 So let's have a quick look at linear momentum. So I just want to show you quickly how to do this. It's very simple. I think let's start down here. We just see linear momentum. That's just a scalar times a vector. That's mass times velocity. That gives us linear momentum, and we know velocity is the first derivative of position. So if mass stays constant, now if you think of a rocket with a chemical type fuel going up, the mass of that rocket is going to change over time. So that mass becomes a function of time, but if mass stays the same, mass is just a scalar, linear momentum is just mass times velocity. So let's have this vector up here. We make a position vector. It's the cosine of 3t in the C coordinate system in the x direction and sine of t in the y direction. And then the v of t, the computer variable v of t is just my r of t. It's first derivative with respect to C in that Cartesian coordinate system. And let's just run that, and indeed we get negative 3 sine of 3 times t plus the cosine of t in the C coordinate system in the x and y directions. Just going to introduce variable m there as a mathematical symbol. In a file now we're just to say p of t is now m times the v of t. Quite simply there, m is just multiplied into both coordinates there. It's just a scalar times a vector, no problem. Just to show you that you can also just write m times the r of t, diff t of c, of course you're just going to get the exact same answer. No problem there. Let's go on to something much more interesting, angular momentum. It is the cross product between the position vector and the linear momentum vector. So it's usually written as capital L as angular momentum in time. It is the cross product of the position vector and momentum vector, or then the position vector and the mass times the first derivative of the position vector. Now how do you do these cross products? I'm going to show you quickly. Let's create two simple vectors. I'm going to have x1 in the x direction, y1 in the y direction and p2 is x2 and y2 in the x and the y directions. If we run that and I take the cross product of p1 and p2. If I run that, there you see the cross product of 2. You'll see that it's only in the z direction. It's x1, y2 minus x2, y1. So it's easy enough. We've seen before how to do the cross product. But let's just scale things up a bit. Let me just show you. If I were to make a matrix x1, y1, x2, y2, which is just combining these two vectors up there, but I make it into matrix form. And I take the determinant of that matrix. Remember that's x1, y2 minus x2, y1. If I run that, off we go. That's where you get x1, y2 minus x2, y1. Now what about scaling things up to three dimensions? If I were to run that, I'm just showing you I've got this. So what I do with my two vectors, which are now in three dimensions, x1, y1, z1, x2, y2, z2, I put a i, j and k for my x, y and z directions on the top. And I get the determinant of this. Look at this. If I were to run that and I get the determinant of that. Now it's not going to split things up nicely for you, but you see there's an i there and there's an i there. So in the i direction, we have y sub 1, z sub 2, minus y sub 2, z sub 1. There's the next two in the j direction and the last two in the k direction. Now here's the exact same two, x1, y1, z1, x2, y2, z2, vectors and I take their cross product. And lo and behold, you'll notice that the determinant and the cross product there is the exact same thing. So my c coordinate system sub x there, that's i, there's my j and there's my k and you'll note that the two is exactly the same. So to take the cross product, you just write it in this form with i, j and k at the top, you take the determinant of that, that would be the same as the cross product. So let's do the cross product between the r of t and p of t of our example above if I were just to run that. Lo and behold, three times the mass of sine of t, times sine of three times t, and the rest there all in the z direction. So this angular momentum of two vectors in the plane will point perpendicularly away in the z direction. What is torque? Well there we have the equation for linear momentum. We start off with that r cross p. If I take the first derivative of that, that's going to be r prime of t times, now look at the second one, just the product rule of the cross product of two vectors. This is the product rule as per normal. So I'm taking the first derivative of the first factor there, cross product, the second one, plus the first one cross product, the derivative of the second one. So it's just that. At the frontier we have r prime of t and cross product r prime of t, remember if you take the cross product of two vectors, similar vectors, or even though this one's multiplied by a scalar that is zero. Because remember you multiply it by the sine of the angle between them, sine of zero is zero. So that term first cross product there is going to fall away. So we have that the rate of change of linear momentum. So imagine something is spinning and I want it to spin faster or slower. I have to apply torque to it. If I apply torque that would be analogous to applying a force in a linear fashion for linear motion. I'm going to apply a twisting force, that's called torque and that's going to give me a rate of change in my angular momentum. It's going to spin faster or slower. So that rate of change in angular momentum, in other words torque equals r cross m r double prime of t. We know that force equals m r double prime of t, force equals mass times acceleration. In other words torque is the cross product of the position vector and the force vector. So we're looking actually at the orthogonal component of one multiplied by the other one. Because if we're just looking at magnitudes, it's the magnitude of this times the magnitude of the second one times the sine of the angle between them. We want the maximum of that, the maximum would be sine of 90 degrees or pi over 2 radians. So if I were just to run it on the example we had above, we had the angular momentum there. If I take its first derivative with respect to t in the c, I'm going to get 8 times the mass times the sine of t times the cosine of 3t in the z direction. Again in the z direction. If I just look at acceleration, that will just be the v of t diff. And if I let f be m times a, of course I'm going to get the force that was that. If I cross r of t and f as I do there, I'm going to get this exactly the same as up there, l of t prime and the cross product r of tf. It's exactly the same. I can test this with a Boolean. Is l t prime equal equal? That means is it equal to? It's going to evaluate that across. If I run that the Boolean value is going to return 2. So I can do the cross product or I can just take the first derivative of the angular momentum. I'm going to leave this last bit. It's quite a bit of fun. I'm not going to do a lecture on it. I'm just going to leave it in the notebook. You can run through it. If you just launch something initially you can look at all Newtons equations. You're going to end up with this position vector here and you can answer a lot of questions if you were to have this position vector to start off with. Good.