Question

A water tank is in the shape of an inverted cone with height twice its radius. Water is being pumped into the tank at a rate of 2m3/min. Find the rate at which the water level is rising when the water is 3m deep.

Answer 1

dh/dt = 8/9π

Explanation:

Given that

Volume of A cone = 1/3πr²h.

Assuming v is the volume of the tank, then dv/dt = 2 m³/min

Assuming h is the height of the water we're looking for dh/dt

To start, we need to find a relationship between dh/dt and dv/dt

We then write v as a function of h, instead of having it as h & r, and then by using similar triangles, we notice that

r/h = 2/4, essentially, r = h/2.

Now, we substitute for this r, in the formula for volume

V = 1/3πr²h

V = 1/3 * π * (h/2)² * h

V = 1/3 * π * h³/4

If we differentiate v, wrt t, we have

dv/dt = π/12 * 3h² * dh/dt

dv/dt = π/4 * h² * dh/dt

Remember that h = 3, so, substitute for h

2 = π/4 * 9 * dh/dt

8/π = 9 * dh/dt

dh/dt = 8/9π