Resources for Rates and Integration
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Questions
12
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Video Tutorials
1
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HSC Questions
3
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Rates and Integration Theory
![The rate of change of a quantity \(Q\) with respect to time is measured by \(\dfrac{d Q}{d t}\). \\ If \(\dfrac{d Q}{d t}\) is given, then \(\displaystyle \int \dfrac{d Q}{d t} d t=Q+C\).\\ \textbf{Example 1}\\ %31002 Water flows into and then out of an empty container at a rate \(R\) litres/min given by \[R = t(5-t)\] \begin{itemize} \item[\bf{i)}]Find the maximum flow rate. \item[\bf{ii)}]Find an expression for the volume, \(V\) litres, of water in the container at time \(t\) minutes assuming that the container is initially empty. \item[\bf{iii)}]Find the total time for the container to fill up. \end{itemize} \textbf{Solution}\\ \begin{multicols}{2} i) \\ $\begin{aligned} R &= t(5 - t)\\&= 5t - t^2\\\frac{dR}{dt} &= 5 - 2t\\\frac{d^2R}{dt^2} &= -2\\\therefore \text{ since } \frac{d^2R}{dt^2} &<0 \quad \text{ a maximum will occur.}\\\text{let } \frac{dR}{dt} &= 0 \text{ to determine the maximum.}\\5 - 2t &= 0\\t &= 2.5\text{ seconds.}\\\therefore R &= 2.5\times(5 - 2.5) = 6.25 \text{ litres/min} \end{aligned}$\\ \columnbreak ii) \\ $\begin{aligned}R &= 5t - t^2\\\therefore V &= \int 5t - t^2 \, dt \\&= \frac{5t^2}{2} - \frac{t^3}{3} + C\\\text{when } V=0,\, t = 0 &\therefore C = 0\\\therefore V &= \frac{5t^2}{2} - \frac{t^3}{3}\end{aligned}$\\ iii) Let \(V = 0\) to determine the time, \\ $\begin{aligned}\frac{5t^2}{2} - \frac{t^3}{3} &= 0\\15t^2 - 2t^3 &= 0\\t^2(15 - 2t) &= 0\\\therefore t = 0 &\text{ or } t = 7.5\end{aligned}$\\ Therefore the total time taken for the container to fill up is 7.5 minutes. \end{multicols}](/media/mvedz0up/23167.png)