Question

Consider a right cone (pointed downwards) that is leaking water. the dimensions of the conical tank are a height of 19 ft and a radius of 6 ft. how fast (in ft/min) does the depth of the water change when the water is 16 ft high if the cone leaks water at a rate of 16 ft³/min?

____ ft/min

Answer 1

The rate at which the depth of the water changes when the water is 16 ft high in the leaking cone is approximately 4.64 ft/min.

Step-by-step explanation:

To find the rate at which the water depth changes, we can utilize the concept of related rates involving the volume of a cone. The formula for the volume of a cone is V = (1/3)πr²h, where V represents volume, r is the radius, and h is the height.

Given:

Height of the cone (h) = 19 ft

Radius of the cone (r) = 6 ft

Rate of water leaking (dV/dt) = 16 ft³/min

We're required to find the rate of change of depth when the water height is 16 ft (dh/dt when h = 16 ft).

First, apply the volume formula for a cone: V = (1/3)πr²h. Then, differentiate it with respect to time (t) to find dV/dt.

V = 1/3 * π * r^2 * h

V = 1/3 * π * 6^2 * h

V = 12πh

Differentiate both sides with respect to time:

dV/dt = 12π * dh/dt

Given dV/dt = 16 ft³/min, when h = 16 ft, substitute the values:

16 = 12π * dh/dt

Now solve for dh/dt:

dh/dt = 16 / (12π) ≈ 4.64 ft/min

Therefore, when the water height is 16 ft, the depth of the water in the cone is changing at a rate of approximately 4.64 ft/min.