec0 tan 0 r i j k C

3 2 C i j k

1 12 2( ) 3 ln sec 2 2sin 1 ln sec2 tan 2t t t t t r i j k



16. 2 2

2 21 422 3

( ) t t ttt t

t dt

r i j k 2 2

2 51 12 22 3

2 1ttt t

t dt i j k

2 2 11 1 12 2 3 3

ln 2 2 5ln 2 tan ;tt t t t t i j k C

12(0) ln 2 5ln 2 r i j C i j k

1

21 ln 2 5ln 2 1 C i j k

2 2 11 1 1 2 12 2 2 2 3 3

( ) 1 ln 1 2 1 5ln 1 tan ttt t t t t r i j k

17. 1( 32 ) 32 ;ddt dt t r k k C 1 1(0) 8 8 32(0) 8 8 8 8 8 8 32 ;d d

dt dt t r ri j k C i j C i j i j k

228 8 32 8 8 16 ;t dt t t t r i j k i j k C 2

2(0) 100 8(0) 8(0) 16(0) 100 r k i j k C k

22 100 8 8 100 16t t t C k r i j k

18. 1;ddt dt t t t r i j k i j k C 1 1(0) 0 0 0d

dt r 0 i j k C 0 C 0

( );ddt t t t r i j k 2 2 2

22 2 2 ;t t tt t t dt r i j k i j k C (0) 10 10 10 r i j k

2 2 20 0 02 22 2 2 10 10 10 10 10 10 i j k C i j k C i j k

2 2 2

2 2 210 10 10t t t r i j k

19. 2 214 2 ;d t t t t t t

dt e e e dt e e e r i j k i j k C

0 0 010 12 2 4d

tdt e e e r i j k C i j k C i j

1 2 3 2 C i j k

22 3 2 2 ;d t t tdt e e e r i j k

2( ) 2 3 2 2t t tt e e e dt r i j k

222 3 2 ;t t te t e t e t i j k C

0 0 02(0) e e e r i j k C

2 3 2 i j k C i j k

2 2 2 C i j k

2( ) 2 2 3 2 2 1t t tt e t e t e t r i j k

20. 21sin cos 4sin cos cos sin 2sin ;d

dt t t t t dt t t t r i j i j k C

210 1cos0 sin 0 2sin 0d

tdt r i j k C i C i

1 2C i

22 cos sin 2sin ;ddt t t t r i j k

( ) 2 cos sin 1 cos2t t t t dt r i j k

1 222 sin cos sin 2 ;t t t t t i j k C

1 22(0) sin 0 cos 0 sin 0 r i j k C 2 j C i k

2 C i j k

12( ) 1 2 sin 1 cos 1 sin 2t t t t t t r i j k

21. 13 ( ) 3 ;ddt t t t t v a i j k v i j k C the particle travels in the direction of the vector

(4 1) (1 2) (4 3) 3 i j k i j k (since it travels in a straight line), and at time 0t it has speed 2

62 2 219 1 1 11 11 11

(0) 3 ( ) 3ddt t t t t

rv i j k C v i j k



2 2 23 6 1 2 1 222 2 211 11 11

( ) ;t t t t t t t r i j k C 2(0) 2 3 r i j k C

2 2 2 23 6 1 2 1 2 1 22 2 2 211 11 11 11

( ) 1 2 3 3 2 3t t t t t t t t t r i j k i j k i j k

22. 12 ( ) 2 ;ddt t t t t v a i j k v i j k C the particle travels in the direction of the vector

(3 1) 0 ( 1) (3 2) 2 i j k i j k (since it travels in a straight line), and at time 0t it has speed 2

2 4 2 214 1 1 6 6 6

(0) 2 ( ) 2ddt t t t t

rv i j k C v i j k

2 2 26 1 2 1 222 26 6 6

( ) ;t t t t t t t r i j k C 2(0) 2 r i j k C

2 2 2 24 1 2 1 2 1 22 2 26 6 6 6

( ) 1 1 2 2 2t t t t t t t t t r i j k i j k i j k

23. 1000 m 21,000m0 1 km (840m/s)(cos 60 )( cos ) (21km) (840 m/s)(cos 60 ) 50 secondsx v t t t

24. (a) 2 2 2

0 0 0(2 ) 4sin 2 sin 2 4 sinv v vg g gR

or 4 times the original range.

(b) Now, let the initial range be 20 sin 2 .vgR Then we want the factor p so that 0pv will double the range

2 20 0( ) 2sin 2 2 sin 2 2 2pv v

g g p p

or about 141%. The same percentage will approximately

double the height: 2 2

0 0( sin ) 2( sin ) 22 2 2 2.pv v

g g p p

25. (a) 02

2 sin 2(500 m/s)(sin 45 )9.8 m/s

72.2 seconds;vgt

2 20

2(500 m/s)9.8 m/s

sin 2 (sin 90 ) 25,510.2 mvgR

(b) 5000 m0 (500 m/s)(cos 45 )( cos ) 5000 m (500 m/s)(cos 45 ) 14.14 s;x v t t t thus,

2 2 21 10 2 2( sin ) (500 m/s)(sin 45 )(14.14s) (9.8 m/s ) (14.14s) 4020 my v t gt y

(c) 220

2(500m/s)(sin 45 )( sin )

max 2 2(9.8m/s )6378mv

gy

26. 2 2 2 21 10 0 2 2( sin ) 32ft (32ft/sec)(sin 30 ) (32ft/sec ) 32 16 16 ;y y v t gt y t t y t t

the ball hits the ground when 20 0 32 16 16 1or 2 2secy t t t t t since 0;t thus,

30 2( cos ) (32ft/sec)(cos30 ) 32 (2) 55.4ftx v t x t

27. (a) 2 20 0

22 2 20 09.8m/s

sin 2 10 m = (sin 90 ) 98 m /s 9.9 m/s;v vgR v v

(b) 2

2(9.9m/s )9.8m/s

6m (sin 2 ) sin 2 0.59999 2 36.87 or 143.12 18.4 or 71.6



28. 60 5 10 m/sv and 40cm 0.4 m;x thus 6

0( cos ) 0.4 m (5 10 m/s)(cos 0 )x v t t 6 80.08 10 s 8 10 s;t also, 21

0 0 2( sin )y y v t gt 6 8 2 8 2 141

2(5 10 m/s)(sin 0 )(8 10 s) (9.8m/s ) (8 10 s) 3.136 10 my or 123.136 10 cm.

Therefore, it drops 123.136 10 cm.

29. 2 20

2(400 m/s)

9.8m/ssin 2 16,000 m sin 2 sin 2 0.98 2 78.5v

gR or 2 101.5 39.3

or 50.7

30. 20 sin 2vgR and maximum R occurs when

20

29.8m/s45 24.5 km (sin 90 )v

2 20 (9.8)(24,500) m /s 490 m/sv

31. The projectile reaches its maximum height when its vertical component of velocity is zero

2 20 0 0 0 0

2sin sin sin ( sin ) ( sin )10 max 0 2 2sin 0 ( sin )v v v v vdy

dt g g g g gv gt t y v g 2

0( sin )2 .v

g To find the flight time we find the time when the projectile lands: 21

0 2( sin ) 0v t gt

10 2sin 0 0t v gt t or 02 sin .v

gt Since 0t is the time when the projectile is fired, then

02 sinvgt is the time when the projectile strikes the ground. The range is the value of the horizontal

component when 2 20 0 0 02 sin 2 sin

0( cos ) (2sin cos ) sin 2 .v v v vg g g gt R x v The range is

largest when 2 1 45 .

32. When marble A is located R units downrange, we have 0

0 0 cos( cos ) ( cos ) .Rvx v t R v t t At that

time the height of marble A is 0 0

221 10 0 02 cos 2 cos( sin ) ( sin ) R R

v vy y v t gt v g

2

2 20

12 cos

tan .Rv

y R g

The height of marble B at the same time 0 cos

Rvt seconds is

2

2 20

21 12 2 cos

tan tan .Rv

h R gt R g

Since the heights are the same, the marbles collide regardless of

the initial velocity 0.v

33. 1( )ddt g dt gt r j j C and 0 0 1 0 0(0) ( cos ) ( sin ) (0) ( cos ) ( sin )d

dt v v g v v r i j j C i j

1 0 0 0 0( cos ) ( sin ) ( cos ) ( sin ) ;ddtv v v v gt rC i j i j 0 0( cos ) ( sin )v v gt dt r i j

210 0 22cos sinv t v t gt i j C and 21

0 0 0 0 22(0) (0)cos (0)sin (0)x y v v g r i j i j C

210 0 2 0 0 0 0 0 0 0 02cos sin cosx y x y x v t y v t gt x x v t i j C i j r i j and

210 0 2siny y v t gt



34. The maximum height is 2

0( sin )2

vgy and this occurs for

2 20 0 sin cos

2 sin 2 .v vg gx These equations

describe parametrically the points on a curve in the -planexy associated with the maximum heights on the parabolic trajectories in terms of the parameter (launch angle) . Eliminating the parameter , we have

4 2 24 2 2 4 2 4 4 2 200 0 0 0 0

2 2 2 2

sin 1 sinsin cos sin sin 222 2 22 2 4 0vv v v v v

g gg g g gx y y x y y

2 4 4 2 40 0 0 0 0

2 2 2

22 2 2

2 416 4 44 4 ,v v v v v

g gg g gx y y x y

where 0.x

35. (a) At the time t when the projectile hits the line OR we have tan ;y

x 0 cos ( )x v t and

210 2sin ( ) 0y v t gt since R is

below level ground. Therefore let 21

02| | sin ( ) 0y gt v t so that

21 10 02 2

00

sin ( ) sin ( )cos ( )cos( )

tangt v t gt v

vv t

10 02cos ( ) tan sin ( )v gt v

0 02 sin ( ) 2 cos ( ) tan ,v vgt

which is the time when the projectile hits the downhill slope. Therefore,

2

0 0 02 sin ( ) 2 cos ( ) tan 2 20 cos ( ) cos ( ) tan sin ( ) cos ( ) .v v v

g gx v

If x is maximized, then OR is maximized: 202 sin 2( ) tan cos 2( ) 0vdx

d g

sin 2( ) tan cos 2( ) 0 tan cot 2( ) 2( ) 90 1 1 12 2 2(90 ) (90 ) of .AOR

(b) At the time t when the projectile hits OR we have tan ;y

x 0 cos( )x v t and

210 2sin ( )y v t gt

2 1100 22

00

sin ( )sin ( )cos ( )cos ( )

tanv gtv t gt

vv t

10 0 2cos ( ) tan sin ( )v v gt

0 02 sin ( ) 2 cos ( ) tan ,v vgt which is the

time when the projectile hits the uphill slope.

Therefore, 0 02 sin ( ) 2 cos ( ) tan0 cos ( ) v v

gx v

202 2sin ( ) cos ( ) cos ( ) tan .v

g If x is maximized, then OR is maximized:

202 cos 2( ) sin 2( ) tan 0 cos 2( ) sin 2( ) tan 0vdx

d g

cot 2( ) tan 0 cot 2( ) tan tan ( ) 2( ) 90 ( ) 90 1 12 2(90 ) of .AOR Therefore 0v would bisect AOR for maximum range uphill.



36. 0 116 ft/ sec, 45 ,v and 0( cos )x v t 369 (116 cos 45 ) 4.50 sec;t t

also 210 2( sin )y v t gt

212(116sin 45 )(4.50) (32)(4.50) 45.11 ft.y

It will take the ball 4.50 sec to travel 369 ft. At that time the ball will be 45.11 ft in the air and will hit the green past the pin.

