Sketching a slope field AP Calculus AB BC 2015 4a
About this video
Using dy/dx = 2x - y from the 2015 AP Calculus AB/BC question 4, the instructor makes a table to find the slope at six points and sketches the slope field. He then finds the second derivative, d^2y/dx^2 = 2 - 2x + y, and shows it is positive in Quadrant II, so all solutions are concave up there.
You'll be able to
- calculate slopes from a differential equation at given points and sketch a slope field.
- find the second derivative from a differential equation in terms of x and y.
- use the sign of the second derivative in a region to decide concavity.
Key words
- Slope field
- A picture of short line segments showing the slope dy/dx at many points.
- Second derivative
- The derivative of the derivative; it tells you how the slope is changing.
- Concave up
- Curving upward like a cup, which happens when the second derivative is positive.
- Quadrant II
- The part of the plane where x < 0 and y > 0.
Transcript
- [Instructor] Consider the differential equation, the derivative Y with respect to X is equal to 2x minus Y. On the axes provided, sketch a slope field for the given differential equation at the six points indicated. And we see one, two, three, four, five, six points.
And so what I can do is, let me set up a table actually. Let me set up a table with three columns, one for X, one for Y, and then one for the derivative of Y with respect to X, which this differential equation tells us in terms of X and Y. And so we can look at the different points.
This point right over here, when X is one and Y is two, X is one, Y is two. While the derivative, they tell us it's 2 times X minus Y. So this is 2x minus Y. So it's going to be 2 times 1 minus 2 which is equal to zero. And so if the derivative there is equal to zero, then if I were to draw a line that indicates the slope at that point, well, I would draw a line with a zero slope.
So it would look something like. Let me draw a little bit. It would look something like that. All right, let's keep going. Let's look at this point where X is equal to zero, Y is equal to two. X is equal to zero, Y is equal to two. The derivative is going to be 2 times X minus Y.
So it's going to be zero minus 2 which is equal to negative 2. So how do we draw a slope of negative 2? Well, it's going to go from the top left to the bottom right. It's going to be pretty steep. So it might look something like that. That looks like a slope of negative.
Let me try to draw it a little bit, a little bit better. So I could draw it. Well, that's pretty good. Maybe I shouldn't use my line tool. Let me just try to draw a reasonable. So if I have a slope of negative 2 as I move one to the right, I move two down. So it should look something like that.
All right, let's keep going. X1, Y1. X is one, Y is one. It's going to be 2 times 1 minus 1. So this is 2 minus 1. It's gonna have a slope of one. And so that would look something like a slope of one, looks something like that, or let me just draw it by hand. That's actually easier.
And then we have X is zero, Y is one. X is zero, Y is one. Slope is gonna be 2 times zero minus 1 which is equal to negative 1. And so now the slope is negative 1 and just like that. Notice it's less negative than up here. It's less deep. And then let's go to X is 1, Y is negative 1, X is 1, Y is negative 1.
So it's gonna be 2 times 1 minus negative 1. So this is 2, 2 plus 1. This equal to positive 3. So here the slope, it's going to be even steeper, but now in the positive sense. So slope of three would look something like that. And then finally we have X is zero, Y is negative 1.
It's gonna be 2 times zero minus negative 1, which is going to be equal to a slope of one, which is going to look like this. It should be the same. It should be parallel to what we have right over there. And we're done. All right, let's do part B now. Find the second derivative of Y with respect to X in terms of X and Y.
Determine the concavity of all solutions for the given differential equation in Quadrant II. Give a reason for your answer. All right, so first let's just find the second derivative. So we already know that DY. Let me give myself a little bit more space actually.
We already know that the derivative of Y with respect to X is equal to 2x minus Y. Now to find the second derivative, I just want to take the derivative of both sides of this with respect to X. So let's do that. So I could take the derivative of the left hand side with respect to X and the derivative of the right hand side with respect to X.
And so what is this going to be? On the left hand side, the notation would just be the second derivative of Y with respect to X. And then over here, let's take the derivative of each of these with respect to X. The derivative of 2x with respect to X is going to be equal to 2 and then minus the derivative of Y with respect to X was just gonna be minus the derivative of Y with respect to X.
And so we could say, okay, we found the second derivative, but remember they're saying in terms of X and Y. Right now I found the second derivative in terms of, well, in terms of a constant and in terms of the first derivative. And so we can substitute our expression for the first derivative back here to have this expression in terms of X and Y.
And so this is all going to be equal to, let me write it over here. The second derivative of Y with respect to X is going to be equal to 2 minus. Well, the derivative Y with respect to X we already know is 2x minus Y. 2x minus Y. And so this is going to be equal to 2 minus 2x plus Y, plus Y.
And then they say, determine the concavity of all solutions for the given differential equation in Quadrant II. Remember, if you're thinking about our coordinate axes, you think about our coordinate axes. So if that's our Y axis, that is our X axis. This is Quadrant I, this is Quadrant II, this is Quadrant III, and this is Quadrant IV.
So they're talking about Quadrant II. So what do we know about X and Y in Quadrant II? We know X is less than zero and we know Y is greater than zero. So Quadrant II, let's write like this. Quadrant, Quadrant II, we know X is less than zero and Y is greater than zero.
And so if that is the case, so if we get, so we have 2 minus 2x plus Y, so this is going to be, this right over here is going to be greater than zero. And negative 2 times a negative value. Well, this is also going to be greater than zero. So Quadrant II, because of this, that means if, this whole expression is going to be positive, negative 2 times a negative number is going to be positive plus a positive plus a positive is going to be positive.
So the second derivative, second derivative with respect to X is positive, positive, which means that our slope is increasing over that interval, which means that we have positive, we could say that our concavity, I always have trouble saying that, concavity, concavity is upwards, upwards.
And if you ever forget whether, okay, a second derivative being positive, is that concavity upwards or downwards? I always just like to draw kind of our the canonical concave upwards or concave downwards. And you can see here that your slope here is turning less negative or you can say it's becoming more positive, it's increasing and it continues to increase when you are concave upwards.
So second derivative positive, you are concave upwards. Second derivative negative, you're going to be concave downwards. I'll get rid of this because you would want to put that on the actual AP test.
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