2015 AP Calculus AB 6a

- [Voiceover] Consider the curve given by the equation Y to the third minus XY is equal to two. It can be shown that the derivative of Y with respect to X is equal to Y over three Y squared minus X. All right? Write an equation for the line tangent to the curve at the point negative one comma one. All right, so we can figure out the equation for the line if we know the slope of the line and we know a point that it goes through so that should be enough to figure out the equation of the line. So the line's going to have a form Y is equal to MX plus B. M is the slope and is going to be equal to DY/DX at that point, and we know that that's going to be equal to. We'll see Y is, when X is negative one, Y is one, that sits on this curve. So Y is one. So one over three Y squared. So three times one squared which is three, minus X, when Y is one, X is negative one, or when X is negative one, Y is one. So X is negative one here. And so this is the same thing as three plus positive one, and so this is equal to one fourth and so the equation of our line is going to be Y is equal to one fourth X plus B. Now we need to solve for B and we know that point negative one comma one is on the line, so we can use that information to solve for B. This line is tangent to the curve. So includes this point and only that point. That's what it has in common with the curve and so why is equal to one when X is equal to negative one, plus B and so we have one is equal to negative one fourth plus B. You add one fourth to both sides, you get B is equal to, we could either write it as one and one fourth, which is equal to five fourths, which is equal to 1.25. We could write it any of those ways, so the equation for the line tangent to the curve at this point is Y is equal to our slope is one fourth X plus and I could write it in any of these ways. I'll write it as plus five over four and we're done at least with that part of the problem.