A cylindrical tank with radius 5m is being filled with water at a rate of $3m^3/min$ How fast is the height of the water increasing?
I know that $$dv/dt = 3$$
And that $$v= \pi r^2 h$$
So I want to find the derivative of h
$$h= \frac {v}{ \pi r^2}$$
$$h' = \frac{3 \pi * 25}{ (\pi * 25)^2 }$$
$h' * v'$
This gives an incorrect answer and I am not sure why.
Best Answer
You want $\dfrac{dh}{dt}$; by the chain rule this is $\dfrac{dh}{dv}\dfrac{dv}{dt}$.
You have $h=\dfrac{v}{\pi r^2}=\dfrac1{\pi r^2}v$, where $\dfrac1{\pi r^2}$ is a constant, so $\dfrac{dh}{dv}=\dfrac1{\pi r^2}$; you don't need the quotient rule for this differentiation. Finally, you have $\dfrac{dv}{dt}=3$, so $$\frac{dh}{dt}=\frac{dh}{dv}\frac{dv}{dt}=\frac3{\pi r^2}\text{ m/min}\;.$$ Since $r=5$ m, the actual rate is $\dfrac3{25\pi}$ m/min.
In a problem like this it's a good idea to use the $\dfrac{dv}{dt}$ notation instead of the $v'$ notation, because you're taking derivatives with respect to more than one variable, and you need to keep track of what that variable is in each calculation.