37. (a) (Assuming that " "x is zero at the point of impact:)

( ) ( ( )) ( ( )) ;t x t y t r i j where ( ) (35cos 27 )x t t and 2( ) 4 (35sin 27 ) 16 .y t t t

(b) 2 2

0( sin ) (35sin 27 )max 2 644 4 7.945 feet,v

gy which is reached at 0 sin 35sin 2732

vgt

0.497seconds.

(c) For the time, solve 24 (35sin 27 ) 16 0y t t for t, using the quadratic formula

235sin 27 35sin 27 25632 1.201sec.t

Then the range is about (1.201) (35cos 27 )(1.201)x

37.453 feet.

(d) For the time, solve 24 (35sin 27 ) 16 7y t t for t, using the quadratic formula

235sin 27 35sin 27 19232 0.254t

and 0.740 seconds. At those times the ball is about

(0.254) (35cos 27 )(0.254) 7.921x feet and (0.740) (35cos 27 )(0.740) 23.077x feet from the impact point, or about 37.453 7.921 29.532 feet and 37.453 23.077 14.376 feet from the landing spot.

(e) Yes. It changes things because the ball won’t clear the net max( 7.945).y

38. 0 0 0 0( cos ) 0 ( cos 40 ) 0.766x x v t v t v t and 2 210 0 02( sin ) 6.5 ( sin 40 ) 16y y v t gt v t t

206.5 0.643 16 ;v t t now the shot went

0

96.383073.833ft 73.833 0.766 sec;vv t t the shot lands

when 20 0

2 148,635 148,63596.3830 68.4740 0 6.5 (0.643)(96.383) 16 0 68.474 46.6ft/sec,v v

y v the

shot’s initial speed

39. Flight time 1sec and the measure of the angle of elevation is about 64 (using a protractor) so that

0 02 sin 2 sin 640321 17.80ft/sec.v v

gt v Then 217.80sin 64max 2(32) 4.00fty and

20 sin 2vgR

2(17.80)32 sin128 7.80ftR the engine traveled about 7.80ft in1sec the engine velocity was

about 7.80 ft/sec

40. (a) ( ) ( ( )) ( ( )) ;t x t y t r i j where ( ) (145cos 23 14)x t t and 2( ) 2.5 (145sin 23 ) 16 .y t t t

(b) 2 2

0( sin ) (145sin 23 )max 2 642.5 2.5 52.655feet,v

gy which is reached at 0 sin 145sin 2332

vgt

1.771 seconds.



(c) For the time, solve 22.5 (145sin 23 ) 16 0for ,y t t t using the quadratic formula

2145sin 23 145sin 23 16032 3.585sec.t

Then the range at 3.585t is about

(145cos 23 14)(3.585) 428.311 feet.x

(d) For the time, solve 22.5 (145sin 23 ) 16 20y t t for t, using the quadratic formula

2145sin 23 145sin 23 112032 0.342t

and 3.199seconds. At those times the ball is about

(0.342) (145cos 23 14)(0.342) 40.860feetx from home plate and (3.199) (145cos 23 14)(3.199) 382.195feetx from home plate.

(e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate.

41. 2

2 ( )d ddtdt

k g P t k r r j and ( ) 1( )( ) ( ) ( ) ( ) ( )P t dt dkt

dt v tt g P t dt kt v t e e v t t dt rQ j Q

1 ,kt gkt kt kt ktek kge e dt ge e j j C j C where 1;g C C apply the initial condition:

0 0 0 0 0( cos ) ( sin ) ( cos ) sind g gtdt k kv v v v r i j j C C i j

0 0cos sin ,d g gkt ktdt k kv e e v r i j 0 0cos sing gkt kt

k kv e e v dt r i j

00 2cos sin ;

ktv gt gkt ek k k ke v

i j C apply the initial condition:

0 0 0 02 2

sin sin2 2(0) cos cosv v v vg g

k k k kk k r 0 i j C C i j

0 02( ) 1 cos 1 sin 1v v gkt kt kt

k k kt e e kt e r i j

42. (a) ( ) ( ) ( ) ;t x t y t r i j where 0.121520.12( ) 1 (cos 20 )tx t e and

20.12 0.12152 32

0.12 0.12( ) 3 1 (sin 20 ) 1 0.12t ty t e t e

(b) Solve graphically using a calculator or CAS: At 1.484t seconds the ball reaches a maximum height of about 40.435feet.

(c) Use a graphing calculator or CAS to find that 0y when the ball has traveled for 3.126 seconds. The

range is about 0.12(3.126)1520.12(3.126) 1 (cos 20 ) 372.311x e feet.

(d) Use a graphing calculator or CAS to find that 30y for 0.689t and 2.305 seconds, at which times the ball is about (0.689) 94.454feetx and (2.305) 287.621feetx from home plate.

(e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the ground when it passes over the fence.

43. (a) ( ) ( ) ( ) ( ) ( ) ( ) ( )b b b b ba a a a a

k t dt kf t kg t kh t dt kf t dt kg t dt kh t dt r i j k i j k

( ) ( ) ( ) ( )b b b ba a a a

k f t dt g t dt h t dt k t dt i j k r

(b) 1 2 1 1 1 2 2 2( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )b ba a

t t dt f t g t h t f t g t h t dt r r i j k i j k

1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )ba

f t f t g t g t h t h t dt i j k



1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )b b ba a a

f t f t dt g t g t dt h t h t dt i j k

1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )b b b b b ba a a a a a

f t dt f t dt g t dt g t dt h t dt h t dt i i j j k k

1 2( ) ( )b ba a

t dt t dt r r

(c) Let 1 2 3 .c c c C i j k Then 1 2 3( ) ( ) ( ) ( )b ba a

t dt c f t c g t c h t dt C r

1 2 3( ) ( ) ( ) ( ) ;b b b ba a a a

c f t dt c g t dt c h t dt t dt C r

2 3 3 1 1 2( ) ( ) ( ) ( ) ( ) ( ) ( )b ba a

t dt c h t c g t c f t c h t c g t c f t dt C r i j k

2 3 3 1 1 2( ) ( ) ( ) ( ) ( ) ( )b b b b b ba a a a a a

c h t dt c g t dt c f t dt c h t dt c g t dt c f t dt i j k

( )ba

t dt C r

44. (a) Let u and r be continuous on [ , ].a b Then 0 0

lim ( ) ( ) lim ( ) ( ) ( ) ( ) ( ) ( )t t t t

u t t u t f t u t g t u t h t

r i j k

0 0 0 0 0 0 0 0( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )u t f t u t g t u t h t u t t u i j k r r is continuous for every 0 in[ , ].t a b (b) Let u and r be differentiable. Then ( ) ( ) ( ) ( ) ( ) ( ) ( )d d

dt dtu u t f t u t g t u t h t r i j k

( ) ( ) ( ) ( ) ( ) ( )df dgdu du du dhdt dt dt dt dt dtf t u t g t u t h t u t i j k

( ) ( ) ( ) ( ) df dgdu dh du ddt dt dt dt dt dtf t g t h t u t u ri j k i j k r

45. (a) If 1( )tR and 2 ( )tR have identical derivatives on I, then 1 1 1 1 2 2 2d df dg dh df dg dhdt dt dt dt dt dt dt R i j k i j k

2 1 2 ,d df dfdt dt dt R 1 2 ,dg dg

dt dt 1 21 2 1( ) ( ) ,dh dh

dt dt f t f t c 1 2 2( ) ( ) ,g t g t c 1 2 3( ) ( )h t h t c

1 1 1 2 1 2 2 2 3 1 2( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,f t g t h t f t c g t c h t c t t i j k i j k R R C where

1 2 3 .c c c C i j k (b) Let ( )tR be an antiderivative of ( )tr on I. Then ( ) ( ).t t R r If ( )tU is an antiderivative of ( )tr on I,

then ( ) ( ).t t U r Thus ( ) ( )t t U R on I ( ) ( ) .t t U R C

46. ( ) ( ) ( ) ( ) ( ) ( ) ( )t t t t td d d d d

dt dt dt dt dta a a a ad f g h d f d g d h d r i j k i j kt t

( ) ( ) ( ) ( ).f t g t h t t i j k r Since ( ) ( ),td

dt ad t r r we have that ( )

ta

d r is an antiderivative of r.

If R is any antidervative of r, then ( ) ( )ta

t d R r C by Exercise 41(b). Then ( ) ( )aa

a d R r C

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ).t ba a

a d t t a d b a 0 C C R r R C R R r R R

47. (a) ( ) ( ( )) ( ( )) ;t x t y t r i j where 0.0810.08( ) 1 (152cos 20° 17.6)tx t e and

20.08 0.08152 32

0.08 0.08( ) 3 1 (sin 20°) 1 0.08t ty t e t e

(b) Solve graphically using a calculator or CAS: At 1.527t seconds the ball reaches a maximum height of about 41.893 feet.

13.3 Arc Length in Space 975


(c) Use a graphing calculator or CAS to find that 0y when the ball has traveled for 3.181 seconds. The

range is about 0.08(3.181)10.08(3.181) 1 (152cos 20° 17.6) 351.734x e feet.

(d) Use a graphing calculator or CAS to find that 35y for 0.877t and 2.190 seconds, at which times the ball is about (0.877) 106.028x feet and (2.190) 251.530x feet from home plate.

(e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that 20y at 0.376t and 2.716 seconds. Then define

0.08(2.716)10.08( ) 1 (152cos 20° ),x w e w and solve ( ) 380x w to find 12.846w ft/sec.

48. 2 2

0 0( sin ) 3( sin )3max max2 4 8

v vg gy y and

203( sin )2 21 1

0 02 8 2( sin ) ( sin )vgy v t gt v t gt

2 2 2 2 2 20 0 0 0 03( sin ) (8 sin ) 4 4 (8 sin ) 3( sin ) 0 2 3 sin 0v gv t g t g t gv t v gt v or

03 sin0 22 sin 0 v

ggt v t or 0 sin2 .v

gt Since the time it takes to reach maxy is 0 sinmax ,v

gt then

the time it takes the projectile to reach 34 of maxy is the shorter time 0 sin

2v

gt or half the time it takes to

reach the maximum height.

