00:01
So part a here, initial conditions, we have w is 2 .2 and r is 1.
00:11
Let's start with dw over dt and just plug those values in.
00:16
So 2 .2 minus 2 .2 times 1, since that's w minus wr, this will be 0.
00:26
So dw over dt at 0 is 0.
00:30
So there's no change there.
00:31
Dr over dt is equal to minus 1 plus 2 .2 times 1.
00:40
Remember, that's minus r plus wr.
00:44
So 2 .2 times 1, minus 1, that's 1 .2.
00:52
Those are both your answers for part a, for part b.
00:57
Now remember, we need those values also for part b.
01:01
So beginning with part b with time is equal to 0 .1.
01:08
So let's start with w now.
01:10
We need to find the population w at 0 .1.
01:14
So let's start with 2 .2, which is the previous population, plus the rate of change at 0, which is 0, times time, which is 0 .1.
01:28
That'll be 0, so worms would remain at 2 .2.
01:35
R, however, r is equal to the initial population first.
01:41
That's 1 plus 1 .2 times 0 .1.
01:48
So that'll give you approximately 1 .1.
01:56
Part c, part c, which is time is 0 .2, we need to consider what we have in part b as well.
02:12
As we did in part b, we used values for part a.
02:15
We want to do the same thing here.
02:17
So time is 0 .2...