Science:Math Exam Resources/Courses/MATH103/April 2005/Question 04

MATH103 April 2005

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Question 04
The level of water in a reservoir changes over the year due to inflow and outflow. If the volume at $\displaystyle t=0$ is $\displaystyle 10^{4}$ cubic metres and the rate of change of volume is $\displaystyle 10\sin(\pi t/10)$ cubic metres per day, where t is the time in days, find the total change in the volume between $\displaystyle t=0$ and $\displaystyle t=5$.

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Hint
Given the rate of change of some process r(t), the total change over the time interval $a\leq t\leq b$ is given by $\int _{a}^{b}r(t)dt$

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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. Given that the rate of change of volume is $10\sin(\pi t/10)$ cubic centimetres per day, the total change in the volume between t=0 and t=5, in cubic centimetres, is ${\begin{aligned}\int _{0}^{5}10\sin(\pi t/10)dt&=\left[10{\frac {-10}{\pi }}\cos(\pi t/10)\right]_{0}^{5}\\&={\frac {-100}{\pi }}\left(\cos(5\pi /10)-\cos(0/10)\right)\\&={\frac {-100}{\pi }}(0-1)\\&=\color {blue}{\frac {100}{\pi }}\end{aligned}}$

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