13.3 ARC LENGTH IN SPACE

1. 22 22cos 2sin 5 2sin 2cos 5 2sin 2cos 5t t t t t t t r i j k v i j k v

2 24sin 4cos 5 3;t t 52 23 3 3sin cost t v

vT i j k and Length 0 0

3dt dt

v

03 3t

2. 6sin 2 6cos 2 5 12cos 2 12sin 2 5t t t t t r i j k v i j k

2 2 2 2 212cos 2 12sin 2 5 144cos 2 144sin 2 25 13;t t t t v

512 1213 13 13cos 2 sin 2t t v

vT i j k and Length 00 0| | 13 13 13dt dt t

v

3. 23 2 1 2 2 1 223 1 1 ;t t t t t r i k v i k v 1

1 1t

t t v

vT i k and

Length 88 3 2 522

3 30 0= 1 (1 )t dt t

4. 2 2 2(2 ) ( 1) 1 ( 1) 1 3;t t t r i j k v i j k v 1 1 13 3 3

vvT i j k and

Length 33

0 03 3 3 3dt t

5. 3 3 2 2cos sin 3cos sin 3sin cost t t t t t r j k v j k

2 22 2 2 2 2 23cos sin 3sin cos 9cos sin cos sin 3 cos sin ;t t t t t t t t t t v



2 23cos sin 3sin cos23 cos sin 3cos sin cos sin , if 0 ,t t t t

t t t tt t t v

vT j k j k and

Length 22 2 2 3 3 3

2 4 20 0 0 03 cos sin 3cos sin sin 2 cos 2t t dt t t dt t dt t

6. 2 2 23 3 3 2 2 2 2 2 2 4 26 2 3 18 6 9 18 6 9 441 21 ;t t t t t t t t t t t r i j k v i j k v

2 2 2

2 2 218 6 9 6 32

7 7 721 21 21t t tt t t

vvT i j k i j k and Length

22 2 31 1

21 7 49t dt t

7. 3 2 1 22 23cos sin cos sin sin cos 2t t t t t t t t t t t t r i j k v i j k

22 2 22cos sin sin cos 2 1 2 1 1 1,t t t t t t t t t t t t v if 0;t

1 22cos sin sin cos1 1 1

tt t t t t tt t t v

vT i j k and Length 2 2

2 20 01 tt dt t

8. sin cos cos sin sin cos sin cos sin cost t t t t t t t t t t t t t r i j v i j

2 2 2cos sin cos sin | |t t t t t t t t t t t i j v if 2 2;t

cos sin cos sint t t tt t t t v

vT i j i j and Length 2 2222 2

1tt dt

9. Let 0P t denote the point. Then 5cos 5sin 12t t v i j k and 0 2 20

26 25cos 25sin 144t

t t dt

00 00

13 13 2 ,t

dt t t and the point is 2 5 sin 2 , 5 cos 2 , 24 0, 5, 24P

10. Let 0P t denote the point. Then 12cos 12sin 5t t v i j k and

0 02 20 00 0

13 144cos 144sin 25 13 13 ,t t

t t dt dt t t and the point is

12sin , 12cos , 5 0, 12, 5P

11. 2 2 24cos 4sin 3 4sin 4cos 3 4sin 4cos 3t t t t t t t r i j k v i j k v

025 5 5 5

ts t d t Length 5

2 2s

12. cos sin sin cos sin sin cos cos cos sint t t t t t t t t t t t t t r i j v i j

2 2 2cos sin cos cos ,t t t t t t t t t t i j v since 2

2 20t tt s t d

Length 22 22 32 2 2 8s s

13.3 Arc Length in Space 977


13. cos sin cos sin sin cost t t t t t t te t e t e e t e t e t e t e r i j k v i j k

2 2 2 2

0cos sin sin cos 3 3 3 3 3

tt t t t t t t te t e t e t e t e e e s t e d e v

Length ln 4 3 340 ln 4 0 3 3s s e

14. 2 2 20

(1 2 ) (1 3 ) (6 6 ) 2 3 6 2 3 ( 6) 7 ( ) 7 7t

t t t s t d t r i j k v i j k v

Length (0) ( 1) 0 ( 7) 7s s

15. 2 2 22 22 2 1 2 2 2 2 2 2 4 4t t t t t t r i j k v i j k v

22 1 t Length 11 2 2 21

2 20 02 1 2 1 ln 1 2 ln 1 2tt dt t t t

16. Let the helix make one complete turn from 0t to 2 .t Note that the radius of the cylinder is 1

the circumference of the base is 2 . When 2 ,t the point P is

cos 2 , sin 2 , 2 1, 0, 2 the cylinder is 2 units high. Cut the cylinder along PQ and flatten. The resulting rectangle has a width equal to the circumference of the cylinder 2 and a height equal to 2 , the height of the cylinder. Therefore, the rectangle is a square and the portion of the helix from 0t to 2t is its diagonal.

17. (a) cos sin 1 cos , 0 2 cos , sin , 1 cost t t t x t y t z t r i j k 2 2 2 2cos sin 1,x y t t a right circular cylinder with the -axisz as the axis and radius 1.

Therefore cos , sin , 1 cosP t t t lies on the cylinder 2 2 1; 0 1, 0, 0x y t P is on the curve;

2 0, 1, 1t Q is on the curve; 1, 0, 2t R is on the curve. Then PQ i j k

and

2 2 1 1 1 2 22 0 2

PR PQ PR

i j ki k i k

is a vector normal to the plane of P, Q, and R. Then

the plane containing P, Q, and R has an equation 2 2 2(1) 2(0)x z or 1.x z Any point on the curve will satisfy this equation since cos (1 cos ) 1.x z t t Therefore, any point on the curve lies on

the intersection of the cylinder 2 2 1x y and the plane 1x z the curve is an ellipse.

(b) 2 2 2 2sin cos sin sin cos sin 1 sint t t t t t t v i j k v 2

sin cos sin 32 22 21 sin

(0) , , ( ) ,t t t

t

i j kv i k i k

vT T j T T j T



(c) cos sin cos ;t t t a i j k n i k is normal to the plane 1x z

cos cos 0t t n a a is orthogonal to n a is parallel to the plane;

2(0) , , a i k a j 32( ) , a i k a j

(d) 21 sin t v (See part (b))

2 20

1 sinL t dt

(e) 7.64L (by Mathematica)

18. (a) cos 4 sin 4 4 4sin 4 4cos 4 4t t t t t r i j k v i j k

2 2 24sin 4 4cos 4 4 32 4 2t t v Length 22

0 04 2 4 2 2 2dt t

(b) 1 1 12 2 2 2 2 2 2 2cos sin sin cost t t t t r i j k v i j k

2 2 2 21 1 1 1 12 2 2 2 2 4 4 2sin cost t v Length

44 2 22 20 0

2 2dt t

(c) 2 2 2cos sin sin cos sin cos 1 1 1t t t t t t t r i j k v i j k v

2 Length 00

2 22 2 2 2dt t

19. PQB QOB t and PQ arc ( )AQ t since PQ length of the unwound string length of arc ( );AQ thus cos sin ,x OB BC OB DP t t t and

sin cosy PC QB QD t t t

20. cos sin sin cos sin cos sin cos ( sin ) cost t t t t t t t t t t t t t r i j v i j

2 2 2cos sin cos sin , 0t t t t t t t t t t t t i j v cos sin cos sint t t tt t t t v

vT i j i j

21. 0 1 0 2 0 3 1 2 3 ,d d ddt dt dtx t u y t u z t u u u u v i j k i j k u so

0 0 0( ) 1

t t ts t dt d d t v u

22. 22 3 2 2 2 2 2 4( ) ( ) 2 3 ( ) (1) (2 ) 3 1 4 9 .t t t t t t t t t t t t r i j k v i j k v 0, 0, 0 0t

and 2, 4, 8 2.t Thus 2 2 2 40 0

( ) 1 4 9 .L t dt t t dt v Using Simpson’s rule with 10n and

2 010 0.2x

13.4 Curvature and Normal Vectors of a Curve 979


0.23 (0) 4 (0.2) 2 (0.4) 4 (0.6) 2 (0.8) 4 (1) 2 (1.2) 4 (1.4) 2 (1.6)L v v v v v v v v v

4 (1.8) (2) v v

0.23 1 4(1.0837) 2(1.3676) 4(1.8991) 2(2.6919) 4(3.7417) 2(5.0421) 4(6.5890) 2(8.3800)

0.234(10.4134) 12.6886 (143.5594) 9.5706

13.4 CURVATURE AND NORMAL VECTORS OF A CURVE

1. 22 2sincosln cos tan 1 tan sec sec sec ,t

tt t t t t t t r i j v i j i j v

since tan12 2 sec sec cos sin ;t

t tt t t vvT i j i j sin cosd

dt t t T i j

2 2sin cos 1 sin cos ;ddt

ddt

ddt t t t t

T

TT N i j 1 1

sec 1 cos .ddt t t T

vκ

2. 2 2 2sec tansecln sec tan tan 1 sec sec sec ,t t

tt t t t t t t r i j v i j i j v

since tan 12 2 sec sec sin cos ;t

t tt t t vvT i j i j cos sind

dt t t T i j

2 2cos sin 1 cos sin ;ddt

ddt

ddt t t t t

T

TT N i j 1 1

sec 1 cos .ddt t t T

vκ=

3. 2 2

22 2 2 222 1 2 1

2 3 5 2 2 2 2 2 1 tt t

t t t t t

vvr i j v i j v T i j

2 21

1 1;t

t t i j

3 3 3 3 2 222 2 2 2

2 2

1 1 1 1111 1 1 1

d dt tdt dt ttt t t t

T Ti j

2 2

11 1

;ddt

ddt

tt t

T

TN i j 2 3 22 2

1 1 1 112 1 2 1

ddt tt t

Tvκ=

4. 2 2 2cos sin sin cos cos sin cos sin ,t t t t t t t t t t t t t t t t t r i j v i j v

since cos sin0 cos sin ;t t t ttt t t i jv

vT i j sin cosddt t t T i j


ddt

ddt t t t t

T

TT N i j 1 1 1

| | 1ddt t t T

vκ

5. (a) ( )1( )

( ) .d xdtx

x Tvκ Now, 2( ) ( ) 1 ( )f x x f x v i j v

1 2 1 22 21 ( ) ( ) 1 ( ) .f x f x f x

vvT i j Thus

3 2 3 22 2

( ) ( ) ( )

1 ( ) 1 ( )( )d f x f x f x

dtf x f x

x

T i j



2 2

3 2 3 2 3 22 2 2

2 2( ) 1 ( ) ( )( ) ( ) ( ) ( )

1 ( )1 ( ) 1 ( ) 1 ( )

f x f x f xd x f x f x f xdt f xf x f x f x

T

Thus 1 2 3 222 2

( ) ( )11 ( )1 ( ) 1 ( )

( ) f x f x

f xf x f xx

κ

(b)

22 2

2 3 2 32

sec2 sec1cos sec1 tan

ln cos sin tan secxdy d y x

dx x dx xxy x x x x

κ

1sec cos ,x x since 2 2x

(c) Note that ( ) 0f x at an inflection point.

