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# Applications of the Derivative Practice Test (page 2)

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1.

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

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

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

Therefore

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

or

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

6. Of course

10000 = V = π · r 2 · h .

We conclude that

We wish to minimize

Thus the function to minimize is

Thus

We find therefore that

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

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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