Question

A chemical substance is draining from a conical filtering system at a rate of 100 cubic centimeters per minute into a cylindrical storage tank below. The conical filter and cylindrical tank each have a diameter of 60 centimeters, and the height of the cone also measures 60 centimeters. At time T, the depth of the chemical substance in the conical filter is 48 centimeters.

a. How fast is the level of the substance in the storage tank rising at time T?

b. How fast is the level in the filtering system falling at time T?

I am lost, can someone point me in the right direction and show me what steps to do?

Answer 1

dV/dt=100 cm^2/min

in the storage tank: V=PI*r^2 h
dV/dt=100=PI*r^2 dh/dt solve for dH/dt. R=30cm

In the cone: you need the radius at the top of the liquid in the cone of height 48 cm. Since it is a proportion
D/60=48/60, so D=48, so r=24 cm
in this cone, r is a function of height, so r=h/2
V=1/3 * PI r^2 h
= 1/2 PI h^3 / 4
V= 1/8 PI h^3
dV/dt= 3/8 PI h^2 dh/dt solve for dh/dt

check my thinking.