2015 AP Calculus BC 2c

- [Voiceover] 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, squared. Plus your y component, dy dt, squared. This right here is the speed. And we need to figure out at 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 gonna square that whole thing. And then plus the y component of the velocity, the rate of change, the rate at which y's changing with respect to time. That's e to the zero point five t, and we're going to square that. So plus e to the zero point five t, and we're going to square that. This right over here is our expression for speed, as a function of time. And we still have to figure out, when does this thing, when does this thing equal nine? So there's a couple of ways, we could just subtract nine, or sorry when does this equal three? When does this 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 zero point five t, and then squaring that, well two 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 nine from both sides, and we get cosine of t squared, squared, I could've written cosine squared of t squared, but this makes it a little bit clearer, I think. Plus e to the t minus nine 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, zero is equal to is equal to, let me just delete all of this actually, just to get it out of the way. Alright, is equal to cosine, and I'm gonna use x as my variable, cosine of x squared, close that parenthesis, and then I wanna square the cosine, the whole thing, plus e to the t. So second e to the, but my variable that I'm gonna 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 nine, minus nine 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, two point one nine six. So we get t is approximately two point one nine six. Did I type that in right? Two point one nine, yep and then round that up. And we are all done.