CHAPTER 05.01: SYSTEM OF EQUATIONS: A real life problem of setting up simultaneous linear equations

In this segment we will talk about a real life problem of setting up simultaneously linear equations. Let’s suppose somebody gives you a velocity of a rocket as a function of time. The rocket is going in an upward motion-a straight line, let’s suppose- And we are given the velocity of the rocket at different times: 5 8 12 seconds. The values given are 106.8, 177.2 m/s, 279.2 m/s. And one of the questions which people might ask you about when somebody gives you time data at (47-49)number of points is-Hey can you find the velocity at some point-Let’s suppose 6 seconds or so or 9 seconds or so, somewhere in between 5 and 12. And one of the ways to do it is to simply draw a straight line between-let’s suppose 2 consecutive points like if somebody is asking you to find the velocity at 6 seconds. You might say ok hey, I am going to take these two points and draw a straight line and find out what the velocity of 6 seconds. Or let’s suppose somebody asked me to find the velocity of 10seconds I can always draw a line between these two data points and be able to figure that out.

But let’s suppose somebody says hey, no you have to do better than a straight line. So if we plot these points on a piece of graph paper. We got velocity here and time here. So velocity at 5 is given as 5, 106.8-that’s the coordinate there- Then it is 8, 177.2. Then at 12 it is 279.2. And what somebody is asking you is, hey, I want you to draw a second order polynomial. I want to have a velocity profile or a velocity curve, which is a 2nd order polynomial which is going through these 3 points. So in that case, if we are going to call that velocity profile or velocity curve to be A T squared plus BT plus C. That is our 2nd order polynomial which is going to approximate the velocity from 5 to 12 seconds. So our velocity profile or what we call a interpolant also will be AT squared plus BT plus C 5 less than or equal to T less than or equal to 12 because you are going to write an interpolant for discrete data, it has to be also, what has to be given is the domain in which that particular interpolant is valid.

So the question is that: If I know, if I can find out what this velocity profile is, then I’ll be able to find the velocity at any point between time equal to 5 and time equal to 12 seconds. So the question rises, hey, how do I find out A B and C? Well the thing we are seeing here is that this velocity profile the 2nd order of the velocity profile, the 2nd polynomial of the velocity profile is going through 3 points. So that is going to help me set up the 3 equations. So what I mean by that is hey, if I say the velocity at 5-velocity at 5 is given by A times 5 squared plus B times 5 plus C. But what is the velocity at 5? It is 106.8. And what is the velocity at 8? It will be A times 8 squares plus B times 8 plus C is equal to 177.2. And what is the velocity at 12? It is A times 12 square plus B times 12 plus C equals 279.2. Now if I make the substitute and make the simplification here, the 1st equation, which I have here, will become 25A plus 5B plus C is equal to 106.8. Then the 2nd equation, which is right here, will become 64A plus 8B plus C is equal to 177.2. The 3rd equation, which is right here, will be 144A plus 12B plus C is equal to 279.2.

So what I have is 3 equations and 3 unknowns. So this is my first equation, this my 2nd equation and this is my 3rd equation. So I have 3 equations and the 3 unknowns are A B and C. SO this is an example of a real life problem of a simultaneous linear equation. So we got 3 equations and we got 3 unknowns. If we are able to solve these 3 simultaneous linear equations we will be able to find the values of A B and C. Once we know the values of A B and C is, we can have the velocity profile or the velocity cure with time equal to 5 to time equal to 12 hence being able to find the approximate value of the velocity at any other time other than 5 8 and 12 which are already given to us. And that is the end of this segment.