Science:Math Exam Resources/Courses/MATH102/December 2017/Question 14 (a)/Solution 1

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