Calculus Help Needed!

Question

1. A tube is being stretched while maintaining its cylindrical shape. The height is increasing at the rate of 2 millimeters per second. At the instant that the radius of the tube is 6 millimeters, the volume is increasing at the rate of 96π cubic millimeters per second. Which of the following statements about the surface area of the tube is true at this instant?

a. The surface area is increasing by 28π square millimeters per second

b. The surface area is decreasing by 28π square millimeters per second

c. The surface area is increasing by 32π square millimeters per second

d. The surface area is decreasing by 32π square millimeters per second

2. A particle moves on the hyperbola xy= 15 for time t≥0 seconds. At a certain instant, x= 3 and dx/dt= 6. Which of the following is true about y in this instant?

a. y is decreasing by 10 units per second

b, y is increasing by 10 units per second

c. y is decreasing by 5 units per second

d. y is increasing by 5 units per second

3. Let x and y be functions of time t such that the sum of x and y is constant. Which of the following equations describes the relationship between the rate of change of x with respect to time and the rate of change of y with respect to time?

a. dx/dt=dy/dt

b. dx/dt=-dy/dt

c. dx/dt+dy/dt= dK/dt, where K is a function of t

d. dx/dt+dy/dt= K, where K is a function of t

Answer 1

1.

Let

r(t) be the radius at a time t,

h(t) be the height at a time t,

V(t) be the volume at a time t,

S(t) be the surface at a time t,

t₁ be the instant of time in question.

To simplify notation we will drop (t) and write instead

r, h, V, S.

Their rates of change are

r', h', V', S'.

Instead of r(t₁ ), h(t₁), V(t₁), S(t₁), r'(t₁), h'(t₁), V'(t₁), S'(t₁)

we will write

r₁, h₁, V₁, S₁, r₁', h₁', V₁', S₁'

We are given

h' = 2 (which also implies h₁' = 2)

r₁ = 6

V₁' = 96π

The volume at a time t is

V = πr²h

The change of volume at a time t is

V' = π(2rr'h + r²h')

At time t₁

V₁' = π(2r₁r₁'h₁ + r₁²h₁')

96π = π(2×6×r₁'h₁ + 6²×2)

Solving for r₁'h₁

r₁'h₁ = 2 (1)

The surface at time t

S = 2πrh

The change of surface at time t is

S' = 2π(r'h + rh')

At time t₁

S₁' = 2π(r₁'h₁ + r₁h₁')

Substituting the given quantities and the result (1)

S₁' = 2π(2 + 6×2)= 28π

The correct answer is a.

Please note that the situation as describes is physically impossible.

Physically r' < 0 and h > 0, therefore the product r₁'h₁ must be negative, not 2.

In other words, at the given instant t₁ the volume could not be increasing as fast

as 96π.

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2.

Let x, y denote the values of the coordinates at an arbitrary time t,

Let x₁, y₁ denote the values of the coordinates at the time instant in question.

Let x', y' denote the derivatives of x, y with respect to time,

Let x₁', y₁' denote the values of x', y' at the time instant in question.

We are given

x₁ = 3 therefore from the equation of the hyperbola y₁ = 5

x₁' = 6

Differentiating the equation of the hyperbola with respect to time:

x'y + xy' = 0

Therefore specifically at the given time instant

x₁'y₁ + x₁y₁' = 0

Substituting actual values

6×5 + 3×y₁' = 0

y₁' = -10

The correct answer is a.

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3.

Since x + y is a constant

dx/dt + dy/dt = 0

Therefore

a. is false in general; it would be true only if both x and y were constant functions of time.

b. is true.

c. is true with K(t) = 42 (or any other constant function K).

d. is true with K(t) = 0.