Science:Math Exam Resources/Courses/MATH102/December 2013/Question C 01

MATH102 December 2013

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[hide] Question C 01
A mollusc grows inside a shell and produces new shell material at its opening (indicated in the image below in bold) The volume of the shell can be approximated as that of a cone of base radius r and height h. Given that the height of the cone grows at a constant rate of 0.1 cm/year, at what rate will the volume of the cone be changing when $h=10$ and $r=1$ cm? Leave your answer in terms of $\pi$ . Note that: $V_{\text{cone}}={\frac {1}{3}}\pi r^{2}h$

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[show] Hint 1
What relationship between r and h can we infer by similar triangles?

[show] Hint 2
Since we are only given the rate of change in the cone's height, try expressing the volume of the shell (cone) entirely in terms of h.

[show] Hint 3
Differentiate with respect to time to relate the rate of change in volume to the rate of change in height (note that both are functions of time).

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[show] 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 of a cone is given by $V={\frac {1}{3}}\pi r^{2}h$ By the physical application of the problem, we can assume that the base radius of the cone changes as the height increases. First, we should express the radius of the cone as a function of its height. By similar triangles, and the fact that r = 1 when h = 10, we find that the following relation holds: ${\frac {r}{h}}={\frac {1}{10}}\quad \iff \quad r={\frac {h}{10}}$ Thus the volume of the shell can be expressed entirely in terms of h: ${\begin{aligned}V&={\frac {1}{3}}\pi \left({\frac {h}{10}}\right)^{2}h\\&={\frac {\pi }{300}}h^{3}\end{aligned}}$ The rate of change in the volume of the cone can be computed by differentiating V with respect to t, giving us ${\frac {dV}{dt}}={\frac {\pi }{100}}h^{2}\cdot {\frac {dh}{dt}}$ Since we are told that the height of the cone changes at a constant rate of 0.1 cm/year, the rate of change of in the cone's volume when h = 10 cm and r = 1 cm is given by ${\begin{aligned}{\frac {dV}{dt}}&={\frac {\pi }{100}}(10{\text{cm}})^{2}\cdot 0.1{\frac {\text{cm}}{\text{year}}}\\&={\color {blue}{\frac {\pi }{10}}\,{\frac {{\text{cm}}^{3}}{\text{year}}}}\end{aligned}}$