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

MATH 180 December 2017

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  •  December 2017  •  

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 ${\textstyle R}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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!

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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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $$C={\frac {\pi }{9}}{\frac {R^{2}}{H^{2}}}\cdot H^{2}\cdot {\dfrac {dh}{dt}}\implies \color {blue}{\dfrac {dh}{dt}}={\frac {9C}{\pi R^{2}}}.$$ .