2015 AP Calculus AB 6b | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.

So, we want to figure out the points on that curve where the tangent line is vertical. Let's just remind ourselves what the slope of a tangent line is, or what it isn't. I guess this may be a better way to think about it.

If you have a horizontal line, so let me draw a horizontal line, if you have a horizontal line like that, well then your slope is zero. You could say your rate of change of y with respect to x is equal to zero.

But what about a vertical line? If you have a vertical line like that, what is your rate of change of y with respect to x? Well, some people might say it's infinity, or you could say it's undefined. It's undefined in some way because at some point, or one way to think about it, you're going to try to divide by zero because you have a huge change in y over no change in x.

Another way to think about it that's a little bit more in line with that is you could say that your change in x with respect to change in y—notice I took the reciprocal—so now we're talking about our change in the derivative of x with respect to y is equal to zero. Because your y can change, but as your y changes, your x does not change.

So, can we use this little insight here on vertical lines to think about the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical? Well, before they told us, they gave us the curve of the equation and they also told us what dy/dx is. Let me just rewrite them again just so we have them there.

So we know that y to the third minus x y is equal to two. This is the equation of our curve. And we know that dy/dx is equal to y over 3y squared minus x.

One thing that we could do is, well, let's just figure out what the derivative of x with respect to y is and set that to be equal to zero. So this is the derivative of y with respect to x. If we take the reciprocal of that, the derivative of x with respect to y—just the reciprocal of this—is going to be 3y squared minus x over y.

If we want this to be equal to zero, like we said right over here, well then that's only going to happen if the numerator is equal to zero. So we can say, okay, at what xy pair does this numerator equal to zero? 3y squared minus x is equal to zero. You can add x to both sides and you get 3y squared is equal to x.

Another way you could have thought about it is: What x and y values does the derivative of y with respect to x become undefined? Well, that's going to become undefined if the denominator here is zero. But when you're dealing with things like undefined, it gets a little bit more hand-wave.

Yeah, I like to just think of this as the rate at which x is changing with respect to y is zero. And so that got us to the same conclusion.

Well, for that to be true, x has to be equal to 3y squared. Of course, the xy pair has to also satisfy the equation for the curve. So why don't we use both of these constraints and then we can solve for x and y?

The easiest thing I can think of doing is let's substitute x with 3y squared because they are the same—that's the second constraint. So if we take our original equation of the curve, we get y to the third minus instead of writing x, I could write 3y squared.

3y squared times y is equal to two. And so we get y to the third minus three y to the third is equal to two. This is negative two y to the third is equal to two. We can divide both sides by negative two and we get y to the third is equal to negative one, or y is equal to negative one.

Negative one to the third power is negative one. Now, if y is negative one, what is x? Well, x is going to be equal to 3 times negative one squared. So negative one squared is just one, so x is going to be equal to three.

So the point on that curve at which the tangent line is vertical is going to be the point three comma negative one, and we are all done.