Solutions To Selected Problems

Chapter 12

Exercises 12.1

  1. 1.

    parametric equations

  2. 3.

    displacement

  3. 5.
    1234-11234xy
  4. 7.
    -55510xy
  5. 9.
    -3-2-1123-1-212xy
  6. 11.
    -3-2-1123-1-212xy
  7. 13.
  8. 15.
  9. 17.

    r(t)=t2+t4=|t|t2+1.

  10. 19.

    r(t)=4cos2t+4sin2t+t2=t2+4.

  11. 21.
    Answers may vary, though most direct solution is r(t)=2cost+1,2sint+2.
  12. 23.
    Answers may vary, though most direct solution is r(t)=1.5cost,5sint.
  13. 25.
    Answers may vary, though most direct solutions are r(t)=t,5(t-2)+3 and r(t)=t+2,5t+3.
  14. 27.
    Answers may vary, though most direct solution is r(t)=2cost,2sint,2t.
  15. 29.

    Answers may vary, though most direct solution is r(t)=Rcos(2πNt),Rsin(2πNt),ht.

  16. 31.

    1,0

  17. 33.

    0,0,1

Exercises 12.2

  1. 1.

    component

  2. 3.

    It is difficult to identify the points on the graphs of r(t) and r(t) that correspond to each other.

  3. 5.

    11,74,sin5

  4. 7.

    1,e

  5. 9.

    (-,0)(0,)

  6. 11.

    r(t)=-sint,et,1/t

  7. 13.

    r(t)=(2t)sint,2t+5+(t2)cost,2=2tsint+t2cost,6t2+10t

  8. 15.
    r(t)=2t,1,0×sint,2t+5,1+t2+1,t-1,1×cost,2,0= -1,cost-2t,6t2+10t+2+cost-sint-tcost
  9. 17.
    246246r(1)xy r(t)=2t+1,2t-1
  10. 19.
    24-22r(1)xy r(t)=2t,3t2-1
  11. 21.

    (t)=2,0+t3,1

  12. 23.

    (t)=-3,0,π+t0,-3,1

  13. 25.
    t=2nπ, where n is an integer; so t=-4π,-2π,0,2π,4π,
  14. 27.

    r(t) is not smooth at t=3π/4+nπ, where n is an integer

  15. 29.

    Both derivatives return 5t4,4t3-3t2,3t2.

  16. 31.
    Both derivatives return 2t-et-1,cost-3t2, (t2+2t)et-(t-1)cost-sint.
  17. 33.

    14t4,sint,tet-et+C

  18. 35.

    -2,0

  19. 37.

    r(t)=12t2+2,-cost+3

  20. 39.

    r(t)=t4/12+t+4,t3/6+2t+5,t2/2+3t+6

  21. 41.

    213π

  22. 43.

    154((22)3/2-8)

  23. 45.

    12

  24. 47.
    As r(t) has constant length, r(t)r(t)=c2 for some constant c. Thus r(t)r(t) =c2 ddt(r(t)r(t)) =ddt(c2) r(t)r(t)+r(t)r(t) =0 2r(t)r(t) =0 r(t)r(t) =0.

Exercises 12.3

  1. 1.

    Velocity is a vector, indicating an objects direction of travel and its rate of distance change (i.e., its speed). Speed is a scalar.

  2. 3.

    The average velocity is found by dividing the displacement by the time traveled — it is a vector. The average speed is found by dividing the distance traveled by the time traveled — it is a scalar.

  3. 5.

    One example is traveling at a constant speed s in a circle, ending at the starting position. Since the displacement is 0, the average velocity is 0, hence 0=0. But traveling at constant speed s means the average speed is also s>0.

  4. 7.

    v(t)=2,5,0, a(t)=0,0,0

  5. 9.

    v(t)=-sint,cost, a(t)=-cost,-sint

  6. 11.

    v(t)=1,cost, a(t)=0,-sint

    0.511.50.511.5v(π/4)a(π/4)xy
  7. 13.

    v(t)=2t+1,-2t+2, a(t)=2,-2

    2462-2-4-6-8v(1)a(1)xy
  8. 15.
    v(t)=4t2+1. Min at t=0; Max at t=±1.
  9. 17.
    v(t)=5. Speed is constant, so there is no difference between min/max
  10. 19.
    v(t)=|sect|tan2t+sec2t. min: t=0; max: t=π/4
  11. 21.
    v(t)=13. speed is constant, so there is no difference between min/max
  12. 23.
    v(t)=4t2+1+t2/(1-t2). min: t=0; max: there is no max; speed approaches as t±1
  13. 25.
    (a) r1(1)=1,1; r2(1)=1,1 (b) v1(1)=1,2; v1(1)=5; a1(1)=0,2 v2(1)=2,4; v2(1)=25; a2(1)=2,12
  14. 27.
    (a) r1(2)=6,4; r2(2)=6,4 (b) v1(2)=3,2; v1(2)=13; a1(2)=0,0 v2(2)=6,4; v2(2)=213; a2(2)=0,0
  15. 29.
    v(t)=2t+1,3t+2, r(t)=t2+t+5,3t2/2+2t-2
  16. 31.

    v(t)=sint,cost, r(t)=1-cost,sint

  17. 33.