6. (a) 2 2 2 2

2 2( ) ( ) yxx y x y

f t g t x y x y x y

vvr i j i j v i j v T i j

22 2

3 2 3 2 3 2 3 2 3 2 22 2 2 2 2 2 2 2 2 2

2 2

;y x yx xy yx xyy yx xy x xy yx y yx xy x xy yxd d

dt dt x yx y x y x y x y x y

T Ti j

3 22 22 2 2 21 1 .yx xy xy yxd

dt x yx y x y

T

v

κ

(b) ( ) ln (sin ) , 0t t t t x t r i j and 2cossinln sin 1, 0; cot , csct

ty t x x y t y t

2

2

3/2 32

csc 0 csccsc1 cot

sint t

ttt

κ

(c) 1 1( ) tan sinh ln cosh tan sinht t t x t r i j and 2cosh 1

cosh1 sinhln cosh t

tty t x

3 2

2 2

sech sech tanh2sinhcosh sech tanh

sech , sech tanh ; tanh , sech |sech | secht t tt

t t tt x t t y t y t t t

κ

7. (a) ( ) ( ) ( ) ( ) ( )t f t g t f t g t r i j v i j is tangent to the curve at the point ( ), ( ) ;f t g t

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0;g t f t f t g t g t f t f t g t n v i j i j ( ) 0; n v n v thus, n and –n are both normal to the curve at the point

(b) 2 2 2( ) 2 2t t tt t e e e r i j v i j n i j points toward the concave side of the curve; nnN and

2

4 44 2 1

1 4 1 44 1

t

t tt e

e ee

n N i j

(c) 2 2

24 4

( ) 4 t tt t

t t t

r i j v i j n i j points toward the concave side of the curve;

nnN and

2

2 222 1

24 41 4t

t tt t

n N i j

8. (a) 3 2 213( )t t t t t r i j v i j n i j points toward the concave side of the curve when 0t and

2t n i j points toward the concave side when 421

10

tt t

N i j for 0t and

421

1 tt

N i j for 0t



(b) From part (a),

2 3 6 2

3 2 3 2 34 4 4 4 4

4 2 2 4 411 1 1 1 1

1 d dt t t t tdt dtt t t t t

t

T Tv T i j i j

4 3 3

4 3 2 3 2 4 44 4

2 1 2 221 1 11 1

; ; 0.ddt

ddt

t t t t t ttt t t t tt t

t

T

TN i j i j N does not exist at 0,t where

the curve has a point of inflection; 0

0ddt t

T so the curvature 0d d dtds ds ds T T

κ= at

10 ddst TN

κ is undefined. Since x t and 3 31 1

3 3 ,y t y x the curve is the cubic power curve which is concave down for 0x t and concave up for 0.x t

9. 2 2 23sin 3cos 4 3cos 3sin 4 3cos 3sin 4 25 5t t t t t t t r i j k v i j k v

2 23 3 3 3 3 3 345 5 5 5 5 5 5 5cos sin sin cos sin cosd d

dt dtt t t t t t T TvvT i j k i j

3 315 5 25sin cos ;

ddtddt

t t T

TN i jκ

10. 2 2 2cos sin sin cos 3 cos sin cos sint t t t t t t t t t t t t t t r i j k v i j v

,t t if 0 cos sin , 0 sin cosddtt t t t t t v T

vT i j i j


ddt

ddt t t t t

T

TT N i j 1 11t t κ

11. cos sin 2 cos sin sin cost t t t t te t e t e t e t e t e t r i j k v i j

2 2 2cos sin sin cos 2 2;t t t t t te t e t e t e t e e v cos sin sin cos2 2

t t t t vvT i j

2 2sin cos cos sin sin cos cos sin

2 2 2 21d dt t t t t t t t

dt dt T Ti j

cos sin sin cos2 2

;ddt

ddt

t t t t T

TN i j 1 1 12 2

1t tddt e e

Tvκ

12. 2 2 26sin 2 6cos 2 5 12cos 2 12sin 2 5 12cos 2 12sin 2 5t t t t t t t r i j k v i j k v

512 12 24 2413 13 13 13 13169 13 cos 2 sin 2 sin 2 cos 2d

dtt t t t TvvT i j k i j

2 224 24 2413 13 13sin 2 cos 2 sin 2 cos 2 ;

ddt

ddt

ddt t t t t

T

TT N i j 1 1 24 24

13 13 169 .ddt T

vκ



13. 3 2 2 4 2 23 2 , 0 1,t t t t t t t t t r i j v i j v since

2 21

1 10 t

t tt

v

vT i j

2

3 2 3 2 3/2 3/2 3 22 2 2 2 2

2 2

11 1 111 1 1 1 1

d dt t tdt dt tt t t t t

T Ti j

2 21

1 1;

ddt

ddt

tt t

T

TN i j 2 3 22 2

1 1 1 111 1

.ddt tt t t t

Tvκ

14. 3 3 2 22cos sin , 0 3cos sin 3sin cost t t t t t t r i j v i j

2 22 2 4 2 4 23cos sin 3sin cos 9cos sin 9sin cos 3cos sin ,t t t t t t t t t t v since 20 t

2 2cos sin sin cos sin cos 1d ddt dtt t t t t t T Tv

vT i j i j

sin cos ;ddt

ddt

t t T

TN i j 1 1 13cos sin 3cos sin1 .d

dt t t t t Tvκ

15. 2 2cosh , 0 sinh 1 sinh cosh cosht t t t ta a a a at a a r i j v i j v

21 1sech tanh sech tanh sechdt t t t ta a dt a a a a a Tv

vT i j i j

2 22 2 41 1 1sech tanh sech sech tanh sech ;

ddt

ddt

d t t t t t tdt a a a a a a aa a

T

TT N i j

21 1 1 1cosh

sech sech .ta

d t tdt a a a av

T

16. 22cosh sinh sinh cosh sinh cosh 1 2 cosht t t t t t t t r i j k v i j k v

21 1 1 1 12 2 2 2 2

tanh sech sech sech tanhddtt t t t t Tv

vT i j k i k

4 2 21 1 12 2 2

sech sech tanh sech sech tanh ;ddt

ddt

ddt t t t t t t

T

TT N i k

21 1 1 122cosh 2

sech sech .ddt t

t t Tvκ

17. 2 2 2 ;y ax y ax y a from Exercise 5(a), 3 22 2

3 22 2 2

1 4( ) 2 1 4a

a xx a a x

κ

5 22 2 232( ) 2 1 4 8 ;x a a x a x

κ thus, ( ) 0 0.x x κ Now, ( ) 0x κ for 0x and ( ) 0x κ for

0x so that ( )xκ has an absolute maximum at 0x which is the vertex of the parabola. Since 0x is the only critical point for ( ),xκ the curvature has no minimum value.



18. cos sin sin cos cos sina t b t a t b t a t b t r i j v i j a i j

sin cos 0 ,cos sin 0

a t b t ab ab aba t b t

i j kv a k v a since 30; ( )a b t v a

vκ

3 22 2 2 2sin cos ;ab a t b t

5 22 2 2 2 2 232( ) sin cos 2 sin cos 2 sin cost ab a t b t a t t b t t

κ

5 22 2 2 2 2 232 sin sin cos ;ab a b t a t b t

thus, ( ) 0 sin 2 0 0,t t t κ identifying points

on the major axis, or 32 2,t identifying points on the minor axis. Furthermore, ( ) 0t κ for 20 t and

for 32 ; ( ) 0t t κ for 2 t and 3

2 2 .t Therefore, the points associated with 0t and t

on the major axis give absolute maximum curvature and the points associated with 2t and 32t on the

minor axis give absolute minimum curvature.

19.

2 2

2 2 22 2

2 2; 0 0a d a b dda daa b a b

a b a b a b

κ κκ since , 0.a b Now, 0dda κ if a b

and 0dda κ if a b κ is at a maximum for a b and 2 2

12( ) b

bb bb

κ is the maximum value of κ.

20. (a) From Example 5, the curvature of the helix ( ) cos sin , , 0t a t a t bt a b r i j k is 2 2 ;aa b

κ also

2 2 .a b v For the helix ( ) 3cos 3sin , 0 4 , 3t t t t t a r i j k and 2 23 3

103 11b

κ

and 44 3 3 12

100 10 10010 10K dt t

v

(b) 2y x x t and 2 2 2, ( ) 2 1 4 ;y t t r t t t t t i j v i j v

2

3 2 3 2 3 22 2 2 2 22 4 16 41 2 2

1 41 4 1 4 1 4 1 4 1 4; ; .d dt t t

dt dt tt t t t t

T TT i j i j Thus 2221

1 41 4ttt

κ

32

2

1 4.

t Then

3 2 2 22

022 2 2 201 4 1 4 1 41 4

1 4 lim limb

at t ta btK t dt dt dt dt

01 1 1 12 20

lim tan 2 lim tan 2 lim tan 2 lim tan 2b

aa b a bt t a b

21. 22 2 22 2sin cos 1 cos 1 cos 1 cos 1;t t t t t r i j v i j v v

2

3 2 3 2 2 22 2 222

sinsincos sin cos sin 111 cos 1 cos1 cos 1 cos 1 cos

; 1.tt d d dt t tdt dt dtt tt t t

i j T T TvvT i j Thus

1 12 1 11 1 1 κ and the center is 2 2

2 2,0 1x y

22. 2 2

2 2 2 2 2 2

21 2 11 2 1 4 1

1 12ln 1 1 ;t t t

t t t t t t t tt t

v

vr i j v i j v T i j

22 2 2

2 2 4 22 2 2

2 1 4 1 164 211 1 1.

t t td dtdt dt tt t t

T Ti j Thus

2 2

2 2 2221 2

1 1 1

d t tdt t t t

Tvκ=



22 1 1

22(1) 2.

κκ The circle of curvature is tangent to the curve at 0, 2P circle has same

tangent as the curve (1) 2 v i is tangent to the circle the center lies on the -axis.y If 1( 0),t t then

22 2 2 11 0 2 1 0 1 2 2ttt t t t t since 1 10 2 2 2t tt t t y on

both sides of 0, 2 the curve is concave down center of circle of curvature is 0, 4

22 4 4x y is an equation of the circle of curvature

23. 2 ( ) 2y x f x x and ( ) 2f x

3 2 3 22 2

2 21 (2 ) 1 4x x

κ

24. 4 3

4 ( )xy f x x and 2( ) 3f x x

2

2

3 2 3 22 63

3 311

x x

xx

κ

25. sin ( ) cosy x f x x and ( ) sinf x x

3 2 3 22 2

sin sin

1 cos 1 cos

x x

x x

κ

26. ( )x xy e f x e and ( ) xf x e

3 2 3 22 2

| |

11

x x

xx

e e

ee

κ

27. ( ) lnf x x 1( ) xf x and 21( )

xf x

12

3 2 3 22 21 11( ) x

x

x

xx

κ 1 23 22 23

232

1 (1) 1 2

1( )

x x x x

xx

κ

2

5 221 2

10x

x

1

2;x

so the maximum curvature is 3 21 22 3

.κ



28. 1( ) xxf x 2

1( 1)

( )x

f x

and 32

( 1)( )

xf x

233( 1)

3 2 3 22 41

2( 1)

2( 1)

( 1) 11

( ) x

x

x

xx

κ 3 2 1 24 2 3 4 33

234

( 1) 1 6( 1) 2( 1) ( 1) 1 4( 1)

( 1) 1( )

x x x x x

xx

κ

2 4

5 24

6( 1) 1 ( 1)

( 1) 10

x x

x

0;x

so the maximum curvature is 12

0 .κ

29. We will use the formula 1 ( )( )

d aa dt

Tv

κ to find the curvature at the point 2,a a .

By Example 4 in Section 13.4, 2( ) 1 4t t v and

3/222 1 4 2 .d t tdt

T i j

At t a this gives

3/22 23/22 2

1 2 2( ) 1 4 1 4( ) 1 4 1 4

d a a aa dt a a

Tv

κ . Thus the radius of

the osculating circle is 3/221 1 4 .2

r a To show that the given formulas for center and radius are correct we

must first show that the distance between 2,a a and 3 2 14 , 32

a a is .r This distance is

22 33 2 2 6 4 2 21 1 14 3 16 12 3 1 4

2 4 2a a a a a a a a

as required. Finally we must show

that the line containing the points 3 2 14 , 32

a a and 2,a a is perpendicular to the tangent line at 2, ,a a

which has slope 2 .a This requires that 2 2

3

13 12 ,24

a a

aa a

which is correct.

30. By Exercise 29, for 1,a the center of the osculating circle is at 74,2

and its radius is 5 5 .

