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

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