Direct Vandermonde Method of Interpolation – Quadratic Interpolation

In this segment, we'll talk about the direct method of interpolation, also called the Vandermonde method of interpolation, and we will concentrate on quadratic interpolation.

So, in quadratic interpolation what that means is that I will need to have exactly three data points to be able to do quadratic interpolation. So, let's suppose we have these three data points, the function is given discretely only at, r is given at three data points. Then, what we can do is we can draw a second order polynomial. So, that could be a second-order polynomial which might look like this: a naught plus a1 x plus a2 x squared. So, the question arises that hey, how do I find a naught a1 and a2? Because once I find those three constants of this polynomial, then I can find the value of the polynomial at any point between x naught and x2. So, that will allow me to find the value of the function at points which are not given to me And of course, how do we find a naught even a2? We find it by saying that hey, the polynomial is going through the three data points you're given to us. That will set up three equations three unknowns, and hence, we can find out these three constants. And of course, this particular polynomial will be only valid between x naught and x2. So, it's best illustrated through an example, so let's go ahead and do that.

So, let's go and look at the example which we're asked to be solved. Here, we are given six data points, but we're doing quadratic interpolation. We're asked to find the value of the velocity at 16 seconds with a second order polynomial. So, what we have to do is we have to choose three data points from the six data points. So, we first have to figure out hey, what are the two closest data points to 16 so that it brackets that t equal to 16. So, we can already find out that 15 and 20 brackets 16, and they're the closest two data points. But we need the third data point. So, should we take 22.5, or should we take 10? But if you look at a how far is 16 from, 10, the absolute value, the modulus value is 6. And how far is 16 from 22.5, it is 6.5. So, we can see that 16 is further away from 22.5 than from 10, so we will choose 10 as our third data point. And that, those are three data points we should be choosing in order to do the second order polynomial interpolation. Now, keep in mind that in this particular problem, we are also asked to find the absolute relative approximate error for the second order polynomial approximation, which we'll get for the velocity at 16 seconds. So, whatever we get from here how can we calculate the absolute relative approximate, to get some sense of how good our answer is? We're also asked to find out the distance covered by the rocket from 11 to 16 seconds. So, what this is a good example of showing that hey, polynomial interpolation or any kind of interpolation is not just about finding the value of the function at the points which is, which are not given to us, but to be able to do the things which we need to do. So, if we want to find the distance covered by the rocket, which will mean that hey since velocity is given, I’ll have to integrate the velocity to be able to find the distance covered. Look at the fourth part here, we are asked to find the acceleration of the rocket at 16 seconds, and that implies that since the velocity is given to us, or velocity will be found by us, that we can find the acceleration as well by differentiating the velocity expression. So, keep in mind that interpolation has several applications, not just about finding the value of the function at a point which is not given to us. The other, more important part of the, not more important, as important part is that hey, how do we choose the data points for our interpolation? That's also an important part of conducting polynomial interpolation. So, let's go and solve these problems one at a time.

So, part a, the velocity expression is a naught plus a1 t plus a sub 2 t squared. And since we have chosen the values between 10 and 20, this particular interpolant, or this particular second-order polynomial will be applicable only between times of 10 and 20. We’re also given that the velocity at 10 is, will be a naught plus a1 times 10, plus a2 times 10 squared, right? And that will be equal to 227.04, because that's the data which is given to us at 10; it is 227.04. The value of the velocity at 15, if we plug 15 into our second order polynomial, we get this. But the value of the velocity at that time is 362.78. Same thing, the value of the velocity at 20 is the following expression, and we get that as our third equation. So, basically what we have is we have three equations three unknowns, and we should be able to find out what the values of a naught a1 and a2 are. So, let's go and write these equations in a cleaner form with coefficients in the front of the unknowns. We'll get this, the second equation will look like this, the third equation will look like this. So, we have three equations three unknowns. What we're going to do is we're going to write them in the matrix form because by now, you should be doing things by using numerical methods. So, we'll have a naught, a1, and a2 and using matrix multiplication we'll have 1, 10 and 100 here. 115 and 225 here, 120 and 400 here, and the right-hand side vector will be as follows. So, we have three equations, three unknowns, which we should be able to solve by any of the methods which you might have learned, such as Gauss elimination Gauss-Seidel method, and so on and so forth.

