Question

A tube is being stretched while maintaining its cylindrical shape. The heiggt 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 96pi cubic millimeters per second. Which of the following statements about the surface area of the tube is true at this instant? (The Volume V of a cylinder with radius r and a height h is V=pi r²h. The Surface Area S of a cylinder, not including the top and bottom is S=2pi rh.

A) The surface area is increasing at the rate of 28pi square mm per second.
B) The surface area is decreasing at the rate of 28pi square mm per second.
C) The surface area is increasing at the rate if 32pi square mm per second.
D) The surface area is decreasing at the rate of 32pi square mm per second.

Answer 1

To find the rate at which the surface area of the tube is changing, we need to use the formulas for the volume and surface area of a cylinder. By differentiating the volume formula, we can find the rate of change of the height of the tube. Then, using the formula for the surface area, we can find the rate of change of the surface area. At the given instant, the surface area is increasing at the rate of 108pi square mm per second.

Step-by-step explanation:

To find the rate at which the surface area of the tube is changing, we need to use the formula for the surface area of a cylinder, which is S = 2pi rh. We are given that the height (h) of the tube is increasing at a rate of 2 millimeters per second, and the volume (V) is increasing at a rate of 96pi cubic millimeters per second. The formula for the volume of a cylinder is V = pi r²h. We can differentiate this equation with respect to time (t) to find the rate of change of the volume:

dV/dt = pi * 2r * dh/dt

Substituting the given values, we have:

96pi = pi * 2(6) * 2

Canceling out the pi and simplifying, we get:

96 = 24 * dh/dt

Solving for dh/dt, we find that the rate of change of the height is dh/dt = 4 millimeters per second. Now, we can use the formula for the surface area (S) and differentiate it with respect to time to find the rate of change of the surface area:

dS/dt = 2pi * (r * dh/dt + h * dr/dt)

Substituting the given values, we have:

dS/dt = 2pi * (6 * 4 + h * dr/dt)

Since we are interested in the rate of change of the surface area when the radius is 6 millimeters, we can substitute this value and solve for dS/dt:

dS/dt = 2pi * (6 * 4 + 6 * dr/dt)

The rate of change of the volume is given as 96pi cubic millimeters per second, so we can solve for dr/dt:

96pi = pi * 6² * 4 + pi * 6 * dr/dt

Canceling out the pi and simplifying, we get:

96 = 36 * 4 + 6dr/dt

Simplifying further, we find that:

dr/dt = 5 millimeters per second

Now, we can substitute the values of h (2 millimeters per second), r (6 millimeters), and dr/dt (5 millimeters per second) into the equation and calculate the rate of change of the surface area:

dS/dt = 2pi * (6 * 4 + 6 * 5)

Simplifying, we get:

dS/dt = 2pi * (24 + 30)

dS/dt = 2pi * 54

dS/dt = 108pi

Therefore, the correct statement about the surface area of the tube at this instant is The surface area is increasing at the rate of 108pi square mm per second (option not given).