2015 AP Calculus AB/BC 4ab | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Consider the differential equation: the derivative of y with respect to x is equal to 2x minus y.

On the axis provided, sketch a slope field for the given differential equation at the six points indicated. We see 1, 2, 3, 4, 5, 6 points.

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.

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, is 2 * x - y. So this is 2 * x - y, so it's going to be 2 * 1 - 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 0, y is equal to 2. x is equal to 0, y is equal to 2. The derivative is going to be 2 * x - y, so it's going to be 0 - 2, which is equal to -2.

So how do we draw a slope of -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... it might look something like that. That looks like a slope of negative... let me try to draw it 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 line. So, if I have a slope of -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. x is 1, y is 1. It's going to be 2 * 1 minus 1, so this is 2 - 1. It's going to have a slope of 1, and so that would look something like a slope of 1. It looks something like that. Or I could just draw it by hand. That's actually easier.

Then we have x is 0, y is 1. x is 0, y is 1. The slope is going to be 2 * 0 minus 1, which is equal to -1. So now the slope is -1, and just like that, notice it's less negative than up here; it's less steep.

Next, let's go to x is 1, y is 1. So it's going to be 2 * 1 minus 1, so this is 2 - 1. This is equal to 1.

So here, the slope is going to be even steeper, but now in the positive sense. A slope of 3 would look something like that.

Finally, we have x is 0, y is 1. It's going to be 2 * 0 minus -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 2. Give a reason for your answer.

All right, so first let's just find the second derivative. We already know that dy/dx 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.

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. 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.

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. Then, minus the derivative of y with respect to x is just going to be -dy/dx.

So when 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 a constant and in terms of the first derivative.

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 the derivative of y with respect to x.

We already know is 2x - y, so this is going to be equal to 2 - (2x - y). This is going to be equal to 2 - 2x + y, which simplifies to 2 + y - 2x.

And then they say to determine the concavity of all solutions for the given differential equation in quadrant 2.

Remember, if you're thinking about our coordinate axes, if you think about our coordinate axes. So if that's our y-axis, that is our x-axis. This is quadrant one, this is quadrant two, this is quadrant three, and this is quadrant four.

So they're talking about quadrant two. So what do we know about x and y in quadrant 2? We know x is less than zero, and we know y is greater than zero.

So, if that is the case, so if we get 2 + y - 2x... well, we know that -2 times a negative value is also going to be greater than zero.

So in quadrant 2, because of this, that means if this whole expression is going to be positive. -2 times a negative number is going to be positive... plus a positive + a positive. It's going to be positive.

So the second derivative with respect to x is positive, which means that our slope is increasing over that interval. This means that we have positive... we could say that our concavity... I always have trouble saying that... concavity. Concavity is upwards.

If you ever forget whether, okay, a second derivative being positive is that concavity upwards or downwards, I always just like to draw the canonical form: concave upwards or concave downwards.

You can see here that your slope is turning less negative or you can say it's becoming more positive, it's increasing.

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 wouldn't want to put that on the actual AP test.