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Reparametrize the curve

$$\mathbf{r}(t)=\left(\frac{2}{t^{2}+1}-1\right) \mathbf{i}+\frac{2 t}{t^{2}+1} \mathbf{j}$$

with respect to arc length measured from the point $(1,0)$ in

the direction of increasing $t$ . Express the reparametrization

in its simplest form. What can you conclude about the

curve?

$$\vec{r}(t(s))=\cos (s) \vec{i}+\sin (s) \vec{j}$$ We can conclude that the curve is a circle.

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Johns Hopkins University

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all right. In this problem, we want to re parameter rise The curve are of tea with respect to our Klink from the 0.1 comma zero, um, in the direction of tea increasing. So the first thing that we, um what did you is? Find the, um, find the derivative of this vector. So we're gonna look at our private T I'm actually what's let's go ahead and express this first in the other vector notation because I think it's a little bit easier to look at and Teoh simplify things for when we go to actually take the derivative. I'm gonna go one step further. And great things is a product instead of quotient. We're gonna write things like this just to make things a little bit easier. Look at work with All right there, You know, is that our FT everyone I find our prime of tea. All right, So what we will have is negative, too. Times to t times t squared, plus one to the power of negative two. That's just using our, um, our chain rule there. And then for our second term, we need to use our product rule. What we're gonna have is two times t squared plus one to the negative First power minus to t times to t times t squared plus one to the power of negative, too. That's just applying our product rules on. And then our chain will to take take the derivative of those two component functions and we can go ahead and simplify that a little bit further. Um, because ultimately we want toe take the magnitude of this vector, and that's gonna be easier to do. The simpler or two component functions are So this 1st 1 will simplify too negative for tea, uh, times T squared plus one to the power of negative, too. And our second component function, the first term will stay the same, and our second term will become 40 squared times t squared plus one to the power of negative too. All right. And we have our premise t we want to find the magnitude of that. So we're gonna look at the magnitude of our prime of tea, which is going to be the square root of the some of our component functions squared. So our 1st 1 um, the component function is negative. Four t. I'm gonna go ahead and put that, um, t squared, plus one quantity squared in the denominator. We're gonna have that component functions squared. They were going to be adding two over T squared plus one minus 40 squared over T squared plus one squared. And that difference is also going to be squared and who want to simplify this a little bit. So that way, we can, um, eventually integrate this function that we get. So I'm gonna start off by squaring both of the, uh, expanding both of the squares that we have. So, our first term in the numerator we have negative four t quantity square or 16 T squared. And in the denominator, we have t squared plus one to the fourth power and then expanding This, uh, the second term out. We're gonna have four over t squared, plus one squared, and then we're gonna have negative 16 looks. I lost my pen for a moment. There. Negative 16 t squared over t squared, plus one to the third power. So that comes from multiplying our first and second term. Um, the two over T squared plus one and the negative 40 squared over t squared, plus one squared, we multiply those, and that's gonna happen twice. So we multiply that result by two. So we end up getting negative 16. He squared over t squared plus one cubed. And then we're gonna add this, uh, second term Squares. It'll be 16 t to the fourth T squared, plus one over T squared, plus one to the fourth. And then we want to be able to condensed this a bit. So what I'm gonna do is get all of my get a common denominator of t squared plus one to the fourth power. So it's still gonna be under the square root sign. Our first term is gonna be okay. This is all going to be over. He squared plus one to the fourth. So I just need to multiply. The numerator is by the proper proper value or proper expression. So our second term, the four over T squared plus one squared. I need to multiply that four by a T squared plus one squared in order for the denominator to become, uh, the same as the first term for our second er For third term, the 16 T squared gets multiplied by t squared, plus one. And then our last term can be left a 16 t to the fourth, because the denominator is already t squared, plus one to the fourth power. All right, we can go ahead and, um, expand all of the things that are numerator. So that way we're able to add are like terms. So we have 16 t squared. Plus, let's distribute that four. Will, it's expand. Um, the T squared plus one squared and multiply it by four. We end up getting four times t to the fourth plus eight times t squared plus four. Then we have expanded their distributing the negative 16 t squared. We end up getting negative 16 t to the fourth, minus 16 t squared, then plus 16 t to the fourth at the end. That's all gonna be over t squared, plus one to the fourth power. All right, we're slowly getting there. Ah, So what we can see is that our next error 16 t squared of negative 16 t squared, add up to be zero, and then our 16 t to the fourth and negative 16 t to the fourth. Also add up to be zero. So what? We get