Question

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 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 height h is V = π.r^2h. The surface area S of a cylinder, not including the top and bottom of the cylinder, is S = 2πrh.

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

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

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

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

Answer 1

The volume V of the cylinder is V = πr^2h. Taking the derivative of both sides with respect to time t, we get:

dV/dt = π(2rh.dr/dt + r^2dh/dt)

We are given that dh/dt = 2 mm/s and r = 6 mm, so we have:

dV/dt = π(2rh.dr/dt + r^2dh/dt) = π(2(6)(dr/dt)(2) + 6^2(2)) = 24π(dr/dt + 2)

We know that dV/dt = 0, since we are only interested in the instant when the volume is increasing at the rate of cubic millimeters per second. Solving for dr/dt, we get:

dr/dt = -2/(24π) = -1/(12π)

Now we can find the surface area S of the cylinder, not including the top and bottom of the cylinder, which is S = 2πrh. Substituting r = 6 mm and h = 2 mm, we get:

S = 2π(6)(2) = 24π square mm

Taking the derivative of S with respect to time t, we get:

dS/dt = 2πr(dh/dt) + 2πh(dr/dt)

Substituting r = 6 mm, h = 2 mm, dh/dt = 2 mm/s, and dr/dt = -1/(12π), we get:

dS/dt = 2π(6)(2) + 2π(2)(-1/(12π)) = 28π square mm per second

Therefore, the correct answer is a. The surface area is increasing by 28π square mm per second.