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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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    I will also add that if you are already doing the things that I suggested, then post the examples you looked at or anything else you have done to try to solve the problem.2012-08-08

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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    No @AustinBroussard, why?! We already had $y=(x+1)^3\left[(x+1)^3-1\right]=(x+1)^3(x(x^2+3x+3))=x(x+1)^3(x^2+3x+3)$Please do pay due attention to degree of polynomials: if you had degree $\,6\,$ at some point, then *unless there was cancelation* the final result *still must* have degree $\,6\,$2012-08-08
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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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    Yes you are right. I realized that after posting that too, but I deleted my answer anyways. Thanks.2012-08-08