Calculate $f(800,200) = $

This means that, if at time $t$, species $a$ population size is $a(t)=800$ and species $b$ population size is $b(t)=200$, then the rate of change in population $a$ is $\diff{a}{t}=$

. At that particular moment, the population size is of species $a$ is

.
On the other hand, if $a(t)=800$ and $b(t)=200$, what is the rate of change in population $b$? $\diff{b}{t}=g(800,200) = $

. At that particular moment, the population size of species $b$ is

.

Notice that the point $(800, 200)$ is on the $a$ nullcline and it is not on the $b$-nullcline. This means you didn't need to calculate $\diff{a}{t}$ but could know immediate that it was

. On the other hand, you could know immediately $\diff{b}{t}$ was not zero.

What you cannot see directly from the nullclines is whether $b$ is increasing or $b$ is decreasing at the point $(800,200)$. So, you didn't waste your time calculating $g(800,200)$. To understand the behavior of the population size $a(t)$ and $b(t)$ at $(a,b)=(800,200)$, though, all we really care about is the sign of $g(800,200)$ so that we could conclude that $b$ is

.

When $(a(t),b(t))=(800,200)$ what direction is the point $(a(t),b(t))$ moving in the phase plane? Since the point is on the $a$-nullcline, $a(t)$ is not changing; $(a(t),b(t))$ cannot move

. Instead, it can only move

. Since $\diff{b}{t}$ is

, the direction the point $(a(t),b(t))$ must be moving is

. To represent this fact, draw a vector pointing

at the point $(800,200)$ on the above phase plane.

##### Hint

$f(800,200) = 2(800)(1-(800+200)/1000) = \ldots$

To draw the vector online, change the step slider in the phase plane above to 4. Purple vectors appear on the nullclines. There is one purple vector on attached to the $a+b=1000$ branch of the $a$-nullcline. Move it near the point $(800,200)$ by dragging the point at its tail to make sure it is at a point $(a,b)$ where $a \gt 0$ and $b \gt 0$. Then drag the point at its head so that it points in the proper direction.

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Another branch of the $a$-nullcline is the line $a=0$. We know if we plug $a=0$ into the expression for $\diff{a}{t}$, i.e., into the function $f(a,b)$, we get $f(0,b)=$

. What happens if we plug $a=0$ into the expression for $\diff{b}{t}$, i.e., input $g(a,b)$? Calculate an expression for $g(0,b)$, which will still be in terms of $b$. $g(0,b) = $

.
Let's try a couple values of $b$. First try $b=1900$: $g(0,1900) = $

. Since this value is

, $b(t)$ must be

. Since we already calculated that $\diff{a}{t}=$

, we know that the trajectory at $(a,b)=(0,1900)$ must move

.

For the second value, let's increase $b$ slightly to $b=2100$: $g(0,2100) =$

. Since this value is

, $b(t)$ must be

. The trajectory at $(a,b)=(0,2100)$ switched to moving

.

What happened when we moved from $b=1900$ to $b=2100$? We are working on the $a$-nullcline, and, when moving from $b=1900$ to $b=2100$ on the $a$-nullcline, we crossed the

, i.e., the curve where $\diff{b}{t}=0$. Only when crossing that curve, can the sign of $\diff{b}{t}$ switch.

You can also read the fact that $\diff{b}{t}$ switches sign this directly from your expression for $\diff{b}{t}$. Along the $a=0$ branch of the $a$-nullcline, $\diff{b}{t}=g(0,b)$, which you calculated above. This expression is

for $0 \lt b \lt $

and becomes

for $b \gt $

.

To document the fact that $a$ is not changing and $b$ is either increasing or decreasing, draw a vector

on the part of the $a$-nullcline where $a=0$ and $0 \lt b \lt 2000$. Draw a vector pointing

on the $a$-nullcline where $a=0$ and $b \gt 2000$.

##### Hint

Online, in the above phase phase, when the step slider is set to 4, there are two purple vectors attached to the $a=0$ branch of the $a$-nullcline. Move one so that its tail is at a point $(0,b)$ with $0 \lt b \lt 2000$, i.e., anywhere on that nullcline above zero and below the point where the $b$-nullcline intersects. (The point $(a,b)=(0,1000)$ where the other branch of the $a$-nullcline intersects is irrelevant; knowing that $\diff{a}{t}=0$ twice doesn't give us any more information.) move the other purple vector above the point where the $b$-nullcline intersects, i.e., move it to a point $(0,b)$ with $b \gt 2000$. Then, rotate each vector so it points in the correct direction.

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