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Alagendram's question via email about approximating change

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
The population of a colony of bacteria of size $\displaystyle \begin{align*} N \end{align*}$ is given by $\displaystyle \begin{align*} N(t) = N_0\,\mathrm{e}^{\frac{k\,t}{100}} \end{align*}$, where $\displaystyle \begin{align*} t \end{align*}$ is the time in hours.

(a) Use calculus to show that a small change $\displaystyle \begin{align*} \Delta\,N \end{align*}$ in the population size in a small interval of time $\displaystyle \begin{align*} \Delta \, t \end{align*}$ satisfies $\displaystyle \begin{align*} \frac{\Delta\,N}{N} \approx \frac{k}{100}\,\Delta\,t \end{align*}$.

(b) In the space of any hour, what is the approximate percentage change in the size of the population?
(a) We should recall that for a small change in x, then $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \approx \frac{\Delta\,y}{\Delta\,x} \end{align*}$, so $\displaystyle \begin{align*} \Delta\,y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\Delta\,x \end{align*}$, so in this case

$\displaystyle \begin{align*} N &= N_0\,\mathrm{e}^{\frac{k\,t}{100}} \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= N_0\,\frac{k}{100}\,\mathrm{e}^{\frac{k\,t}{100}} \\ \\ \Delta\,N &\approx \frac{\mathrm{d}N}{\mathrm{d}t}\,\Delta\,t \\ &= N_0\,\frac{k}{100}\,\mathrm{e}^{\frac{k\,t}{100}}\,\Delta\,t \\ \\ \frac{\Delta\,N}{N} &\approx \frac{N_0\,\frac{k}{100}\,\mathrm{e}^{\frac{k\,t}{100}}\,\Delta\,t}{N} \\ &= \frac{N_0\,\frac{k}{100}\,\mathrm{e}^{\frac{k\,t}{100}}\,\Delta\,t}{N_0\,\mathrm{e}^{\frac{k\,t}{100}}} \\ &= \frac{k}{100}\,\Delta\,t \end{align*}$

(b) In the space of one hour, $\displaystyle \begin{align*} \Delta\,t = 1 \end{align*}$, so that means

$\displaystyle \begin{align*} \frac{\Delta\,N}{N} &\approx \frac{k}{100}\,\Delta\,t \\ &= \frac{k}{100}\cdot 1 \\ &= \frac{k}{100} \\ \Delta\,N &\approx \frac{k}{100}\,N \end{align*}$

So the change in N is approximately k% in the space of one hour.