So, when we solve these three equations three unknowns, we get the values of the unknowns to be as follows. And now, what that would mean is that the velocity expression which we had, which was a naught plus a1 t plus a2 t squared will have these coefficients a1, a naught a1 a2 evaluated for you. And again, keep in mind that this particular interpolant is only valid between 10 and 20. So, if we want to find the value of the velocity at 16, all we have to do is to substitute t equal to 16 in there, and this value here turns out to be equal to 392.19 meter per second. So, let's go and do the next part now.

In this part, the question is asking hey, what is the absolute relative approximate error for the value which you just obtained? So, the question arises that hey what does it mean we have a previous approximation? Yes, we do. What we got was the value of the velocity at 16, we got as 392.19 just in part a. This was a second order polynomial, right? But, if you remember we, we got the value of the velocity to be approximately equal to 393.70 in the first order polynomial in a previous lesson. So, we can use that as our previous approximation. So, now we can calculate our absolute relative approximate error as current approximation minus the previous approximation, divided by the current approximation. The current approximation is the second order polynomial, and then if we want to find in terms of percentage, we get 0.38410%. So, how does it help us? It helps us to figure out that hey, how good our answer is when we are using the second order polynomial, and maybe we'll get a better answer when we use a third order polynomial, and so on and so forth. So, here we get epsilon a to be 0.38410, which is less than 5%. So, we know at least one significant digit is correct; it is less than 0.5%, so we know that at least two significantly so correct. So, what that means that, most probably, probably what we can do is we can trust 3 and 9 in this answer right here. Let's go and do the next part.

In this part, we are supposed to calculate the distance covered by the rocket between 11 seconds to 16 seconds, distance covered. We gotta recognize that we already have found the velocity expression to be this one, and it is valid between 10 and 20. So, we could integrate this velocity expression from 11 to 16 to be able to find out what the distance covered between 16 and 10 seconds is. So, to be just nothing but integrating the velocity expression from 11 to 16, that’s what will give us the location at 11, and location at 16 the difference between the two. So, all we have to do is, we have to conduct the integration just by using our integral calculus knowledge. So, I substitute the velocity expression here, and I apply my integral calculus knowledge. So, that's what the integral will be. And the lower limit will be 11 and upper limit will be 16. So, if we put these limits in here, we'll get 1604.3 meters. So, that is the distance covered by the rocket from 11 to 16 seconds. Approximation of it, because we are approximating the velocity by a second order polynomial. So, that's what it turns out to be.

Let's go for the last part now. The last part is asking that hey, what is the acceleration at 16? Now, we can say that the acceleration expression will be simply taking the derivative of the velocity, and since we know the value, the expression for the velocity between 10 and 20, we should be able to take its derivative to find the acceleration at 16. So, we know the velocity expression to be as follows from the first part of the problem, and again, keeping in mind that the interpolant is valid between 10 and 20. So, let's take the derivative of the velocity and use our differential calculus knowledge. And this derivative will give us 17.733 plus 2 times 0.37660 t. So that is the acceleration expression, but again, this expression is also valid between 10 and 20 now, but we want to find the acceleration 16. So, all we need is to substitute the value of time here, and that will allow us to find the value of the acceleration at 16 seconds, and that turns out to be 29.784 with the proper units of meters per second squared. So, what I want to summarize here, is that here we took a second-order polynomial to find out what the, how the function approximately behaves between three data points which are given to us. Our function is only given that three, is given at three discrete data points, and we want to be able to find out how does it behave between those, those points. But we also are able to do integration, to do the things which we want to do. We're able to differentiate the expression for the second order polynomial to do the things which we want to do, and the things which we want to do is, we want to find the distance covered from the velocity, we want to find the acceleration from the velocity, and we are able to do that as well. And that's the end of this segment.