Hi everyone and welcome to our lecture on velocity, speed and acceleration. These are notions that we've seen before in single variable calculus. We know the derivative is associated of the position function is associated with velocity. Speed is its absolute value and acceleration of course is usually the second derivative. But now I want to think of a curve that lives in space. So I have three dimensions and I'm flying around in space, some rocket ships and particle. And what does it mean when I talk about speed, acceleration and velocity in this context. So remember, C will always be defined by a position vector, we'll call it RFT. And this will be a vector with three components, x of t, y of t, and z of t. So the position the location of space with xyz coordinates is given in terms of time. So now I'd like to define the velocity vector which we will denote as v, so v of t, and this will be the derivative of the position vector. So related to the derivative, what this means, instead of a single derivative, I have to take the derivative of each of the component functions. So instead of taking one derivative, I have to take three. Now if we can do a once we can do a three times. And then if I want to talk about the speed of some particle moving in space, well then this becomes instead of the absolute value of the velocity well now as a vector, we say that it's the magnitude or the norm. This is the speed. And this is the same formula as always it is the square root of the components squared and we add them up. So the sum of the components squared. And all these formulas work, of course, in two dimensions, we just have a z component of 0. And then last but not least, if I wanted to talk about the acceleration of a particle along this curve at introduce a new vector a of t. And of course, this is the derivative of the velocity vector or the second derivative of the position vector. And this is once again a vector. It's important to realize when you're working with vectors and when you're not. x double prime of t, y double prime of t, and z double prime of t. Add these to your ever growing list of formulas. But again, this should look very familiar. It mimics the single variable case for a function f of x. All right, let's do an example just to go through and practice some of these new notions. So let me give you a curve c, defined by the position function. And we'll pick easy derivatives just to go through but you can imagine that this can get tougher if I start asking for more complicated functions and need derivatives using product or quotient rules. So I have my position function. This defines some curve in space. I don't know what this looks like, but we can graph it on a calculator or a computer and try to get a picture. But it's some curve in space. So let's go through and just find our new components and practice working this out. So our velocity as a vector, v (t) is the vector 2t derivative of t squared + 1, 3t squared, 2t. And again, I could ask you as a function of t, the velocity changing over time, you give me different values of t and I plug it in, and I give you different velocities. If I want the speed, well this becomes the magnitude of the velocity. And as a formula that's not going to be too exciting, but it's each component squared. So 2t squared plus 3t squared whole thing squared plus 2t squared. I don't think this is going to clean up too nicely, but if you want it to just simplify it a little bit, you're going to get 4t squared for the first component and 4t square for the second one. So that's 8t squared plus 9t to the fourth. It is what it is. I don't think we're going to get much more into that. And then just for practice, here's the acceleration vector. This is our vector a of t. And of course, this is the derivative of the velocity vector. So I go to my velocity vector and I just take some derivatives of each component. It's important to realize though, each time, we're giving back a vector, and here we get 2 60 and 2 again. And you can imagine questions they'll ask you to plug in different values of t, t equals zero t equals one, t equals pi. Whatever it is, and you can answer of course, was the velocity, speed and acceleration, just by knowing calculus and of course. The position of the parameterization of the curve. So now we're going to use these tools and head over to physics for a minute and sort of re derive some classic results in physics. mimicking a little bit what Newton did. So to do this, I'm just going to recall Newton's second law of motion. I think we've all seen this. This is force equals mass times acceleration. When we first learned this, maybe young as children, it's F equals ma. F, of course is the force measured in Newtons, m is mass and a is acceleration. Normally, though is presented in scalar form, we want the vector version of this. And so remember, forces a vector has a magnitude and a direction. Everything's changing with respect to time. And now it's mass times the acceleration vector. So is the more adult more grown up version of this think of it in terms of vectors. And this is just scalar multiplication we can move the mass in masses a scalar. We multiply it by a vector as usual. So now let's take a very common situation in physics. Let's imagine I have some particle, it's moving. Let's keep it in the xy plane. To keep things simple. We're going to have an orbit. We're going to go in the positive or counterclockwise direction. Let's do a perfect circle. And we're going to go a radius, so my radius here is a. We know the parameterization of a circle, so c is a circle of radius a. And this is defined by r(t) is of course the vector a cosine of t, a sine of t. So we can parameterize a circle, and one lap around is from 0 to 2 pi. Let's go ahead and see what's going on with the other pieces. So velocity, what's going on with the velocity of an object moving in a circle. This is a constant speed. Derivative of a cosine