Science:Math Exam Resources/Courses/MATH110/December 2016/Question 02 (c)

MATH110 December 2016

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Question 02 (c)
The volume ${\textstyle V=(4/3)\pi r^{3}}$ of a spherical balloon changes with its radius ${\textstyle r}$. What is the rate of change of the balloon’s volume with respect to its radius when the radius is ${\textstyle r=2{\text{m}}}$?

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 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
According to the question, we need to find ${\textstyle V'(2)}$.

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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. By the Hint, we need to find ${\textstyle V'}$ and put ${\textstyle r=2{\text{m}}}$. Since $$V(r)={\dfrac {4}{3}}\pi r^{3},$$ we have $$V'(r)={\dfrac {4}{3}}\pi \cdot 3r^{2}=4\pi r^{2}.$$ When ${\textstyle r=2}$, $$V'(2)=4\pi 2^{2}=16\pi .$$ So the rate of change is ${\textstyle 16\pi }$. Answer: the rate of change is ${\textstyle {\color {blue}16\pi ({\text{m}}^{3}/{\text{m}}}).}$