I'm trying to solve this problem:
"Sand being emptied from a hopper at the rate of $10$ cubic feet per second forms a conical pile whose height is always twice its radius. At what rate is the radius of the pile increasing when its height is $5$ feet?"
My Attempt
The volume of a cone is given by $$V=\frac{1}{3}\pi r^2h$$ We know that $\frac{dV}{dt}=10$, $h=5$, $h=2r$, and we are solving for $\frac{dr}{dt}$.
Implicit differentiation gives $$\frac{dV}{dt} = \frac{2\pi}{3}r\frac{dr}{dt}\frac{dh}{dt}$$ But how can we proceed from here? Any help would be greatly appreciated.
You forgot to use the product rule. IMO it is best to think of every variable as a function of time, as this it would be the most accurate interpretation. We would then have
$$V(t)=\frac{1}{3}\pi r^2(t)\cdot h(t)$$ $$\implies \frac{d}{dt}[V(t)]=\frac{d}{dt}\left[\frac{1}{3}\pi r^2(t)\cdot h(t)\right]$$ $$\implies V'(t) = \frac13\pi\left(2r(t)\cdot r'(t)\cdot h(t)+r^2(t)\cdot h'(t)\right)$$
When $h(t_0)=5$ for some $t_0$, then we also know that $r(t_0)=\frac52$ and $V'(t_0) = 10$. We want to find $r(t_0)$. Substituting this information in makes our equation $$10 = \frac13 \pi \left( 2\cdot \frac52\cdot r'(t_0)\cdot 5 + \left(\frac52\right)^2 \cdot h'(t_0)\right)$$
We also know that $$h(t)=2r(t)\Longleftrightarrow h'(t) = 2r'(t)$$ $$10 = \frac13 \pi \left( 2\cdot \frac52\cdot r'(t_0)\cdot 5 + \left(\frac52\right)^2 \cdot 2r'(t_0)\right)$$ $$\implies r'(t_0) = \frac4{5\pi}$$