Question

derivative application....

Water is flowing into a spherical tank with 6 ft radius at constant rate of 30pi cu ft per hr. When the water is h feet deep, the volume of water is given by
V=((pih^2)/3)*(18-h)
What is the rate at which the depth of the water in tank is increasing at the moment when the water is 2 ft deep?

Answer 1

\[\frac{dV}{dt} = \frac{dV}{dh}*\frac{dh}{dt}\]
\[\rightarrow \frac{dh}{dt} = \frac{30}{dV/dh}\]
take derivative and then plug in h=2

Answer 2

Can you please explain how you would take the derivative?

Answer 3

\[V = \frac{\pi}{3}h^2 (18-h) = 6 \pi h^2 - \frac{\pi}{3} h^3\]
you take derivative of polynomial
\[\frac{d}{dx} x^n = n x ^{n-1}\]

Answer 4

Thanks so much!

Answer 5

yw :)