2015 AP Calculus BC 2b | AP Calculus BC solved exams | AP Calculus BC | Khan Academy

• [Voiceover] Part b. "For zero is less than t is less than one, there is a point on the curve at which the line tangent to the curve has a slope of two." The line tangent to the curve has a slope of two. "At what time is the object at that point?" So, the slope of the tangent line is two. That means that the rate of change of y with respect to x is equal to two.

Well, they don't directly give us the derivative of y with respect to x, but they do give us the derivative of x with respect to t. That's the derivative of x with respect to t. And they give us the derivative of y with respect to t. The x component of the velocity function is the rate of change of x with respect to time, and the y component of the velocity function is the rate of change of y with respect to time.

And using those two, we can figure out the rate of change of y with respect to x. If you were to take the derivative of y with respect to t and divide it by the derivative of x with respect to t, derivative of x with respect to t, well, if you, for the sake of, I guess conceptually understanding it, if you view the differentials the way that you would view traditional numbers and fractions, well, the dt's would cancel out and you'd be left with dy divided by dx.

Or a little bit more formally, you could go to the chain rule and you'd say, all right, the derivative of y with respect to t is equal to the derivative of y with respect to x times the derivative of x with respect to t. This comes straight out of the chain rule. So this is the chain rule right over here. That is the chain rule.

And then if you divide both sides by the derivative of x with respect to t, you're going to get that original expression right over here. Well, how is this useful? Well, we know what the derivative of y with respect to t is. We know the derivative of x with respect to t is. We know them as functions of t, and then we can set them equal to two and then use our calculators to solve for t. So let's do that.

The derivative of y with respect to t, that is, e to the point .5t. So we have e to the 0.5t and then we divide it by the derivative of x with respect to t, so that's going to be the x component of the velocity function. So cosine of t squared.

And so this is the derivative of y with respect to x and we need to figure out at what t does this equal to two. Or if we want to simplify this, and in our calculator, we need to set this up so it's some expression, you know, some function of x is equal to zero. So let me rearrange this equation. So I have a bunch of things equaling zero.

So let's see, I could just subtract two from both sides, or actually, what I could do is I could multiply both sides times cosine of t squared. And so I'll have e to the 0.5t is equal to two, cosine of t squared. Then I could subtract this from both sides, and I will get e to the 0.5t minus two, cosine of t squared is equal to zero.

And now I could use the solver on my calculator to figure this out. So let's get the calculator out, and let's go to Math. Whoops, let me make sure it's on. So, Math. Let me go all the way down to the solver. So, select that. My equation is zero is equal to, so I'm gonna say, e to the 0.5, and the variable that I'm gonna solve for, I'm gonna use x instead of t. The same thing, I'll get the same answer.

e to the 0.5x. All right. And then I'm gonna have minus two times cosine of x squared is equal to zero. All right, so there you have it. Zero is equal to e to the 0.5x minus two, cosine of x squared. Click Enter.

And then we put our initial guess. And they tell us that for t is between zero and one. So maybe a good guess would be right in between. So let's put .5 there. And then we press Alpha, at least, on this calculator, and then you see that little blue Solve there. That will actually solve it. Let it munch on it for a little bit.

And I get t is equal to 0.840. 0.840. t is approximately 0.840. And we are done.