match-the-parametric-equations-with-the-graphs-labeled-i-vi-give-reasons-for-your-choices-do-not-use

Question

Match the parametric equations with the graphs labeled I-VI. Give reasons for your choices. (Do not use a graphing device.)

(a) $ x = t^4 - t + 1 $, $ \; y = t^2 $
(b) $ x = t^2 - 2t $, $ \; y = \sqrt{t} $
(c) $ x = \sin 2t $, $ \; y = \sin(t + \sin 2t) $
(d) $ x = \cos 5t $, $ \; y = \sin 2t $
(e) $ x = t + \sin 4t $, $ y = t^2 + \cos 3t $
(f) $ x = \dfrac{\sin 2t}{4 + t^2} $, $ \; y = \dfrac{\cos 2t}{4 + t^2} $

Aparna Shakti · Accepted Answer

So the problem says mass. The parametric equations with the graphs labeled 1-6. Give reasons for the choice is do not use a graphic device. So we are not using graphing device but we know the method for one thing that we can put the equals to zero or infinity. Then we&#x27;ll see what will be the value of X and Y. Right. So here are few or sanitary equations. Right? So we will see here that what all military equations are there. So this is a solitary right? This is our military. You can say, Okay, I&#x27;m sorry. This is 4th 1. Right? So this is arson. It read This is our military. And we can say this is also a solitary. Right? So this is how we find out. Right? So yeah, this is also a signatory. Like so what what can we do now? This is for sure that the part A and B does not contain any sinus or it&#x27;ll wave or sign costs. I know towns. Right? So it will not have that value. Right? The 2nd, 3rd, 4th and 6th values excluded from it. Right. So now we will solve simply that we can just observe that what is happening in this problem? Yeah. Right. So here we can see this equation parametric equation. Right? Uh, as we put the equals to infinity. Right? So far X and y goes to infinity. Right? If we put equals to infinity here. So X will be infinity and I will be also infinity. So The graph must contain 10 value. Anyone raped. So no. Which is only possible. Which is only possible in cyst because here we can see if we&#x27;ve seen minute lee This is 10 value rate. This is 10. You can see here Here there is 10 coordinated. Right. Can we say this? So the best matches first graph or you can see graph fine. All right. So now moving forward, we can see here X equals two T square minus duty. These this equation is given to us. This is not a solitary. We have already discussed that one and five are not oscillate story. Right? So, we can see here even we can put uh t question infinity as I have told. So, yeah, the equestria infinity. So, this is for sure. After that after conclusion from above. We can also directly right that the answer is one. But we&#x27;ll see how. So, the equation passes through 00. When we put the equals two in the media. All right. So, this will be 00. Right. So, we are getting the origin. So, best matches. I was thinking bro. What? This is very easy. Yes. No. Now we are moving forward for part C. We can see a people Tequesta by beto then the point should be then point point should be 11 is attained. Right? But it should be it can be minus five to also because the values of uh sine wave varies from minus five to plus five. Two. Yes. So here there could be -1 -1. There would be too cruel conditions. Right so, so here we can see the symmetry here that this is the second graf, right. Second graph if you see keenly the coordinates are 11 year. Right, and here it is, you can see minus one. Oh I&#x27;m sorry, this is minus one plus one and this is minus one minus one. We can see here. So that&#x27;s right option. You can see Graph two is correct. So best matches here, mhm, mhm Drop of doom here, thank you. Now moving forward, we are given with next to equations for X and Y in terms of T. So here We will put equals to zero right at the point when when the equals to zero, then the point should be 10 should be attending. So how could be possible? How could it be possible? It would be possible for graph six white. Because I just see you. Oh 10. It is telling. Right, So here only in draft six, we can see Right here you can see right, this is one zero point. Right, so 10 point is attained in glass sixth. So best matches mhm, graph six, Okay, now moving forward, mm here we are given with value of, you can say the equation for X and Y in terms of T again. So if we put the T infinity. Right, so X and y will also go to infinity but the difference is what? But the difference is one only that it is possible. A Tory in behavior. So let us see you. Okay, so let us compare for E and F. Part. So you Here we&#x27;ll see 00 cannot be the point in the draft for the Efta part. Right? So whatever is the graph for 00, let us see which one is that? Okay? Mhm. So four? Yeah. Okay. So option you can see Graph four is the answer for this? Best match. Right? Best matches Deborah. Mhm. Now, for the last part we are given with two equations X and Y. Here. Now 00 cannot be the uh tradition. Right? Because here&#x27;s the scientist is given and this is called given is given. So if one is zero then another should be something is that should be a constant. Right? So it would be constant, not one year rate. So how in which condition it would be possible? So yes, there is no symmetric kind of things that zero and one only exists. So the value is burying in part today. Right. You can see in Graph three, that&#x27;s why in part we have taken graph four, but not graph three. So best matches here. Rough. Yeah, I hope you understand the concept and this is very easy problem. Thank you for watching. Yeah,

