exponential function?

Wesker · #1 $Old$ July 8th, 2009, 01:24 AM

The number of elk after t years in a state park is modeled by the function
P(t) = 1216/(1+75e^-.03t)
a) What was the initial population of elk?
b) When will the number of elk be 750?
c) What is the maximum number of elk possible in the park?

a) The initial would be 1216/100 = 12.16 right?
I need help with b) and c). How do I find e?
Thanks a bunch!

alexmahone · #2 $Old$ July 8th, 2009, 01:50 AM

a) Initial population of elk is when $t=0$ .

Initial population = $\frac{1216}{1+75*1}=\frac{1216}{76}=16$

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b) $P(t)=750$

$\frac{1216}{1+75e^{-.03t}}=750$

$1+75e^{-0.03t}=\frac{1216}{750}$

$1+75e^{-0.03t}=1.621$

$75e^{-0.03t}=0.621$

$e^{-0.03t}=\frac{0.621}{75}$

$-0.03t=ln \frac{0.621}{75}$

$-0.03t=-4.8$

$t=160$

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c) To find the maximum number of elk possible in the park, put $t=\infty$

Maximum number of elk possible in the park is $\frac{1216}{1+75e^{-\infty}}=\frac{1216}{1+0}=1216$

Prove It · #3 $Old$ July 8th, 2009, 01:53 AM

Quote:

Originally Posted by Wesker $View Post$

The number of elk after t years in a state park is modeled by the function
P(t) = 1216/(1+75e^-.03t)
a) What was the initial population of elk?
b) When will the number of elk be 750?
c) What is the maximum number of elk possible in the park?

a) The initial would be 1216/100 = 12.16 right?
I need help with b) and c). How do I find e?
Thanks a bunch!

a) The initial population is found when $t = 0$ .

So $P(0) = \frac{1216}{1 + 75e^{-0.03\cdot 0}}$

$= \frac{1216}{1 + 75\cdot 1}$

$= \frac{1216}{76}$

$= 16$ .

b) We need to solve for $t$ when $P(t) = 750$ .

c) You need to find $\frac{dP}{dt}$ and solve for $t$ when $\frac{dP}{dt} = 0$ .

Then substitute this value of $t$ back into your original equation to find $P$ .

Grandad · #4 $Old$ July 8th, 2009, 01:56 AM

Hello Wesker

Quote:

Originally Posted by Wesker $View Post$

The number of elk after t years in a state park is modeled by the function
P(t) = 1216/(1+75e^-.03t)
a) What was the initial population of elk?
b) When will the number of elk be 750?
c) What is the maximum number of elk possible in the park?

a) The initial would be 1216/100 = 12.16 right?
I need help with b) and c). How do I find e?
Thanks a bunch!

$e$ is the exponential number $= 2.718...$ , the base of natural logarithms.

For (a), you need to give $t$ the value $0$ . So, noting that $e^0 = 1$ :

$P(0) = \frac{1216}{1 + 75}= 16$ . ( $12.16$ elk didn't sound right, did it? I wonder what part of an elk the $0.16$ bit was. The antlers, perhaps?)

(b) We need to find the value of $t$ for which $P(t) = 750$ .

So $750 = \frac{1216}{1+75e^{-0.03t}}$

$\Rightarrow 1+75e^{-0.03t} = \frac{1216}{750} = 1.6213...$

$\Rightarrow e^{-0.03t} = \frac{0.6213...}{75}=0.008284$

Now here's the bit you might not understand if you don't know about natural logarithms. We take logs of both sides, and get:

$-0.03t = \ln(0.0082844) = -4.79337...$

$\Rightarrow t = 159.8...$

So the population will be $750$ after just under $160$ years.

(c) As $t \rightarrow \infty, e^{-0.03t} \rightarrow 0$ . (Do you understand that bit?) So the population has a maximum value given by

$\frac{1216}{1+0} = 1216$ .

Grandad