Solutions To Selected Problems

Chapter 4

Exercises 4.1

  1. 1.

    T

  2. 3.

    3 ft/min

  3. 5.
    (a) 5/(2π)0.796cm/s (b) 1/(40π)0.00796 cm/s (c) 1/(4000π)0.0000796 cm/s
  4. 7.
    (a) 64.44 mph (b) 78.89 mph
  5. 9.
    Due to the height of the plane, the gun does not have to rotate very fast. (a) 0.073 rad/s (b) 3.66 rad/s (about 1/2 revolution/sec) (c) In the limit, rate goes to 7.33 rad/s (more than 1 revolution/sec)
  6. 11.
    (a) 30.59 ft/min (b) 36.1 ft/min (c) 301 ft/min (d) The boat no longer floats as usual, but is being pulled up by the winch (assuming it has the power to do so).
  7. 13.
    (a) 0.63 ft/sec (b) 1.6 ft/sec About 52 ft.
  8. 15.
    (a) The balloon is 105ft in the air. (b) The balloon is rising at a rate of 17.45ft/min. (Hint: convert all angles to radians.)

Exercises 4.2

  1. 1.

    T

  2. 3.

    2500; the two numbers are each 50.

  3. 5.

    There is no maximum sum; the fundamental equation has only 1 critical value that corresponds to a minimum.

  4. 7.

    Area = 1/4, with sides of length 1/2.

  5. 9.

    The radius should be about 3.84cm and the height should be 2r=7.67cm. No, this is not the size of the standard can.

  6. 11.

    The height and width should be 18 and the length should be 36, giving a volume of 11,664in3.

  7. 13.

    5-10/393.4 miles should be run underground, giving a minimum cost of $374,899.96.

  8. 15.

    The dog should run about 19 feet along the shore before starting to swim.

  9. 17.

    The largest area is 2 formed by a square with sides of length 2.

  10. 19.

    A length of 2 in and height of 2.5 will give a cost of 60 ¢.

Exercises 4.3

  1. 1.

    T

  2. 3.

    F

  3. 5.

    Answers will vary.

  4. 7.

    Use y=x2; dy=2xdx with x=2 and dx=0.05. Thus dy=.2; knowing 22=4, we have 2.0524.2.

  5. 9.

    Use y=x3; dy=3x2dx with x=5 and dx=0.1. Thus dy=7.5; knowing 53=125, we have 5.13132.5.

  6. 11.

    Use y=x; dy=1/(2x)dx with x=16 and dx=0.5. Thus dy=.0625; knowing 16=4, we have 16.54.0625.

  7. 13.

    Use y=x3; dy=1/(3x23)dx with x=64 and dx=-1. Thus dy=-1/480.0208; we could use -1/48-1/50=-0.02; knowing 643=4, we have 6333.98.

  8. 15.

    Use y=sinx; dy=cosxdx with x=π and dx-0.14. Thus dy=0.14; knowing sinπ=0, we have sin30.14.

  9. 17.

    dy=(2x+3)dx

  10. 19.

    dy=-24x3dx

  11. 21.

    dy=(2xe3x+3x2e3x)dx

  12. 23.

    dy=2(tanx+1)-2xsec2x(tanx+1)2dx

  13. 25.

    dy=(exsinx+excosx)dx

  14. 27.

    dy=1(x+2)2dx

  15. 29.

    dy=(lnx)dx

  16. 31.

    1-6x

  17. 33.

    1-x2

  18. 35.

    22/3+x24/3

  19. 37.

    dV=±0.157

  20. 39.

    ±15π/8±5.89in2

  21. 41.
    (a) 297.8 feet (b) ±62.3 ft (c) ±20.9%
  22. 43.
    (a) 298.9 feet (b) ±8.67 ft (c) ±2.9%
  23. 45.

Exercises 4.4

  1. 1.

    F

  2. 3.

    x0=1.5, x1=1.5709148, x2=1.5707963, x3=1.5707963, x4=1.5707963, x5=1.5707963

  3. 5.

    x0=0, x1=2, x2=1.2, x3=1.0117647, x4=1.0000458, x5=1

  4. 7.

    x0=2, x1=0.6137056389, x2=0.9133412072, x3=0.9961317034, x4=0.9999925085, x5=1

  5. 9.

    roots are: x=-5.156, x=-0.369 and x=0.525

  6. 11.

    roots are: x=-1.013, x=0.988, and x=1.393

  7. 13.

    x=±0.824,

  8. 15.

    x=±0.743

  9. 17.

    The approximations alternate between x=1 and x=2.

  10. 19.

    f(x)=x2-16.5 and x0=4 yield x1=6516=4.0625 and x2=844920804.0620192.

  11. 21.

    f(x)=x3-63 and x0=4 yield x1=191483.97916667 and x23.9790572.

  12. 23.
    (a) xn- (b) x1 is undefined (c) xn2 (d) x1 is undefined (e) xn6
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