# Exponential function

• Apr 17th 2008, 02:15 PM
Snowboarder
Exponential function
The spread of head lice at a particular school was traced back to a sports trip to a neighbouring school. The spread amongst the students at the school could be modelled by the exponential functionL(t) = 4e^(8t - t^2)k (^ - square) , where t represents the number of weeks after the sports trip, and k is the rate of growth

How many students were initially infected? Justify your answer.
• Apr 17th 2008, 06:26 PM
Mathstud28
Quote:

Originally Posted by Snowboarder
The spread of head lice at a particular school was traced back to a sports trip to a neighbouring school. The spread amongst the students at the school could be modelled by the exponential functionL(t) = 4e^(8t - t^2)k (^ - square) , where t represents the number of weeks after the sports trip, and k is the rate of growth

How many students were initially infected? Justify your answer.

In the equation $y=Ce^{kt}$ $...C$ is always the initial value...since when t=0 $y=Ce^{0k}=C$
• Apr 17th 2008, 06:40 PM
Snowboarder
oki here is another one:

After 2 weeks, it was found that 56 students had become infected.
Calculate the rate of growth, k (to 4 dp).
• Apr 17th 2008, 09:52 PM
earboth
Quote:

Originally Posted by Snowboarder
oki here is another one:

After 2 weeks, it was found that 56 students had become infected.
Calculate the rate of growth, k (to 4 dp).

1. Do yourself and do us a favour: If you have a new problem please start a new thread. Otherwise no member of the forum can see that you need some help.

2. Plug in t = 2 into the given equation of the function:

$L(t) = 4e^{(8t-t^2) \cdot k}~\buildrel {t = 2} \over \longrightarrow~ L(2)=56= 4e^{(16-4) \cdot k}$

$56 = 4e^{12k}~\iff~14=e^{12k}~\iff~\ln(14)=12k$ ... $\iff~ \boxed{k=\frac1{12} \ln(14) \approx 0.21992 \approx 22\%}$
• Apr 18th 2008, 12:13 AM
Snowboarder
thank you earboth .

Does anybody know when the outbreak was at its worst?
When the outbreak was effectively over. Include any assumptions.
• Apr 18th 2008, 01:59 AM
earboth
Quote:

Originally Posted by Snowboarder
thank you earboth .

Does anybody know when the outbreak was at its worst?

1. Calculate the first derivative. Since you have to use the chainrule one of the factor is (8-2t). This is the only factor which can become zero.
Therefore you'll get t = 4 if you calculate L'(t) = 0.

Quote:

When the outbreak was effectively over. Include any assumptions.
You have first to define a state called "outbreak is over". Assume the outbreak has passed if only one person is actually infected:

$1 = 4e^{(8t-t^2) \cdot \frac{\ln(14)}{12}}$

I haven't the time yet to do the necessary calculatations in detail but you get $t \approx -0.72267~ or~ t\approx 8.72267$

The first value (measured in weeks!) gives the start of the infection and the second value marks the actual end of the infection.