Is there a quicker way to find this differential $\frac{dh}{dt}$

Just as some context, this question involved revolving the curve $y=\frac{1}{4}x^2$ around the y-axis for $y\in[0,25]$. The first part was just to find this volume which I had no problems with. Howver, the second part involves filling this container full with water and then letting it leak out. It was said that the rate at which the volume of water was exiting was $$ \frac{dV}{dt}=-kh$$ Where $k$ is some constant and $h$ is the height of the water at any time $t$. It is required to show that $$\frac{dh}{dt}=\frac{-k}{4\pi}$$ I got this result in the following method but it feels like there may be a shortcut I can take within my method. We know that $$V=\pi\int(x^2(y))dy$$ But $y$ is the height so $$V=4\pi\int(h)dh$$ $$V=2\pi h^2+c$$ Then differentiate W.R.T $t$ $$\frac{dV}{dt}=4\pi h\frac{dh}{dt}$$ Which indeed gives us the correct answer. However, it feels I have gone backwards here as I have integrated and then differentiated? I'm thinking we can take $$V=4\pi\int h dh$$ And then differentiate to get $$\frac{dV}{dt}=4\pi\frac{d}{dt}(\int hdh) $$ But can we simplify the RHS without having to compute this integral? I know it is not a major time saver but just interested in different methods. Thanks.

Solution 1:

As you have written in your comment, you can simplify $\frac{d}{dt}\int hdh$ because, by the Chain Rule, this is the same as $$\frac{d}{dh}\int h dh\left(\frac{dh}{dt}\right)=h\frac{dh}{dt}$$

Alternatively, $$V=\pi\int x^2dy\implies \frac{dV}{dy}=\pi x^2=4\pi y$$

But at time $t$, $y=h$, so $\frac{dV}{dh}=4\pi h$.

Therefore, $$\frac{dh}{dt}=\frac{dh}{dV}\cdot\frac{dV}{dt}=\frac{1}{4\pi h}\cdot(-kh)=-\frac{k}{4\pi}$$

Solution 2:

Since we're working with a specified height, we should be using a definite integral (to avoid constants of integration).

Since $y=\frac{1}{4}x^2$, $x=2\sqrt{y}$. Therefore, the volume when the height is $h$ is $$ V=\int_0^h \pi (2\sqrt{y})^2dy=4\pi\int_0^hydy. $$ Then, differentiating both sides with respect to $t$ gives (using the fundamental theorem of Calculus/differentiating under an integral), $$ \frac{dV}{dt}=4\pi h\frac{dh}{dt}. $$

Since $\frac{dV}{dt}$ was given, we have that $$ -kh=4\pi h\frac{dh}{dt}. $$ From there, we solve to get $$ \frac{dh}{dt}=-\frac{k}{4\pi}, $$ as desired.