# Calculate the Rate of Change of the Volume of a Frustum of a Right Circular Cone

I have a question for calculus which I'm having trouble with. I've calculated the volume of the frustum with the formula. However, I'm having difficulty deriving this equation in order to find the rate of change of the volume. Would appreciate advice, here is the problem.

"A reservoir containing water has the shape of a frustum of a right circular cone of altitude 10 feet, a lower base of radius 10 feet, and an upper base radius 15 feet. How fast is the volume of the water increasing when the water is 6 feet deep and rising at a rate of 2 ft./hr?"

• Correct altitude $h=10,\,$ or $h+H== 10 \, ?$ – Narasimham Oct 23 '17 at 2:21

You can simply model it by finding the rate at which the volume of the subtracted cone changes. This is true because the Volume of the whole reservoir remains constant.

Implicit differentiation! Take the derivative of both sides with respect to $t$ like so:

\begin{align*} V=\frac{\pi}{3} h \left( R^2+Rr+r^2 \right) &\implies \frac{d}{dt}(V) =\frac{\pi}{3}\frac{d}{dt} \left( h R^2+ hRr+ hr^2 \right) \end{align*} $$\implies \frac{dV}{dt} = \frac{\pi}{3} \left( \frac{dh}{dt} R^2 + 2Rh\frac{dR}{dt} + \frac{dh}{dt}Rr+h\frac{dR}{dt}r+hR\frac{dr}{dt} + \frac{dh}{dt}r^2+2rh\frac{dt}{dt} \right)$$

Essentially implicit differentiation is doing the chain rule over and over but treating everything in the expression as an unknown function of $t$, hence why you're seeing $\frac{d}{dt}$ everywhere.