Question

A piece of rubber tubing maintains a cylindrical shape as it is stretched. At the instant that the inner radius of the tube is 2 millimeters and the height is 20 millimeters, the inner radius is decreasing at the rate of 0.1 millimeter per second and the height is increasing at the rate of 3 millimeters per second. At this instant, what is the rate of change, in cubic millimeters per second, of the volume of the tube? (The volume V of a cylinder with radius r and height h is V-2h.) B) 20m 80 O 80 D. 84?

Answer 1

Answer:

the rate of change in volume is dV/dt = 4π mm³/s = 12.56 mm³/s

Step-by-step explanation:

since the volume V of a cylinder is related with the height H and the radius R through:

V = πR²*H

then the change in time is given by the derivative with respect to time t

dV/dt = (∂V/∂R)*(dR/dt) + (∂V/∂H)*(dH/dt)

the change in volume with radius at constant height is

(∂V/∂R) = 2*πR*H

the change in volume with height at constant radius is

(∂V/∂H) = πR²

then

dV/dt = 2π*R*H *(dR/dt) + πR²*(dH/dt)

replacing values

dV/dt = 2π* 2 mm * 20 mm  * (-0.1 mm/s) + π (2 mm) ²* 3 mm/s = 4π mm³/s

dV/dt = 4π mm³/s = 12.56 mm³/s