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### Course: AP®︎/College Calculus BC>Unit 7

Lesson 9: Logistic models with differential equations

# Worked example: Logistic model word problem

Finding the carrying capacity of a population that grows logistically. Also finding the population's size when it's growing the fastest.

## Want to join the conversation?

• In a previous video Sal stated the formula

dn/dt = Rn (1 - N/K)

in the above video he then used the formula

Nr(K-N).

How are these interchangeable?

Many thanks.
• Proof that dN/dt = Nr * (1 - N/K) is the same as dN/dt = Nc * (K-N), where c = r/K:

dN/dt = Nr * (1 - N/K)

We know that (1 - N/K) = 1/K * (K-N)

dN/dt = Nr * 1/K (K - N)

dN/dt = Nr/K * (K-N)

r and K (carrying capacity) are just constants. So, let's make r/K = c, a constant.

dN/dt = Nc * (K-N)

Now, we've shown that the two forms are equivalent.

But note that in the second, we have Nc * (K-N), while in the first we have Nr * (1 - N/K). c is NOT equal to r. In fact, c = r/K.

Sal Khan used dP/dt = kP (a - P) which is the same as the second form, dN/dt = Nc * (K-N). Here, P is N, a is uppercase K, and lowercase k is c.

Ask if you have questions! And let me know if I made any errors. Thanks!
• At . The population's size is the half-way between the initial population(P) and maximum population(a). The initial population is 700, not 0. Shouldn't the middle point be shifted by 350?
• The initial population is 700, but this is where t=0. What Sal did was finding the vertex of dP/dt, which is a function of P, not t. The vertex is halfway between the points where dP/dt is equal to 0. dP/dt is equal to 0 when P is equal to 0. This is not the same thing as the initial population. Therefore the middle point is not shifted over.
• We would all appreciate it if Sal could explain why he uses an equation that is completely different from the ones he teaches in the surrounding videos.
The cognitive dissonance is real. It's like we're driving down the road all happy and learning... and up pops a brick wall out of nowhere and we just smash into it
• Is it possible to find the fastest growth by finding the derivative of the logistic equation, and then locating the inflection point?
• Yes. The fastest growth would occur when the derivative is maximized. To maximize the derivative, we find where it's derivative is 0, i.e. where the second derivative is 0, i.e. an inflection point.
• At , when Sal says dpt, shouldn't it say dp/dt?
• yes it should be dp/dt, the rate of change of population with respect to time
• How do you know the max growth is halfway between zeros ?
• 𝑑𝑃∕𝑑𝑡 is a quadratic function (when graphed over 𝑃), which is symmetric about the vertex.
• So will the population ever reach the max capacity or will it just get very close to the max?
• Can we take second derivative to find the max?
Let y = dP/dt = P x 48000/4000 - P^2/4000
y'= 48000/4000 - 2P/4000 = 0
2P/4000 = 48000/4000
2P = 48000
P = 24000
• That works too! Understand why it does though.

When you find the derivative of $\frac{dP}{dt}$, what you're doing is finding critical points of the function. Now, when you equate it to zero and find the value of P, it is the value for which $\frac{dP}{dt}$ is a maximum or minimum. You can do a second derivative test (or, as our function itself is a derivative, this would technically be a third derivative test) to see if the value of P is a maximum or minimum. As it should come to be a maximum, P = 24,000 would be the maximum of the function $\frac{dP}{dt}$. So, at P = 24,000, $\frac{dP}{dt}$ is maximum, which is exactly what we needed to show!
• for the second part of the question, could we find the answer (P) by equating the double derivative of the equation to 0? if so, how would the calculations go?
• Yep, exactly! Like Sal says, there are many ways to find where the maximum of our derivative graph is. One way that we know is with calculus: the maximum of the derivative should be where its derivative hits 0, going from positive to negative. To do this, we'd have to first find the second derivative, and then set it equal to 0 in order to solve for P:
dP/dt = P/4000 (48000 - P) = 12P - P^2/4000
d^2P/dt^2 = 12 - P/2000
0 = 12 - P/2000
P = 24000
As you can see, we'd get the same answer no matter which method we use.
• I assume he simplified $kP(1-\frac{P}{a})$ to $kP(\frac{a-P}{a})$ and then took $\frac{k}{a}$ = k. Though yes, I do agree that he should've touched upon that a bit before using it