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

Find the total distance traveled by the particle from time t equals zero to t equals one.

Now let's remember, they didn't say find the total displacement; they said find the total distance traveled by the particle. So if something goes to the right by one and then goes up by one, their distance is two. Actually, if they go back—if they go back over here, actually let me draw a straighter line—if they then go back to the original starting point, then this distance right here would be what? The square root of the square root of two. The displacement would be zero; they got back to where they started. But the distance would be 1 plus 1 plus the square root of 2.

So how do we figure out the total distance in this scenario right over here? Let me erase this since it's not relevant to the problem. It's just to remind ourselves we're talking about distance.

Well, distance is equal to rate times time, or it's equal to speed times time. So in a very small amount of time, if you want to figure out the distance, well, you could take your speed, which is the magnitude of your velocity function. If you took your speed—if that's your speed right over there—and if you multiply it by a little small change in time, that's going to give you your infinitesimally small change in distance over that infinitesimally small change in time.

If you wanted to find the total distance over a non-infinitesimal change in time, well then you can integrate. You can integrate those little changes in time, in this case from t equals 0 to t equals 1. This is going to be the expression for the total distance.

Well, what is this going to be? Well, this is going to be equal to the integral from 0 to 1. In the last part of the problem, we already have an expression for our speed. Our speed we saw was this business right over here. So it's going to be equal to the square root—give myself a little bit more space to work with—it's going to be equal to the square root of the x component of the velocity, cosine of t squared, plus the y component of our velocity squared. The rate of change of y with respect to time squared, so plus e to the 0.5 t—that's the rate of change of y with respect to time, or the y component of velocity—we're going to square that.

This is our expression for our speed as a function of time, and then you multiply that times dt. We're going to integrate all this, and this is going to give us our total distance. Lucky for us, we can use our calculator in this part of the AP exam, so let's evaluate this.

Let me—oops, let me turn it on and let me clear out. We're just going to evaluate. Let's go to math and function integral right over there; that's for evaluating definite integrals. We want to evaluate the square root—I'll put an open parenthesis—I'm going to have a lot of parentheses here, but let's see if we can do it.

So the square root of cosine of, and I'm going to use x as my variable of integration, cosine of x squared. So that's that parenthesis; I need another parenthesis squared. Plus e to this business squared—well, this we already saw before. This is the same thing as e to the t, right? If we raise something to the 0.5 t and we raise that to the second power, 2 times 0.5 t is just t.

Let me—I can do that if I like. I could just type in all of this business if you'd like. So plus second e to the—my variable of integration here is x, e to the x. So close that, and then I close my square root. See, I do that right? This closes around the x squared; that closes with that. Yep, okay, that looks right. And then my variable of integration is x, and I'm integrating from 0 to 1.

Now let's let the calculator munch on it a little bit, and I get approximately 1.595. So this is approximately 1.595, the total distance traveled by the particle from time equals zero to t equals one, which is kind of neat that we can do these things just from, you know, the information that they gave us at the beginning of the problem.