HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Applications of calculus to the physical world

Exponential Growth and Decay

2 unit course

The exponential function can be used to show the growth or decay of a given variable, including the growth or decay of population in a city, the heating or cooling of a body, radioactive decay of radioisotopes in nuclear chemistry, and amount of bacteria in a culture.

The exponential growth and decay formula is $N=N$₀e^kt

where:
$N$₀ is the first value of N (where $t=0$)
$t$ represents time in given units (seconds, hours, days, years, etc.)
$e$ is the exponential constant ($e=2.718281828...$), and
$k$ is the growth ($k+ve$) or decay($k-ve$) constant.

Differentiation can be used to show that the rate of change (with respect to time, $t$) of $N$ is proportional (∞) to $N$. if:
$N=N$₀e^kt,
then the derivative of $N$ can be shown as:
dN $=kN$₀e^kt
dt
$=kN$, substituting $N=N$₀e^kt.

(note the derivative of e is the variable $k$ of the power of e times $e^{kx}$ and $N,t$ are constant.)

3 Unit applications

not yet complete

The variable of a given application can be proportionate to the difference between the variable and a constant. An example of this is the internal cooling of a body as it adjusts to the external room temperature.

dN = $k(N-P)$
dt
$=kN$₀e^kt$-kP$

where $P$ = the external constant (e.g., the external room temperature)

using natural logarithms, $log$_e$x$, we can find any variable when given certain information.
Example:
A cup of boiling water is initially $100$^oC. The external room temperature is $24$^oC. after 10 minutes, the temperature of the water is $74$^oC. find
(i) k
(ii)how many minutes it takes for the temperature to equal 30 degrees.

(i)$74=24-100$e^10k
$50=-100$e^10k

${\frac {50}{-100}}=e^{10k}$

$log$_e$1-log$_e$(-2)=10k$

$k={\frac {log_{e}1-log_{e}(-2)}{2}}$

= .34567359... (store in memory)

(ii) 30=24-100e^(.34657359t)

incomplete 10th august '08