is four t to the fourth power plus eight t to the second power plus four over t squared plus one to the fourth power, which, if we want we can separator numerator and denominator Taking the square root of the denominator gives us t squared plus one squared. And then our numerator we actually see that we've got, um, see, you can factor four out and then we're left with t squared plus one squared, which we can take the square root of that and we'll get two times t squared, plus one over t squared plus one squared, which, seeing the common terminar numerator and denominator ends up giving us too over t squared plus one. Okay, so we found the magnitude of the derivative of the vector are. And now what we need to do with that is determined what the arc length is. So we're gonna have s or s of tea, right? Cause our arc length of the function of the parameter t, and we need to figure out what our bounds of integration are going to be. So if we looks, let's get back in a full screen here. So if we take a look at what were originally given were given this 0.1 common zeros. We need to figure out what value of tea corresponds to that. And we can do so by just picking one of our two one of our two component functions. Andi setting a equal Teoh the correct coordinates. So, for example, our first component function represents the x coordinate. So if you look at two over t squared plus one minus one, we know from the point that were given that that is equal to one. So now what I can do is solve this expression for tea or the equation for tea. So on the left hand side, I'll keep my two over t squared plus one, and then add one to both sides getting to, um, if I multiply both sides by t squared plus one and divide both sides by two. I end up getting one equals t squared plus one, which you can simplify that or sulfur t Here, subtract one from both sides. We get t squared equals zero, which taking the square root of that just yields t equals zero. So we know that we're starting at this value of the parameter teeth. So that means that when we set up our integral to calculate arc length, we're going from zero. And we were told that tease increasing. So we're gonna go from zero to t and we're going to integrate the magnitude of the derivative of that original vector were given. So we integrate from zero to t integrating T two times t squared plus one to the negative one Power. Oh, I'm gonna use my dummy variable you here. So that way we don't get it confused with tea, and this integral ends up being okay two times inverse tangent of you. It goes from zero to t using fundamental theorem of calculus. We end up getting two times inverse tangent of tea and, uh, that minus two times inverse tangent of zero, which this term ends up being second term ends up being zero. So we just end up with two inverse tangent T. And we're trying Teoh Express the, uh, the repairman authorization of the curve in terms of our Klink. So what I want to do is be able to plug in ah, value that's equal to t. That is in terms of s So I take this resulting equation here, and what I'm gonna do is solve it for TB. So if I divide both sides by to get s ever two equals inverse tangent of T, and then I can take tangent of both sides. So what I'm gonna get is t equals tangent of s over two. All right, so we know what quantity we need to use to repair amateur eyes curve. So let's let's start there. We're gonna have to do some simplification work because our problems says to express it in the simplest form possible, and that's gonna be able to tell us something about the curve. So if we look at the the vector that we were originally given, we have to over t squared plus one minus one, and to tea over a T squared plus one. So we know here that are X component is to over t squared plus one minus one. And if I plug the, uh, this equation that we got here in terms of s, what we're gonna get is two over tangent squared s ever two plus one minus one. Now, that is definitely not the simplest form in which we can write this. So what I'm gonna do is use some of the Trigana metric identities that you might want to review. Teoh, go ahead and simplify this down even further. It turns out it's gonna actually get quite concise once we've worked with it a little bit. So the first thing that I'm going to do is use the half angle formula. Teoh change. Put the denominator in terms of CO sign on and reduce that, uh, reduce the power of the trick function there and also put it in trig functions in terms of s instead of s over two. We're going to be a little bit easier to do with. So we're gonna have in the denominator for that tangent squared. Um, of s over two is gonna have one minus co sign of s over one plus co sign of ass. And then we carry down the rest of our function here. So that was applying the half angle formula to this component here. And now that we've got that, we can reduce er complex fraction here. So, in the denominator, if I want to actually be able to add that plus one, I'm gonna put it in terms of one plus co sign s over one plus co sign s So we end up with is the following. So all I did there was multiply this one term by one plus co sign over at one plus co sign of s over one plus co sign of s that we could get that common denominator. And what we see is that our negative co sign of s and plus co sign of s add up to be zero simplifying that quite a bit. So this is going to turn into to times one plus co sign of s over two minus one. Our twos are going Teoh, simplify a little bit and we're gonna be left with one plus co sign of s minus one are ones air going toe Also add up to be zero and we're just left with co sign of s. So we were able