rate of change - cones

Question

A tank of water in the shape of a cone is leaking at the rate of 2 ft³/hour. The base radius of the take is 5ft. and the height of the take is 14ft. at what rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft?

Answer 1

The height of the cone is 14 ft and its radius is 5 ft.

Here water is leaking (negative value), so the rate of change of volume dv/dt = - 2 ft³/hour.

The volume of cone is .

Differentiate implicitly with respect to t to obtain the related - rate equation

.

The above equation represent the rate of change of V is related to the rates of change of both h and r.

Here to find rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft.

So, the value of h is convert into the r values in the volume equation.

Here two cones are available, first one is original cone and second one is shape of cone with water height 6 ft.

Two similar triangles then ratios of any two sides will be equal.

.

Volume :

Differentiate implicitly with respect to t.

If h = 6 ft then and dv/dt = - 2.

.

The rate of change of the radius of the top of the water in the tank changing when the depth of the water is 6ft is dr/dt = - 0.04949 ft/hour.