00:01
We're going to do chapter 11, section 2, problem 13, matching the given slope field equations with the slopes in figure 11 .27, which we can see here, plotted in the negative 5 to 5 cartesian coordinates.
00:14
So a is the equation y prime is equal to 1 plus y squared.
00:27
So starting with that, we should always see positive values.
00:31
Negative squareds are positive, positive squareds are positive, so it should always be positive slope values.
00:36
So with that, we can knock out anything that has negative slopes.
00:39
So that would be this one.
00:41
There's some negative slopes there.
00:44
We have negative slopes here and negative slopes here and here.
00:49
So that already knocked us into just this section right here.
00:53
Let's continue.
00:54
If y is equal to zero, we should see a slope of one, which we do see in this section right here.
00:59
So that's good.
01:01
And then anytime it goes up, one plus one squared is two, and the slopes are just going to get steeper and steeper.
01:05
So this is the answer to a.
01:17
Okay, part b, we have y prime is equal to x.
01:24
And here we should see negative and positive slopes.
01:27
We should see as it gets further and further away from the zero position in x, the y axis, they should become larger and more negative.
01:36
And we shouldn't see too much iteration throughout the pattern.
01:39
It should be pretty much consistent.
01:41
So what we can see here, excuse me, as x gets big, we want a high, steep slope.
01:48
As x is zero, so at the y position, we want zero slope.
01:54
So let's identify where we have zero slope along the y.
01:58
There's zero slope on y here, and there is zero slope on y here.
02:06
Now, there's not here, here, or here.
02:09
So those are out.
02:10
And as x is very large positive, we want very large positive slopes, not like those.
02:16
And x is very large negative.
02:18
We want very large negative slopes, unlike those.
02:20
So this is the answer to b.
02:30
Problem c.
02:33
Y prime is equal to sine of x.
02:39
Now we should always anticipate with the sign functions, seeing some type of cyclical pattern.
02:45
So that would immediately leave us towards problem number four...