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Rates and Differentiation Theory
![The rate of change of a quantity \(Q\) with respect to time \(t\) is measured by \(\dfrac{d Q}{d t}\)\\ \textbf{Example}\\ The surface area in \(\mathrm{cm}^2\) of a balloon being inflated is given by \[ S=t^3-2 t^2+4 t-1\] Where \(t\) is time in seconds.\\ Find the rate of increase in the balloon's surface area after 4 seconds.\\ \textbf{Solution}\\ $\begin{aligned} S & =t^3-2 t^2+4 t-1 \\ \frac{d S}{d t} & =3 t^2-4 t+4 \\ \text { when }t=4 \quad \frac{d S}{d t}&=3(4)^2-4 \times 4+4 \\ & =36 \mathrm{~cm}^2 / \mathrm{sec} \end{aligned}$\\](/media/5vepbdyd/23166.png)
NSW Y12 Maths - Advanced Motion and Rates Rates and Differentiation
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Rates and Differentiation - Video - Rates of change application
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