▶️ Watch on 3Speak - YouTube - Telegram - Summary - Notes - Sections playlist - Vector Functions playlist - MES Links --- In this video, I show that we can obtain the velocity of a moving object by...
Transcript
integrating acceleration and velocity vectors, and we'll illustrate this with an example. So the vector integrals that were introduced in my earlier video, there's a link to that, can be used to find position vectors, one velocity, or acceleration vectors are known as in the next example. And again, here's the recap. This is a thumbnail of that video, and then there is a thumbnail of the section in this video here, although all these will be in the link in over here. And so basically, Yes, to recap, so just recall the derivative of the position vector, you could use this definition of derivative, and you could also do it by components. So the derivative of this equals derivative of each, X, Y, Z, or I, J, K components, like that, and same thing with integral. So the integral is going to be the same thing, integral of the position vector is equal to the integral of each component. So there's the I, there's the J, there's the K, yeah, this I, J, K. and then that equals to this one or here, capital R of T, that's the anti -derivative, where anti-derivative is equal to just RFT. So that's just what the anti-derivative is, and then you evaluate it just like regular integrals. And yeah, here's the thumbnail for the other one, and again, this is the same setup as normal integrals, and there's the, you would sum up into infinite rectangles like this, and yeah, and then you do that for each component. So you do some of infinite rectangles for the i component, the j, the k, and then basically as the limit, as that n approaches to infinity, as the amount of rectangles and so on, you get the integral. And again, there's this whole setup here. Anyway, so let's illustrate this with example three, and that's integrating to get velocity and position vectors. So a moving particle starts at an initial position of R right here. R of 0 equals to 1-0 with initial velocity. And so with an initial velocity of v vector of 0 is equal to i, negative j plus k, again, these are standard basis vectors. Again, in a coordinate format, or this would be like this, 1, negative 1, 1, yes, that's what it is. Anyways, just keep it like that. It's pretty straightforward. Alright, so in its acceleration, it's given by AFT equals to 4TI, plus 6TJ plus K, and now we're asked, find its velocity and position at time T. All right, so let's take a look at the solution. So we were given the acceleration, and now we're going to move backwards. We're going to go get the velocity and the previous examples. Notice what we had, we were given the position, and then we take the derivative to get the velocity, and then derivative again to get the acceleration, but now we're going to reverse way. Now we're going to go acceleration to velocity, then velocity to position, and that's using integration. Alright, so let's take a look at the integrals of that. So let's say we're given. So since we know that A of T, that acceleration is equal to the derivative of the velocity, prime of T, then, yeah, so we have that, so then, this is going to be V, yeah, V of T, the velocity is equal to the integral. of the acceleration. Yeah, because we have this derivative to take the integral to get the regular velocity. So we're going to get, now this is going to be the integral of, and then we have the acceleration, it's going to be 4. I'll put a bracket like this, 4T, I should put a normal bracket, not a square bracket, so 4TI plus 6TJ plus K, like that, and this is going to be now, this is going to be DT, like that. All right, here, just fix that up. So now take the integral of that, this is going to equals to all of each of these components. We're going to get, now this is going to be a integral of 4T, that's what we, t squared, and then move a t divided by 2 there, so it's going to be 4 2. Because if you take the derivative, put the 2 down, it would be 4, then T, power 1. So 2, ti, like that, plus, And now we have this one, t squared, then divided by two. So this is going to be, yeah, let's delete that, this is going to be t squared divided by two, so six is going to be three. Yeah, six divided by two like that. The next one is going to be below, plus just t, like that. T, and this is going to be, yeah, this is, we forgot the J, J component, and then plus T. K component, and now we have to have an integration vector constant. We're going to call this vector C, because this is the, if you just have a vector like this, and then the derivative of, that's going to be zero. So you have to, so you have to have that a constant like that. All right, and now we're going further, so to determine the value of the constant vector C, we plug in our given value for the initial velocities of V, you have V vector 0. So this one right here. And we know that the initial value we're given is, Initial velocity at 0, t equals 0 is equal to i, negative j, and then plus k. So let's write this out. So we're going to have, we're going to have V of 0 equals 2, let's write this out, i negative j k. i vector, negative j vector plus k, vector like that. Now this equals 2, and now we just plug in t equals 0 for all this. and then, so we're going to have a, this is going to be a zero, so if a zero, let's put a zero like that, plus there's another zero, plus another zero, so yeah, plus C. So in other words, yeah, so in other words, C, the C constant, your vector constant of integration is equal to just the initial velocity. Fee of zero, and this just equals to I, negative j, plus k, like that. Yeah, so thus, our velocity right here, we just plug in the value for K there, so thus, now put it better thus. Thus, V, yes, so that's what we have is a V of T is equal to, to plug these in, so 2T squared I plus 3T squared