Michael M. answered • 05/06/21

Math, Chem, Physics, Test Prep Tutoring with Michael ("800" SAT math)

First get an equation for the volume. V = (1/3)πr^{2}h

The height is 12m when the radius is 4m, therefore the height of the water is always going to be 3 times the radius of water due to similar triangles.

V = (1/3)π(h/3)^{2}h = (π/27)h^{3}

Next take the derivative of both sides with respect to time

dV/dt = (π/27)(3h^{2})dh/dt

dV/dt = (π/9)(h^{2})dh/dt

dV/dt = (π/9)(h^{2})(h - 12)

Lastly, solve by plugging in h = 3

b)

Use the equation for a volume of a cylinder with area of base = 400π

V = 400πh

Take the derivative of both sides with respect to time:

dV/dt = 400π(dh/dt)

The rate at which the water is flowing into the cylinder is going to be negative of the rate that the water was flowing out of the cone. So plug that into dV/dt and solve for dh/dt.