t, of course is negative a sine of t, and then a cosine of t for the derivative of the second component, okay, all good. Let's get acceleration of an object moving at a constant speed in a circle. And this becomes minus a cosine of t and then minus a sine of t. So now let's put this all together. With Newton's second law, so we go up and we look at what is the force of an object moving in a circle at a constant speed. So remember the force is the mass times the acceleration. Let's use our acceleration vector here and we have a cosine of t minus a sine of t. Let's bring out the negative. So since that appears in both components I can factor it out completely and I get a cosine of t, and a sine of t. All right, I know it's a lot of derivatives and there's lots of sine and cosines but stare at this for a minute. What did you just discover? What did you just notice? What did you just find? Where have you seen a cosine t and a sine of t before this is the position vector. So the force that this object experience is -m times the position vector. Let's draw a picture to understand what the position vector is doing. If I'm at any position, the position vector is the vector from the origin to that spot. So here's RFT. And we just said for an object moving in a circle at a constant speed, there's a force on this object proportional to negative the mass. So it's going in the opposite direction, negative RFT. So it's pushing it back in the opposite direction. For us, that means towards the center. What is this call this force has a name, you know this force. If you're moving in a circle, what's the force that pushes an object towards the center? And it turns out to be proportional to the mass of the object. If you said centripetal force, you're absolutely correct. This is centripetal force. Give yourself a little pat on the shoulder. You just discovered centripetal force using Newton's second law and the rules of calculus. That's a great example. Here's another one that I think we sort of discovered in our childhood. But we also see it pretty early on in any physics course. So this is going to be the notion of a projectile. So projectile, you fire something out of a cannon out of a gun, you throw it, whatever it is, and you have a fixed initial velocity. And it's at some angle to the ground, we'll call it alpha, and then you let gravity take over. And it turns out it travels in an arc and it lands and hits the ground somewhere else away. The only force acting on this object is gravity itself. The only thing you can control is the angle off the ground. And the initial velocity upon which this thing is fired. We're going to place our cannon or gun or whatever at the origin, we'll keep it simple. And my question to you is, and you probably already know the answer, if you ever try to throw a ball pretty far is what is the angle that maximizes the distance traveled? Again of a projectile here, which just means the only force acting on the object is gravity. I think if I asked people and I said like, what angle do you throw it? I think most people will tell me 45 degrees. And that turns out to be correct, but let's go through the steps and show that, but I think we sort of discovered this almost accidentally. But now it's amazing to see like the math behind it, that makes us work. So when we work with projectiles the only force we have is that of gravity. So now F equals ma really becomes F equals mg. And g is pretty well known constant in physics 9.8 meters per second squared. So g is a constant is fixed. Since we're thinking of things now as a vector, especially in terms of time, I want to think of this as gravity, it's pointing down. So there's no force in the x component. So my x coordinate is zero, and then my due to the downward nature, we're going to really call it minus mg. This tells me remember that forces mass times acceleration. If I factor out the m, I can get my acceleration vector and this becomes the vector is 0 comma minus g. Let's go through the process now in reverse where I last time I had position I got velocity and I had acceleration I got force. Let's walk through that in reverse. So now I have force I can get from that the acceleration vector. From the acceleration vector, I can find the velocity vector. So a little integration on both parts here all with respect to t. So the what function has derivative zero? Well, that's just some constant. I'm going to call it c1, what function has negative g? Well, that's negative g t, plus some other constant of integration c2. So c1 and c2 are just some real numbers. Let's keep going. I now have my position vector R of t, and I have to integrate again. So I have velocity and I want my position, I gotta integrate. When I do that, I get c1 of t plus a new constant of integration d1. Then minus g t squared over two plus c2 t plus d2. And here we go again, do you want to do to our other constants of integration. So every time we integrate, we do have to factor in for that plus c. But since I have different components, we want to just be careful and make sure we're doing incorrectly. Okay? The good news is it's not as bad as it looks. So for example, let's use the the model here to situate this. Remember that initial position is at the origin, that's 00. That tells me that my position at zero is so if I plug in t equals zero, I get basically just what's left d1 and d2. So I get d1 and d2. If I plug in and since I on purpose, put the cannon at the origin. This just tells me that d1 and d2 are 00. So, not as bad as it looks these constants of integration turned out to be 0 as long as you place your cannon at the origin which is pretty natural thing to do. The other thing I want to think about here is I have to start, I have a question about angles and I need to introduce my angle somewhere. But the initial velocity that is going to correspond to this is the length of this vector. This is the length of the vector, and the initial