Quadratic Spline Interpolation Application

In this segment, we'll talk about quadratic spline interpolation. We'll look at an example of how to apply the theory of quadratic spline interpolation. So, let's recall what we learned in our quadratic spline interpolation theory, you can always go into the description to find the full playlist of quadratic spline interpolation.

Here, what we have is that we are given n plus one data points, and what we are doing is that we are making quadratics go through two consecutive data points. And the bottom line in quadratic spline interpolation becomes can we find the coefficients of these quadratics? So, if we have n plus one data points, we are going to have n such quadratics, and in each quadratic, we're going to have these unknowns and hence, it becomes a problem of having 3n unknowns and wanting to find three n equations so that we can solve them simultaneously. So, let's go and look at look at it through an example.

So, here we are given velocity as a function of time for six data points, and we are asked to find what the value of the velocity will be 16 seconds, but we're also asked to find the acceleration 16 seconds, the distance covered from 11 to 16 seconds. So, it is a it is an application which goes beyond interpolation itself, but let's concentrate first on trying to find the quadratic splines themselves.

So, here is the data, which was given to us, six data points and I’m showing to you here in a graph these particular six data points. So, what we are doing is that we are going to take one spline, this is the first quadratic, the second quadratic, the third quadratic, the fourth quadratic, and the fifth quadratic. So, you will have five quadratics which we have to find in order to be able to develop the quadratic spline interpolation. So, let's make a note of it, we have five quadratics which we have to find, and there will be 15 unknowns because each quadratic has three unknowns.

So, since we have six data points, we'll have five quadratics, and these are the quadratics which are going to exist in the domains given. So, let's suppose we have the first quadratic which is a1t squared plus b1 t plus c1; it will exist for the domain of 0 to 10. Then the next one will go from 10 to 15, and so on and so forth. So, as you can see from here, that we have five quadratics and each of the quadratics have three unknowns. So, that makes it the 15 unknowns which we have to find. So, once we've established that we have to find 15 unknowns, and what domains each of those quadratics are valid for, let's go and set up the equations.

So, the first part of the theory, which we talked about in the previous videos, that each spline goes through two consecutive data points. So, if you look at this particular spline right here, which was termed as a1 t squared plus b1 t plus c1, and it is from 0 to 10 as mentioned here. This is going through this data point where the velocity is 0. and the same spline is going through this point right here at t equal to 10 where the velocity is 227.04. So, that gives us the two equations for saying that each spline is going through two consecutive data points. So, let's go and see what the aggregate number of equations we get by applying this principle that each spline goes through two consecutive data points.

So, we can see that these two equations which we just derived, we get them or not derived, we found, was by saying that hey, the first spline is going through 0 and 10. And if we look at this one, that is for this spline going through 10 and 15. As you can see here, the second spline which has a2 b2 and c2 as its coefficients, but they are going through two points 10 and 15 and using the corresponding velocities for 10 and 15 there. So, that's the second spline. The third spline is going between 15 and 20, that gives us this equation. Then the fourth spline is going from 20 to 22.5, and we get this, these two equations. And then the last spline is going from 22.5 to 30, and that gives us this set of equations. So, basically, since each spline is going through two consecutive data points and we have uh five splines, we'll get five times two, ten equations from there. So, we are five to go, because we need 15 equations. We have been able to show that hey, these, this particular part of the algorithm tells us each spline goes through two consecutive data points, gives us 10 equations.

We also know that the derivatives are continuous at the interior data points, so we have six data points but four of them are interior right here. So, if you look at this particular interior data point right here, it has this spline that is this one right here, and then it has this spline the second spline which is right here. They both go through the common interior data point which is at t equal to 10. So, what that means is the derivative at t equal 10 of the first spline will be same as the derivative of the second spline at the same point, at t equal to 10. So, derivative of the first spline is 2 a sub 1 t plus b1, and the derivative of the second spline is 2 a sub 2 t plus b2. By plugging in the value of 10, we are able to find out what the equation is, and then we take all of these to the left-hand side because these are the unknowns. Because whenever you set up equations, you always put the unknowns on the left-hand side and knowns to the right-hand side. So, since a2 and b2 are unknown, that's why we will take them to the left-hand side and that gives us the equation which comes from, that the derivatives are continuous at the interior data points. Since we have three more interior data points like that, we'll get three more equations, for a total of four equations coming from the fact the derivatives are continuous at the interior data points. So, these four equations, which we are getting from that the derivatives are continuous at the interior data points from the splines, which go through the common data points. So, at t equal to 10, we have the first and the second spline going at t equal to 10, they are, they are, they have a common point there. So, the derivative will be the same, which we just derived this equation. So, similarly you're going to get another equation for t equal to 15, another one for t equal to 20, and another one for t equal to 22.5. So, this gives us four equations. We already had 10 equations from saying that hey, each spline goes through two consecutive data points. Now, we have four equations which are coming from that the derivatives are continuous at the interior data points from the, from the splines, which are going through the common interior point, so we have 14 equations. So, what that means is that we're still looking for one more linear equation to be able to set up our 15 equations and 15 unknowns. So, how we're going to do that?