j, and then plus t k and then plus this is going to be well i negative j plus k and these are just a component so we just add these up accordingly yes we're going to add up the k components here with this oh not that one with this one with this so we're going to get a this equals to well put a bracket two t squared plus one so just factor out the i Like that, this is going to be i component, and now the likewise here, this is going to be plus 3 t squared minus the j1, so minus 1, to factor out that, and this is going to be the j component, and then lastly, it's going to be t plus 1. So it's going to be t plus 1, okay, like that. So yeah, that's what we have, let's bold this whole thing, and for our velocity. So that is our velocity vector. Yeah, so now the next setup is, now we just do the next one. So since, to get the position vector, since R of T is equal to, actually not R of T, the other way around, since the velocity of T is equal to the derivative of the position vector, yeah, so since we have that, so again, pretty much the same things before, likewise, the R of the position vector, It's going to equal to the integral of v of t. DT, like that. So, yeah, make sure to put a DT there. Let's see if I put a DT over here earlier, yeah, this one, A, DT. Here, here, I just fix that up, yes, added the DT there, all right. So again, I'll just leave it like that. So now this is going to be now, yeah, this is going to equal to, well, integral now of this whole setup there. So we're going to have a, I'll put a square bracket, yeah, square bracket, now we have 2T square plus 1, and then this is going to be, yeah, it's going to be the I component, and then this is going to be plus. 3T squared minus 1, this is the J component, and then plus 2T plus 1 K component, and then put this whole thing bracket, and this DT now. Like that, all right, so now let's just evaluate this, the integral of each component, so this 2T squared, well, that's going to be t cubed, then we divide by 3, well, it's going to be 2, 2, 3. And now the next one's going to be, the integral of 1 is just T, like that, so that's what we have, this is going to be, I'll make this less dark, 1. Yeah, I mean, plus T, I'll just fix everything up, so this is going to be I. Now the next one's going to be here. This is going to be, well, t cubed, and then divided by two. So it's going to be three, I mean, divided by three, yeah, so three over three is just going to be cancelled, so we just leave a t cubed. Like that, yeah. And now the next one's will be well minus t, like that, this is j. And now plus, next one's going to be well, t squared over two, like that. 2 squared over 2, and then plus t. And this is the K component. Yes, that's what we have. And actually, we need to include the integration concept. So I'll put it plus D vector, because we already took C before. And I'll just move this over. I'll move the whole thing over, actually. This whole thing, I'm going to shift it over just to fit my little box here. Because I have this line here, just to make it nicely here, this line. All right, so that's what we have. All right, so likewise, using our given value for R of zeros or just the initial position at T-go-0 to solve for the integration constant vector, we get, and again, that's if you scroll back up here, we're given this, so the initial position is 1 -0, all right, so let's put that in. So, this is going to be R of 0 at T-0 equals to 1-0. Yeah, now this just equals to, in our iJK format, this equals to i, and then all the j and k is zero, and now this equals to now plug in zero into all this, so these are all t, so those have become zero there, plus another is zero, this t-cube minus t, uh t, plug in zero for everything, this one is zero, and then plus, well, d vector. So again, notice what happens, d, is equal to the D vector is equal to, well, just the initial position vector, R of 0, which equals to I, like that. Yeah, so then we just add I to this one here. So all we do is add of one there. So then, so thus, the position vector, R of T, is equal to, well, the whole thing, I'm just going to move my mouse here, so I'm going to put it like this, 2 over 3, T cubed plus T, and then plus the 1, the I. So that's all, plus T plus 1, so that's only difference that we have there. And then plus, and then T cubed minus 1, I mean, minus T, plus, and then last one, T squared over 2, plus T, and this is going to be our K. Yeah, yeah, so that's pretty much it, we could box this whole thing in. Make a better box. Perfect. All right, going further. So the expression here for this position vector, R of T, that we obtained in example 3, was used to plot the path of the particle in the figure below for, yeah, for zero, yes, so zero is less than T is less than equal to 3. So we're T between 0 and 3. like that. So here's the figure from my calculus book, and I'm going to graph with the GOG, but before we get to that, so we have, we're given the initial point 100, and now this last point here, so that looks, this looks like, it's like 8, I think, so we'll just look at this point right here, at this point, so at t equals 3, so at t equals the 3, let's just look at our actual position vector, that's this part right here, and what we get is, well, we're gonna get a, Yeah, so at this, we're going to get a R of T. Now, R of 3 is equal 2, well, this is going to be 2 over 3, and then this is going to be 3 cubed, that's 27. Because this is just 3 times 3 times 3 is, I mean, 3 times, yeah, this is going to be 3 times 3 times 3. So this is going to be 9 times 3 is 27. So 3 times 3 is 27. So 3 times 3 is 9 times it by 3 is 27. All right, so it's going to be plus 3 plus 1. And this is going to be our i component. All right, for the j component, that's just going to be plus, and then it's going to be, well, t cubed minus t, t, that's going to be 3 cubed, minus 3, and this is going to be our j component. Then lastly, the k component is going to be, well, 3 squared, divided by 2, and then plus 3. This is our K component like that. Yeah, so let's just evaluate this. So this is going to be, well, the top part's going to be, this is going to, we can cancel one of these out, so cancel the 3, with, this becomes a 2, so in other words, 2 times 3 to power of 2, that's 9. So 2 times 9 is 18, so 18 plus 3 plus 1, so in other words, 18 plus 4 is 22. So 22I. And again, notice here, 22, there's a 22 there. Yeah, so it goes to X on the 22 side. And then the next setup is, yes, so the next one is going to be the J component. 