# HSC Extension 1 and 2 Mathematics/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 ${\displaystyle N=N}$0ekt

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

Differentiation can be used to show that the rate of change (with respect to time, ${\displaystyle t}$) of ${\displaystyle N}$ is proportional (∞) to ${\displaystyle N}$. if:
${\displaystyle N=N}$0ekt,
then the Derivative of ${\displaystyle N}$ can be show as:
dN ${\displaystyle =kN}$0ekt
dt
${\displaystyle =kN}$, substituting ${\displaystyle N=N}$0ekt.

(note the derivative of e is the variable ${\displaystyle k}$ of the power of e times ${\displaystyle e^{kx}}$ and ${\displaystyle N,t}$ 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 = ${\displaystyle k(N-P)}$
dt
${\displaystyle =kN}$0ekt${\displaystyle -kP}$

where ${\displaystyle P}$ = the external constant (e.g., the external room temperature)

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

(i)${\displaystyle 74=24-100}$e10k
${\displaystyle 50=-100}$e10k

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

${\displaystyle log}$e${\displaystyle 1-log}$e${\displaystyle (-2)=10k}$

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

= .34567359... (store in memory)

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

incomplete 10th august '08