to reduce the X component function significantly down to a more concise expression. So I want to keep that in mind that's gonna be needed for when we write the final expression of our curve. Now we also need Teoh figure out the second component function so we were given that we, uh the y component function is to t over t squared plus one again, we're going, Teoh use this t equals tangent s over to you to plug into that function. So we end up with to times tangent of s over to over Tangent Squared s over to plus one and we're going Teoh, simplify. This is well using some of our trig identities. Now, keep in mind, there may be several different ways of doing this. Um, but I I'll show you the one that I choose. So first thing I'm gonna do is similarly to the X component function. I'm gonna use my half angle formula, Teoh Deal with the denominator. And what that's gonna give me is one minus co sign of s over one plus co sign of s. And similarly, I'm gonna turn this plus one term into one plus co sign of s over the same denominator. What that's gonna give me is two times tangent of us over to our negative co sign of s and plus co Sign of s are going to, uh, going to add up to be zero. So we're left with two over one plus co sign of s which ends up being Tangent s over two times one plus co sign of s. All right, We're starting to get there, but we're not quite there yet. So what I'm gonna do now is use the half angle formula for the tangent s ever to term. So using the half angle formula again, or one of the half angle formulas more specifically so the that tangent term is going. Teoh, change to sign of s Over to over. Co sign s over to That's all Times one plus co sign of s and I am going Teoh. Let's see. Ah, that was not the half angle for their so tangent of acidity That just is equal. Teoh sign of best silver two overcoats song invested s ever to. So now we're using the half angle formula to do the next step. My apologies for that. So our sign of s over to you never co sign s over two eyes going to stay the same. What we're doing is we're using the half angle formula to deal with one plus co sign of s terms and what we get again. A current uto review your trig identities before tackling this problem. Or at least, um, if you're unfamiliar with what some of these trig identities are. But we end up getting to co sign squared of s ever to for that that underlying term in the previous line. And we noticed that we've got co sign s ever to now, and our numerator and denominator so ends up simplifying to two times sign X over to times co sign s over two. All right, we are nearly done. Uh, what we can look at is the product to some formula. Um and I'm gonna go ahead and write that one off to the side because it's a little bit less straightforward than some of the others. So the product is some formula. One of them can be written as co sign of an angle you times or sign of an angle. You Times Co sign of an angle V, which is 1/2 time sign of U plus V plus sign of U plus V. We're sorry U minus fee, and we see that we have something in terms of two times. Sign oven angle times CO sign of angle. So if I devise Ah, multiply both sides by two we're gonna get it is two times sign of you Times co sign a V is equal to sign of U plus V plus sign of U minus fee. Um, so if I consider u and V to be the same, we see that this expression is going to become sign of s over two plus s over to plus sign of s ever to minus s over to this first term becomes sign of s, the second term is gonna become sign of zero, which is just zero. So we now have the simplified perversion of So this is the simplest way that we can write X and Y, um, we have that X X component can be written in terms of arc length as co sign of s and the, uh, why component can be written in terms of arc length. A sign of s. So what we're left with is a re premature ization of our curve are of tea of s as our X component function co sign of s times, vector I plus sign of s times, Vector J. And that is going to be the repair amateur ization of this curve that we were given in terms of our F t now written in terms of the arc length s. Now the problem also asks what we can tell about this curve. And one of the things that we can look at is, since this is two dimensional, it's not too bad to graph. Um, so let's just draw simple coordinate playing out here and let this be one one negative 11 So let's say that I have t equals zero. Actually, I'm gonna make a little table for this. So t x Why? So let's say that we have the Value T equals zero Rx component function is gonna be co sign of zero, which is going to be one. And our why component function is going to be sign of zero, which is gonna be zero. So that means that for t of zero, our first point is one comma zero. So we have a point here on our curve because we're looking at sign and co sign. Let's look it, um, increments of pi over to So say that I have pi over two for my value of tea, my parameter, that is gonna give me co sign Pi over two, which is zero for my X component and sign of pi over two for my y component, which is one. So that means that zero comma one is going to be our point there. We can continue this pattern, right? So if I plug in pie for a parameter t, we get negative one and zero for X and Y components. And if I plug three pi over four in for my parameter end up getting zero and negative one, so have a point here. And as soon as we go through, um, go through the interval from 0 to 2 pi r values are going to start cycling back around. So what we can see is that our curve is actually going to be a circle so we can conclude that the curve represented by R T or the vector are is a circle