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

Part C: Find the time at which the speed of the particle is three.

So let's just remind ourselves what speed is. It's the magnitude of velocity. If you have the x, actually let me draw it this way. If you have the x dimension of, or the x component of a velocity right over there, so this is the rate of which x is changing with respect to time.

And you have the y component of the velocity. If you have the y component of the velocity, let's say it looks something like that—that is dy/dt—then the speed is going to be the magnitude of the sum of those two vectors. So this right over here, the magnitude of this vector right over here, is going to be the speed.

Well, what's the magnitude of that? Well, the Pythagorean theorem tells us it's going to be the square root of your x component of velocity squared, so (dx/dt) squared, plus your y component (dy/dt) squared. This right here is the speed, and we need to figure out what time this thing is equal to three.

So let's figure that out. The square root of—what's the x component of our velocity? Well, they told us over here the x component of our velocity is (cos(t))^2. So (cos(t))^2 we're going to square that whole thing, and then plus the y component of the velocity, the rate at which y is changing with respect to time, that's (e^(0.5t)) and we're going to square that.

So plus (e^(0.5t))^2. This right over here is our expression for speed as a function of time, and we still have to figure out when this thing equals 3.

So there are a couple of ways we could just subtract 3 from both sides and input this into our solver, or we could begin to simplify this a little bit. We could square both sides, and you would get (cos(t))^2 + (e^(0.5t))^2 = 9. So now we can subtract 9 from both sides.

And we get (cos(t))^2 + (e^(0.5t))^2 - 9 = 0. Now, once again in this part of the AP exam, we can use our calculators. So let's use our calculators to solve for—in this case, t—but I'll do everything in terms of x.

So the equation 0 = (cos(x))^2 + (e^x) - 9 = 0. We already have this set equal to zero, and so we click enter. Then we could just use our previous answer as our initial guess, and we click—we have to do this little blue solve there.

So I click alpha solve, let the calculator munch on it a little bit, and it gets t is equal to where x is equal to—but this is really t: 2.196. So we get t is approximately 2.196. Did I type that in right? 2.19? Yup. And round that up, and we are all done.