##### Step-by-step procedure to directly compute the directions along the nullclines

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

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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Pick another point $(a,b)$ on the same branch of the $a$-nullcline. The values of $a$ and $b$ don't matter, as long as $a+b=1000$ and both $a$ and $b$ are positive. Calculate $f(a,b)$ and $g(a,b)$. For that point $(a,b)$, $\diff{a}{t}=f(a,b) =$

and $\diff{b}{t}=g(a,b)$ is

. The population size $a(t)$ is

and the population size $b(t)$ is

. When the solution $(a(t),b(t))$ of the dynamical system passes through the point $(a,b)$ that you chose, the solution $(a(t),b(t))$ must be moving

.

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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Repeat this process for the $b$-nullcline. Since $\diff{b}{t}=0$ for points on the $b$-nullcline, there can be no vertical movement along the $b$-nullcline. Instead, any trajectory $(a(t),b(t))$ must cross on the $b$-nullcline moving

. This horizontal movement can change from leftward to rightward movement only at the

. The $b=0$ branch of the $b$-nullcline is cut into two pieces by the $a$-nullcline (we don't care about negative values of $a$ or $b$). So, you need to test two points, one on each side, to determine the sign of $\diff{a}{t}$. The $a+b=2000$ branch of the $b$ nullcline has only one piece that we care about, so just test one point there. Draw three arrows at the locations you tested to show the direction of movement at the different pieces of the $b$-nullcline.

##### Hint

Online, with the step slider set to 4 in the above phase plane, there are two purple vectors attached to the $b=0$ branch of the $b$-nullcline. Move them so that one is on either side of the point where the $a$-nullcline intersects this branch of the $b$-nullcline (the point where the other $b$-nullcline branch intersects is irrelevant since we already know $\diff{b}{t}=0$ along the nullcline). To determine the direction of each vector, you can just pick a value of $(a,b)$ on that part of the nullcline and test the sign of $\diff{a}{t}$.

There is also one purple point attached to the $a+b=2000$ branch of the $b$-nullcline. Move that point, if necessary, to make sure both coordinates of the point at its tail are positive. You can pick any point along that nullcline (with $a \gt 0$ and $b \gt 0$) to check the sign of $\diff{a}{t}$ along that nullcline. Rotate the vector to point in the correct direction.

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