Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 08 (a)/Solution 1

Denote the depth of the water in the bottom tank by $h$ and the radius of the cone filled with water with the depth of $h$ by $r$. Then, using the ratio between the height and the radius as in the picture on the right,

we get ${\frac {R}{3H}}={\frac {r}{h}}\implies r={\frac {R}{3H}}h$.

Observe that at the depth $h$ of water, the volume of the water is given by

$$V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi \left({\frac {R}{3H}}h\right)^{2}h={\frac {\pi }{27}}{\frac {R^{2}}{H^{2}}}h^{3}.$$

By the Hint, the rate of change of $V$ over time is constant. i.e., ${\frac {dV}{dt}}=C$. Taking a derivative on both side of the equation with respect to $t$, we have

$$C={\dfrac {dV}{dt}}={\frac {\pi }{27}}{\frac {R^{2}}{H^{2}}}{\dfrac {d(h^{3})}{dt}}={\frac {\pi }{9}}{\frac {R^{2}}{H^{2}}}h^{2}{\dfrac {dh}{dt}}.$$

In the second equality, the derivative only hits $h^{3}$ because $R$ and $H$ are fixed constants. (i.e., independent of time.)

Therefore, at the depth $h=H$, the rate is

$$C={\frac {\pi }{9}}{\frac {R^{2}}{H^{2}}}\cdot H^{2}\cdot {\dfrac {dh}{dt}}\implies \color {blue}{\dfrac {dh}{dt}}={\frac {9C}{\pi R^{2}}}.$$

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