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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB/BC 4ab

Slope field for differential equation.

## Video transcript

- [Voiceover] 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 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. Well, the derivative, they
tell us is two times x minus y. So this is 2x minus y. So it's gonna be two times one minus two, 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. It would look something like, let me draw it 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 two times x minus y. So it's going to be zero minus two, which is equal to negative two. So how do we draw a slope of negative two? Well, it's going to go from the top left to the bottom right, it's
going to be pretty steep. 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, so if I have a slope of negative two as I move one to the right, I move two down. So it should look something like, something like that. All right, let's keep going. x one, y one. x is one, y is one. It's gonna be two times one minus one. So this is two minus one. It's gonna have a slope of one. And so that would look something like, a slope of one looks something like that. 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 two
times zero minus one, which is equal to negative one. So now the slope is negative one. And just like that,
notice it's less negative than up here, it's less steep. And then let's go to x is one, y is negative one. x is one, y is negative one. So it's gonna be two times
one minus negative one. So this is two plus one. This is equal to positive three. So here, the slope, it's
going to be even steeper, but now in the positive sense. So a slope of three would
look something like that. And then finally, we have x
is zero, y is negative one. It's gonna be two times
zero minus negative one, 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 two. 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 wanna 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 two, and then minus the derivative
of y with respect to x, well, it's just gonna be
minus the derivative of y with respect to x. So we could say, okay, we
found the second derivative, but remember, they're
saying in terms of x and y. Right now I've 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 two 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 two minus 2x plus y. Plus y. And then they say, determine the concavity of all solutions for the
given differential equation in quadrant two. 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. What do we know about x
and y in quadrant two? We know x is less than zero, and we know y is greater than zero. So quadrant two, let's write like this. Quadrant two. 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 two minus 2x plus y. So this is going to be,
this right over here is gonna be greater than zero. And negative two times a negative value, well, this is also going
to be greater than zero. So quadrant two, because of this, that means if this is, this whole expression
is going to be positive, negative two times a negative
number is gonna 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
con-kuh, our concavity, I always have trouble saying that. 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 could 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 are going to be concave downwards. I'll get rid of this 'cause
you wouldn't put that on the actual AP test.