This is the question: Water is being poured into a tank at $10ft^3/min$. Find the rate at which the water level is increasing when the depth of the water is ${1\over 3}ft$ Here is a picture of the tank with it's measurements
So to find the rate at which the water level is increasing I started with the formula for a prism $V=Bh$ and then I used implicit differentiation, as well as the product rule.
$${d\over dt}[V]={d\over dt}[Bh]$$ $${dv\over dt}=(h){d\over dt}B+(B){d\over dt}h$$ $${dv\over dt}=(h){dB\over dt}+(B){dh\over dt}$$ Now I replaced these terms: $h=20$; ${dB\over dt}={1\over 3}$; ${dh\over dt}=10$ (because the rate that the water is being poured is $10ft^3/min$); and $B=4$
$${dv\over dt}=(20)({1\over 3})+(10)({4})$$ $${dv\over dt}=({20\over 3})+(40)$$ $${dv\over dt}={140\over 3}$$ $${dv\over dt}\approx46.67ft^3/min$$
Did I do everything right?
Here is an approach. Note that the full volume of the prism is $V=BH$, where $B=80$ft$^2$ and $H=\sqrt{12}$ft.
Now, let's say that at time $t$, the water has reached a height of $h(t)$. Then, the volume of water in the prism at $t$ is
$$\begin{align} v(t)&=BH\left(\frac{h(t)}{H}\right)^2\\\\ &=\frac{B}{H}h^2(t) \end{align}$$
The rate of increase in the volume of water $\frac{dv(t)}{dt}$ is given by
$$\frac{dv(t)}{dt}=\frac{B}{H}2h(t)\frac{dh(t)}{dt}$$
We are given the rate of increase in the volume of water $\frac{dv(t)}{dt}$ is $10$ft$^3$ per minute. Thus, at time $\tau$ when the water height is $h(\tau)=\frac13$foot, the rate of increase in height $h'(\tau)$ is
$$\begin{align} h'(\tau)&=\frac12\frac{H}{Bh}v'(\tau)\\\\ &=\frac12\frac{\sqrt{12}}{(80)\frac13}10\\\\ &=\frac3{16}\sqrt{12} \end{align}$$