This question is incomplete, the complete question is;
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? (The volume V of a cylinder with radius r and height h is V=πr2h. The surface area Sof a cylinder, not including the top and bottom of the cylinder, is S=2πrh.)
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:
Surface Area is increasing at the rate of 28π mm²/sec ( Option A )
Explanation:
Given the data in the question;
let r be the radius and h be the height of the tube
so volume V = πr²h
dv/dt = πr² dh/dt + 2nrh dr/dt
⇒ 96n = π × 36 × 2 + 2π × 6h dr/dt
⇒ 24π = 12nh dr/dt ⇒ h.dr/dt = 2
Area Surface S = 2nrh
ds/dt = 2π ( r.dh/dt + h.dr/dt )
= 2π ( 6×2 + 2)
= 2π (14)
= 28π
therefore Surface Area is increasing at the rate of 28π mm²/sec ( Option A )