2 A

parametrization of this circle is 5 5 7 5 5( ) 4 cos , ( ) sin .2 2 2

x y


Maple: with( plots );

r : t - [3*cos(t),5*sin(t)];

1o : 0;

hi : 2*Pi;

t0 : Pi/4;



P1: plot( [r(t)[], t 1o..hi] ):

display( P1, scaling constrained, title "#31(a) (Section 13.4)" );

CURVATURE : (x,y,t) - simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/

(diff(x,t)^2+diff(y,t)^2)^(3/2));

kappa : eval(CURVATURE(r(t)[],t),t t0);

UnitNormal : (x,y,t) - expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) );

N : eval( UnitNormal(r(t)[],t), t t0 );

C : expand( r(t0) N/kappa );

OscCircle : (x-C[1]^2+(y-C[2])^2 1/kappa^2;

evalf( OscCircle );

P2 : implicitplot( (x-C[1])^2+(y-C[2])^2 1/kappa^2, x -7..4, -4..6, color blue ):y

display( [P1,P2], scaling constrained, title "#31(e) (Section 13.4)" );

Mathematica: (assigned functions and parameters may vary) In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot". Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with

the word, "Cross". However, the Cross command assumes the vectors are in three dimensions. For the purposes of applying the cross product command, we will define the position vector r as a three

dimensional vector with zero for its z-component. For graphing, we will use only the first two components. Clear[r, t, x, y]

_r[t ] {3 Cos[t], 5 Sin[t] }

t0 π/4; tmin 0; tmax 2π;

_r2[t ] {r[t][[1]], r[t][[2]]}

pp ParametricPlot[r2[t], {t, tmin, tmax}];

_mag[v ] Sqrt[v.v]

_vel[t ] r'[t]

_speed[t ] mag[vel[t]]

_acc[t ] vel'[t]

3_curv[t ] mag[Cross[vel[t],acc[t]]]/speed[t] //Simplify

_unittan[t ] vel[t]/speed[t]//Simplify

_unitnorm[t ] unittan'[t] / mag[unittan'[t]]

ctr r[t0] (1/ curv[t0]) unitnorm[t0] //Simplify

{a,b} {ctr[[1]], ctr[[2]]}

To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve.

Graphics`ImplicitPlot`

2pc ImplicitPlot[(x a)^2 (y b)^2 1/curv[t0] , {x, 8, 8}, {y, 8, 8}]

radius Graphics[Line[{{a, b}, r2[t0]}]]

Show[pp, pc, radius, AspectRatio 1]

13.5 Tangential and Normal Components of Acceleration 987


13.5 TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION

1. 2 2 2cos sin sin cos sin cosa t a t bt a t a t b a t a t b r i j k v i j k v

2 2 0;dT dta b a v 2 2 2cos sin cos sina t a t a t a t a a a i j a

2 22 20 (0)N Ta a a a a a a a a T N N

2. 2 2 21 3 2 3 3 3 3 1 ( 3) 19 0;dT dtt t t a r i j k v i j k v v

2 2 0 (0) (0)N Ta a a 0 a a T N 0

3. 1 22 2 2 2 2 212( 1) 2 2 2 1 2 (2 ) 5 4 5 4 (8 )Tt t t t t t a t t

r i j k v i j k v

1 22 4 439

4 5 4 (1) ;Tt t a

2 (1) 2 (1) 2 a k a k a

22 2 2 2 5 2 5204 43 9 3 3 32 (1)N Ta a a a T N

4. 2cos sin cos sin sin cos 2t t t t t t t t t t t t r i j k v i j k

2

1 22 2 2 2 2 512 5 1

cos sin sin cos 2 5 1 5 1 (10 ) tT

tt t t t t t t t a t t

v

(0) 0;Ta 2 22sin cos 2cos sin 2 (0) 2 2 (0) 2 2 2 2t t t t t t a i j k a j k a

22 2 22 2 0 2 2 (0) (0) 2 2 2 2N Ta a a a T N N

5. 2 222 3 3 2 2 2 21 13 3 2 1 1 2 1 1t t t t t t t t t t t r i j k v i j k v

4 2 22 2 1 2 1 2 2 (0) 0;T Tt t t a t a 2 2 2 (0) 2 (0) 2t t a i j k a i a

2 2 2 22 0 2 (0) (0) 2 2N Ta a a a T N N

6. cos sin 2 cos sin sin cos 2t t t t t t t te t e t e e t e t e t e t e r i j k v i j k

2 2 2 2cos sin sin cos 2 4 2 2 (0) 2;t t t t t t t tT Te t e t e t e t e e e a e a v

cos sin sin cos sin cos cos sin 2t t t t t t t t te t e t e t e t e t e t e t e t e a i j k

222 sin 2 cos 2 (0) 2 2 (0) 2 2 6t t te t e t e i j k a j k a

22 2 26 2 2 (0) 2 2N Ta a a a T N



7. 2 2cos sin sin cos sin cos 1t t t t t t r i j k v i j v

2 24 2 2sin cos ;t t v

vT i j T i j cos (sin )ddt t t T i j

2 2 2 24 2 2cos sin 1 cos sin ;

ddt

ddt

ddt t t t t

T

TT N i j N i j

4sin cos 0 ,cos sin 0

t tt t

i j kB T N k B k the normal to the osculating plane; 2 2

4 2 2 r i j k

2 22 2, , 1P lies on the osculating plane 2 2

2 20 0 ( 1) 0 1x y z z is the

osculating plane; T is normal to the normal plane 2 2 2 22 2 2 2 0 ( 1) 0x y z

2 22 2 0 0x y x y is the normal plane; N is normal to the rectifying plane

22 2 2 2 2 2

2 2 2 2 20 ( 1) 0 1 2x y z x y x y is the rectifying

plane.

8. 2 2cos sin sin cos sin cos 1 2t t t t t t t r i j k v i j k v

1 1 1 1 12 2 2 2 2

sin cos cos sinddtt t t t Tv

vT i j k i j

2 21 1 12 2 2

cos sin cos sin ;ddt

ddt

ddt t t t t

T

TT N i j thus 1 1

2 2(0) T j k and (0) N i

1 1 1 12 2 2 2

(0) 0 ,

1 0 0

i j kB j k the normal to the osculating plane; (0) 1, 0, 0P r i lies on the

osculating plane 1 12 2

0 1 0 0 0 0x y z y z is the osculating plane; T is normal to the

normal plane 1 12 2

0 1 0 0 0 0x y z y z is the normal plane; N is normal to the

rectifying plane 1 1 0 0 0 0 0 1x y z x is the rectifying plane.

9. By Exercise 9 in Section 13.4, 3 3 45 5 5cos sint t T i j k and sin cost t N i j so that

3 3 34 4 45 5 5 5 5 5cos sin cos sin .

sin cos 0

t t t t

t t

i j kB T N i j k Also 3cos 3sin 4t t v i j k

3sin 3cos 3cos 3sinddtt t t t aa i j i j and 3cos 3sin 4

3sin 3cos 0t tt t

i j kv a



2 22 212cos 12sin 9 | | 12cos 12sin ( 9) 225.t t t t i j k v a Thus

2 2

3cos 3sin 43sin 3cos 0

4 9sin 9cos3cos 3sin 0 36 4225 225 225 25

t tt t

t tt t

10. By Exercise 10 in Section 13.4, cos sint t T i j and sin cos ;t t N i j thus

2 2cos sin 0 cos sin .sin cos 0

t t t tt t

i j kB T N k k Also cos sint t t t v i j

sin cos cos sin cos sin sin sin cos cosddtt t t t t t t t t t t t t t aa i j i j

cos 2sin 2cos sin .t t t t t t i j Thus cos sin 0sin cos cos sin 0

t t t tt t t t t t

i j kv a

222 2 4cos cos sin sin sin cos .t t t t t t t t t t t t t k k v a

Thus 4 4

cos sin 0cos sin sin cos 02sin cos 2cos sin 0 0 0

t t t tt t t t tt t t t t

t t

11. By Exercise 11 in Section 13.4, cos sin sin cos2 2

t t t t T i j and cos sin sin cos2 2

;t t t t N i j Thus

2 2 2 2cos sin sin cos cos 2cos sin sin sin 2sin cos cos2 22 2

cos sin sin cos2 2

0

0

t t t t t t t t t t t t

t t t t

i j kB T N k

1 sin (2 ) 1 sin (2 )2 2 .t t

k k Also, cos sin sin cost t t te t e t e t e t v i j

sin cos cos sin cos sin sin cost t t te t t e t t e t t e t t a i j

2 sin 2 cos 2 cos sin 2 sin cos .dt t t tdte t e t e t t e t t ai j i j

Thus 222 2 4cos sin sin cos 0 2 2 4 .

2 sin 2 cos 0

t t t t t

t t

e t t e t t e e e

e t e t

i j k

v a k v a

Thus

4

cos sin sin cos 0

2 sin 2 cos 0

2 cos sin 2 sin cos 0

40

t t

t t

t t

t

e t t e t t

e t e t

e t t e t t

e



12. By Exercise 12 in Section 13.4, 512 1213 13 13cos 2 sin 2t t T i j k and ( sin 2 ) (cos 2 )t t N i j Thus

5 5 512 12 1213 13 13 13 13 13cos 2 sin 2 cos 2 sin 2 .

sin 2 cos 2 0

t t t t

t t

i j kB T N i j k

Also, 12cos 2 12sin 2 5 24sin 2 24cos 2t t t t v i j k a i j and 48cos 2 48sin 2ddt t t a i j

12cos 2 12sin 2 5 120cos 2 120sin 2 28824sin 2 24cos 2 0

t t t tt t

i j kv a i j k

2 2 2 2 2 2 2 2120cos 2 120sin 2 ( 288) 120 cos 2 sin 2 288 97344.t t t t v a Thus

12cos 2 12sin 2 524sin 2 24cos 2 048cos 2 48sin 2 0 5 ( 24 48) 10

97344 97344 169

t tt tt t

13. By Exercise 13 in Section 13.4, 2 2

11 1

tt t

T i j and 2 21

1 1t

t t N i j so that

2 2

2 2

11 1

11 1

0 .

0

tt t

tt t

i j kB T N k Also, 2 2 2d

dtt t t av i j a i j i so that

2 02 1 0 0 02 0 0

t tt

14. By Exercise 14 in Section 13.4, cos sint t T i j and sin cost t N i j so that

cos sin 0 .sin cos 0

t tt t

i j k

B T N k Also, 2 23cos sin 3sin cost t t t v i j

2 2 2 23cos sin 3sin cos 3cos sin 3sin cosdd d d d d ddt dt dt dt dt dt dtt t t t t t t t

aa i j i j

2 2

2 2

2 2

3cos sin 3sin cos 0

3cos sin 3sin cos 0 0 0

3cos sin 3sin cos 0

d ddt dt

d d d ddt dt dt dt

t t t t

t t t t

t t t t



15. By Exercise 15 in Section 13.4, sech tanht ta a v

vT i j and tanh secht ta a N i j so that

sech tanh 0 .

tanh sech 0

t ta a

t ta a

i j k

B T N k Also, 21 1sinh cosh sinhdt t t

a a a dt aa av i j a j j so

that

2

1

1

1 sinh 0

0 cosh 0 0 0

0 sinh 0

ta

ta a

taa

16. By Exercise 16 in Section 13.4, 1 1 12 2 2

tanh secht t T i j k and sech tanht t N i k so that

1 1 1 1 1 12 2 2 2 2 2

tanh sech tanh sech .

sech 0 tanh

t t t t

t t

i j kB T N i j k Also, sinh cosht t v i j k

cosh sinh sinh coshddtt t t t aa i j i j and sinh cosh 1

cosh sinh 0t tt t

i j kv a

22 2 2 2sinh cosh cosh sinh sinh cosh sinh cosh 1.t t t t t t t t i j k i j k v a Thus

2 2 2 2 2

sinh cosh 1cosh sinh 0sinh cosh 0 1 1sinh cosh 1 sinh cosh 1 2cosh

.

t tt tt t

t t t t t

17. Yes. If the car is moving along a curved path, then 20 and | | 0 .N T Na k a a v a T N 0κ

18. constant 0dT Ndta a v v a N is orthogonal to T the acceleration is normal to the path

19. 0 0dT dta a v a T v v is constant

20. ( ) ,T Nt a a a T N where (10) 0d dT dt dta v and 2 100 0 100 .Na v a T Nκκ κ Now, from

Exercise 5(a) Section 13.4, we find for 2( )y f x x that 3 2 3 2 3 22 2 2

( ) 2 21 (2 ) 1 41 ( )

;f x

x xf x

κ also,

2( )t t t r i j is the position vector of the moving mass 2

2 11 4

2 1 4 2 .t

t t t

v i j v T i j

At (0, 0): (0) , (0) T i N j and (0) 2 (100 ) 200 ;m m m F a N jκ κ

At 2 2 2 21 1 13 3 3 3 32, 2 : 2 2 2 , 2 , T i j i j N i j and 2

272 κ

2 2 400 2200 200127 3 3 81 81(100 )m m m m m F a N i j i jκ



21. 0 0 0 0.x At y Bt z Ct A B C r i j k v i j k a 0 v a 0 κ Since the curve is a plane curve, 0.

