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Applications of the Derivative Practice Test (page 2)

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Answers:

  1.  

    SOLUTIONS TO EXERCISES Chapter 3

    Fig. S3.1

  2. Figure S3.2 shows a schematic of the imbedded cylinder. We see that the volume of the imbedded cylinder, as a function of height h, is

    V ( h ) = π · h · (3 − h /2) 2.

    Then we solve

    SOLUTIONS TO EXERCISES Chapter 3

    SOLUTIONS TO EXERCISES Chapter 3

    Fig. S3.2

    The roots of this equation are h = 2, 6. Of course height 6 gives a trivial cylinder, as does height 0. We find that the solution of our problem is height 2, radius 2.

  3. We know that

    V = l · w · h

    hence

    SOLUTIONS TO EXERCISES Chapter 3

  4. We know that v 0 = −15. Therefore the position of the body is given by

    p ( t ) = −16 t 2 − 15 t + h 0 .

    Since

    0 = p (5) = −16 · 5 2 − 15 · 5 + h 0 ,

    we find that h 0 = 475. The body has initial height 475 feet.

  5. We know that

    SOLUTIONS TO EXERCISES Chapter 3

    Therefore

    SOLUTIONS TO EXERCISES Chapter 3

    At the moment of the problem, dh/dt = 3, r = 5, h 12/(5π). Hence

    SOLUTIONS TO EXERCISES Chapter 3

    or

    SOLUTIONS TO EXERCISES Chapter 3

    We conclude that dr/dt = −75π/8 microns per minute.

  6. Of course

    10000 = V = π · r 2 · h .

    We conclude that

    SOLUTIONS TO EXERCISES Chapter 3

    We wish to minimize

    SOLUTIONS TO EXERCISES Chapter 3

    Thus the function to minimize is

    SOLUTIONS TO EXERCISES Chapter 3

    Thus

    SOLUTIONS TO EXERCISES Chapter 3

    We find therefore that

    SOLUTIONS TO EXERCISES Chapter 3

    or SOLUTIONS TO EXERCISES Chapter 3 Since the problem makes sense for 0 < r < ∞, and since it clearly has no maximum, we conclude that SOLUTIONS TO EXERCISES Chapter 3 , SOLUTIONS TO EXERCISES Chapter 3 .

  7. We calculate that g ′( x ) = sin x + x cos x and g ″( x ) = 2 cos xx sin x . The roots of these transcendental functions are best estimated with a calculator or computer. Figure S3.7 gives an idea of where the extrema and inflection points are located.

  8. We know that v 0 = −5 and h 0 = 400. Hence

    p ( t ) = −16 t 2 −5 t + 400.

    The body hits the ground when

    0 = p ( t ) = −16 t 2 −5 t + 400.

    Solving, we find that t ≈ 4.85 seconds.

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