CHAPTER 05.04: LAGRANGE METHOD: Cubic Interpolation: Example: Part 2 of 2

So what does that turn out to be?  It turns out to be as follows, so it turns out to be v of t is L0 of t, so L0 of t is t minus t1, divided by t0 minus t1, so I'm just repeating, t minus t2, divided by t0 minus t2, t minus t3, divided by t0 minus t3, times the velocity at t0. Then the second, there's four parts, so the second part is L1 t times the value of the velocity at t1. So that'll be t minus t0, divided by t1 minus t0, times t minus t2, divided by t1 minus t2, so as you're seeing here that we are skipping t1, whether it's here or here, because that's the j not equal to i part, and again, it will be t minus t3, divided by t1 minus t3, times the velocity at t1.  Similarly, we have L2 t now, so it'll be t minus t0, divided by t2 minus t0 here, times t minus t1, divided by t2 minus t1, and then we're going to again skip the 2, because it will become 0 here otherwise, so that's the way the formula works, I'm not saying that's why we are doing that, but we have t minus t3 now, skipping t2 here, divided by t2 minus t3, times the value of the velocity at t2. It might seem to be algebraically complicated, but again, if you follow the principles of what each of the Ls are, then it is not.  So again, if you look at L4 of . . . L3 of t, it'll be t minus t0, divided by t3 minus t0, so t3 will be in all the denominators, so it'll be t3 minus t1 here, and then t3 minus t2 here, so if you want to look at it that way, but t3 minus t3 will not be there because i is not equal to j.  Then at the top, it'll be t minus t0, t minus t1, t minus t2, and t minus t3 will not be there, because it is L3, and so that's the way it flows.  So all I have to do now is to substitute the values of t0, t1, t2, t3, and so on and so forth, and I'm not going to show you those, because it's simply just substituting those values, so I'll just write down what values which you have to substitute in here. So going back to what we had said, t0 is 10, the value of the velocity of t0 is equal to 227.04, so wherever there's t0, you're going to substitute 10, wherever there's velocity of t0, which is 227.04, and then wherever there is t1, which is 15, you're going to substitute the velocity at t1, which is 362.78, wherever there is t2, you're going to substitute 20, and your velocity at t2 to be 517.35, and wherever there is 22.5, you're going to substitute the velocity of t3 to be 602.97.  So that's what you're going to do, because, and you can see that everything is here, everything is in this formula here.  Let me go and look back at the top of the formula here.  So if you look at this particular formula, everything which we have written down here, t0, t1, t2, t3, the velocity at t0, velocity at t1, velocity at t2, velocity at t3, and just the t is not there, because the Lagrangian polynomial which you are trying to find is a function of time, but since we are interested in finding the value of the velocity at 16, that's what we're going to substitute for time, t equal to 16.  Since we are finding the velocity at 16, we're going to substitute t equal to 16 in there.  So if you go ahead and do those substitutions, what's going to happen is the velocity at 16 will turn out to be as follows, you get -0.0416, this will be L0 of t, times 227.04, so if you look at this quantity here, it's nothing but L0 at 16, and the plus, you'll have 0.832 times the value of the velocity at t1, which is 362.78, and this is nothing but the L0 at . . . sorry, this is nothing but L1 at 16, then plus 0.312 times the value of velocity at t2, which is 517.35, and again, this is L2 at 16, then plus -0.1024 times the value of the velocity of 602.97, which is the velocity at t3, and this is nothing but L3 at 16.  So these are the weights which you are calculating for the time, t equal to 16.  If you add all the weights, you are going to get 1, and these are the weights which are given to each of the individual velocities which are given to you, and this turns out to be 392.06 meters per second.  So that's how the Lagrangian interpolation works for this example for the cubic interpolation part.  And that is the end of this segment.