MHB Find the Derivative of a Function: Step-by-Step Guide for Beginners

[yli]

Hi, so I am trying to find the derivative of this function.
\[
\frac{dP}{dt}=rP(K-P),
\]
r and K are positive constants describing the natural growth rate and carrying capacity of the population, respectively. I was trying to find the derivative, and I suppose that I am supposed to apply the chain rule, but I seems like I'm doing everything wrong. The derivative of the outside function I think is r. Next, the derivative of the inner function is (0-1), I think. But this comes to -r, which doesn't seem to make sense. Any guidance is greatly appreciated.

 

[MarkFL]

If you are differentiating w.r.t $t$, then using the product rule, you would have:

$$\frac{d^2P}{dt^2}=\left(r\d{P}{t}\right)\left(K-P\right)+\left(rP\right)\left(-\d{P}{t}\right)$$

Now, factor...:)

 

[yli]

Would it end up being rK+2rP? Or did I factor wrong?

 

[MarkFL]

Would it end up being rK+2rP? Or did I factor wrong?

We see that $$r\d{P}{t}$$ is common to both terms, so we have:

$$\frac{d^2P}{dt^2}=\left(r\d{P}{t}\right)\left(K-P\right)+\left(rP\right)\left(-\d{P}{t}\right)=r\d{P}{t}\left(K-P-P\right)=r\d{P}{t}\left(K-2P\right)$$

 

[yli]

So if I were to graph dp/dt as a function of P, how would I find the critical values? Would I just set this derivative to 0? Also, if a new constant is introduced into this, how would I find the derivative? The equation would change to
\[\frac{dP}{dt}=rP(K-P)-H.
\]

 

[S.G. Janssens]

So if I were to graph dp/dt as a function of P, how would I find the critical values? Would I just set this derivative to 0? Also, if a new constant is introduced into this, how would I find the derivative? The equation would change to
\[\frac{dP}{dt}=rP(K-P)-H.
\]

Could it be that you are trying to take the derivative of the right-hand side of the equation for $\frac{dP}{dt}$ with respect to $P$ instead?

Namely, it seems to me that you are trying to do a linear stability analysis of the equilibria of this differential equation. For this, you would differentiate the one-dimensional vector field $P \mapsto rP(K - P)$ (or: $P \mapsto rP(K - P) - H$) using the product rule and evaluate the resulting derivative at the equilibria. A strictly negative value would indicate linear stability, while a strictly positive value would indicate linear instability.

In addition, for autonomous one-dimensional differential equations it would be more instructive to also draw the vector field (vertical) as a function of $P$ (horizontal). Equilibria then correspond to points of the graph on the horizontal axis and the direction of the flow - hence, the (in)stability of the equilibria - is directly inferred from the graph. (Can you see how?)

 

[HallsofIvy]

You have "rP(K- P)" and talk about an "inner function" and an "outer function". Are you think of "P(K-P)" as "P times K- P" (as others responding have) or as the composition, "P of K-P"?

 

[yli]

I think that Krylov has gotten what I was attempting to do, I was confused about it myself. In the case of drawing the graph of \[\frac{dP}{dt}=rP(K−P)−H.\] with respect to P, could someone tell me what would change? For the first graph, I got an upside down U-shape, and the question that I am referring to says that 0<H< $\frac{rk^2}{4}$. I know that $\frac{rk^2}{4}$ is the $\frac{dP}{dt}$ value of the global maximum of the function.

 

[S.G. Janssens]

I think that Krylov has gotten what I was attempting to do, I was confused about it myself. In the case of drawing the graph of \[\frac{dP}{dt}=rP(K−P)−H.\] with respect to P, could someone tell me what would change? For the first graph, I got an upside down U-shape, and the question that I am referring to says that 0<H< $\frac{rk^2}{4}$. I know that $\frac{rk^2}{4}$ is the $\frac{dP}{dt}$ value of the global maximum of the function.

By subtracting $H$, you perform a vertical translation of the graph, lowering its top. By comparing $H$ with the value of the global maximum for the case without harvesting, you compare the maximal harvest-free growth rate with the harvesting rate. If the latter is too large, you will "reap more than you sow", so to speak.

 

[MarkFL]

I think that Krylov has gotten what I was attempting to do, I was confused about it myself. In the case of drawing the graph of \[\frac{dP}{dt}=rP(K−P)−H.\] with respect to P, could someone tell me what would change? For the first graph, I got an upside down U-shape, and the question that I am referring to says that 0<H< $\frac{rk^2}{4}$. I know that $\frac{rk^2}{4}$ is the $\frac{dP}{dt}$ value of the global maximum of the function.

What I would do here, if we are to look at the stability of the equilibria, is write:

$$rP(K-P)-H=-\left(rP^2-rKP+H\right)$$

The roots, by the quadratic formula, are:

$$P=\frac{rK\pm\sqrt{r^2K^2-4rH}}{2r}$$

We require the discriminant to be non-negative:

$$r^2K^2-4rH\ge0$$

Since $r$ is positive, we may divide through by $r$:

$$rK^2-4H\ge0$$

$$rK^2\ge4H$$

And we find:

$$0<H\le\frac{rK^2}{4}$$

Can you explain why the smaller root is unstable while the larger is stable?

edit: I previously had the above reversed with regards to the stability of the roots because I failed to recall that I had factored out a negative sign. This is cleared up in the following posts, but I didn't want to OP to read this last post on the page, and it have bad information in it, and cause the OP any additional grief. :)

 

[S.G. Janssens]

Can you explain why the smaller root is stable while the larger is unstable?

Could it be the other way around?

 

[MarkFL]

Could it be the other way around?

Oops...I failed to recall that I factored out a negative sign...so yes it is the other way around. (Bandit)

 

[S.G. Janssens]

No worries. I recommend that yli tries to follow up on your post #10 (with this small correction).

 

[MarkFL]

No worries. I recommend that yli tries to follow up on your post #10 (with this small correction).

Yes, and for clarity that would be:

Can you explain why the smaller root is an unstable equilibrium while the larger root is stable.

 

[HallsofIvy]

That isn't in general true. If you have dy/dt= ky(1- y) then equilibrium points are points where dy/dx, so ky(1- y), is 0. ky(1- y)= 0 for y= 0 and y= 1. Which equilibrium is stable and which is unstable depends on the sign of k.

If k> 0 then, for y< 0, y is negative, 1- y is positive, so ky(1- y) is "positive*negative*positive" so dy/dt is negative, y is decreasing. If y starts less than 0, it moves away from 0 not toward it. For 0< y< 1 k, y, and 1- y are all positive so dy/dt is positive, y is increasing. If y starts between 0 and 1, it moves away from 0 and toward 1. If y> 1, k and y are positive but 1- y is negative so dy/dt is negative, y is decreasing. If y starts larger t it moves toward 1.

That is, if y is close to 0, whether less than or larger, y moves away from 0. 0 is an unstable equilibrium. if y is close to 1, whether less than or larger than 1, y moves toward 1. 1 is a stable equilibrium.

But if k< 0, the sign in each interval is reversed. Using the same argument about whether y moves toward or way from an equilibrium point shows a reversed direction. 0 is now stable equilibrium and 1 is an unstable equilibrium.

 

[S.G. Janssens]

Which equilibrium is stable and which is unstable depends on the sign of k.

In general you are right, but this is a specific model where $K$ is a carrying capacity, see post #1. Hence it is stipulated to be positive.