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 Integrating: - 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 rate..." 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: -ln(ABS(u)) 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 equation: 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.