3 cubed is 27 minus 3 is 24, then, 27 minus 3 is 24. Yeah, J. So we have, this is 22, and this is, yeah, 24, so 24, and 22 on the other side, so the Y is 22. So that is that. So the next one should be around 8 -ish or so. So it's going to be now. plus 3 squared is going to be 9 divided by 2 is going to be 4 .5 plus 3 is 7.5. All right, so it was, it was close to 8. And you can see right here, it goes to 7, and then there's a 0.5, 7.5. Yeah, so that is the height of that. So right here. So there it is. That is that part here. It's fascinating stuff. Yeah, let's leave it out. So I just wanted to put that at t equals 3 there. All right, and now let's also graph this in the Geoguber 3D Graphen calculator, just so we have the equation there, we can look around it and view it in 3D and so on, and again, it's pretty cool to work with it. And again, here is the setup here, so I have the variable A equals 3, just so I could move along the curve. So here's the curve R equals, and this one here, I can vary this A equals 3, I can move it around, back and forth, and so on. and now we have the curve, this is going to be 2 over 3 t cubed plus t plus 1, and then t cubed minus t, and then this is 1, 1 half t squared plus t. And again, that's just this whole setup there. And then I have the T from 0 to 100, and so on. So now we have the starting point A. I just put A equals to R of A. Well, it's not starting point, it's just a point on it that I'm going to take the vector from. And initially, this is the A, this is that T goes 3. And again, notice that T equals 3, what we have is, well, 22, 24, 7.5. So we have this 22, 24, 7.5, so exactly what we just solved there. That's for this A, where A is 3. Yeah, so now what we have is velocity. I have the velocity vector, starting from A, and then goes to A plus, now we have the velocity. vector there, so 2A squared plus 1, and 3a squared minus 1, and a plus 1, a where a is just our T, but we're just manipulating the A. So this is going to be, again, that's just the velocity one, going back up here. So 2T squared plus 1, 3T squared minus 1, and T plus 1. We're just replacing T with A. All right, and now the last one is acceleration, that's just equals to, again, vector from A to A to A plus, 4A, 6A, 1. and that's right here, going with this velocity, I mean, the acceleration, we're given this. Acceleration is this, 4T6T1, in other words, 4A, 6A, and then 1. Yeah, so that's what we have, and then there's a graph like that, so you can see, this is, it looks exactly like that setup we had over here, so it looks exactly like that, and let's just play around with it, and it's already opened, I'm going to go over here, so here it is, I'm not actually save this, As I used it, yeah, so anyways, I had to fix some stuff up. Anyways, so that's how it is. And you can see this, now if you go to R, A, A, 0, I go to 0, I'll move this, I don't like this thing, make it 0. All right, so it starts up at 0, enter. Yeah, so, it's like, I made a 0, it doesn't do it. Okay, so 0, yes, so 0 is the beginning, and notice at 0 is 1-0. So again, that's exactly what we had. what we should have, and the notice that the velocities change. I mean, the velocity and acceleration vectors, again, absolutely amazing, and it goes follows all the way along, go to the three, and then we have the values cut, about 22, 24, 7.5, and zoom out, and there it is, yeah, absolutely amazing stuff, so again, play around with this, and yeah, that's what I'll actually... I like doing this because you literally see, and you could validate that your equations are right because you can literally see a graph in 3D, like that's pretty amazing stuff there. I'm going to close this and leave. All right, epic stuff. And yeah, so to summarize all this, well, in general, vector integrals allow us to recover velocity when acceleration is known, in position when velocity is known, and we could write this out as, yeah, we could write this as the velocity, so if we have accelerations, velocity, VFT is equal to, and again, remember how our acceleration constant for velocity, this part right here, the C, the C just equals to the initial velocity, V of 0. And then likewise for the position vector, the integration constant was just equals to the initial position there. Yeah, this initial position at t equals zero, because all these vanish. So typically, that's what the case we have. So in general, this is what we usually have, and this is going to be equals to v. at T0, or the starting position. So, I'll put that instead of T equals 0, sometimes you start off at different positions. So T not plus, yeah, so at a specific point there, and then this is going to be from T0 to T of A, and the acceleration, we'll use a different variable U for the DU, just so we can, for the integration, just change the variable there. And then for the position vector, This equals to when the velocity is known, this is going to be equals to the initial position at T-0, because we're going to be taking integral from T-0 to that. So typically it's T-quil 0, but not always. So we're going to go T-0 to T, and now we're going to have the V-E-V-V-V-U, just so we have a different variable. So we have the T there, we're going to plug the T inside. So yes, absolutely fascinating stuff there.