Jeffrey Russell · Answer

we&#x27;re going to match the equations on the right hand side of the screen to the grafts on the left hand side of the screen. We&#x27;ll start with X equals T to the fourth minus T plus one. And why is equal t squared? Why can never be negative so we can rule out any graphs that have a negative Why? So we&#x27;re left with these three graphs. Additionally, it can&#x27;t be this bottom left draft because X and Y are both full of no meals, meaning we wouldn&#x27;t get this sort of squiggly behavior like this. And we especially wouldn&#x27;t have a loop near the origin. So that leaves of these two graphs, however, noticed. The top left graph has an endpoint implying that some for some values of T either X or Y or not defined. But that&#x27;s not true of these two equations. So the only graph which in courts one of these equations. It&#x27;s a spot in the middle one, so I&#x27;ll draw line pointing at the graph. Now consider X is equal to T squared minus duty, and wise equals the square root of team again. Why can&#x27;t be negative? So we&#x27;ll rule out these three graphs with a negative y and again, these Air Poland on wheels. So we can&#x27;t have this sort of squiggly behavior with the loop in this hour. So that only leaves this graph and also notice, even if we haven&#x27;t already taken this graph since T will not give the find values for why, If he is negative, we couldn&#x27;t have it be this graph. Anyway, this graph matches up. So let&#x27;s point at this graph. Now we consider X equal sign of duty and why equals sign of T plus sign of duty. So first off, this is bounded between negative one on one and this bottom left graph is unbounded, so it can&#x27;t be this bottom left graph moving on. We also can&#x27;t have it be the top right graph because this just continuously lose inwards towards the centre. The X and Y values both get smaller and smaller, and these equations aren&#x27;t going toe have that sort of behavior. So it&#x27;s down to these two raps notice that why is equal to sign of T plus X. That means that when t is small, why is equal to sign of X and when word small values of X sign of X is equal to X So at small values of T, where X is also small just by definition of acts here, why is going to be about equal to X? This graph This top middle graph displays that behavior, but this bottom middle graph doesn&#x27;t so it can only be the top middle graph moving on. We have X is equal to co sign a five T and Y is equal sign of two T. So again, it&#x27;s not going to be the bottom left graph because the bottom left graph is unbounded and X and wire bounded by Negative Woman one. So what&#x27;s going to be one of these two? But we can&#x27;t have them both getting infinitely small together like we see here. And additionally, this bottom right graph, uh, because its co sign of five t and sign of two t the periods of X and Y are going to take a while to align, and we see that behavior in this bottom right graph. So this set of equations course once upon right graph now we&#x27;ll look at X is equal to t plus sign of 14 and why is equal to t squared plus co sign of three t. So this is unbounded because asked, he gets larger, the magnitude of X and Y are going to get larger. And why is this going to get very positive? So it can&#x27;t be this top right graphic assistant founded the bottom left graph matches and that also explained the squiggles here and the loop. It&#x27;s because they&#x27;re defined partially by triggered a metric functions. So the set of equations force ones that left and clearly by process of elimination, all that salt is a top right graph. But looking closer asked, he gets larger and larger. In these two equations, X is equal a sign of two t over four plus t Square. And why is he gonna co sign up to 3/4 60 Square X and Y get smaller and smaller Ste goes to infinity and additionally instance, we have signed a two team coastline of two t. We&#x27;re going to have a looping like in this top right graph, as we see here. So these two equations match the top right graph and we&#x27;ve matched all the equations to their corresponding graph

Wen Zheng · Answer

the problem is match the Parametric Equations sect Wells. He reasons for your choices. Your axe is equal to a scientist square. Why&#x27;s the court scientist squire say? Is it t So we have X plus y is it would want he is from make to infinity. To infinity, Andi is a projection of security. Um, Axe. A plane is actually what you call sign Teo Squire. And how Why&#x27;s it playing? That&#x27;s why I called you Fine, please. Now let&#x27;s look happens of roofs on the number two and three, the candy. Any members from neck to reinvent into infinity and then compiled. It&#x27;s a projection punk. I see playing on. Why&#x27;s it fine? Because they only number three thirty five. The conditions says graph. I number the ring.

