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

MATH102 December 2017

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[hide] 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.)

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

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[show] Solution

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: ${\textstyle V(t)={\color {blue}{\dfrac {10\pi }{3}}\left(40t-t^{2}\right)}.}$