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Write each pair of equations as a single equation in $x$ and $y$.
a)$\begin{cases} x=t+1 &\\ y=t^2-t & \\ \end{cases}$ b)$ \begin{cases} x=\sqrt[3]{t}-1 &\\ y=t^2-t & \\ \end{cases}$ c)$\begin{cases} x=\sin t &\\ y=\cos t & \\ \end{cases}$

All I want to know is what the question is asking me to do. Please do not give me the answer to any of these, if needed please make up an example. After that, I will edit with my steps to see if I am doing this correctly.

Edit: Now, I know this has come up before, but can someone please tell me the difference between $\arcsin$ and $\sin^{-1}$. Or are they the same?

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    $\begingroup$ What the question is asking you to do is this: Right now each of (a) (b) and (c) express x and y as functions of t. The problem would like you to just have one expression relating x and y. One way to do this is to solve the x or y expression for t, and then substitute into the other. Example: $x = t + 1$, $y = t - 1$. Then from the first equation $t = x - 1$, so substituting into the second $y = (x - 1) - 1 $, so $y = x-2$. $\endgroup$
    – Jason Knapp
    Aug 7, 2012 at 23:56
  • $\begingroup$ @AustinBroussard The method is the same, but the solution differs, because Jason solved a different problem ;) $\endgroup$
    – M Turgeon
    Aug 8, 2012 at 0:07
  • $\begingroup$ @MTurgeon oh goodness. I was reading a different problem! $\endgroup$
    – Austin Broussard
    Aug 8, 2012 at 0:08
  • $\begingroup$ "...can someone please tell me the difference between..." - see this $\endgroup$
    – J. M. ain't a mathematician
    Aug 8, 2012 at 0:30
  • $\begingroup$ @AustinBroussard Welcome to math.SE! I see that you have posted quite a few questions here. In order to help you, I would like to offer a couple of suggestions. First, if you don't understand what a homework exercise is asking, you should look at the examples in the chapter. Also, you should look at your notes from class for similar problems. These can help you get started. Finally, when you ask questions here and there is something you don't understand about the problem, we can more easily answer your question if you tell us which words in the exercise confuse you. Good luck with your HW! $\endgroup$
    – Code-Guru
    Aug 8, 2012 at 0:58

3 Answers 3

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$$(a)\,\,\,t=x-1\Longrightarrow y=t^2-t=(x-1)^2-(x-1)=(x-1)(x-2)\Longrightarrow y=(x-1)(x-2)$$ $$(b)\,\,\,x=\sqrt[3] t-1\Longrightarrow t=(x+1)^3....etc.$$ Can you now continue by yourself?

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  • $\begingroup$ Is the next part of $(b)$, $=(x+1)^6-(x+1)^3$? $\endgroup$
    – Austin Broussard
    Aug 7, 2012 at 23:55
  • $\begingroup$ Indeed, and that equals $\,(x+1)^3\left((x+1)^3-1\right)\,$, if you love to factor stuff. For the last one think of Pythagoras Theorem... the trigonometric version. $\endgroup$
    – DonAntonio
    Aug 7, 2012 at 23:57
  • $\begingroup$ I have to factor $(x+1)(x+1)(x+1)\Big((x+1)(x+1)(x+1)-1\Big)$??? $\endgroup$
    – Austin Broussard
    Aug 8, 2012 at 0:01
  • $\begingroup$ Oh, dear god: not at all, unless threatened with a bazooka or something like that! Why would you do such a thing? Leave it as it is unless specifically asked otherwise $\endgroup$
    – DonAntonio
    Aug 8, 2012 at 0:03
  • $\begingroup$ That's why I asked! So what do I do after $\,(x+1)^3\left((x+1)^3-1\right)\,$? $\endgroup$
    – Austin Broussard
    Aug 8, 2012 at 0:07
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For (c), think about a very familiar trig identity.

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Sorry, I posted the answers before without fully reading your question. Anyways, this type of question is called parametric. What you want to do is solve for t for one equation and then substitute that into the other equation. For the first one for example, t = x - 1 in the first equation. Now plug that t into the second equation.

Exact same thing for b (remember you want to get t ALONE)

For c, a hint is this: cos(arcsin x) or sin(arccos x) is always sqrt(1-x^2)

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    $\begingroup$ Your solution for (c) is lacking: $\,\sqrt{1-x^2}\,$ only gives the upper semicircle of the unit circle centered at the origin. You still need to express the lower semicircle. $\endgroup$
    – DonAntonio
    Aug 8, 2012 at 0:02
  • $\begingroup$ Yes you are right. I realized that after posting that too, but I deleted my answer anyways. Thanks. $\endgroup$
    – Sidd Singal
    Aug 8, 2012 at 0:05

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