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You are filling up a cylindrical tank with a radius of 5m with water at a rate of 3 cm^3/min. Unbeknownst to you the tank has a hole and is leaking at a rate of 1 cm^3/min.
How fast is the height of the water increasing?
The volume of a cylinder V = πr^(2)h.

Respuesta :

Answer:

dh/dt = 3/25π m/min

Step-by-step explanation:

Radius = 5m

Rate (dV/dt) = 3 cm^3 / min

Leaking rate = 1 cm^3 / min

Volume = πr^2h

Volume = π(5)^2h

V= 25πh

Differentiate volume implicitly with respect to time

dV/dt = 3 cm^3/min

3 = 25π(dh/dt)

dh/dt = 3 m^3/min /25πm^2

dh/dt = 3/25πm/min

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