2011 Calculus AB Free Response #1 parts b c d | AP Calculus AB | Khan Academy

Alright, now let's 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 6.

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. 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. We want to find the average velocity for this time period, so we're going to 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 6. This is going to be equal to the integral from zero to 6 of 2 sine of e to the T over 4 power plus 1 DT, and then all of that divided by 6. Where did I get this from? Well, they tell us what our velocity as a function of time is; it's that right over there.

So we can get our calculator out to solve this part. On our calculator, we would hit math, and then we would want to do number 9, which is the definite integral. So I hit 9, and we are going to go from 0 to 6 of... this is going to be 2 sine of e to the... so let me do second e to the, and I'll just use the variable X instead of T because it's easier to type in. X divided by 4 power, so that's my sine. Then let me close the parentheses for the sine plus 1, and then I'm integrating here with respect to X just because it was more convenient.

I am going to get that, and then I divide by 6. When you're taking an AP calculus exam, it's important to round to three decimal places unless they tell you otherwise; that's what they expect. So it's approximately equal to 1.949, approximately 1.949.

Now, let's 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 want to figure out distance, what you want to 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. Then we can type this; actually, let me just write it out since you would want to do this if you're actually taking the test.

So it's integral from zero to six of the absolute value of 2 sine of e to the T over 4 plus 1, close the absolute value, DT, which is going to be approximately equal to... 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 2 sine of e to the... I'll use X again just because it's easier to use on the calculator. Then I'll close the parentheses.

Whoops, let me make sure I do this right. So let me close the parentheses on the sine, and then I have plus 1. And there you go, and then I'm integrating with respect to X. Approximately 12.573.

Now, let's 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 I'm going to 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 let's just say 2 sine of e to the T over 4 plus 1 is equal to zero. There are multiple ways you can solve this, but because we're allowed to use a calculator, we might as well do that.

We hit math, and then we hit B right over there. I'll just hit enter, equation solver, so I could just say look at the left side of my equation is 2 sine of... we are going to go e to the... I'll just use X as what I'm solving for because it's easier to type in the calculator and close that parentheses plus 1. The right-hand side is equal to zero, and now I just have to give an initial guess. I'll just go right in between that interval between T equals zero and T equals six, so I'll just put a 3 right over there.

Now, I just need to solve it, and I can do that by hitting alpha and then solve. You see it got us to 5.196. So T is approximately equal to 5.196. But remember, they're not asking us for at what time does the particle change direction; they want to 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 T equals 5.196. The displacement we're integrating the velocity function, so that's going to be 2 sine of e to the T over 4 plus 1 DT.

Now, this is going to be our change in position, but where did we start? Well, they say at X equals 0, or when X equals 1 at time zero, our position was 2. 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 2 plus, and then I go to math, definite integral from 0 to 5.196, and I've typed this in many times already: 2 sine of e to the X divided by 4 power, and then close that parenthesis plus 1.

Then I'm integrating with respect to X, and so this gives us 14.135. Approximately 14.135 is this position and let me make sure that I put those parentheses there in the right place, and we are done.