 All right, we're now going to solve part two of our example problem of calculating the acceleration of a fluid particle going through a nozzle. If you recall back, this was the problem statement that we had and what we were looking for is ways of computing the acceleration of a fluid particle going through a nozzle. We were given the velocity field here and we did the first approach using the substantial or the material derivative to compute the acceleration and the second part is to do this from the Lagrangian perspective and so that's what we're going to do now in this part of the problem. Part two using the Lagrangian technique where we follow a particle. So what we're going to do is we're going to begin by assigning the particle position xp with the new variable f and that function or that variable will be a function of time and therefore we can say that the velocity of the particle is equal to the derivative of f with respect to t. And so what we can do is we can write out dxp by dt equals df by dt and that's equal to the particle velocity. Well, we know from the problem statement what the particle velocity is. It's given right here. So let's pull that into our expression. We then get v1 times 1 plus x over l equals v1 times 1 plus f over l where I've now made the substitution for x in putting f as we've denoted x as being little f. Now what we're going to do, we're going to separate the variables. We're going to get the apps on one side and the time on the other side. So we have an expression like that. Now we are going to want to integrate that. So let's write out the expression again. And we were told that this is t equals 0, x equals 0. So at time 0 our particle is at the beginning of the nozzle, at position x equals 0. And we're trying to find at time t the particle is at position f and we want to know what the acceleration is and so that's what we're going to work for in this problem. So we're going to integrate the equation above and the way that we're going to go about doing this integral is we're going to use the substitution method. And so what we will do, we'll sub in little u for 1 plus f over l. And then we can write du by df is 1 plus, it's 1 over l. And from that we can write du is equal to df over l and df is ldu which we can then substitute that into our integral. So what we're going to do, we're going to take df and we're going to sub it in here and we know from our substitution u is there. So let's make that substitution. And then what we're going to have to do is adjust the limits of the integration. So let's do that before we make the substitution. The limits of integration for 0, u will equal 1. And at f, u will equal, that would be 1 plus f over l. So those will be the limits of our integral when we make the substitution. So what we have, we have integral from 1 to 1 plus f over l, du over u is equal to integral 0 to t, this is an easy integral to evaluate. It basically works out to be, you can pull v1 out and we have v1t minus 0, so that is equal to v1t. So the right hand side is easy. The left hand side, du over u, if you look back at your integral tables, you'll recognize that that is a natural logarithm. And did I forget something here? Just a moment, we have du by df, du is df by dl, df is ldu. So df, I forgot the l here, I apologize for that. There should be an l, and the l could be pulled out because that's a constant. But what we get is we get l multiplied by du over u integrated is natural logarithm of u, and that is evaluated in the limits from 1 to 1 plus f over l, which we can then re-express as being l times ln of 1 plus f over l minus ln of 1. And if you check ln of 1, it turns out to be 0. And that will be equal to v1t on the right hand side. Finally what we're left with is l, natural logarithm of 1 plus f over l equals v1t. So that's an interesting equation, but it's not very useful. What we want to be able to do is we want to be able to isolate f and express it as a function of time. And then that would tell us where our particle is within the nozzle. So let's go ahead and rearrange that and see what we get. So that's what we get for the particle position expressed in Lagrangian. So as a function of particle location and as a function of time, I should say, as you'll notice we have t as being the variable. And if you recall, we are looking for the acceleration of the particle, or the acceleration of the particle is just going to be the second derivative of the particle position with respect to time. And so let's evaluate that and we get this expression. So this then tells us what the particle acceleration is as a function of time. It is expressed as the, through the Lagrangian method, versus, versus if we go back, I think it was, where was it, right here. This was the acceleration that we obtained using the Eulerian. And that is where you just determine the acceleration at a given point in space. With the Lagrangian, what we're doing is we're determining acceleration at a point in time. And consequently, that would be for a particle starting at the inlet at time t equals zero. So there's an example of using both the Eulerian and the Lagrangian for calculating the acceleration of a particle. Eulerian is the more clear and more straightforward method. And that's typically what we'll be using throughout this course, is the Eulerian approach. But the purpose of the example is just to show you the difference between the two methods.