Science:Math Exam Resources/Courses/MATH110/December 2015/Question 08 (b)

MATH110 December 2015

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Question 08 (b)

If a spherical tank of radius 2 metres is filled up to h metres with water, then the volume of water $V$ in the tank is given by the following function: $V(h)=\pi h^{2}\left(2-{\frac {h}{3}}\right),$ where $h$ is measured from the deepest point in the tank. (b) As the water is poured into the tank, at what rate is the volume of water in the tank changing with respect to the height of the water when the water is 1 m deep? What are the units of this quantity?

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
We are looking for the rate of change in volume with respect to the height, this means the derivative of volume with respect to height: ${\frac {dV}{dh}}=V'(h)$ at $h=1$ m .

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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. Rate of change in volume with respect to the height means the derivative of volume with respect to height: ${\frac {dV}{dh}}=V'(h)$ at $h=1$ m . ${\begin{aligned}V(h)=\pi h^{2}\left(2-{\frac {h}{3}}\right)&\Rightarrow {\frac {dV}{dh}}=V'(h)=\pi \cdot 2h\left(2-{\frac {h}{3}}\right)+\pi h^{2}\cdot \left(-{\frac {1}{3}}\right)\\&\Rightarrow V'(1)=2\pi \cdot {\frac {5}{3}}-{\frac {\pi }{3}}=\color {blue}3\pi (m^{3}/m)=3\pi (m^{2})\end{aligned}}$