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# 2015 AP Calculus BC 2c

## Video transcript

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 so 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 you have the Y component of the velocity let's say it looks something like that that is dy DT then the 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 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 the rate at which X is changing with respect to time to time squared plus your Y component dy DT squared this right here is the speed and we need to figure out what time is this thing equal to three so let's figure that out so 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 cosine of T squared so cosine of T squared we're going to square that whole thing and then plus the Y component of the velocity the rate the rate of change it with the rate at which Y is changing with respect to time that's e to the 0.5 T and we're going to square that so plus e to the 0.5 T and we're going to square that this right over here is our expression for speed as a function of time and we just have to figure out when does this thing when does this thing equal 9 so there's a couple of ways we could just subtract 9 or sorry when does this equal 3 when does the equal three and so we could try to we could just subtract three 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 cosine let me write it this way you could get cosine of T squared squared plus e to the 0.5 T and then squaring that well two point times zero point five is just one so this is the same thing as e to the t is equal to nine and now we can subtract 9 from both sides and we get cosine of T squared squared I could have writen cosine squared of T squared but this makes it a little bit clearer I think plus e to the t minus 9 is equal to zero and now once again in this part of the AP exam we can use our calculators so let's use our calculators to solve let's use it to solve for in this case T but I'll do everything in terms of X so the equation 0 is equal to is equal to let me just delete all of this actually just to get it out of the way all right it's equal to cosine and I'm going to use X as my variable cosine of x squared close that parentheses and then I want to square the cosine the whole thing plus e to the T so second e to the but my variable that I'm going to solve for is X I'm just replacing all the T's with X just for inputting into the calculator e to the X minus 9 minus 9 is equal to zero is equal to zero we already have this set equal to zero and so we click enter and then we could 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 or X is equal to but this is really T to point one nine six so we get T approximately two point one nine six did I type that in right 2.19 yep and round that up and we are all done
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