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

MATH110 December 2015

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Question 08 (c)

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. (c) Over time, both the volume of water and the height of water in the tank change. Suppose you observe that the height of water is increasing at a rate of 0.1 m/min when the water in the tank is 1 m deep, how fast (in m 3 /min) is the volume increasing at that instant?

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
In this part $h$ and $V$ are both functions of time $t$ and we are looking for the rate of change in the volume with respect to time i.e. ${\frac {dV}{dt}}$ . The volume function is given with respect to $h$ as $V(h(t))$, so we should use chain rule to find its derivative with respect to $t$ .

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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. The volume function is $V(t)=V(h(t))$ which is a composition of two function, so in order to find ${\frac {dV}{dt}}=V'(t)$ we apply the chain rule as the following: $V'(t)=V'(h(t))\cdot h'(t)$ . We have already found $V'(h(t))$ in part (b) as ${\begin{aligned}V'(h(t))&={\frac {dV}{dh}}=\pi \left(2h(2-{\frac {h}{3}})-{\frac {h^{2}}{3}}\right),\quad {\text{so}}\\V'(t)&=\pi \left(2h(2-{\frac {h}{3}})-{\frac {h^{2}}{3}}\right)\cdot h'(t)\end{aligned}}$ The rate of change in the height is given by $h'(t)=0.1$ m/min at the moment of $h=1$, so at that moment, we have ${\begin{aligned}V'(t)=\pi \left(2(2-{\frac {1}{3}})-{\frac {1}{3}}\right)\cdot 0.1=3\pi \cdot 0.1=\color {blue}0.3\pi \ m^{3}/min\end{aligned}}$ .