MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 35

Applications of exponential functions

If the bacteria in a culture increase continuously at a rate proportional to the number present, and the initial number is $N_0$ find the number at time $t\text{.}$

Idea: Change in bacteria is proportional to the amount of bacteria $$\frac{dB}{dt}=kB\text{.}$$ What could $B$ be? $$\frac{dB}{B}=kdt\text{.}$$ So $$\int \frac{1}{B}dB=\int kdt\text{.}$$ So $$\text{ln}\hspace{0.17em}B=kt+c\text{.}$$ So $$B=e^{kt+c}=e^c{e}^{kt}=C{e}^{kt},$$ where $C$ is a constant. At $t=0,$ $B=N_0=C{e}^{k·0}=C\text{.}$ So $C=N_0\text{.}$ So $$B=N_0{e}^{kt}\text{.}$$

A roast turkey is taken from an oven when its temperature reaches ${185}^{\circ }$ F and is placed on a table in a room where the temperature is ${75}^{\circ }$ F. It cools at a rate proportional to the difference between its current temperature and the room temperature.

(a)	If the temperature of the turkey is ${150}^{\circ }$ F after half an hour what is the temperature after 45 minutes?
(b)	When will the turkey have cooled to ${100}^{\circ }$ F?

Idea: Change in temperature is proportional to current temperature $-$ room temperature. $$\frac{dT}{dt}=k(T-R)\text{.}$$ So $$\frac{dT}{T-R}=kdt\text{.}$$ So $$\int \frac{dT}{T-R}=\int kdt\text{.}$$ So $$\text{ln}(T-R)=kt+c\text{.}$$ So $$T-R=e^{kt+c}=e^c{e}^{kt}=C{e}^{kt}$$ where $C$ is a constant. So $$T=C{e}^{kt}+R\text{.}$$ At $t=0,$ $$T=185=C{e}^{k·0}+75=C+75\text{.}$$ So $$C=185-75=115\text{.}$$ So $$T=115e^{kt}+75\text{.}$$ At $t=\frac{1}{2},$ $$T=115e^{k\frac{1}{2}}+75+150\text{.}$$ So $$e^{k\frac{1}{2}}=\frac{150-75}{115}=\frac{75}{115}\text{.}$$ So $$\frac{1}{2}k=\text{ln}\left(\frac{75}{115}\right)\text{.}$$ So $$k=2\text{ln}\left(\frac{75}{115}\right)\text{.}$$ So $$T=115e^{2\text{ln}\left(\frac{75}{115}\right)t}+75\text{.}$$ At $t=\frac{3}{4},$ $$\begin{array}{ccc}{\displaystyle T}& =& {\displaystyle 115e^{2\text{ln}\left(\frac{75}{115}\right)\frac{3}{4}}+75}\\ & =& {\displaystyle 115e^{\frac{3}{2}\text{ln}\left(\frac{75}{115}\right)}+75}\\ & =& {\displaystyle 115{\left(e^{\text{ln}\left(\frac{75}{115}\right)}\right)}^{\frac{3}{2}}+75}\\ & =& {\displaystyle 115{\left(\frac{75}{115}\right)}^{\frac{3}{2}}+75\text{.}}\end{array}$$ If $T=100$ then $115e^{2\text{ln}\left(\frac{75}{115}\right)t}+75=100\text{.}$ So $$e^{2\text{ln}\left(\frac{75}{115}\right)t}=\frac{100-75}{115}=\frac{25}{115}\text{.}$$ So $$2\text{ln}\left(\frac{75}{115}\right)t=\text{ln}\left(\frac{25}{115}\right)\text{.}$$ So $$t=\frac{\text{ln}\left(\frac{25}{115}\right)}{2\text{ln}\left(\frac{75}{115}\right)}\text{.}$$

The majority of naturally occurring rhenium is ${}_{75}^{187}\text{Re},$ which is radioactive and has a half life of $7\times {10}^{10}$ years. In how many years will 5% of the earth's ${}_{75}^{187}\text{Re}$ decompose.

