In this video, I show how to obtain the curvature of parametric equations that involve integrals, by first utilizing the fundamental theorem of calculus to obtain the derivative of the position vec...
Transcript
the solution to problem four, so curvature from parametric equations, and we're asked find the curvature of the curve with parametric equations, so x component is equal to the integral from zero to t of sine of one half pi theta, theta, and y is equal to the integral from zero to t of cosine, one half pi. theta squared and then d theta and then the solution so where else to find the curvature so recall the equation of curvature and the unit tangent vector from my earlier videos linked to the notes on Hive and yeah so there is there is the setup right here so definition is this so it's equal to the magnitude of the derivative of the unit tangent vector, this capital T, in terms of arc length, a derivative in terms of arc length, where arc length is just this setup here, how long it's going. So basically, and then the unit tangent vector is equal to the derivative of the position vector divided by the magnitude of the position vector, so that the magnitude just becomes one. So if you look at the tangent, it's going to be similar like this, here's tangent like this, and then you're taking derivative with respect this arc length. So as you're moving across this, so as you're moving, you're going to get a unit tangent vector and then it's unit tangent vector, and now we're looking at the rate change of it. So this is going to be a low curvature. This right here is going down and then going up. This is a high curvature or a kappa, so, and this one is low. This one is low, well, so basically small, and then if it's flat, it just goes zero, and so on. And then in my earlier video, I derived it in terms of the parameter T, and that equals to, well, we're just going to look at this one right here. So this is going to be, yeah, it's going to equal to the magnitude of the derivative of the unit tangent vector divided by the magnitude of the position vector in terms of T. So again, and then you could also, this is the one in terms of just R and its derivatives. But for this case, we're just going to use this because we're going to get this equals to one, the magnitude of the derivative, and we'll illustrate that right as we go. So by the fundamental theorem of calculus, what we have, so we're given this, X and Y. But in our case, what we want is, well, we want this right here. We want this derivative of it. So we don't actually want this by itself, and this is in parameter's T. So I'm just going to write this out. So we have the position vector is equal to, so there's going to be this whole integral there. So integral 0t sine pi, I mean, 1 1⁄2 squared over d theta. And then we're going to have over here, 0, t, and then cosine, 1 1.5, theta, squared over d theta, like that, and then this close. All right, yeah, so now this whole setup is the x, and this is going to be the y component. And, yeah, so now the next setup is, well, we just take, yeah, the integral using fundamental theorem of calculus. What we end up having is we take the indigrant. of each one. So there's an integral, it's going to be the derivative of the integral, it's going to be the integrand, and then this is from 0 to t, so we're going to replace theta with t. So this equals 2 by the fundamental theorem of calculus, and again, I'm going to recap that after I finish this problem. So it's going to be now just sign 1 1.5 pi t squared like that, and then this is going to be cosine, 1.5 pi t squared, and then like that. So we close that up. So now that we have this, the derivative of the position vector, the next setup is, well, we need, to get the curvature, we need this derivative of the unit tangent vector. So first of all, let's get the unit tangent vector, is just going to be derivative of the position vector divided by the magnitude of the position vector. So we're going to have to solve for this part right here. So the magnitude is going to be, So we're going to have the magnitude of this derivative of the position vector is equal to using just our basic distance formula in 3D, square root, and then the square of the components. So we're going to have a sine squared, 1 1.5 pi t squared, and then plus the other x, the y component, 1 half pi t squared, like that, so we're going to square root. All of that. All right, and now this is just a Pythagorean identity, so again, we have a sine squared plus co-squared, so this equals to 1. Yeah, so this whole thing is going to be equal to all square root of 1, just going to be plus or minus 1, so we always take the positive, so I also put again, every time that this is a square root, you can have a plus or minus 1, but for real purposes, distance is going to be a magnitude 1, so it's going to be 1, yes, it's just going to be square root of 1, so this whole thing is equal to 1. Like that, so that equals to 1. So what that means is, so what we get is thus, we have the unit tangent vector of T is equal to the position vector derivative divided by the magnitude of the unit, this one right here. And again, that's just this part right here, there's the magnitude of that. Yeah, so then that's going to equal to, well, this equals to, just, this bottom part of part is equal to 1. So put it better like that. Let's just move it down a bit, yeah, so this just equals to just the derivative of the unit tangent vector. So we're going to need this for our curvature, which is Kappa is equal to, this right here, the magnitude of the derivative of the unit tangent vector, divided by the magnitude of the derivative of the position vector, and this one's already 1. So this one's already equals to 1, but this part, we already know it's equal to 1 for our case, so we just need to worry about the top part. So the top part is going to be, and I'll write, and... the magnitude, or first thing is the derivative, and this derivative right here of the unit tangent vector equals