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

MATH 180 December 2017

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[hide] Question 08 (a)

Consider two tanks of water, one above the other, as pictured below. The top tank, which is initially full, is a cylinder of radius ${\textstyle R}$ and height ${\textstyle H}$. The bottom tank, which is initially empty, is a cone of radius ${\textstyle R}$ and identical volume as the top tank. Suppose water drains at a constant rate ${\textstyle C}$ from the top tank to the bottom tank. (a) Find the rate at which the depth of water in the bottom tank is increasing when it is equal to ${\textstyle H}$. Your answer should be in terms of ${\textstyle C}$ and ${\textstyle R}$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Since water drains at a constant rate $C$, it means that the rate of change of the volume of water over time is the constant.

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[show] Solution

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