velocity v0 is the length of my velocity vector, add zero. And if I plug in zero to my velocity vector, what do I get back? I just get back c1 and c2. So this turns out to be the magnitude of c1 and c2. So these constance of integration are coming from my initial speed, and we have a formula for this. Remember, it's the sum of the components squared. So these two numbers are sort of given with the problem. I want to think about these numbers though have to introduce data some way somehow. So let's think about them in terms of their sort of breakdown their x component and their y component. So let's break down the initial velocity In terms of a vector, as c1 and c2. But I want to think about more as what are they in terms of sine and cosine of theta. So the length, usually we denote it as r, but in this case it's v0. So normally it's r cosine theta. Now it's v0 cosine, I call it alpha for no good reason. And then I also have v0 sine of alpha. So those are the x and y components. This corresponds to c1 and c2. That allows me to do the best I can to get a better picture of my position vector. So we're going to go back and plug this into my position vector c1 is V naught cosine of alpha times t. We said d1 was 0- gt squared over 2 + c2, which is v naught, sine of alpha, and all of that times t. These formulas may look familiar, if you've studied a little bit of physics or if you've seen projectiles before, this gives you the parametric equations that you might be pretty familiar with. They're usually denoted as x of t or just x equals v naught. Cosine alpha times t and y coordinate or yft is given by an easily switch the order on these things v zero, sine of alpha, all of this times t minus one half gt squared. So hopefully this looks familiar. All right, so we're going to use these two parametric equations of position to answer our question, because really, at the end of the day, it does involve position where I want the distance traveled. So the next observation to make is well the distance traveled when I hit the ground of course, my y component is zero. Let's use that I ran out of room on this slide. So let's go to a new slide. So I want to set my y coordinate to zero. So friendly reminder that y is equal to v naught sine of alpha times t minus one half gt squared, and I set that equal to zero. So I'll put the picture back here for a second. I have my initial velocity, gravity takes over my Cannonball hits the ground. So where it has the ground, the x component is the distance that I want. So really DS like the x component, and of course the y component is zero. And I want to maximize this x component I want to maximize this distance. So y equals 0, so let's stare at this equation. If you notice, I could factor out a t from both pieces and I get v naught sine alpha minus 1 half gt equals 0. And this gives me two solutions where t is 0. And of course we're v the naught sine alpha minus one half, gt zero. This should make sense when t is 0. I kind of haven't started yet. I don't know I'm starting at the origin and of course my y value zero. So this other time and again this is a formula you might recognise when I rearrange the time when the object hits the ground if you solve for t. So do a little algebra and solve you get 2v naught sine alpha over g. This is another formula that might look familiar to you if you study projectile motion, just solve for t in the equation above. So this is the time when the object hits the ground. I take this time from the y component, and I plug it back into the x component. So friendly reminder, x was v0 cosine of alpha times t. So if I substitute my value of t in here, I get 2 v0 sine of alpha over g. I'm going to rearrange a little bit here. So we have v naught squared over g. Remember v naught is the initial speed, whatever the power of my cannon is or how hard I throw, that's fixed that's coming in from the problem. G is 9.8 meters per second squared. It's also constant. So this thing is just some number. And then I have times two, sine alpha, cosine alpha. This is a common double angle identity. I don't know if you remember it or not but you can look it up or trust me on this, but this is a double angle identity that you can rewrite as sin of 2 alpha. So appreciate this for a moment. The x component is completely determined by alpha, where this thing lands is determined by alpha is the only thing you can kind of control the g is fixed and v note is also fixed as well. So where you aim the cannon will determine where this thing lands. So if I want to maximize x, well that means that I have to maximize. The only thing I control sine of two theta, and this is a sine function, we know the sine function. When does it max out? What's its amplitude? That's it one. So I want sine of two theta or sine of two alpha two equal one. When does that happen? You can easily in a circle to see this, or the graph either ones fine. That's when two alpha has to be at pi over two. So for sine t equal one, the angle inside is pi over two, rearrange a little bit and you get alpha equals pi over four. This is in radians and of course, perhaps more common number is 45 degrees. This is amazing, this is beautiful, this took two slides to get. Again, I think we know this intuitively but just to see sort of all the math and all the calculus in play, this makes me happy. I really like going through these calculations. Putting it all together. It's, I think it's beautiful. I think it's amazing. And hopefully you do too. So great job on this one if you follow this before, if you follow all the work here and if you see these formulas before it's really nice just to see where they all come from. And again, things are easy and easy to find with the right tools. Of course, we've developed a lot of calculus to get to here but when you see it all come together like this. I really just hope you appreciate it. I really do. So great job on this video, and we'll see you next time.