to the second derivative of the position vector, and this equals to, let's put this for completeness with D over DT, like this, of, this whole setup there, so sign 1⁄2 pi t squared, and then cosine, 1 1⁄2 pi t squared, all right, and then just box this in like that, all right. So now I just shrunk it and move it in here. So now if we take the derivative of this, this just equals 2. Yeah, so the second derivative of the position vector. So derivative is going to be cosine. I'm going to put it like this, so we'll have a cosine, and we're going to have a t squared, and then using chain rule, we're going to take the derivative of the inside. So this inside is going to be, well, one half pi, uh, one half times pi is constant. The two goes down, then we have a t there. So what we're going to be left with is, yeah, so the two cancels with the two rows, we're left with a pi t. So pi t times it by cosine, uh, one half. And this other one's going to be the exact same thing, but now we have a negative, uh, uh, sign. So we have a negative sign. 1.5 pi t squared like this, and then this, this puts a 2 down using chain rule, get it out of there, so 2 is cancel, we're going to left with pi t. Pi t times that is the derivative, and now we can just factor that whole thing out, so we're going to get a, the derivative of the unit tangent vector is equal to, so pi t, again, this is just, factor that out. and now we're just going to be left with a cosine, 1.5 pi t squared minus, and now this whole setup in there, it's just going to be a negative sign, 1.5 pi t squared, like that. All right, but now we want the magnitude of that too, because number of the derivative, yeah, the magnitude for the curvature, we need to have the magnitude of this derivative of the unit's henture vector, and that is just equals 2, and I'm going to go right here, the magnitude of this, like that is going to equal to, yeah, so this T there, we need to, yeah, that needs to be inside both. Yeah, so just to put that in the components when we sum this all up, we're going to get a pi, T squared, now this one's going to be, well, and then we're also going to factor it out, because we're going to put them in both, so factor that out at the beginning, so we don't need to worry about that. Now we're going to get a cosine squared, one-half pi t squared, and then this other part's going to be well plus, now because we're doing the, squaring it, so the negative becomes positive, so we're going to have a sign. squared, 1-half pi t squared like that, and then, yeah, now it's just a magnetic number, so we don't need the brackets anymore, it's not a vector. All right, so what we end up happening is, well, this is, again, the Pythagran identity, so this equals to 1, it's a co -squared, plus sine squared. Yeah, so this whole thing is equal to square root, and then we're going to have a pi t squared times by 1, like that. So, yeah, so what that means is, well, yeah, that, that's just gone. So I'll just put it out of there. Yeah, here, I just left it out of there. Yeah, I don't know why my calculus book just has the one there, but yeah, they didn't have this other part, anyways, we got rid of the one. So now this one square root of a square, just going to equal to a plus or minus, but we also consider the plus. For this case, because, yeah, we just want a positive number. So we're going to get now, this is going to be just pi t, like that. So this equals to pi. T. Yeah, but actually, since we want the magnitude, we don't want a negative sign here, because this is just a magnitude distance and so on. So we want the positive, the absolute value of T like that. Yeah, just so in case we ever go a T negative and so on. So anyways, use positive value there. So thus, I'll write thus, curvature Kappa is equal to the magnitude of the derivative of the unit tangent vector in terms of T, the magnitude of the derivative of the position vector in terms of T, which this part, yeah, this just equals to now pi, yeah, pi t over 1, so this equals to pi, and then the magnitude, yes, we want to always positive, so that pi t, I got. Yeah, pi times the magnitude, the absolute value of T. And yeah, so that's the curvature. And now, just to quickly recall my earlier videos on part one and part two of the fundamental theorem of calculus, and that's for the derivative we had here, this part right here, derivative, well, that's this integral, I mean the integral, and the derivative is just take the integrant and put in T there. So again, that's from the fundamental theorem of calculus. So this says if, yeah, this is part one, if F is continuous on the interval A and B, then we're going to get integral. So G of X, if G of X is equal to integral from A to X of F of X, then a derivative, is this equal to, uh, yeah, derivative of X is just equal to FX. Yeah, so this is just going to be FX. We take the integrand and then replace the T with X. There. And then a fundamental theorem part two, that's for the actual opposite values there. So that's when you plug these in to find the difference in there. So when you're actually finding it. This one is just the indefinite integral there. Yeah, this is just up to point X. Now we're doing a point A and B, so we're going to take the difference. And so on, but again, the derivative of this, yeah, the derivative of the capital F, and again, F is any antiderivative of F. It just equals to F here. So that's the integral there. All right, and then here's the, in a summary version of this, fundamental theorem of calculus, suppose F is continuous on A and B, closed interval, so that includes A and B, so if G of X equals to integral from A to DT, DT, then derivative G prime of X equals to F, and so, and then there's for definite integrals to actually solve them, evaluate them out, same thing, capital of just entity derivative of F, and where the derivative of this F is equal to, just this part here, anyways. Yeah, that's pretty much all calculus is, just connecting integrals and, and derivatives. Just fascinating stuff.