Wen Zheng · Answer

The problem is much. The parametric equations with graphs give reasons for your choices. Axe. Is that going to cost nineteen? Why sick or to a society? See is they could tow sign Tootie. So we have X Square plus y square. Is he control one? It&#x27;s a protection. It&#x27;s a plane, Um is equal to zero X, like plain is a circle. And then we also have, say, its founding. I nicked one on one. Say it&#x27;s from neck to want to whine Now let&#x27;s no cat the grafts. So only number one satisfied There&#x27;s two conditions. Sounds. A graph is number one.

Wen Zheng · Answer

the problem, ese. Match the parametric equations with the laughs. Your reasons for your choices. Mexico to Kasai Inching. Why&#x27;s it called society andan? See you seek with one over one past his squire. So we have square past y squared. It&#x27;s a good one. The thing is, Richards and cereal and more of what So from a square plus y square. One way? No. How&#x27;s the points on this crew of Ah, almost sunder? I square past my square. Here&#x27;s why. On sea is founded by zero and one. So let&#x27;s see. Look at the graph here. So only one, three, four and six the points on the roof off. Oh, it&#x27;s a sign. Do I swear? That&#x27;s why squarely could what? But only six say is Bonnie by zero and one. So Graff is six.

Wen Zheng · Answer

the problem is much of the parametric equations. Ways well, offs your reasons for your choices. You&#x27;re actually able to tee times society wise because teen, see if he could He&#x27;d hands scientist on peace. Gretel and zero. So we&#x27;re half square us wise. Last three square. It&#x27;s the continent e squared. So this is why square this question is a corn. So it means that how&#x27;s the points on this curve? Uh um a chord. I squired past a square in court. Why score now? The fool cat, the grafts. So only number to satisfy this condition. So we have. The graph is number two.

Aparna Shakti · Answer

The problem is much the parametric equations. Basic graphs labeled one too. Six The reasons for your choices. So let&#x27;s look at the grafts from one to six. Hey, we have what he is able to cereal actually Goto one. And why you could do you know Andi what he goes to negative Negative. The affinity or positive in twenty have access codes to infinity. Why cost two in quantity? Now that&#x27;s tech the grafts from one to six only number five satisfies this two conditions. This is Paul Hot Bean. The first half why is greeted and zero then that&#x27;s check. One, two, three, four and six Only one in the floor satisfy satisfies this condition and then, Well, kind of number four. So for number four ax can be when we were small So you numbers war axe Andi were where is small But well, our expression actually is equal to squared minus tootie Act&#x27;s cannot be where small sacks cannot go to next collective infinity so number for it&#x27;s not correct self or B to graph is number one I see So we&#x27;LL have What axe is the court to zero? Why you got to sign? He since one zero. Should he is the cultural a pie? It&#x27;s over half T is equal to Cape High over too. So why is the society why come Vico to zero one on the next one? So why Ojai&#x27;s three while use the next attack? Shoo! Three on dh one, two, three and six. Oh, no, True three, four, six. It&#x27;s only number two satisfies this condition. This is number two for part of the when axe equal to zero with half after is a little cape. Hi Us high over two. So he called you a pie or a fall us high over ten. It&#x27;s over half. Why you got to sign Hootie. This is too k pi over Paul plus high over. So why high five news then look at is the grafts Only number six satisfies this condition Number six What? So Max goes to Conectiv Finerty? If he goes to negative twenty only number four satisfies this conditions. This number four the last one This here We have one too. Wall five six number far off is number three

