Science:Math Exam Resources/Courses/MATH104/December 2011/Question 04

MATH104 December 2011

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Question 04

A conical tank of height of 5 metres and top radius 4 metres is filled with water and then drains into a cylindrical container of height 5 metres and radius 4 metres. If the water level in the conical tank drops at a constant rate of 0.5 metres per minute, at what rate does the water level in the cylindrical tank rise when the water level in the conical tank is 3 metres? The volume of a cone of radius r and height h is $V=(1/3)\pi r^{2}h$. 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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Write down everything you know about the problem. Try drawing a picture.

Hint 2
In order to solve this problem, you need to differentiate both volume formulas with respect to t. However, before you do so, you should re-write the formula for the volume of a cone in terms of h, with no r.

Hint 3
Using similar triangles, we can determine that the relationship between the radius and height of water in the first tank is given by the ratio ${\frac {r}{h}}={\frac {4}{5}}$.

Hint 4
What is the relationship between the volume of water, and its rate of change, in the first and second tank?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We begin by assembling all the information we know about the problem: The volume of water in the conical tank, which we will denote as $V_{1}$ is given by the formula $V_{1}=1/3\pi r^{2}h_{1}$ where $h_{1}$ is the height of water in the tank. The volume of water in the cylindrical tank, which we will denote as $V_{2}$ is given by the formula $V_{2}=\pi r^{2}h_{2}$ where $h_{2}$ is the height of water in the tank. Due to similar triangles, we know that the ratio of radius/height within the conical tank is always constant, thus for the conical tank $r/h=4/5$. The rate of change of the volume of water in the first tank is $dV_{1}/dt$. Because the water is pouring from the first tank into the second one, the change in the volume of water in the second tank, or $dV_{2}/dt$ will be the same as $-dV_{1}/dt$. The height of the water in the conical tank is dropping by .5 m/min, which is given by the derivative $dh_{1}/dt=-0.5$. We are looking for the change of the water's height in the second tank, which is given by $dh_{2}/dt$. We note that for this problem, we are mostly concerned with change in height and change in volume, not change in radius. Therefore, we will rewrite our two volume formulas in terms of h as follows: First, we use fact #3 and solve for r in terms of h, or $r={\frac {4}{5}}h$. We can then rewrite the formula given in #1 as a function of $h_{1}$. $V_{1}(h_{1})={\frac {1}{3}}\pi \left({\frac {4}{5}}\right)^{2}(h_{1})^{3}$. For the volume formula given in #2, the radius doesn't change (because the tank has straight sides) and equals 4. So we have a volume formula as a function of $h_{2}$. $\displaystyle V_{2}(h_{2})=\pi (4)^{2}h_{2}$. We now have two volume formulas in terms of h. Because the question is asking about change over time, we will implicitly differentiate with respect to t. This gives: ${\frac {dV_{1}}{dt}}(h_{1})=\pi \left({\frac {4}{5}}\right)^{2}(h_{1})^{2}{\frac {dh_{1}}{dt}}$ for the first tank and $\displaystyle {\frac {dV_{2}}{dt}}(h_{2})=\pi (4)^{2}{\frac {dh_{2}}{dt}}$ for the second tank. If we consider the first formula, we realize from #5 that we know $dh_{1}/dt=-0.5$ and furthermore, the statement of the question tells us that $h=3$. We plug these into the first formula to get ${\begin{array}{ccl}{\frac {dV_{1}}{dt}}(3)&=&\pi \left({\frac {4}{5}}\right)^{2}(3)^{2}(-0.5)\\&=&-{\frac {72}{25}}\pi \end{array}}$ Now we remember from #4 that $dV_{2}/dt=-dV_{1}/dt$. So $dV_{2}/dt=(72/25)\pi$. Plugging that into our second volume formula we get ${\frac {72}{25}}\pi =\pi 16{\frac {dh_{2}}{dt}}$ Finally, we know from #6 that we're trying to find $dh_{2}/dt$. Thus we simply solve the previous equation for $dh_{2}/dt$ to get: ${\frac {dh_{2}}{dt}}={\frac {9}{50}}$ Which is our answer.

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