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NSW Y12 Maths - Extension 1 Differential Equations Modelling First Order

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Modelling First Order Theory

There are many models of first order differential equations, such as:\\ The exponential growth and decay model \(N=N_0 e^{kt}\) where \(N_0\) is the initial value of \(N\) when \(t=0\) and \(k\) is the growth or decay constant.\\ In practical applications the differential equation \(\dfrac{d N}{d t}=k(N-P)\) is used, where the rate of change is proportional to the difference between \(N\) and a constant \(P\). (where \(k\) is a constant).\\ \textbf{Example}\\ %Question 16065 The rate of cooling of a metal object, initially at \(T = 125^\circ C\), is given by \(\dfrac{{dT}}{{dt}} = - 0.64(T - 30)\), where \(t\) is in hours. Find how long it would take for the object to cool to \(66^\circ C\) \\ \textbf{Solution}\\ $\begin{aligned} \frac{dT}{dt}&=-0.64(T-30) \\ \frac{dt}{dT}&=\frac{1}{-0.64(T-30)} \\ \int dt&=\frac{1}{-0.64} \int \frac{1}{T-30} d T\\ -0.64(t+c) &=\log _{e}(T-30) \\ T-30 &=Ae^{-0.64 t} \\ T &=30+Ae^{-0.64 t} \\ 125 &=30+A \rightarrow A=95 \\ T &=30+95 e^{-0.64t} \\ 66 &=30+95 e^{-0.64t}\\ \frac{36}{95}&=e^{-0.64 t} \\ -0.64 t&=\log_{e}\frac{36}{95} \\ \therefore t&=1.5 \text { hours } \end{aligned}$\\

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