Aparna Shakti · Answer

So if we want to match up these parametric equations with these graphs, we could try to, like, sketch each of these and go from there. But that is going to take way too much time. So what we can do instead is use um, the fact that some of these have certain properties because, like, notice over here, um, for why, why or not? I for one, our output values are y values I should say are always positive or zero. So if we have some function over here where it ever outputs a negative for why we can throw that one out. So this is what I&#x27;m going to use to kind of help us eliminate some things. So let&#x27;s look over here at a A. We know that why has to be greater than or equal to zero. So kind of like what I was saying. We can go ahead and throw out anything where it has a negative output. So that&#x27;s 234 Ah, five is good. And then six. So it&#x27;s between one and two now. Now, if we have where x outputs a negative value, um, then we would have won. And if For some reason, all of our X values are always going to be positive. Then we can go ahead and use this one over here. Um, but something that I&#x27;ll actually do instead is just look at t 00 and compare these. So let&#x27;s just do that. So for T 0 to 0 here, we&#x27;re going to get X. Is he good at one? And then why is he going to zero? So over here, X is equal to one. Well, why definitely is that equals zero so we can rule out one. So that means a matches with five. So let me get rid of all these X is really quick. I mean those back, and then I&#x27;ll just remove these really quickly. Okay, so we have that now, since we know that for one, the UAE has to be positive and all the other ones have a negative output for why? We just need to come over here and look from B F to see if one of these has such an output. And it looks like that is going to be be because we know here, the square root of white has to be greater than equal zero and then sign can output a negative again sign cannot put a negative t squared plus co sign three t. Um, A I think if we plug in, like, a small negative number, that should output a negative. But if you were to just plug that in, I&#x27;m sure you could see that this has some negative output. Um, and then over here, uh, co sign is negative for some value, so that would be negative. So that that looks like it would be the only one. So we can go ahead and just take this one and match it up over here. Okay, so So we got the easy ones out of the way to start Now, we have to work a little bit harder. Um, for these other ones, possibly. Well, the next one that kind of jumps out to me is graph three here. Um, you may not be able to see here because it doesn&#x27;t look too good from what I&#x27;ve drawn. Um, but it&#x27;s kind of just keep swirling and swirling around, um, the origin. So that kind of looks like as we just keep plugging in. Larger and larger values are just smaller and smaller values. This is just going to keep getting closer and closer to the origin. So one thing we might try to look at is the limit as he goes to infinity for some of these and you might notice down here from F if we were to take the limit as t goes to infinity of this X is going to go to zero, and y is going to go to zero. Um, but notice all these other ones, They don&#x27;t have anything like that. So I would make the guess that f and you can do a little bit more to also check this. But I would think this is, like the best approach to kind of connect f with three moments that below again. Okay, Yeah. Um, now. So let&#x27;s go ahead and just compare what we have left for these functions. Well, for C and D, notice that all of these outputs for both X and Y have to be trapped between negative one and one. So that means these two are just going to keep swirling around and around. So these are going to be trapped between negative one and one for all of its outputs for X and Y. So the only one that doesn&#x27;t have this kind of property between what we have left is over here for four. So that means we can take four and throw this below it like that. Um, Now we just need to figure out some way to distinguish between two and six between C and D. I think the best way to maybe distinguish between these is to just try to plug in zero into each of them. Because notice if we were to just plug in zero, one of these will not have a point, I think because, um, or we could try setting them equal to zero as well. But let&#x27;s just start with trying to plug in zero and then if plugging in zero gives us something nice and we can just kind of use that. So let&#x27;s see, when t is equal to zero, that would give us we&#x27;ll sign up zero, which is X zero. And over here we have Why is zero okay? I mean, that doesn&#x27;t help us because both of those just have the origin. Um, but now what about if I do t 0 to 0 here. Well, X is going to be one. And then why is going to be zero? And so if we look at X is equal to one over here. Well, only one of these looks like it actually has access. You could have one. Why is he going to zero? So we can go ahead and match D step six and then two with C. So these would be the way you would want to match these. Now, you could, um, come up with a different kind of rationale or how to get this. Um, this was just the best way I would think of going about it without actually trying to, like, graph a ton of points. Um, just start with some of the easy things, you know, normally, like seeing some of these only output positive values. Um, for certain things, maybe some of them only have negatives. You might have some kind of like swirling effect, So you might think, Oh, this is going to go to positive or negative infinity, Um, Or as he goes to positive or negative Infinity. Then this is going to close in around a certain point. In the case of three. And then sometimes you might have where all your values have to be bounded between two numbers, like in the case of two and six and then you can use all that kind of help you eliminate some of the information and even then, a lot of times just plug it in like a good point. Um will also help alleviate that, because you can see we use t 003 times to kind of help us eliminate. Um, some of them Yeah. I mean, you could have done this in a different way. This is just the way I feel was the most straightforward to get these answers.

Problem

Graph the curve $ x = y - 2 \sin \pi y $.

Problem 28 Hard Difficulty

Answer

(a) Graph V $\\$
(b) Graph I$\\$
(c) Graph II $\\$
(d) Graph VI $\\$
(e) Graph IV $\\$
(f) Graph III

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(a) Graph V $\\$(b) Graph I$\\$(c) Graph II $\\$(d) Graph VI $\\$(e) Graph IV $\\$(f) Graph III

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(a) Graph V $\\$
(b) Graph I$\\$
(c) Graph II $\\$
(d) Graph VI $\\$
(e) Graph IV $\\$
(f) Graph III

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