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 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 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 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.

Explanation:

Step1: Differentiate the volume formula

Given \(V=\pi r^{2}h\), differentiate with respect to time \(t\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\). So \(\frac{dV}{dt}=\pi(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt})\). We know \(\frac{dV}{dt} = 96\pi\), \(r = 6\), and \(\frac{dh}{dt}=2\). Plug in these values: \(96\pi=\pi(2\times6\times h\times\frac{dr}{dt}+6^{2}\times2)\). Divide both sides by \(\pi\): \(96 = 12h\frac{dr}{dt}+ 72\). Then \(12h\frac{dr}{dt}=96 - 72=24\), so \(h\frac{dr}{dt}=2\).

Step2: Differentiate the surface - area formula

Given \(S = 2\pi rh\), differentiate with respect to time \(t\) using the product rule. \(\frac{dS}{dt}=2\pi(r\frac{dh}{dt}+h\frac{dr}{dt})\). Substitute \(r = 6\), \(\frac{dh}{dt}=2\), and \(h\frac{dr}{dt}=2\) into the formula. \(\frac{dS}{dt}=2\pi(6\times2 + 2)\). First, calculate inside the parentheses: \(6\times2+2=12 + 2=14\). Then \(\frac{dS}{dt}=2\pi\times14=28\pi\). Since \(\frac{dS}{dt}>0\), the surface area is increasing.

Answer:

A. The surface area is increasing by \(28\pi\) square millimeters per second.