Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 5.00 inches at the top and a height of 6.00 inches. At the instant when the water in the container is 4.00 inches deep, the surface level is falling at a rate of 0.7 in./sec. Find the rate at which water is being drained from the container.

The rate at which water is being drained from the container is;

dV/dt = 24.4 in³/s

I've attached an image showing this inverted cone with the corresponding letters to represent the height and radius.

We are given;

Radius of Cone; R = 5 inches

Height of cone; H = 6 inches

h = 4 inches

Height rate of change; dh/dt = 0.7 in/s

Looking at the image attached, we can use similar triangle property to derive the relationship;

r/R = h/H

Where r is the radius at the instant where dh/dt = 0.7 in/s

Thus;

r/5 = 4/6

r = (5 × 4)/6

r = 3.33 inches

Now, from r/R = h/H

We can write with initial values of cone and differentiate;

r/5 = h/6

6r = 5h

6(dr/dt) = 5(dh/dt)

Since dh/dt = 0.7 in/s

Thus;

dr/dt = (5/6) × 0.7

dr/dt = 0.5833

Volume of cone is given by;

V = ⅓πr²h

We can find the rate at which the rate at which water is being drained from the container using partial differentiation on the volume equation.

Thus;

dV/dt = ⅓π((2rh × dr/dt) + (r² × dh/dt))

Plugging in the relevant values, we have;

dV/dt = ⅓π((2 × 3.33 × 4 × 0.5833) + (3.33² × 0.7))

dV/dt = 24.4 in³/s

Read more about rate of change of cylinder volume at; link

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