    Displacement: 0,0,6π; distance traveled: 213π22.65ft; average velocity: 0,0,3; average speed: 133.61ft/s

  18. 35.

    Displacement: 0,0; distance traveled: 2π6.28ft; average velocity: 0,0; average speed: 1ft/s

  19. 37.

    At t-values of sin-1(9/30)/(4π)+n/20.024+n/2 seconds, where n is an integer.

  20. 39.
    (a) Holding the crossbow at an angle of 0.013 radians, 0.745 will hit the target 0.4s later. (Another solution exists, with an angle of 89, landing 18.75s later, but this is impractical.) (b) In the .4 seconds the arrow travels, a deer, traveling at 20mph or 29.33ft/s, can travel 11.7ft. So she needs to lead the deer by 11.7ft.
  21. 41.

    The position function is r(t)=220t,-16t2+1000. The y-component is 0 when t=7.9; r(7.9)=1739.25,0, meaning the box will travel about 1740ft horizontally before it lands.

Exercises 12.4

  1. 1.

    1

  2. 3.

    T(t) and N(t).

  3. 5.
    T(t)=4t20t2-4t+1,2t-120t2-4t+1; T(1)=4/17,1/17
  4. 7.

    T(t)=costsintcos2tsin2t-cost,sint. (Be careful; this cannot be simplified as -cost,sint because cos2tsin2tcostsint, but rather |costsint|.) T(π/4)=-2/2,2/2

  5. 9.

    (t)=2,0+t4/17,1/17; in parametric form,

    (t)={x=2+4t/17y=t/17

  6. 11.

    (t)=2/4,2/4+t-2/2,2/2; in parametric form,

    (t)={x=2/4-2t/2y=2/4+2t/2

  7. 13.

    T(t)=-sint,cost; N(t)=-cost,-sint

  8. 15.
    T(t)=-sint4cos2t+sin2t,2cost4cos2t+sin2t; N(t)=-2cost4cos2t+sin2t,-sint4cos2t+sin2t
  9. 17.
    (a) Be sure to show work (b) N(π/4)=-5/34,-3/34
  10. 19.
    (a) Be sure to show work (b) N(0)=-15,25
  11. 21.

    T(t)=152,cost,-sint; N(t)=0,-sint,-cost

  12. 23.
    T(t)=1a2+b2-asint,acost,b; N(t)=-cost,-sint,0
  13. 25.
    aT=4t1+4t2 and aN=4-16t21+4t2 At t=0, aT=0 and aN=2; At t=1, aT=4/5 and aN=2/5. At t=0, all acceleration comes in the form of changing the direction of velocity and not the speed; at t=1, more acceleration comes in changing the speed than in changing direction.
  14. 27.
    aT=0 and aN=2 At t=0, aT=0 and aN=2; At t=π/2, aT=0 and aN=2. The object moves at constant speed, so all acceleration comes from changing direction, hence aT=0. a(t) is always parallel to N(t), but twice as long, hence aN=2.
  15. 29.
    aT=0 and aN=a At t=0, aT=0 and aN=a; At t=π/2, aT=0 and aN=a. The object moves at constant speed, meaning that aT is always 0. The object “rises” along the z-axis at a constant rate, so all acceleration comes in the form of changing direction circling the z-axis. The greater the radius of this circle the greater the acceleration, hence aN=a.

Exercises 12.5

  1. 1.

    time and/or distance

  2. 3.

    Answers may include lines, circles, helixes

  3. 5.

    κ

  4. 7.

    s=3t, so r(s)=2s/3,s/3,-2s/3

  5. 9.
    s=13t, so r(s)=3cos(s/13),3sin(s/13),2s/13
  6. 11.

    κ=|6x|(1+(3x2-1)2)3/2;

    κ(0)=0, κ(1/2)=19217172.74.

  7. 13.

    κ=|cosx|(1+sin2x)3/2;

    κ(0)=1, κ(π/2)=0

  8. 15.

    κ=|2costcos(2t)+4sintsin(2t)|(4cos2(2t)+sin2t)3/2;

    κ(0)=1/4, κ(π/4)=8

  9. 17.

    κ=|6t2+2|(4t2+(3t2-1)2)3/2;

    κ(0)=2, κ(5)=19139413940.0004

  10. 19.

    κ=0;

    κ(0)=0, κ(1)=0

  11. 21.

    κ=313;

    κ(0)=3/13, κ(π/2)=3/13

  12. 23.

    maximized at x=±254

  13. 25.

    maximized at t=1/4

  14. 27.

    radius of curvature is 55/4.

  15. 29.

    radius of curvature is 9.

  16. 31.
    x2+(y-1/2)2=1/4, or c(t)=1/2cost,1/2sint+1/2
  17. 33.

    x2+(y+8)2=81, or c(t)=9cost,9sint-8

  18. 35.

    Let r(t)=x(t),y(t),0 and apply the second formula of part 3.

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