Science:Math Exam Resources/Courses/MATH104/December 2013/Question 03

MATH104 December 2013

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Question 03
Two cylindrical tanks are being filled simultaneously at exactly the same rate. The smaller tank has a radius of 5 metres, and the water rises at a rate of 0.5 metres per minute. The larger tank has a radius of 8 metres. How fast is the water rising in the larger tank? It may be helpful to know that the volume of a cylinder of radius R and height H is $V=\pi R^{2}H$.

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
Both tanks are being filled at the same rate. What does that tell you about the rate of change of the volume of water in each tank relative to each other?

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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. To do this question, we start by letting $h_{1}$ and $h_{2}$ denote the height of the water level in the smaller and larger cylinder respectively. We also let $r_{1}=5$, $r_{2}=8$ be the radii of the tanks and $V_{1},V_{2}$ be the respective volumes of water within each tank. Since we are told that each tank is being filled at the same rate, this implies that ${\frac {dV_{1}}{dt}}={\frac {dV_{2}}{dt}}$ (i.e. The tanks have an equal rate of change of volume of water per unit time). Since the radius of the tank is not changing with time, but the height of the water level is changing with time, we can differentiate the expression for volume of a cylinder with respect to time giving us ${\frac {dV_{1}}{dt}}=\pi r_{1}^{2}{\frac {dh_{1}}{dt}},\quad {\frac {dV_{2}}{dt}}=\pi r_{2}^{2}{\frac {dh_{2}}{dt}}.$ We want to find how fast the height of water in the larger tank (i.e. tank 2) is changing, hence we are looking for the value of $dh_{2}/dt$. Using the above equations we can write $\pi r_{1}^{2}{\frac {dh_{1}}{dt}}=\pi r_{2}^{2}{\frac {dh_{2}}{dt}}.$ Isolating for $dh_{2}/dt$ gives ${\begin{aligned}{\frac {dh_{2}}{dt}}&={\frac {r_{1}^{2}}{r_{2}^{2}}}{\frac {dh_{1}}{dt}}={\frac {5^{2}}{8^{2}}}(0.5)={\color {blue}{\frac {25}{128}}{\frac {\text{m}}{\text{min}}}}.\end{aligned}}$ Thus, the water level in the larger tank is rising at a rate of (25/128) metres per minute.

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