Science:Math Exam Resources/Courses/MATH102/December 2017/Question 14 (a)

MATH102 December 2017

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Question 14 (a)

The males of the magnificent frigatebirds have a bright red throat pouch that gets inflated like a balloon to attract females during mating season. Suppose that the rate of change of the volume during inflation is $$V'(t)={\frac {100\pi }{15}}\left(20-t\right)$$ measured in ${\textstyle \mathrm {cm} ^{3}/\mathrm {min} }$ for ${\textstyle 0\leq t\leq 20}$. At ${\textstyle t=0~\mathrm {min} }$, the pouch is empty. (a) Find the function ${\textstyle V(t)}$ for the volume at time ${\textstyle t}$ during inflation. (Hint: consider a simple antiderivative.)

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
The antiderivative of ${\textstyle x^{n}}$ is ${\textstyle {\frac {1}{n+1}}x^{n+1}+c}$.

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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, the antiderivative of ${\textstyle (20-t)}$ is $$20t-{\frac {1}{2}}t^{2}+c,$$ where $c$ is an arbitrary constant. Since $$V'(t)={\dfrac {100\pi }{15}}(20-t),$$ this implies that $$V(t)={\dfrac {100\pi }{15}}\left(20t-{\frac {1}{2}}t^{2}\right)+C,$$ where ${\textstyle C}$ is a constant. As we are given that ${\textstyle V(0)=0}$, we have $$0={\dfrac {100\pi }{15}}\left(20\cdot 0-{\frac {1}{2}}\cdot 0^{2}\right)+C=C.$$ Hence, $$V(t)={\dfrac {100\pi }{15}}\left(20t-{\frac {1}{2}}t^{2}\right)={\dfrac {10\pi }{3}}\left(40t-t^{2}\right).$$ Answer: ${\textstyle V(t)={\color {blue}{\dfrac {10\pi }{3}}\left(40t-t^{2}\right)}.}$