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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 10

Lesson 4: Slope fields

# Worked example: forming a slope field

Given a differential equation in x and y, we can draw a segment with dy/dx as slope at any point (x,y). That's the slope field of the equation. See how we determine the slopes of a few segments in the slope field of an equation.

## 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.