Determining the average rate of change of parametric planar motion functions over a given interval

0:00- [Instructor] Particle follows the path of the parametric function, f of t through the points A, which is 1, -1 and B, 3, 4. Let f of t equal all of this business. Determine the average rate of change of x of t from A to B. Pause this video, have a go at that before we do it together.

0:19Okay, so we can see that x of t is defined by this. So let me write that down again. x of t is equal to -3t + 13 over 2t -7. Now the average rate of change from A to B, I could write this this way. The average rate of change is going to be our change in x of t over the corresponding change in t.

0:47So we could call it x of t2 - x of t1, all of that over t2-t1. Now, when we look at these points, 1, -1 and 3, 4, let's just make this the, let's say that we hit this point. We hit this point at t2 and we hit this point at t1. So we know what x of t2 is. x of t2, when we're at t2, x is at three.

1:21So this is going to be equal to three minus and actually let me color code it so we know where things are coming from. x of t2 is going to be three. So it's going to be three minus x of t1, which is this value right over here, when we're at t1, x is equal to one.

1:41So that's our numerator. The numerator was pretty straightforward to come up with. But now we have to figure out what t2 and t1 actually are. So to do that, we have to say, all right, what t value will make x equal three and what t value would make x equal to one?

1:59So for to figure out for three, we could say, all right, three is equal to -3t+13 over 2t-7. We could multiply both sides of this equation by 2t-7. And let's see, 3 times 2t is 6t and then 3 times -7 is -21. Did I do that? Yep. 6T, sorry, not, yeah, sorry. 6t-21 is equal to -3t+13.

2:34Now I could add 3t to both sides, and then I could also, let see, this is the 13. Let me make sure I get that right. If I add 3t to both sides, I get 9t on this side. And if I add 21 to both sides, I'm just skipping a one step just for the sake of space. Then let's see, if I do add 21 over here, I'm going to get 234.

2:599t is equal to 34. Did I do that right? And so t, and maybe I should say t1, maybe I'll put t1 here to know which t we're dealing with. So t1 is equal to 34/9. Did I do that right? Yeah, that looks right. Alright, so t1 is 34/9. Actually no, this is t2. This is t2 right over here.

3:24I was solving, so let me be very careful. Get rid of those ones. These are t2s. This is equal to 34/9. So this is going to be 34/9. And then I'm going to subtract. Now we could solve for t1 and I don't know, maybe let me do it over here. When at t1, x is equal to one, so we're gonna get one is equal to -3 times t1 + 13 over 2 times t1-7.

3:59Multiply both sides by this denominator. We're going to get 2 times t1-7 is equal to -3 times t1 plus 13. So we'll just now do the same thing. Add 3t to both sides. And I'm gonna go over here, add 3t to both sides. We get 5t on the left, and if I add seven to both sides is equal to 20, divide both sides by five.

4:23We get t is equal to four. This was t1 is equal to four. So this right over here is going to be equal to four. And if we wanted to write it as a over nine, four is the same thing as 36/9. I just did that to simplify this. And now we can actually figure out this average rate of change.

4:40In the numerator, we have a two. In the denominator here we're gonna have a 2/9, right? Or no, we're gonna have a -2/9. So this is -2/9. So this business, let me make a little dividing line here. This is going to be equal to 2 times -9/2. And everything cancels out nicely and we get our average rate of change is -9.