Resources for Exponential Growth and Decay
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Questions
11
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Video Tutorials
1
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HSC Questions
7
With Worked SolutionClick Here
Exponential Growth and Decay Theory
![Exponential growth or decay can be written as \(\dfrac{d Q}{d t}=k Q\) where \(k\) is the growth or decay constant.\\ \(\dfrac{d Q}{d t}=k Q\) is solved by the equation \(Q=A e^{k t}\) when \(t=0 \quad Q=A e^0=A\) \(\therefore A\) is the initial amount.\\ $\begin{aligned} \text { i.e. } \quad Q & =A e^{k t} \\ \frac{d Q}{dt} & =A k e^{k t} \\ \frac{d Q}{d t} & =k A e^{kt} \\ \therefore \frac{d e}{d t} & =k Q \end{aligned}$\\ \begin{multicols}{2} \textbf{Example}\\ An isotope of carbon, \(C_{14}\) decays at a rate proportional to the mass present. The rate of change is given by \[\frac{dM}{dt} = -KM\] Where \(K\) is a positive constant and \(M\) is the mass present. \begin{itemize} \item[\bf{i)}]Show that \(M = M_0 e^{-Kt}\) is a solution to this equation. \item[\bf{ii)}]The half life of \(C_{14}\) is 5730 years. This means it takes 5730 years for 10 grams of \(C_{14}\) to decay to 5 grams. Find the value of \(K\) correct to 3 significant figures. \item[\bf{iii)}]At an archaeological dig, a piece of pottery was discovered which had one-third of the original \(C_{14}\) remaining. Calculate the age of the pottery. \end{itemize} \columnbreak \textbf{Solution}\\ \textbf{i)} \\ $\begin{aligned} M &= M_0e^{-Kt} && (1)\\ \frac{dM}{dt} &= -KM_0e^{-Kt} && (2)\;\;\text{ Differentiating (1) }\\\therefore \frac{dM}{dt} &= -KM && \text{(sub (1) into (2))} \end{aligned}$\\ \textbf{ii)} \\ $\begin{aligned}\frac{1}{2} M_0 &= M_0e^{-K\times 5730}\\\frac{1}{2} &= e^{-K\times 5730}\\\therefore -K\times 5730 &= \ln\left(\frac{1}{2}\right)\\K\times 5730 &= \ln 2\\K &= \frac{\ln 2}{5730}\\&= 0.000121\end{aligned}$\\ \textbf{iii)} \\ $\begin{aligned}\frac{1}{3}M_0 &= M_0 e^{-0.000121t}\\\frac{1}{3} &= e^{-0.000121t}\\-0.000121 t &= \ln \left(\frac{1}{3}\right)\\t &= \frac{\ln 3}{0.000121}\\&= 9079 \text{ years.}\end{aligned}$\\ \end{multicols}](/media/aikaognc/23168.png)