Related Rates: Conical Pile Height Growth with Sand Conveyor

[courtrigrad]

Sand falls from a conveyor belt at the rate of 10 \frac{ft^{3}}{min} onto a conical pile. The radius of the base of the pile is always equal to half the pile's height. How fast is the height growing when the pile is 5 ft high?

So r = \frac{1}{2} h. That means when h = 5 , r = 2.5. We want to find \frac{dh}{dt}. I know the volume of a cone is: \frac{1}{3}\pi r^{2}h.
\frac{dV}{dt} = \frac{1}{3}\pi r^{2} h \frac{dh}{dt}. So would I just do:

10 = \frac{1}{3}\pi (2.5)^{2}(5) \frac{dh}{dt} and solve for \frac{dh}{dt}?

Thanks

 

[mezarashi]

Your general approach is correct. You must express the volume of the cone as a function of the height of the cone only. So that you have only 2 variables, something like:

V = f(h), then differentiating you get dV/dt = df(h)/dt

10 = \frac{1}{3}\pi (2.5)^{2}(5) \frac{dh}{dt}

This is an incorrect differentiation. Because r and h are functions of t, you must use the product rule on the right hand side. It is easier however if you replace r by 0.5h as given in the question so that you only have 1 variable. You are differentiating with respect to time. All variables that change with respect to time must be treated accordingly. For reference, you should be getting

\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}