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## Sketching slope fields

# Worked example: forming a slope field

AP.CALC:

FUN‑7 (EU)

, FUN‑7.C (LO)

, FUN‑7.C.1 (EK)

## Video transcript

- [Instructor] In drawing
the slope field for the differential equation, the
derivative of y with respect to x is equal to y minus 2x, I would place short line segments at select points on the xy-plane. Complete the sentences. At the point (-1,1) I would draw a short segment of slope blank, and like always pause this
video and see if you can fill out these three blanks. The short segments that
you're trying to draw to construct this slope field, you figure out their slope based on the differential equation
so you're saying when x is equal to -1 and y is equal to 1, what is the derivative
of y with respect to x, and that's what this
differential equation tells us. So for this first case, the
derivative of y with respect to x is going to be equal to
y, which is 1 - 2 times x, x is negative 1, so this is going to be -2
but you're subtracting it so it's gonna be +2, so the derivative of y with
respect to x at this point is going to be 3, so I would
draw a short line segment or a short segment of slope 3, and we keep going, at the point (0,2). Well let's see, when x is 0 and y is 2 the derivative of y with
respect to x is going to be equal to y which is 2 - 2 times 0, well that's just going to be 2. And then last but not
least, for this third point the derivative of y with
respect to x is going to be equal to y which is 3
- 2 times x, x here is 2, 2 times 2, 3 - 4 is equal to -1, and that's all that problem asks us to do, now if we actually had
to do it, it would look something like, I'll try
to draw it real fast, so let's see, let me make sure I have space for all of these points here, so that's my coordinate axis and I want to get the point (0,2), actually I want to go all the way to (2,3) so let me get some space here, so 1, 2, 3, and then 1, 2, 3, and then we have to go
-1 and once we might go right over here, and
so for this first one, this exercise isn't asking
us to do it but I'm just making it very clear how we
would construct the slope field, so the point (-1,1), a
short segment of slope 3, so slope 3 would look something like that, then at the point (0,2) a slope of 2, (0,2), the slope is going to
be 2 which looks something like that, and then at the point (2,3), at (2,3), a short segment of slope -1, so (2,3) a slope segment of slope -1, it would look something like that, and you would keep doing
this at more and more points, if you had a computer to do
it that's what the computer would do, and you would draw
these short line segments to indicate what the
derivative is at those points and you get a sense of,
I guess you would say the solution space for
that differential equation.

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