k8lyn911
  • k8lyn911
Calculus III Homework Help: Eliminate the parameter to find a Cartesian equation of the curve. x=t-1, y= t^3 +1, -2 <= t <= 2 I'm just trying to get a basic understanding of how to do this kind of problem so I can type it up in Mathematica. If anyone knows the format for that, I'd really appreciate some help.
Mathematics
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
Eliminate the parameter? :o Hmm I guess you could solve for t in your x equation,\[\large\rm x=t-1\qquad\to\qquad \color{royalblue}{t=x+1}\]and then sub that into the other equation,\[\large\rm y=(\color{royalblue}{t})^3+1\]And then also get some boundaries for x using your first equation using the old t boundaries. Mathematica I don't know though. :c
k8lyn911
  • k8lyn911
Okay. How do I go about finding the new limits? Like how t ranges from -2 to 2. In the examples I've looked at, those change without any explanation. Would I plug -2 and 2 into one of these equations?
zepdrix
  • zepdrix
\(\large\rm x=t-1\) t ranges from -2 to 2, Which means x will range from: \(\large\rm x=(-2)-1\) to \(\large\rm x=(2)-1\) ya? :)

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k8lyn911
  • k8lyn911
Oh, that helps so much! Thank you!
k8lyn911
  • k8lyn911
Is the Cartesian equation y = 2 + 3 x + 3 x^2 + x^3 ?
zepdrix
  • zepdrix
You expanded out the cube? Mmmm ya! That looks correct! :)
k8lyn911
  • k8lyn911
This is just how the answer came out in Mathematica. Our assignment is use Mathematica to find the answers, so I'm mostly here to double check them. I'm so sleep deprived at this point that nothing seems to make sense anymore.
zepdrix
  • zepdrix
Oh I see :) lol

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