Exponential Growth function

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becky0307

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In 1996, an outbreak of a disease infected 14 people in a large community. By 1997, the number of those infected had grown to 40. Find an exponential growth function that fits the data.

would it be:
N(t) = 14e1.85t, where t is the number of years after 1996

value at end of period was 40 subtract 14 and divide by 14 to get growth percentage
 
where is 1.85 coming from?

You need to solve for k in this instance, knowing N(t), No, and t. Knowing k, will allow you to set up an equation which describes the number N(t) at any given time.

Let me know if this helps or makes sense.
 
I thought I was solving K by taking the beginning number of people was 14 next year it was 40,
like this:
y = a(1+r)^n, where y is the # of infected people in 1997, a the number of infected people in 1996, r the growth rate, and n the number of years.

40 = 14(1+r)^1
40 = 14(1+r)
40 = 14+14r
14r = 26
r = 26/14
r = 13/7 or 1.857
 
Hmm, nope

The starting equation is N(t) = Noe^kt right?

now, the initial conditions state that No = 14 and after t = 1 year the final infections are N(t) = 40. Solving the starting equation for k yields:

k = Ln(Nt/No)/t

do you see how I got this? Plug and chug.
 
tutor_joel said:
Hmm, nope

The starting equation is N(t) = Noe^kt right?

now, the initial conditions state that No = 14 and after t = 1 year the final infections are N(t) = 40. Solving the starting equation for k yields:

k = Ln(Nt/No)/t

do you see how I got this? Plug and chug.
Click to expand...

tutor_joel is right, however I'll put it in tex form since I want to answer a question but there are no more un-answered ones!

\(\displaystyle N_t = N_0e^{kt}\)

therefore...

\(\displaystyle 40 = 14e^k\)

and...

\(\displaystyle ln\frac{40}{14} = k\)
 
Yeah Burger, just working on the La TeX notation and increasing my posts :)

So,the exponential growth function would be:

\(\displaystyle N(t) = 14e^{Ln\left(\frac{40}{14}\right)t}\)

or just make a decimal out of

\(\displaystyle Ln\left(\frac{40}{14}\right)\)
 
tutor_joel said:
...or the \(\displaystyle e^{Ln}\) simply cancels out yeilding

\(\displaystyle N(t) = 14\left(\frac{40}{14}\right)t} = 40t\)
Click to expand...

Careful, the t is not part of the natural log, its better to leave it as

\(\displaystyle N_t = 14e^{ln(\frac{40}{14})t}\)

and do no simplifying.
 
tutor_joel said:
yeah, you're right, I thought that looked weird. I was just doing the math and not thinking the problem.

How about

\(\displaystyle N(t) = 14\left(\frac{40}{14}\right)e^t} = 40e^t\)

much more presentable
Click to expand...


No no no!

\(\displaystyle e^x + e^y = e^{xy} \ \ does \ \ NOT \ \ equal \ \ e^xe^y\)

Therefore...

\(\displaystyle e^{ln(\frac{40}{14})*t} \ \ does \ \ NOT \ \ equal \ \ e^{ln(\frac{40}{14})}e^t\)
 
I would like to see somebody reduce 40/14 to 20/7.

Oh, I just did. 8-)

Despite what I would like to see, here's how I would report function N:

N(t) = 14 e^(1.0498t)
 
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