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# 2015 AP Calculus BC 2d

## Video transcript

fine the total distance traveled by the particle from time T equals zero to T equals one now let's remember they didn't say you'd 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 and actually if then 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 and 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 one plus one plus the square root of two so how do we figure out the total distance in this scenario right over here and 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 Magnus magnitude of your velocity function if you took your speed if that's your speed right over there and if you multiplied it by a little small change in time that's going to give you your your your infinitesimally small change in distance over that infinitesimally small change in time and if you wanted to change your you want if you want to find the total distance over a non infinitesimal change in time well then you can integrate and you can integrate those little changes in time in this case from T equals zero to T equals one 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 both the X component of the velocity cosine of T squared weird let me put a little other parentheses there squared plus the y component of our velocity squared the rate of change of Y with respect to x 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 this is our expression for our speed as a function of time and then you multiply that times DT and we're gonna integrate all this and this is going to give us our total this is going to give us our total distance and lucky for us we can use our calculator in this part of the AP exam so let's evaluate this so let me let me let me turn it on and let me clear out and so we're just going to evaluate let's go to math and function integral right over there that's for evaluating definite integrals and we want to evaluate the square root I'll put an open parenthesis oh yeah I'm gonna 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 gonna use X as my variable of integration cosine of x squared so that's that parenthesis I need another parenthesis squared squared plus e to this business squared well this we already saw before this is this is the same thing as e to the t right if we raise something to the 0.5 t and then we raise that to the second power two times 0.5 t is just t so i mean i can do that if i like i could just type in all of this business if you like so plus second e to them my variable of integration here is x e to the x so close that and then i close my square root so you don't 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 and now let's let the calculator or munch on it a little bit and I get approximately one point five nine so this is approximately one point five nine five the total distance traveled by the particle of 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 at the beginning of the problem
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