22. 20 0 0Na vκ κ (since the particle is moving, we cannot have zero speed) the curvature is zero so the particle is moving along a straight line

23. From Example 1, tv and Na t so that 2 22 1 1, 0Na t

N tta t t

vv

κκκ

24. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows:

2

( ) ( ) 0( ) ( ) 0( ) ( ) 0( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0

f t g tf t g tf t g td

dtf t g t f t g t f t g t f t g t

a

v ar i j v i j a i j i j

25. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ;t f t g t h t f t g t h t r i j k v i j k 0 ( ) 0 ( )h t h t C v k ( ) ( ) ( )t f t g t C r i j k and ( ) ( ) ( ) ( ) 0, ( ) 0a f a g a C f a g a r i j k 0 and 0 ( ) 0.C h t

26. From Example 2, 2 2sin cosa t a t b a b v i j k v

2 21 sin cos ;

a ba t a t b

vvT i j k

2 21 cos sind

dt a ba t a t

T i j

cos sin ;ddt

ddt

t t T

TN i j 2 2 2 2 2 2 2 2 2 2 2 2sin cos sin cos

cos sin 0

a t a t b b t b t aa b a b a b a b a b a b

t t

i j kB T N i j k

2 22 2 2 2 2 2 2 21 1 1cos sin ,d d b d b b

dt dt dt a ba b a b a b a bb t b t

B B B

vi j N N

which is consistent with the result in Example 2.


Maple: with( LinearAlgebra ); r : t*cos(t) | t*sin(t) | t ; t0 : sqrt(3); rr : eval( r, t t0 ); v : map( diff, r, t ); vv : eval( v, t t0 ); a : map( diff, v, t ); aa : eval( a, t t0 ); s : simplify(Norm( v, 2 )) assuming t::real; ss : eval( s, t t0 ); T : v/s; TT : vv/ss ;

13.6 Velocity and Acceleration in Polar Coordinates 993


q1: map( diff, simplify(T), t ): NN : simplify(eval( q1/Norm(q1,2), t t0 )); BB : CrossProduct( TT, NN ); kappa : Norm(CrossProduct(vv,aa),2)/ss^3; tau : simplify( Determinant(< vv, aa, eval(map(diff,a,t),t t0) >)/Norm(CrossProduct(vv,aa),2)^3 ); _a t : eval( diff( s, t ), t t0 ); _a n : evalf[4]( kappa*ss^2 );

Mathematica: (assigned functions and value for t0 will vary) Clear[t, v, a, t] _mag[vector ]: Sqrt[vector.vector] _Print["The position vector is", r[t ] {t Cos[t], t Sin[t], t}] _Print["The velocity vector is", v[t ] r'[t]] _Print["The acceleration vector is",a[t ] v'[t]] _Print["The speed is",speed[t ] mag[v[t]]//Simplify] _Print["The unit tangent vector is", utan[t ] v[t]/speed[t] //Simplify]

3_Print["The curvature is",curv[t ] mag[Cross[v[t],a[t]]] / speed[t] //Simplify]

2_Print["The torsion is", torsion[t ] Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]] //Simplify] _Print["The unit normal vector is", unorm[t ] utan'[t] / mag[utan'[t] //Simplify] _Print["The unit binormal vector is", ubinorm[t ] Cross[utan[t],unorm[t]] //Simplify] _Print["The tangential component of the acceleration is",at[t ] a[t].utan[t] //Simplify] _Print["The normal component of the acceleration is",an[t ] a[t].unorm[t] //Simplify] You can evaluate any of these functions at a specified value of t. t0 Sqrt[3] {utan[t0], unorm[t0], ubinorm[t0]} N[{utan[t0], unorm[t0], ubinorm[t0]}] {curv[t0], torsion[t0]} N[{curv[t0], torsion[t0]}] {at[t0], an[t0]} N[{at[t0], an[t0]}]

To verify that the tangential and normal components of the acceleration agree with the formulas in the book: at[t] speed'[t] //Simplify

2an[t] curv [t] speed[t] //Simplify

13.6 VELOCITY AND ACCELERATION IN POLAR COORDINATES

1. 2 0,ddt 2 0;r r r 2 (2) 2 2 ;r r v u u u u

2( ) (0 ( )(2) ) (( )(0) 2(2)(2))rt a u u 4 8r u u



2. 2 2 ,ddt t t 22 2 2 322

2 2 4 4 3

2 2 221 ;t t t ttr r r

2 2 2

2 221( ) ;t t t

r rt t v u u u u 3

2

3 2

22 2 21 1( ) 2 2t t t

rt t t t

a u u

3 3 2 3

3 2

2 2 2t t t t tr

u u

3. 3 0,ddt 1 cos sin 3 sin 3 cos 9 cosd d

dt dtr a r a a r a a

3 sin 1 cos (3) 3 sin 3 1 cosr ra a a a v u u u u

29 cos 1 cos (3) 1 cos 0 2 3 sin (3)ra a a a a u u

9 cos 9 9 cos 18 sin 9 2cos 1 18 sinr ra a a a a a u u u u

4. 2 2,ddt t sin 2 cos 2 2 4 cos 2 4 sin 2 2 4 cos 2d d

dt dtr a r a ta r ta a 216 sin 2 4 cos 2 ;t a a 4 cos 2 sin 2 (2 ) 4 cos 2 2 sin 2r rta a t ta ta v u u u u

2 216 sin 2 4 cos 2 sin 2 (2 ) sin 2 (2) 2 4 cos 2 (2 )rt a a a t a ta t a u u

2 2 216 sin 2 4 cos 2 4 sin 2 2 sin 2 16 cos 2rt a a t a a t a u u

2 220 sin 2 4 cos 2 2 sin 2 16 cos 2rt a a a t a u u

2 24 cos 2 5 sin 2 2 sin 2 8 cos 2ra t a t u u

5. 2 0,ddt 22 2 4a a a a ad d

dt dtr e r e a a e r a e a a e

2 (2) 2 2a a a ar ra e e a e e

v u u u u

2 2 24 (2) (0) 2 2 (2) 4 4 0 8a a a a a a ar ra e e e a e a e e a e

a u u u u

24 1 8a are a a e

u u

6. 1 ,t t te e e (1 sin ) cos sinr a t r a t r a t

cos 1 sin cos 1 sint tr ra t a t e a t a e t

v u u u u

2sin 1 sin 1 sin 2 cost t t

ra t a t e a t e a t e

a u u

2sin 1 sin 1 sin 2 cost t tra t a e t a e t a e t

u u

2sin 1 sin 1 sin 2cost tra t e t a e t t

u u

2sin 1 sin 2cos 1 sint tra t e t a e t t

u u

7. 2 2 0,t 2cos 4 8sin 4 32cos 4r t r t r t 8sin 4 2cos 4 (2) 8 sin 4 4 cos 4r rt t t t v u u u u

13.6 Velocity and Acceleration in Polar Coordinates 995


232cos 4 2cos 4 (2) 2cos 4 0 2 8sin 4 (2)rt t t t a u u 32cos 4 8cos 4 0 32sin 4 40 cos 4 32 sin 4r rt t t t t u u u u

8. 2

0 0

0 0

( 1) ( 1)20 01 ;r v GM e GM e

GM r re v v

Circle: 0

00 GMre v

Ellipse: 0 0

200 1 GM GM

r re v

Parabola: 0

201 GM

re v

Hyperbola: 0

201 GM

re v

9. 22GM GM GM

r rvr v v which is constant since G, M, and r (the radius of orbit) are constant

10. ( ) ( ) ( ) ( ) ( ) ( )1 1 1 1 12 2 2 2( ) ( ) ( ) ( ) ( ) ( ) ( )t t t t t t t t tA

t t t t tA t t t t t t t t r r r r r rr r r r r r r

( ) ( ) ( ) ( )1 1 1 1 12 2 2 2 20

( ) lim ( ) ( ) ( )d dt t t t t tdAt dt t dt dtt

t t t t

r rr r r rr r r r r r

11. 22 2 4 2 4 0 02 2 2 20 0 0 0 0 0

22 2 22 4 41 1 1 1r va a a

r v GMr v r vT e T e

(from Equation 5)

2 4 22 4 2 2 2 4 22 4 2 4 0 00 0 0 0 0 0 0 0 0 0

2 2 2 2 2 2 2 2 2 2 00 0 0 0 0

4 22 22 44 4 222 4

a GM r vr v r v GMr v r v GM r va aGM r GM GMr v G M r v G M r G M

a

2 4 1 224 a GMa (from Equation 10)

2 3 22

32 4 4a T

GM GMaT

12. For each of the planets listed we form the ratio 2 3

2/ .

(4 ) / ( )T a

GM The values we obtain are

Mercury 1.00225 Venus 1.00288 Mars 1.00252 Saturn 1.00019

These values are all close to 1, so they support Kepler’s third law.

13. Solve Kepler’s third law for a and double this result: 1/32

102

(365.256 days)2 29.925 10 m(4 ) / ( )GM

14. Solve Kepler’s third law for a and double this result: 1/32

102

(84 years)2 573.95 10 m(4 ) / ( )GM



15. Assuming Earth has a circular orbit with radius 6150 10 km, the rate of change of area is

26

9 2150 10 km

2.24 10 km /sec.365.256 days

16. Solving Kepler’s third law for T we find 2 3104 77.8 10 m 11.857 years.TGM

17. Solving Kepler’s third law for the mass M of the body around which Io is orbiting we find

3103

2 2 272 2

0.042 10 m4 4 1.876 10 kg.

(1.769 days)aM

T G G

18. To solve this we need a value for the mass of Earth, which is approximately 245.972 10 kg.M Solving Kepler’s third law for the orbital radius we get

1/31/3 1/3 1/32 2 2 6 24 54 4 2.36055 10 sec 5.972 10 kg 3.831 10 km.a T GM G

Since Earth’s radius is about 6371, the orbit of the moon is about 383,143 6371 376.772 km from the surface, assuming a circular orbit for the moon.

