Science:Math Exam Resources/Courses/MATH110/April 2017/Question 03 (b)

MATH110 April 2017

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Question 03 (b)

Cosmos had a brief career as an artist. His masterpiece was a sculpture made of ice in the shape of a circular cone (of volume $V={\frac {\pi }{3}}r^{2}h$, where $r$ and $h$ are the base radius and height of the cone). Soon after its completion, the sculpture started to melt. Assume that the sculpture was melting in such a way that the diameter of its base was always $3$ times its height. When the ice was melting at $0.5{\text{m}}^{3}/{\text{min}}$ and the sculpture was $2{\text{m}}$ high, (b) at what rate was the radius of the base of the sculpture changing?

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 already know the speed ${\textstyle h'}$, what is the relation between ${\textstyle h'}$ and ${\textstyle r'}$?

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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. We know that ${\textstyle 2r=3h}$ always holds, differentiating both sides with respect to ${\textstyle t}$ gives $$2r'=3h'.$$ Thus ${\textstyle r'={\frac {3}{2}}h'={\frac {3}{2}}\times {\frac {1}{18\pi }}=\color {blue}{\frac {1}{12\pi }}}$