Applications of the Derivative Practice Test

Review the following concepts if needed:

Applications of the Derivative Practice Test

Sketch the graph of f ( x ) = x /[ x ² + 3], indicating all local maxima and minima together with concavity properties.
What is the right circular cylinder of greatest volume that can be inscribed upright in a right circular cone of radius 3 and height 6?
An air mattress (in the shape of a rectangular parallelepiped) is being inflated in such a way that, at a given moment, its length is increasing by 1 inch per minute, its width is decreasing by 0.5 inches per minute, and its height is increasing by 0.3 inches per minute. At that moment its dimensions are ℓ = 100″, w = 60″, and h = 15″. How is its volume changing at that time?
A certain body is thrown straight down at an initial velocity of 15 ft/sec. It strikes the ground in 5 seconds. What is its initial height?
Because of viral infection, the shape of a certain cone-shaped cell is changing. The height is increasing at the rate of 3 microns per minute. For metabolic reasons, the volume remains constantly equal to 20 cubic microns. At the moment that the radius is 5 microns, what is the rate of change of the radius of the cell?
A silo is to hold 10,000 cubic feet of grain. The silo will be cylindrical in shape and have a flat top. The floor of the silo will be the earth. What dimensions of the silo will use the least material for construction?
Sketch the graph of the function g ( x ) = x ·sin x . Show maxima and minima.
A body is launched straight down at a velocity of 5 ft/sec from height 400 feet. How long will it take this body to reach the ground?
Sketch the graph of the function h ( x ) = x /( x ² − 1). Exhibit maxima, minima, and concavity.
A punctured balloon, in the shape of a sphere, is losing air at the rate of 2in. ³ /sec. At the moment that the balloon has volume 36 cubic inches, how is the radius changing?
A ten-pound stone and a twenty-pound stone are each dropped from height 100 feet at the same moment. Which will strike the ground first?
A man wants to determine how far below the surface of the earth is the water in a well. How can he use the theory of falling bodies to do so?
A rectangle is to be placed in the first quadrant, with one side on the x -axis and one side on the y -axis, so that the rectangle lies below the line 3 x + 5 y = 15. What dimensions of the rectangle will give greatest area?
A rectangular box with square base is to be constructed to hold 100 cubic inches. The material for the base and the top costs 10 cents per square inch and the material for the sides costs 20 cents per square inch. What dimensions will give the most economical box?
Sketch the graph of the function f ( x ) = [ x ² − 1]/[ x ² + 1]. Exhibit maxima, minima, and concavity.
On the planet Zork, the acceleration due to gravity of a falling body near the surface of the planet is 20 ft/sec. A body is dropped from height 100 feet. How long will it take that body to hit the surface of Zork?

Answers:

Fig. S3.1
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.
We know that

V = l · w · h

hence
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.
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.
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 , .
We calculate that g ′( x ) = sin x + x cos x and g ″( x ) = 2 cos x − x 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.
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.