A paper cup has the shape of a cone with height of 8cm and a radius of 4cm at the top. Water is poured into the cup at a rate of 2cm^3/s. How fast is the water level rising when the water level is 4cm deep?

Answer 1

At any water level, the radius of the surface is 1/2 the height. h = 2r
So, when the water has depth h, the volume of water is
v = 1/3 πr^2 h = 2/3 πr^3
dv/dt = 2πr^2 dr/dt
when y=4, r=2, so
2 = 2π*2^2 dr/dt
dr/dt = 1/(4π)
since h = 2r, dh/dt = 2 dr/dt = 1/(2π) cm/s

or, you could have made v a function of h:
v = 1/3 π (h/2)^2 h = 1/12 π h^3
dv/dt = 1/4 πh^2
2 = π/4 (4^2) dh/dt
dh/dt = 1/(2π) cm/s