CHAPTER 13 PRACTICE EXERCISES

1. ( ) 4cos 2 sint t t r i j

4cosx t and 22

16 22 sin 1;yxy t

4sin 2 cost t v i j and

4cos 2 sin ;t t a i j

(0) 4 , (0) 2 , (0) 4 ; r i v j a i

4 4 42 2 , 2 2 , 2 2 ; r i j v i j a i j 2 216sin 2cost t v

2 2

14sin cos ;16sin 2cos

dT dt

t tat t

v at 20: 0, 0 4, 0 4 4 ,T Nt a a a a T N N

242 2;Naκ

v at 4 27 7 49

4 3 9 38 1: , 9 ,T Nt a a

4 27

3 3 , a T N 24 227

Naκ v

Chapter 13 Practice Exercises 997


2. ( ) 3 sec 3 tan 3 sect t t x t r i j and 22 2 2 2 2

3 33 tan sec tan 1 3;yxy t t t x y

23 sec tan 3 sect t t v i j and

2 3 23 sec tan 3 sec 2 3 sec tan ;t t t t t a i j

(0) 3 , (0) 3 , (0) 3 ; r i v j a i 2 2 43sec tan 3sect t t v

2 3 4

2 2 46sec tan 18sec tan

2 3sec tan 3sec;d t t t t

T dt t t ta

v

at 20: 0, 0 3,T Nt a a a

0 3 3 , a T N N 23 1

3 3Naκ

v

3. 22 2

2 23 2 3 2 3 2 3 22 2 2 21 111 1

1 1 1 1 .ttt t

t t t t t t

r i j v i j v

We want to maximize 22

2

1: d t

dt t

vv and

222

10 0 0.d t

dt tt

v For 0,t

222

10;t

t

for 0,t

222

max10t

t

v occurs when max0 1t v

4. cos sin cos sin sin cost t t t t te t e t e t e t e t e t r i j v i j

cos sin sin cos sin cos cos sin 2 sin 2 cos .t t t t t t t t t te t e t e t e t e t e t e t e t e t e t a i j i j

Let be the angle between r and a. Then

2 2

2 2 2 21 1 2 sin cos 2 sin cos

cos sin 2 sin 2 coscos cos

t t

t t t t

e t t e t t

e t e t e t e t

r ar a

21 10

22cos cos 0te

for all t

5. 3 4 v i j and 5 15 3 4 0 25 25;5 15 0

i j k

a i j v a k v ×a 3 32 2 25 1

553 4 5 κ v×a

vv

6.

3 22

3 2 3 2 5 22 2 2 232

1+1 1 1 2y x x x x x x xdκ

dxy

κ = e e e e e e e

3 2 5 2 5 2 5 22 3 2 2 2 2 2 21 3 1 1 1 3 1 1 2 ;x x x x x x x x x x xe e e e e e e e e e e

2 2 1 1 12 2 2

0 1 2 0 2 ln 2 ln 2 ln 2 ;x xdκdx e e x x y therefore κ is at a

maximum at the point 12

ln 2,



7. dydxdt dtx y r i j v i j and .dx

dty y v i Since the particle moves around the unit circle 2 2 1, 2 2 0 ( ) .dy dy dydx x dx x

dt dt dt y dt dt yx y x y y x Since dxdt y and ,dy

dt x

we have y x v i j at 1, 0 , v j and the motion is clockwise.

8. 3 2 2139 9 3 .dy dydx dx

dt dt dt dty x x x If ,x y r i j where x and y are differentiable functions of t,

then .dydxdt dt v i j Hence 4 4dx

dt v i and 2 21 13 3 (3) (4) 12dy dx

dt dtx v j at (3, 3).

Also, 22

2 2d yd x

dt dt a i j and 2 2

2 2

2 22 13 3 .d y dx d x

dtdt dtx x Hence

2

22 2d xdt

a i and 2

22 22 1

3 3(3)(4) (3) ( 2) 26d ydt

a j at the point ( , ) (3, 3).x y

9. ddtr orthogonal to 1 1 1

2 2 20 ( ) ,d d d ddt dt dt dt K r r rr r r r r r r r a constant. If ,x y r i j where x

and y are differentiable functions of t, then 2 2 2 2 ,x y x y K r r which is the equation of a circle centered at the origin.

10. (a)

(b) cos sint t v i j

2 2sin cos ;t t a i j

(0) v 0 and 2(0) ;a j

(1) 2v i and 2(1) ; a j

(2) v 0 and 2(2) ;a j

(3) 2 ;v i and 2(3) a j

(c) Forward speed at the topmost point is (1) (3) 2 ft/sec; v v since the circle makes 12 revolution per

second, the center moves ft parallel to the -axisx each second the forward speed of C is ft/sec.

11. 2 2 21 10 0 2 2sin 6.5 (44 ft/sec)(sin 45 )(3sec) 32 ft/sec (3sec) 6.5 66 2 144y y v t gt y

44.16 ft the shot put is on the ground. Now, 20 6.5 22 2 16 0 2.13secy t t t (the positive

root) (44 ft/sec)(cos 45 )(2.13sec) 66.27 ftx or about 66 ft, 3 in. from the stopboard

12. 22

02

(80 ft/sec)(sin 45 )sinmax 0 2 (2)(32 ft/sec )

7 ft 57 ftvgy y

13. 0 cosx v t and

21 10 02 2

00

sin sin210 2 coscos

sin tanv t gt v g ty

x vv ty v t gt

0 02 sin 2 cos tan10 0 2cos tan sin ,v v

gv v gt t which is the time when the golf ball hits the

upward slope. At this time 0 02 sin 2 cos tan 2 2 220 0 0cos sin cos cos tan .v v

g gx v v v Now



2 2 20 0sin cos cos tan2

cos cosv vx

gOR OR

2 20 0

22 cos 2 coscos tan sin cos cos sinsin

cos cos cosv v

g g

20

22 cos

cossin( ) .v

g

The distance OR is maximized when x

is maximized: 202 cos 2 sin 2 tan 0vdx

d g

2 2 4cos 2 sin 2 tan 0 cot 2 tan 0 cot 2 tan 2

14. (a) 0 cos 40x v t and 2 210 026.5 sin 40 6.5 sin 40 16 ;y v t gt v t t 5

12262 ftx and

5 262.4167

0 012 cos 400 ft 262 cos 40 or

ty v t v

and 2262.4167cos 40

0 6.5 sin 40 16t

t t

2 14.1684 3.764sec.t t Therefore, 0262.4167 cos 40 (3.764sec)v

262.4167

0 0cos 40 (3.764sec)91 ft/secv v

(b) 220

(91) sin 40sinmax 0 2 (2)(32)6.5 60 ftv

gy y

15. 2 2 222cos 2sin 2sin 2cos 2 2sin 2cos 2t t t t t t t t t r i j k v i j k v

22 1 t 2 24/4 2 2 2

4 16 4 160 0Length = 2 1+t 1 ln 1 1 ln 1dt t t t t

16. 22 23 2 1 2 1 23cos 3sin 2 3sin 3cos 3 3sin 3cos 3t t t t t t t t t r i j k v i j k v

33 3 2

0 09 9 3 1 Length = 3 1 2 1 14t t t dt t

17. 3 2 3 2 1 2 1 24 4 1 2 2 19 9 3 3 3 31 1 1 1t t t t t r i j k v i j k

2 2 21 2 1 2 1 2 1 22 2 1 2 2 1

3 3 3 3 3 31 1 1 1 1t t t t v T i j k

2 2 13 3 3(0) ; T i j k 1 2 1 2 21 1 1 1

3 3 3 3 31 1 (0) (0)d d ddt dt dtt t T T Ti j i j

1 12 2

(0) ; N i j 2 2 1 1 1 43 3 3 3 2 3 2 3 21 12 2

(0) (0) (0) ;

0

i j kB T N i j k

1 2 1 21 1 1 13 3 3 31 1 (0)t t a i j a i j and 2 2 1 2 2 1

3 3 3 3 3 31 13 3

(0) (0) (0)

0

i j kv i j k v a



23

3 32 21 1 4

9 9 9 3 31(0) ;κ v a

vi j k v a

2 2 13 3 31 13 3

1 21 13 186 6

2 22

3

003 2 3 21 1 1 1 1

6 6 6 6 61 1 (0) (0)t t

v aa i j a i j

18. sin 2 cos 2 2 sin 2 2 cos 2 cos 2 2 sin 2 2t t t t t t t te t e t e e t e t e t e t e r i j k v i j k

2 2 2sin 2 2 cos 2 cos 2 2 sin 2 2 3t t t t t te t e t e t e t e e v

1 2 1 2 2 2 1 23 3 3 3 3 3 3 3sin 2 cos 2 cos 2 sin 2 (0) ;t t t t v

vT i j k T i j k

2 4 2 4 2 4 23 3 3 3 3 3 3cos 2 sin 2 sin 2 cos 2 (0) (0) 5d d d

dt dt dtt t t t T T Ti j i j

2 43 3

2 53

1 25 5

(0) ;

i j

N i j 52 1 2 4 23 3 3 3 5 3 5 3 51 25 5

(0) (0) (0) ;

0

i j kB T N i j k

4 cos 2 3 sin 2 3 cos 2 4 sin 2 2 (0) 4 3 2t t t t te t e t e t e t e a i j k a i j k and

(0) 2 2 (0) (0) 2 1 2 8 4 10 64 16 100 6 54 3 2

i j kv i j k v a i j k v a and

36 5 2 5

93(0) 3 (0) ;κ v

4 cos 2 8 sin 2 3 sin 2 6 cos 2 3 cos 2 6 sin 2 4 sin 2 8 cos 2 2t t t t t t t t te t e t e t e t e t e t e t e t e a i j k

2 cos 2 11 sin 2 11 cos 2 2 sin 2 2 (0) 2 11 2t t t t te t e t e t e t e i j k a i j k

2

2 1 24 3 22 11 2 80 4

180 9(0)

v a

19. 2

4 42 2 41 1 1 4

2 17 171 11 (ln 2) ;

t

t tt t t e

e et e e e

r i j v i j v T i j T i j

4 2

3 2 3 24 42 2 32 8 4 1

1717 17 17 17 171 1(ln 2) (ln 2) ;

t t

t t

d de edt dte e

T Ti j i j N i j

1 417 17

4 117 17

(ln 2) (ln 2) (ln 2) 0 ;

0

i j kB T N k 22 (ln 2) 8te a j a j and (ln 2) 4 v i j



(ln 2) (ln 2) 1 4 0 8 80 8 0

i j k

v a k v a and 817 17

(ln 2) 17 (ln 2) ;κ v

2

1 4 00 8 00 16 024 (ln 2) 16 (ln 2) 0te

v a

a j a j

20. 3cosh 2 3sinh 2 6 6sinh 2 6cosh 2 6r t t t t t i j k v i j k

2 2 1 1 12 2 2

36sinh 2 36cosh 2 36 6 2 cosh 2 tanh 2 sech 2t t t t t vvv T i j k

5 8117 2 2 17 2

(ln 2) ; T i j k 22 22 2

sech 2 sech 2 tanh 2ddt t t t T i k

2 22 8 28 8 15 128 240 128 2402 217 17 17 172 2 289 2 289 2 289 2 289 2

(ln 2) (ln 2)d ddt dt T Ti k i k

8 1517 17(ln 2) ; N i k 15 8 15 81 1

17 2 2 17 2 17 2 2 17 28 15

17 17

(ln 2) (ln 2) (ln 2) ;