Idea: Change in ${}_{75}^{187}\text{Re}$ is proportional to existing amount of ${}_{75}^{187}\text{Re}\text{.}$ $$\frac{dR}{dt}=kR\text{.}$$ So $\frac{dR}{R}=kt\text{.}$ So $\int \frac{dT}{R}=\int kt\text{.}$ So $\text{ln}\hspace{0.17em}R=kt+C\text{.}$ So $R=e^{kt+c}=e^c{e}^{kt}=C{e}^{kt}$ where $C$ is a constant. When $t=0$ the amount is $R_0\text{.}$ So $R_0=C{e}^{k·0}=C\text{.}$ When $t=7\times {10}^{10}$ the amount is $\frac{1}{2}{r}_0\text{.}$ So $$\frac{1}{2}{R}_0=R_0{e}^{k·7·{10}^{10}}\text{.}$$ So $$\frac{1}{2}=e^{k·7·{10}^{10}}\text{.}$$ So $$\text{ln}\hspace{0.17em}\frac{1}{2}=k·7·{10}^{10}\text{.}$$ So $$k=\frac{\text{ln}\left(\frac{1}{2}\right)}{7\times {10}^{10}}\text{.}$$ So $$R=R_0{e}^{\frac{\text{ln}\left(\frac{1}{2}\right)}{7\times {10}^{10}}t}\text{.}$$ We want to know when $R=.05R_0\text{.}$ $$.05R_0=R_0{e}^{\frac{\text{ln}\left(\frac{1}{2}\right)}{7\times {10}^{10}}t}\text{.}$$ So $$\frac{1}{20}=e^{\frac{\text{ln}\left(\frac{1}{2}\right)}{7\times {10}^{10}}t}\text{.}$$ So $$\text{ln}\left(\frac{1}{20}\right)=\frac{\text{ln}\left(\frac{1}{2}\right)}{7\times {10}^{10}}t\text{.}$$ So $$t=\frac{7\times {10}^{10}\text{ln}\left(\frac{1}{20}\right)}{\text{ln}\left(\frac{1}{2}\right)}=\frac{7\times {10}^{10}\text{ln}\left(\frac{1}{20}\right)}{\text{ln}\left(\frac{1}{2}\right)}\text{.}$$

If you buy a $200,000 home and put 10% down and take out a 30 year fixed rate mortgage at 8% per year compute how much your payment would be if you paid it all off in one big payment at the end of 30 years.

Idea: Change in money is $.08$ of its current amount. $$\frac{dM}{dt}=.08M\text{.}$$ So $$\frac{dM}{M}=.08dt\text{.}$$ So $$\int \frac{dM}{M}=\int .08dt\text{.}$$ So $$\text{ln}\hspace{0.17em}M=.08t+c\text{.}$$ So $$M=e^{.08t+c}=e^c{e}^{.08t}=C{e}^{.08t}$$ where $C$ is a constant. So $$M=C{e}^{.08t}\text{.}$$ At time $t=0$ we owe $\mathrm{200,000}-\mathrm{20,000}=\mathrm{180,000}\text{.}$ $$\mathrm{180,000}=C{e}^{.08·0}=C\text{.}$$ So $$M=\mathrm{180,000}{e}^{.08t}\text{.}$$ After 30 years we owe $$M=\mathrm{180,000}{e}^{.08·30}=\mathrm{180,000}{e}^{.24}\text{dollars.}$$

If you borrow $500 on your credit card at 14% interest find the amounts due at the end of 2 years if the interest is compounded

(a)	annually,
(b)	quarterly,
(c)	monthly,
(d)	daily,
(e)	hourly,
(f)	every second,
(g)	every nanosecond,
(h)	continuously.

You owe:

