Differentiating implicitly and determining concavity AP Calculus AB BC 2015 4b
About this video
This AP Calculus free-response walkthrough sketches a slope field for dy/dx = 2x - y at six points by computing the slope at each one. It then differentiates the equation again to get the second derivative, 2 - 2x + y, and uses the signs of x and y in Quadrant II to show every solution there is concave upward.
You'll be able to
- compute slopes from a differential equation and sketch them as a slope field.
- find the second derivative in terms of x and y by differentiating and substituting dy/dx.
- use the sign of the second derivative in a region to determine concavity.
Key words
- differential equation
- An equation that gives the derivative of y, here in terms of x and y.
- slope field
- A picture made of short line segments showing the slope dy/dx at many points.
- second derivative
- The derivative of the derivative; it tells you whether the slope is increasing or decreasing.
- concave upward
- The curve bends up like a cup because its slope is increasing (second derivative positive).
- Quadrant II
- The top-left part of the coordinate plane, where x is negative and y is positive.
Transcript
- [Instructor] Consider the differential equation, the derivative of 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 3 columns, 1 for x, 1 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 1 and y is 2, x is 1, y is 2. Well, the derivative, they tell us it's 2 times x minus y. So this is 2x minus y. So it's gonna be 2 times 1 minus 2, which is equal to 0. And so if the derivative there is equal to 0, then if I were to draw a line that indicates the slope at that point, well, I would draw a line with a 0 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 0, y is equal to 2. x is equal to 0, y is equal to 2. The derivative is going to be 2 times x minus y.
So it's going to be 0 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, and 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 1 to the right, I move 2 down. So it should look something like that.
All right, let's keep going. x1, y1. x is 1, y is 1. It's gonna be 2 times 1 minus 1. So this is 2 minus 1. It's gonna have a slope of 1. And so that would look something like a slope of 1, looks something like that, or let me just draw it like, draw it by hand, that's actually easier.
And then we have x is 0, y is 1. x is 0, y is 1. Slope is gonna be 2 times 0 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 steep. 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 is equal to positive 3. So here the slope, it's going to be even steeper, but now in the positive sense. So a slope of 3 would look something like that. And then finally we have x is 0, y is negative 1.
It's gonna be 2 times 0 minus negative 1, which is going to be equal to a slope of 1, 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? Well, 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 of 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 0, and we know y is greater than 0. So Quadrant II, let's write like this. Quadrant, Quadrant II, we know x is less than 0 and y is greater than 0. And so if that is the case, so we have 2 minus 2x plus y.
So this is going to be, this right over here is going to be greater than 0. And negative 2 times a negative value. Well, this is also going to be greater than 0. 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, you know, the canonical concave upwards or concave downwards. And you can see here that your slope, 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 'cause you wouldn't want to put that on the actual AP test.
Video © Khan Academy, used under CC BY-NC-SA. Watch on Khan Academy. Puzzicle is not affiliated with Khan Academy.