Science:Math Exam Resources/Courses/MATH100/December 2015/Question 08 (b)

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MATH100 December 2015

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Question 08 (b)
A tank of water in the shape of an inverted circular cone of height 16 m and diameter 20 m at the top is being filled with water at a rate of 2 m³ per minute. (b) Find the rate of change of the height of the water in the tank when the height of the water is 10 m.

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
Find the relation between the radius of the circle at the top of the water and the height of the water.

Hint 2
Recall the volume formula for a cone.

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Solution
Let $r$ be the radius of the circle at the top of the water and $h$ be the height of the water. Then, using the picture in part (a) with a property of similar triangles gives the ratio between $r$ and $h$ as ${\frac {r}{h}}={\frac {10}{16}}\implies r={\frac {5}{8}}h.$ On the other hand, by the volume formula for a cone, the volume $V$ of water can be written as $V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}h\right)^{2}\cdot h={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot h^{3}.$ Therefore, by differentiating with respect to $t$ on both sides, we obtain ${\frac {dV}{dt}}={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot 3h^{2}\cdot {\frac {dh}{dt}}=\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot h^{2}\cdot {\frac {dh}{dt}}.$ Since the volume changes at a rate of $2\,\mathrm {m} ^{3}/\mathrm {min}$ , the rate of change of the height of the water when the height is $10\,\mathrm {m}$ is $2\,\mathrm {m} ^{3}/\mathrm {min} =\pi \left({\frac {5}{8}}\right)^{2}\cdot (10\,\mathrm {m} )^{2}\cdot {\frac {dh}{dt}}{\bigg \|}_{h=10}\implies {\frac {dh}{dt}}{\bigg \|}_{h=10}={\frac {2\cdot 8^{2}}{5^{2}\cdot 10^{2}\pi }}\,\mathrm {m} /\mathrm {min} =\color {blue}{{\frac {2^{5}}{5^{4}\pi }}\,\mathrm {m} /\mathrm {min} }.$