04 8logapps2

# Exponential Modelling

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Exponential functions are used to model many physical occurances including growth of cells, population growth, continuously compounded interest, radioactive decay and the rate of cooling (see Newton's Law of Cooling).

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Let A be the quantity at time t. Then $A = A_0e^{kt}$,

• where A0 is the initial quantity (a constant)
• where k is the rate constant of the equation.
• where $t \geqslant 0$

Note:
… … Growth: k > 0
… … Decay: k < 0

Note:

• Most functions involved with modelling have a domain of $t \geqslant 0$
• and many have a co-domain of $A \geqslant 0$
• so take care to only draw the appropriate part of the graph (usually 1st quadrant)

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#### Example 1

The population of wombats in an area is given by:

… … $W = 100e^{0.03t}$

… … … where W is the number of wombats
… … … t is the time in years after 1 January, 1998.

a) … Sketch W against t for $t \in [0, \; 30]$

b) … Find the time taken for the population of wombats to double.
… … … State the year and the week.

Solution

a) … Sketch W against t for $t \in [0, \; 30]$

… … … Note that t is always positive and W is always positive
… … … so only draw the 1st Quadrant.

… … … Note the asymptote plays no part in this portion of the graph
… … … so it needn't be marked in.

… … … Note the labelling of the axes is W and t (not x and y)

… … … Note that significant points are identified and labelled
… … … (in this case the endpoints of the domain).

… … … Note that the initial number of Wombats is $W_0 = 100$
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b) … Find the time taken for the population of wombats to double.
… … … State the year and the week.

… … … This can be found algebraically on the calculator
… … … by solving for t when $W = 200$

OR

… … … To find graphically, sketch W = 200 on the same graph
… … … and locate the point of intersection.

… … … Point of intersection at $t = 23.1049$

… … … 23 years after 1998 is 2021

… … … $0.1049 \times 52 = 5.4548$

… … … ie population has not yet doubled by the 5th week

… … … so population has doubled by the 6th week of 20121
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