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Numerade Educator



Problem 22 Hard Difficulty

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of differential equation
$ \frac {dP}{dt} = c \ln (\frac {M}{P})P $
where $ c $ is a constant and $ M $ is the carrying capacity.
(a) Solve this differential equation.
(b) Compute lim $ _{t \to \infty} P(t). $
(c) Graph the Gompertz growth function for $ M = 1000, P_0 = 100, $ and $ c = 0.05, $ and compare it with the logistic function in Example 2. What are the similarities? What are differences?
(d) We know from Exercise 13 that the logistic function grows fastest when $ P = M/2. $ Use the Gompertz differential equation to show that the Gompertz function grows fastest $ P = M/e. $


a) $P(t)=M e^{-A e^{-e t}}$
b) $M$
c) see graph
d) $P=\frac{M}{e}$


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Video Transcript

Hey, it's clear. So when you're married here, so we get dp over DT is equal to see l end of em overpay times p to get one over p till end. Um overpay DP is equal to see 18 We're gonna integrate full side I'm gonna get negative. The U is equal to one over PDP How we got this is that we make you be equal to Ln of them overpay So we got do you It's equal to the derivative of em overpay Oh for m overpay which we get to be negative one for a p d. P Now we're just gonna substitute It's again negative won over you We integrate this BU is equal to the integral of CD T to get negative l and of the l on of overpay is equal to C T plus d. We know that m iss bigger than P and Ellen of a number bigger than one. It's gonna be positive, so we just don't need the absolute value signs now get out. End of l and M of P for p This sequel to negative C T minus d put the base to the base of E get into the negative ct off my minus. D you get Ellen of M minus Ellen of you? Because this is just subtract it. If it's the quotient, we get negative. C t minus d we end up getting l n of m minus. Eleanor, that's equal to a e the negative C t Elton of m minus. Hey, e to the negative ct. It's equal to Elena. P get a p volume to be m e to the negative a e the negative ct. Since we make you be equal to E. Ellen of you now for part B, we found this equation from part Aye. So we just take the limit. Us t goes to infinity of em he to the negative b e the negative C t. Which is equal to, um eat negative e the negative infinity which becomes to m e zero, which is equal to em for part C. We just draw a graph. So one graph. Let's say you are, um, logistic function first gonna look something like this and then in red we have our other function Gompertz function, which looks something like this. So we see some similarities. They have the same initial population in the same equilibrium, But they have different points of inflection for part D. We know from part, eh? That d p over d t is equal to Ste e p Elton of em overpay. So we need to find pee when dp over DT is maximum. That's d over dp dp over dp is equal to zero So yet d over dp of seed p l end of, um over P is equal to zero. Then I'll work right here. So we got l and of em overpay times d c p over dp plus c p times d l end of em over a p all over DP is equal to zero Then we get Ln of em over peep time See plus tp my times one over m overpay times negative M Overpay square is equal zero get l N of m o ver p times C minus C is equal to zero We find for Pee Wee Get em over E That is our answer