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

Consider the curve given by the equation (y^3 - xy = 2). It can be shown that the derivative of (y) with respect to (x) is equal to (\frac{y}{3y^2 - x}).

All right, write an equation for the line tangent to the curve at the point ((-1, 1)).

So, we could 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.

The line's going to have a form (y = mx + b). (m) is the slope and is going to be equal to (\frac{dy}{dx}) at that point. We know that that's going to be equal to, let's see, (y) is 1 when (x) is -1.

So, (y = 1), so (\frac{1}{3y^2}) - (x), when (y = 1), since (x = -1), we can substitute this in. So this is (\frac{1}{3 \cdot 1^2}) which is (3 - (-1)).

So, this is the same thing as (3 + 1) and so this is equal to (\frac{1}{4}).

And so, the equation of our line is going to be (y = \frac{1}{4}x + b).

Now we need to solve for (b) and we know that the point ((-1, 1)) is on the line. So we can use that information to solve for (b).

This line is tangent to the curve, so it includes this point and only that point. That's what has in common with the curve.

So, when (y = 1) when (x = -1 + b), and so we have (1 = -\frac{1}{4} + b).

You add (\frac{1}{4}) to both sides and you get (b) is equal to, we could either write it as (1) and (\frac{1}{4}) which is equal to (\frac{5}{4}) 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 = \frac{1}{4}x + \frac{5}{4}) and we're done, at least with that part of the problem.