Question

gasoline is pouring into a vertical cylindrical tank of radius 2 feet. when the depth of the gasoline is 8 feet, the depth is increasing at 0.1 ft/sec. how fast is the volume of gasoline changing at that instant?

Answer 1

The volume of gasoline in a cylindrical tank is increasing at a rate of 0.4π cubic feet per second when the depth of gasoline is 8 feet and the depth is increasing at 0.1 ft/sec.

Step-by-step explanation:

The question involves calculating how fast the volume of gasoline is changing in a cylindrical tank as the depth increases. This is a related rates problem, where we are given the rate of change of the depth of gasoline (dh/dt) and asked to find the rate of change of the volume of gasoline (dV/dt).

The volume V of a cylinder with radius r and height h is given by V = πr^2h. The radius r of the tank is 2 feet. When the depth h of gasoline is 8 feet, the depth is increasing at the rate dh/dt = 0.1 ft/sec. To find dV/dt, we differentiate V with respect to t:

dV/dt = d(πr^2h)/dt = πr^2(dh/dt)

We then plug in the values: dV/dt = π(2^2)(0.1) = 0.4π cubic feet per second.

So the volume of gasoline is increasing at a rate of 0.4π cubic feet per second at the instant when the depth is 8 feet.