This is a rates of change question:

A rectangular water tank has a square base with sides of length $0.55m$. Water is poured into the tank so that the fluid volume in the tank increases at a constant rate of $0.2m^3$ per hour. How fast is the water leveraging in metres per hour?

I've worked out the volume of the water tank $V=0.55 ^3 =0.166375m ^3$. From the question, $dv/dt=0.2m^3$ p/hr but I do not know what to do next. Any help will be beneficial.


Let's write the formula for the volume of water in the tank as a function of depth $h$ and length $l$:

$$V = hl^2.$$

Now, we have that both volume $V$ and depth $h$ are functions of time:

$$V(t) = h(t)l^2.$$

We know that $dV/dt = 0.2 m^3/hr$. We also know that

$$\frac{dV(t)}{dt} = \frac{d}{dt}\left(h(t)l^2\right) = l^2\frac{dh(t)}{dt}.$$

It should be straightforward from here.


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