(a)	$500+500\left(.14\right)=500(1+.14)$ after one year. $500(1+.14)(1+.14)$ after two years.
(b)	$500+500\left(\frac{.14}{4}\right)=500(1+\frac{.14}{4})$ after one quarter. $500{(1+\frac{.14}{4})}^2$ after two quarters. $500{(1+\frac{.14}{4})}^8$ after two years (8 quarters).
(c)	$500+500\left(\frac{.14}{12}\right)=500(1+\frac{.14}{12})$ after 1 month. $500{(1+\frac{.14}{12})}^8$ after two years (24 months).
(d)	$500+500\left(\frac{.14}{365}\right)=500(1+\frac{.14}{365})$ after 1 day. $500{(1+\frac{.14}{365})}^{2·365}$ after two years $\text{(}2·365$ days).
(e)	$500+500\left(\frac{.14}{365·12}\right)=500(1+\frac{.14}{365·12})$ after 1 hour. $500{(1+\frac{.14}{365·12})}^{2·365·12}$ after two years.
(f)	$500+500\left(\frac{.14}{365·12·3600}\right)=500(1+\frac{.14}{365·12·3600})$ after $1$ second. after two years.
(h)	$$\begin{array}{ccc}{\displaystyle \underset{x\to \infty }{\text{lim}}500{(1+\frac{.14}{n})}^{n·2}}& =& {\displaystyle \underset{x\to \infty }{\text{lim}}500{\left({(1+\frac{.14}{n})}^n\right)}^2}\\ & =& {\displaystyle 500{\left(e^{.14}\right)}^2}\\ & =& {\displaystyle 500e^{.28}}\end{array}$$ after two years, since $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}{(1+\frac{.14}{n})}^n}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}{\left(e^{\text{ln}(1+\frac{.14}{n})}\right)}^n}\\ & =& {\displaystyle \underset{x\to \infty }{\text{lim}}{e}^{n\text{ln}(1+\frac{.14}{n})}}\\ & =& {\displaystyle \underset{x\to \infty }{\text{lim}}{e}^{n\left(\frac{1+\frac{.14}{n}}{\frac{.14}{n}}\right)\left(\frac{.14}{n}\right)}}\\ & =& {\displaystyle \underset{x\to \infty }{\text{lim}}{e}^{.14\frac{\text{ln}(1+\frac{.14}{n})}{\frac{.14}{n}}}}\\ & =& {\displaystyle e^{.14·1}}\\ & =& {\displaystyle e^{.14}\text{.}}\end{array}$$

A sample of a wooden artefact from an Egyptian tomb has a $\raisebox{1ex}{${}^{14}C$}\!\left/ \!\raisebox{-1ex}{${}^{12}C$}\right.$ ratio which is 54.2% of that of freshly cut wood. In approximately what year was the old wood cut? The half life of ${}^{14}C$ is 5720 years.

Idea: The change in ${}^{14}C$ is proportional to the existing amount. $$\frac{d{}^{14}C}{dt}=k$ {}^{14}C$\text{.}$$ So $$\frac{d$ {}^{14}C$}{{}^{14}C}=kdt\text{.}$$ So $$\int \frac{d{}^{14}C}{{}^{14}C}=\int kdt\text{.}$$ So $$\text{ln}{}^{14}C=kt+c\text{.}$$ So $${}^{14}C=e^{kt+c}=e^{kt}{e}^c=K{e}^{kt},$$ where $K$ is a constant. Suppose that at $t=0$ the amount of ${}^{14}C$ is ${}^{14}{C}_0\text{.}$ then $${}^{14}{C}_0=K{e}^{k·0}=K\text{.}$$ So $${}^{14}C={}^{14}{C}_0{e}^{kt}\text{.}$$ The half life of ${}^{14}C$ is 5720 years. So, at $t=5720$ $$\frac{1}{2}{}^{14}{C}_0={}^{14}{C}_0{e}^{kt}={}^{14}{C}_0{e}^{k5720}\text{.}$$ So $$\frac{1}{2}=e^{k5720}\text{.}$$ So $$\text{ln}\left(\frac{1}{2}\right)=k·5720\text{.}$$ So $$k=\frac{\text{ln}\left(\frac{1}{2}\right)}{5720}\text{.}$$ So $${}^{14}C{}^{14}{C}_0{e}^{\frac{\text{ln}\left(\frac{1}{2}\right)}{5720}t}\text{.}$$ Now there is 54.2% of the original ${}^{14}C\text{.}$ So $$\left(.542\right){}^{14}{C}_0={}^{14}{C}_0{e}^{\frac{\text{ln}\left(\frac{1}{2}\right)}{5720}t}\text{.}$$ So $$.542=e^{\frac{\text{ln}\left(\frac{1}{2}\right)}{5720}t}\text{.}$$ So $$\text{ln}\left(.542\right)=\frac{\text{ln}\left(\frac{1}{2}\right)}{5720}t\text{.}$$ So $$t=\frac{\text{ln}\left(.542\right)5720}{\text{ln}\left(\frac{1}{2}\right)}\text{.}$$

Notes and References

These are a typed copy of Lecture 35 from a series of handwritten lecture notes for the class MATH 221 given on December 4, 2000.

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