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Exponential Growth of a Population

Date: 05/03/98 at 09:23:28
From: Caren
Subject: exponetial growth (AP Calculus AB)

P(t) represents the number of wolves in a population at time t years 
when t > 0 or t = 0. The population P(t) is increasing at a rate 
directly proportional to 800 - P(t), where the constant of 
proportionality is k. I need to find:

   If P(0) = 500, what is P(t) in terms of t and k?
   If P(2) = 700, find k.
   Find the limit of P(t) from t = 0 to infinity.

I am really not sure how to start this. I've tried to maneuver the 
first part, but I can't figure out how to get something in terms of t 
if the t = 0 as suggested in P(0) = 500. Please help.

Date: 05/03/98 at 10:58:19
From: Doctor Anthony
Subject: Re: exponential growth (AP Calculus AB)

If P(0) = 500, what is P(t) in terms of t and k?

To find P(t), start with the differential equation:

   dp/dt = k(800 - p)         where p = P(t)

   dp/(800 - p) = k*dt   


   - ln(800 - p) = k*t + constant

At t = 0, p = 500, so:

   - ln(300) = constant

   ln(300) - ln(800 - p) = kt

   ln[300/(800 - p)] = kt
   300/(800 - p) = e^(kt)

   (800 - p)/300 = e^(-kt)           

   800 - p = 300e^(-kt)

   p = 800 - 300e^(-kt)       (Equation 1)

This is the population at time t in terms of k and t. 

Putting t = 2, we get:

   700 = 800 - 300e^(-2k)
   300e^(-2k) = 100

   e^(-2k) = 1/3

   -2k = ln(1/3)

   2k = ln(3)     

   k = ln(3)/2 = 0.5493 

Finally, when t -> infinity, e^(-kt) -> 0, so Equation 1 shows:

   p(infinity) = 800

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 05/03/98 at 11:52:24
From: Doctor Jaffee
Subject: Re: exponential growth (AP Calculus AB)

Hi Caren,

You have presented a very good question. If you take the AP exam on 
May 15, you are likely to see something very similar on it. I think I 
can help you out.

First of all, if I were solving this problem, I would want to express 
an initial equation in terms of k, t, and p, where p is the same as 
P(t). The problem begins by establishing that "P(t) is increasing at a 

Generally, when the problem refers to a rate of change, that can be 
expressed as a derivative. So, the left side of the equation will be 
P'(t), or dp/dt. Then the problem says "...directly proportional to 
800 - P(t)." That means that 800 - P(t) is being multiplied by some 
constant, in this case, k. So, the right side of the equation says 
k(800 - P(t)), or k(800 - p)

Our equation now is:

   dp/dt = k(800 - p)

This is a good example of a separable differential equation. We want 
to get anything with p or dp on one side of the equation, and anything 
with t or dt on the other side. Constants can be on either side. If we 
treat dp and dt as the numerator and denominator, respectively, of a 
fraction, we can multiply both sides of the equation by dt, then 
divide both sides of the equation by 800 - p. The result is:

   (1/(800 - p))dp = k dt

We can now find the indefinite integral of both sides. On the left 
side, the substitution method would work.  

Let u = 800 - p, and du/dp = -1. Thus, dp = -du. When you integrate 
the left side, you get:


and then when you resubstitute 800 - p for u, you get:

   -ln(800 - p)

where p < 800. On the right side of the equation, you get kt + C, 
where C is the constant of integration.

We now have -ln(800 - p) = kt + C. Let's multiply both sides by -1 
and get ln(800 - p) = -kt + C. (Notice that it doesn't look like I 
multiplied the C by -1, but that's because a constant times -1 is just 
another constant, so the C in the second equation actually means the 
negative of C in the previous equation.)

Now, let's transform this logarithmic equation to an exponential 

   e^(-kt + C) = 800 - p

which becomes

   e^(-kt + C) - 800 = -p

Then multiply both sides by -1 and get:

   800 - e^(-kt + C) = p   or   800 - e^(-kt)*e^C

But, since e^C is also a constant, let's rename it C. Our final 
result is:

   p = 800 - Ce^(-kt)

Phew, we finally have a general equation. Now, P(0) = 500. So, 
substitute 0 for t and 500 for p in the equation:

   500 = 800 - Ce^0

Since e^0 = 1, C turns out to be 300. We now have:

   p = 800 - 300e^(-kt)

Since P(2) = 700, we substitute 2 for t and 700 for p, and we get

   700 = 800 - 300e^(-2k)

If you subtract 800 from both sides, then divide by -300, you get:

   1/3 = e^(-2k)

Take the natural logarithm of both sides and get:

   ln(1/3) = -2k

But ln(1/3) = -ln(3), so dividing both sides by -2, you obtain:

   (1/2) ln 3 = k

Substituting that into the general equation, you should eventually 
come up with p = 800 - 300*3^(-t/2). I'll leave it to you to work out 
the details of that last step, and I'll leave it up to you to finish 
the problem. I hope I've given you a good start. And finally, if 
you're taking it, good luck on the AP exam!

-Doctor Jaffee, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
High School Calculus

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