Nonlinear Models - need verification of answer

[hank]

This is a problem that doesn't have an answer in the back of the book, so I'm just hoping someone can verify I got the right answer.

The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially, N(0) = 500, and it is observed that N(1) = 1000. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000.

Here's what I know:
logistic equation: aP / (bP +(a - bP)e^-at)

dN/dt = kN(50,000 - N) = 50,000kN - kN^2 //took this from an example
a = 50,000k, b = k, P = N(0) = 500

//So now, I plug and chug

N(t) = 50,000k * 500 / (500k + (50,000k - 500k)e^-50,000kt)
= 25,000,000k / (500k + 45,500ke^-50,000kt)
= 25,000,000 / (500 + 45,500e^-50,000kt) //divide top and bottom by k

//time to find k

N(1) = 1,000 = 25,000,000 / (500 + 45,500e^-50,000k(1))
500 + 45,500e^-50,000k = 25,000
e^-50,000k = 0.5385
-50,000k = ln (.5385)
k = 0.000012381 //roughly

//So, the answer is...

N(t) = 25,000,000 / (500 + 45,500^-0.619t)

How's that?

 

[Ctayb14]

Re: Topic Review

The numbers are a little off. N(t) should equal 25,000,000 / (500 + 49,500e^-50,000k(t)). Everything else is correct.

N(1) = 1000 = 25,000,000 / (500 + 49,500e^-50,000k(1))
500 + 49,500e^-50,000k = 25,000
e^-50,000k = 0.4949
-50,000k = ln (.4949)
k = 0.000014068

Thus, the answer should be N(t) = 25,000,000 / (500 + 49,500^-50,000(0.000014068)(t))
=25,000,000 / (500 + 49,500^-0.7034t)

Hopes this helps!