So, the way we want to do it, was going to say that hey maybe the first spline is linear, or the last spline is linear. So, we can make this assumption that first spline is linear, or we can say last spline is linear, and that's going to allow us to set up the 15th equation because we don't have any other choice here, or a better choice. Now, the question is that hey, should I choose the first spline to be linear or should I choose the last spline to be linear will depend on which one is the better assumption to use. So, we can already see that hey, the difference here is 7.5 of the x values, of the t values, and the difference here is 10 so far as the distance between the two points is concerned. So, it's better to choose the last spline to be linear, because the distance between the two last points is less than the distance between the first two points. So, that's why we chose a sub five equal to zero, that is the last spline is linear.

So, we can see that we can go and recall that we have 15 equations, and we have 15 unknowns and let's recap. The 15 equations came from: 10 equations came from that each spline goes through two consecutive data points, four equations came from that the derivative is the same at the common points of the two splines which go through that point, that gives us four more equations, that makes it 14, and then we had to somehow find out what the 15th equation would be. And in this case, we chose that the last spline is linear, giving us a sub 5 equal to 0.

So, here what we are doing is that we are taking all those 15 equations and putting them in the matrix form because you want to solve them in the matrix format to be able to make any sense out of it. So, you can also see that this coefficient matrix right here is very sparse. In fact, you can make it more banded; banded meaning that there'll be more zeros in the, in this area, and more zeros in this area. If you rewrite these equations where, for example, this number here or this equation is written somewhere close to, let's suppose, this equation right here. We're not done that because what we did was, we took the first 10 equations, wrote them here and then we wrote the next four equations and then we wrote the last equation. But if they are written in the proper format, you're going to get a banded coefficient matrix: meaning there are lots of zeros in the off-diagonal areas here, lots of zeros here in the off-diagonal areas here because there are some special algorithms written for solving equations which are banded in nature so far as the coefficient matrix is concerned. But we're not getting into that, we're simply saying we have 15 equations 15 unknowns, let's go and solve it by any method which is available to us.

So, we found out what the 15 unknowns are, because we have 15 equations 15 unknowns. So, here what we are doing is that since each quadratic is of the format, this one. So, we are just writing those coefficients of the quadratics here, and since we had six data points hence, five quadratics. So, we'll have five such values, five such rows here for the values of a b and c. Now what we're gonna do is, we're going to write it in the quadratics format which we assumed, and let's go and see what we're going to get.

So, here we have the, the coefficients which we found out by solving 15 equation, 15 unknowns. This is there just for reference, so you can see that hey, this will be my first quadratic, that's my second, third, fourth and fifth quadratic. And keep in mind also, that each quadratic is only valid for a certain domain. So, that's very important to understand that hey, that this is not like polynomial interpolation where it is valid between the minimum value of x and the maximum value of x. That's not the case here because each quadratic is, has, has its own domain in which it is valid. So, this is the graph of the quadratic polynomial interpolation, quadratic spline interpolation which you're seeing here, and you can maybe not visually see it here, but you can also see that the slopes seem to be continuous right here at the interior data points based on what we have found out. Let's go and solve the problem at hand, which is finding the value of the velocity at 16.

So, part a of the problem was to find the value of the velocity at 16. Now, we have these five quadratics here, so we cannot just simply pick and choose which quadratic needs to be used. So, since it is 16, we know that this is the proper quadratic to use because this quadratic is valid between 15 and 20. So, having said that, we will go ahead and take this quadratic right here, and put the value of t equal to 16 seconds in there. So, as to be able to find out what the value of the velocity is at16.

The next part is asking you to find the acceleration at 16. Again, what we have to do is we have to find the proper quadratic to use, and since 16 is between 15 and 20, we will use this quadratic right here. But since we are trying to find the acceleration at 16, what we have to do is we have to take the derivative of the velocity expression, or the velocity quadratic and then put t equal to 16 in there. So, let's see how we can go about doing that.

So, we chose the proper quadratic; we'll take the derivative of the quadratic, one derivative of it, and this becomes our acceleration expression now, for which is valid in the same domain as the velocity expression between 15 and 20. And we want to find the acceleration at 16, all we have to do is to put t equal to 16 in there, and we get an acceleration of 32.261 meters per second squared.

Let's look at the last part, which is saying hey, can you calculate the distance covered by the rocket from 11 to 16 seconds? So again, you may have to you will have to choose the proper quadratic or quadratics in order to do this problem. So, you can see that hey, if I want to go from 11 to 16, part of it is in this domain, and part of it is in this domain. So, that means that you'll have to use this spline as well as this, sorry you'll have to use this quadratic as well as this quadratic of the quadratic spline. So, that means that this has to be broken from 11 to 12, 11 to 15 for the first spline, and then from 15 to 16 for the next spline. So, this will be, this velocity here will be coming from the second spline here, because that's the value between 11 and 15, and this one will be coming from the quadratic, which is valid between 15 and 20, but of course we'll integrate only from 15 to 16. So, let’s go and see what we get from there.

So, we already chose the proper two quadratics, we need two quadratics for here. We broke it up into two integrals: one going from 11 to 15 and one from 15 to 16 so that we can put the proper quadratics in there. So, what we do is we put the first quadratic for the integral going from 11 to 15, and the second quadratic from 15 to 20. When I say first quadratic meaning, the first one which is in the domain of 11 to 16, it is the second, second quadratic and the third quadratic which we are using if you look at all the quadratics. So, you get the point; so, here we're going to integrate this function and integrate this function by using our integral calculus knowledge, and we get that the location difference between 11 and 16 seconds, which would give us the measure of the distance covered from 11 to 16 seconds turns out to be 1590.7 meters. And that is the end of this segment.