Question: There is a fountain at the cross of Main Mall and University Blvd in UBC. After observation, we found that the fountain is leaking water at a constant rate of $2m^{3}/hour$. At what rate is the depth of the water in the fountain changing when the depth of the water is 0.25m? (The fountion’s radius is 2m, and the height is 0.5m)
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Solution: $r=2m$, $h=0.5m$, $\frac{r}{h}=\frac{4}{1}$ so that $r=4h$.
$$V={\pi}{r^2}{h}$$
$$V={\pi}{(4h)^2}{h}$$
$$V=16*{\pi}{h^2}{h}$$
$$V=16*{\pi}{h^3}$$
$$\frac{dv}{dt}=3*16*{\pi}{h^2}*\frac{dh}{dt}$$
$$2=48*{\pi}*(0.25)^{2}*\frac{dh}{dt}$$
$$\frac{dh}{dt}=\frac{2}{3\pi}$$
Thus, when the depth of the fountain is 0.25m, the rate of increasing change is $\frac{2}{3\pi}m^{3}/hour$.