2011 Calculus AB Free Response #1 (b, c, & d)

- [Instructor] Alright, now lets tackle part B. Find the average velocity of the particle for the time period from zero is less than or equal to t is less than or equal to six. So our average velocity, that's just going to be our change in position, which we could view as our displacement, divided by our change in time. Well, what is our change in position going to be? Well they don't give us our position function but they do give us our velocity function. And so to figure out the average velocity, we could figure out our displacement, which is going to be equal to the integral of our velocity. The integral of v of t, dt. And we wanna find the average velocity for this time period. So we're gonna go t equals zero to t equals six and then we're going to divide that by the amount of time that goes by. Well our change in time is going to be equal to six and so, this is going to be equal to the integral from zero to six of two sine of e to the t over four power plus one, dt and then all of that divided by six. Where did I get this from? Well they tell us, what our velocity as a function of time is, its that right over there. So we can get our calculator out to solve this part. So on our calculator we would hit math and then we would wanna do number 9 which is definite integral, so I'll hit nine. And we are going to go from zero to six of, this is going to be two sine of e to the.. So let me do, second e to the And I'll just use the variable x instead of t cause its easier to type in. X divided by four power, so that's my sine and then let me close the parentheses for the sine, plus one and then I'm integrating here with the respect to x just cause it was more convenient. And I am going to get that, and then I divide by six. Divide by six. And I get And when you're taking an AP Calculus exam its important to round to three decimal places unless they tell you otherwise, that's what they expect. So its approximately equal to one point nine four nine. Approximately one point nine four nine. Now lets do part C. So part C Find the total distance traveled by the particle from time t equals zero to t equals six. So you might say, hey didn't we just figure that out? No, this is displacement. And to remember the difference between distance and displacement, if I have something that starts here, it goes over there and then it comes back to right over there. The distance traveled is the total path length so it would be that, plus this. While the displacement would just be this right over here. So if you wanna figure out distance, what you wanna do is take the integral of the absolute value of velocity, you could think about it as you're taking the integral of the speed function. So this is just going to be the integral from zero to six of the absolute value of v of t, dt. And then we can type this, actually let me just write it out, since you would wanna do this if you're actually taking the test. So its the integral from zero to six of the absolute value of two sine of e to the t over four, plus one. Close the absolute value, dt. Which is going to be approximately equal to... So once again, we hit math, definite integral. We're going from zero to six of... Now we'll hit math again, and then we'll go to number, we see absolute value. The absolute value of two sine of e to the, I'll use x again just cause its easier to use on the calculator. And then I'll close the parentheses on... Whoops, let me make sure I do this right. So let me close the parenthases on the sine and then I have plus one and then there ya go. And then I'm integrating with respect to x. Approximately twelve point five seven three. Now lets do part d. I'll do it right over here. Part d. For t between zero and six the particle changes direction exactly once. Find the position of the particle at that time. So the way that I'm gonna tackle it is first I need to think about at what time does the particle change direction and then, I can figure out its position by taking the integral of the velocity function. So how do I think about where that particle changes direction? Well you're going to change direction if your velocity is at zero and right before that your velocity was positive and then right after that your velocity is negative or the other way around, right before that your velocity was negative and then your velocity right after that is positive. So we just have to figure out when does v of t equal zero? So lets just say two sine of e to the t over four, plus one, is equal to zero. And there's multiple ways you can solve this, but because we're allowed to use the calculator we might as well do that. So we hit math. And then we hit B right over there, I'll just hit enter. Equation solver. So I could just say, look the left side of my equation is two sine of, we are going to go e to the, I'll just use x as what I'm solving for cause its easier to type in, in the calculator. And close that parentheses, plus one. The right hand side is equal to zero. And now I just have to give an initial guess and I'll just go right in between that interval between t equals zero and t equals six, so I'll just put a three right over there. And now I just need to solve it and I can do that by hitting alpha and then solve. And then you see, it got us to five point one nine six. So, t is approximately equal to five point one nine six. But remember, they're not asking us for at what time does the particle change direction. They wanna know, what's the position of the particle at that time. Well we know how to find our change in position, our change in position is going to be our displacement from time equals zero to time five point one nine six and the displacement, we're integrating the velocity function. So that's going to be two sine of e to the t over four, plus one, dt. Now this is going to be our change in position, but where did we start? Well they say at x equals zero, or at time zero our position was two. So this is our change, we would start here, this would give us our total position. Where we started plus our change. And so this is going to be approximately equal to... We are going to have two plus and then I go to math, definite integral, from zero to five point one nine six and I've typed this in many times already. Two sine of e, of e, to the x divided by four power and then close that parentheses. Plus one, and then I'm integrating with respect to x and so this gives us, fourteen point one three five. Approximately fourteen point one three five, is its position. And let me make sure that I put those parentheses there in the right place. And we are done.