0

i j kB T N i j k

17 15 51 458 8 2 212cosh 2 12sinh 2 (ln 2) 12 12t t a i j a i j i j and 15 17

8 8(ln 2) 6 6 6 v i j k

45 51 45 514 4 4 4

51 452 2

6 (ln 2) (ln 2) 6 135 153 72 153 2

0

i j ki j k v a i j k v a and 51

4(ln 2) 2v

3514

153 2 328672

(ln 2) ;κ

45 514 4

51 452 2

2

60

45 51 032

86724sinh 2 24cosh 2 (ln 2) 45 51 (ln 2)t t

v a

a i j a i j

21. 2 22 3 3 4 4 6cos 3 6 4 8 6sint t t t t t t t r i j k v i j k

2 2 2 2 23 6 4 8 6sin 25 100 100 36sint t t t t t v

1 22 21

2 25 100 100 36sin 100 200 72sin cos (0) (0) 10;d dTdt dtt t t t t t a

v v

22 2 26 8 6cos 6 8 6cos 100 36cos (0) 136t t t a i j k a a

2 2 2136 10 36 6 (0) 10 6N Ta a a a T N

22. 2 22 2 2 22 2 1 1 4 2 1 1 4 2 2 8 20t t t t t t t t t t r i j k v i j k v

1 221

2 2 8 20 8 40 (0) 2 2;d dTdt dtt t t a

v v 2 24 2 4 2 20 a j k a

22 2 20 2 2 12 2 3 (0) 2 2 2 3N Ta a a a T N



23. sin 2 cos sin cos 2 sin cost t t t t t r i j k v i j k

22 2 1 12 2

cos 2 sin cos 2 cos sin cos ;t t t t t t vvv T i j k

2 221 1 1 12 2 2 2

sin cos sin sin ( cos ) sin 1d ddt dtt t t t t t T Ti j k

1 12 2

sin cos sin ;ddtddt

t t t T

TN i j k 1 1 1 12 2 2 2

1 12 2

cos sin cos ;

sin cos sin

t t t

t t t

i j kB T N i k

sin 2 cos sin cos 2 sin cos 2 2

sin 2 cos sin

t t t t t t

t t t

i j ka i j k v a i k

3 32 1

224 2 ;κ

v a

vv a cos 2 sin cost t t a i j k

2

cos 2 sin cossin 2 cos sin

cos 2 2 sin (0) cos 2cos 2 sin cos4 0

t t tt t t

t t tt t t

v a

24. (5cos ) (3sin ) ( 5sin ) (3cos ) ( 5cos ) (3sin )t t t t t t r i j k v j k a j k 25sin cos 9sin cos 16sin cos ;t t t t t t v a 0 16sin cos 0 sin 0t t t v a or cos 0t

20, ort

25. 12 2 2 2 22 4sin 3 0 ( ) 2(1) 4sin ( 1) 0 2 4sin sint t t t t

r i j k r i j

2 6 3t t (for the first time)

26. 2 3 2 2 4 31 214 14 14

( ) 2 3 1 4 9 (1) 14 (1) ,t t t t t t t t r i j k v i j k v v T i j k

which is normal to the normal plane 31 214 14 14

( 1) ( 1) ( 1) 0 or 2 3 6x y z x y z is an equation

of the normal plane. Next we calculate (1)N which is normal to the rectifying plane. Now, 2 6t a j k

376 19

7 1414(1) 2 6 (1) (1) 1 2 3 6 6 2 (1) (1) 76 (1) ;

0 2 6к

i j ka j k v a i j k v a

2

2

1 22 4 31 222 1411

( ) 1 4 9 8 36 ,ds d sdt dt tt

t t t t t

v so 2

2

2d s dsdtdt

к a T N

22 3 19 14 8 922 117 7 714 14 7 14 2 19

2 6 14 i j kj k N N i j k

8 9117 7 7( 1) ( 1) ( 1) 0x y z or or 11 8 9 10x y z is an equation of the rectifying plane. Finally,

14 1 1 172 19 14 19

(1) (1) (1) 1 2 3 3 3 3( 1) 3( 1) ( 1) 011 8 9

x y z

i j kB T N i j k or

3 3 1x y z is an equation of the osculating plane.

Chapter 13 Additional and Advanced Exercises 1003


27. 11(sin ) ln(1 ) (cos ) (0) ;t t

te t t e t r i j k v i j k v i j k (0) (1, 0, 0) r i is on the line

1 , ,x t y t and z t are parametric equations of the line

28. 2 cos 2 sin 2 sin 2 cost t t t t r i j k v i j k

4 4 42 sin 2 cos v i j k i j k is a vector tangent to the helix when 4t the

tangent line is parallel to 4 ;v also 4 4 4 42 cos 2 sin r i j k the point 41, 1, is on the

line 1 , 1 ,x t y t and 4z t are parametric equations of the line

29. 2 2 2 20 cosx v t and 2 22 2 2 2 2 2 2 21 1

0 02 2siny gt v t x y gt v t

30. 2 2 2 2 2 2 2 22

2 2 2 22 2

22 2 2 2 2 2 2 x y x y x x x x y y y yx x y yx x y yddt x y x yx y

s x y x y s x y

3 22 222 2 2 2 2 2

2 2 2 2 2 2 2 2 2

2 2 2 2 1x yx y y xx y y xx y y x x x y y x yкx y yxx y x y x y x y s

x y s

31. 1 1 12

ds sa a ds a a as a к since a > 0

32. (1) DO OTOT SOSOT TOD

20 6380 638006380 6380 437 6817

y y

0 5971 km;y

(2) 263805971

2 1 dxdyVA x dy

2 2

6817 2 2 63805971 6380

2 6380y

y dy

6817 681759715971

2 6380 2 6380dy y

2 7 216,395,469 km 1.639 10 km ;

(3) percentage visible 2

216,395,469 km4 (6380 km)

3.21%

CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES

1. (a) ( ) ( cos ) ( sin ) ( sin ) ( cos ) ;d ddt dta a b a a b rr i j k i j k

2 2 2 2 2 2 2 22 2 42 2

22 2gz gb gb gbd d d d

dt dt dt dta b a b a b a bgz a b

rv

(b) 2 2 2 2 2 22 2 21 22 ;gb gb gbd d

dt a b a b a bdt t C

2 2

21 20 0 0 2 gba b

t C t

2

2 22;gbt

a b

2 2

2 22gb ta b

z b z

(c) 2 2( ) ( sin ) ( cos ) ( sin ) ( cos ) ,gbtd ddt dt a b

t a a b a a b

rv i j k i j k from part (b)

2 2 2 2 2 2

( sin ) ( cos )( ) ;a a b gbt gbt

a b a b a bt

i j kv T



2 2

2 2

2( cos ) ( sin ) ( sin ) ( cos )d d d

dtdt dta a a a b r i j i j k

2 2 2 2

2( cos ) ( sin ) [( sin ) ( cos ) ]gbt gb

a b a ba a a a b

i j i j k

2 2 2 22 2 2 2 2 2

2( sin ) ( cos ) ( cos ) (sin )a a b gb gbt gb gbta b a ba b a b a b

a a

i j k i j T N (there is

no component in the direction of B).

2. (a) ( ) cos sin cos sin sin cos ;d ddt dta a b a a a a b rr i j k i j k

2 2 2 2

1 2 22 2 2 22 gbd d ddt dt dt a a b

gz a a b

rv

(b) 1 2 1 2 1 22 2 2 2 2 2 2 2 2 2 2 20 0 0 0t t td

dts dt a a b dt a a b d a a u b du v

2 2

22 2 2

0 0,a b

aa u du a a c u du

where 2 2 22 2 2 2

2 20

lna b u ca

c s a c u u c u

2 2 2 2 2 22 ln lna c c c c c

3. 0 02

(1 ) (1 ) ( sin )1 cos (1 cos )

;e r e r edre d e

r

02

(1 ) ( sin )0(1 cos )

0 0 (1 ) ( sin ) 0 sin 0e r edrd e

e r e

0 or . Note that 0drd when sin 0 and 0dr

d when sin 0. Since sin 0 on 0

and sin 0 on 0 , r is a minimum when 0 and 0(1 )01 cos 0(0) e r

er r

4 (a) 12( ) 1 sin 0 (0) 1f x x x f and 1 1

2 2(2) 2 1 sin 2f since sin 2 1; since f is continuous on [0, 2], the Intermediate Value Theorem implies there is a root between 0 and 2

(b) Root 1.4987011335179

5. (a) x y v i j and (cos ) (sin ) ( sin ) (cos )rr r r r x v u u i j i j v i and

cos sin cos sin ;r r x r r y v i v j and

sin cos sin cosr r y r r v j

(b) (cos ) (sin ) cos sin cos sin (cos ) sin cos (sin )r r x y r r r r u i j v u

by part (a), ;r r v u therefore, cos sin ;r x y (sin ) (cos ) u i j

sin cos cos sin ( sin ) sin cos (cos )x y r r r r v u by part (a)

;r v u therefore, sin cosr x y

6. 2 2

2 2

2( ) ( ) ( ) ( ) ;dr d d r d d

dt dt dtdt dtr f f f f

cos sin sin cosdr d dr d dr drdt dt dt dt dt dtr r r

v u u i j

1 2 1 22 2 22 2 ;dr d d

dt dt dtr f f v ,x y y x v ×a

where cosx r and sin .y r Then sin cosdx d drdt dt dtr

2 2 2

2 2 2

2( 2sin ) ( cos ) ( sin ) (cos ) ;d x d dr d d d r

dt dt dtdt dt dtr r ( cos ) (sin )dy d dr

dt dt dtr

Chapter 13 Additional and Advanced Exercises 1005


2 2 2

2 2 2

2(2cos ) ( sin ) ( cos ) (sin ) .d y d dr d d d r

dt dt dtdt dt dtr r Then, after much algebra v a

222 2

2 2 3 22 2

3 2 3 222 22 2 f f f fd d dr d d r d dr ddt dt dt dt dt dtdt dt f f

r r r f f f f к

v av

7. (a) Let 2r t and 3 1drdtt and

2 2

2 23 0.d d r ddt dt dt The halfway point is (1, 3) 1;t

(1) 3 ;dr dr rdt dtr

v u u v u u 2 2

2 2

22d r d d dr d

rdt dt dtdt dtr r

a u u

(1) 9 6r a u u (b) It takes the beetle 2 min to crawl to the origin the rod has revolved 6 radians

22 26 6 62 2 41 13 3 3 9 90 0 0

( ) ( ) 2 4L f f d d d

2 66 637 12 ( 6)2 2 21 1 19 3 3 2 20 0 0

( 6) 1 ( 6) 1 ln 6 ( 6) 1d d

1637 ln 37 6 6.5 in.

8. (a) cos cos sin ;x r dx dr r d sin sin cos ;y r dy dr r d thus 2 2 2 2 2 2cos 2 sin cos sindx dr r dr d r d and 2 2 2 2 2 2 2 2 2 2 2 2 2 2sin 2 sin cos cosdy dr r dr d r d ds dx dy dz dr r d dz

(c) r e dr e d ln 8 2 2 2 20

L dr r d dz

ln8 2 2 20

e e e d

ln8ln80 0

3 3e d e

8 3 3 7 3

(b)

9. (a) cos sin 0sin cos 0

r a

i j ku u k right-handed frame of unit vectors

(b) ( sin ) (cos )rdd u i j u and ( cos ) (sin )d

rd u i j u

(c) From Eq. (7), r r rr r z r r r r r z v u u k a v u u u u u k

2 2rr r r r z u u k

(d) 2

2( ) ( ) ( ) ( ) ( ) ;d d d ddt dt dtdt

t t m t m m m m m L r r LL r v v r v v r a r a

3 3 3 ( )c d c cdtm m m

L

r r rF a r a r a r r r r 0 L constant vector

t 000sin2 2 1 2cos2 - media.pearsoncmg.com · 972 chapter 13 vector-valued functions and motion in...

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