Find the rectangular version of the parametric curve given by x=(t+3)^3, y=(t+2)^2

Question

mathmale · Answer

the trick here is to eliminate the parameter, t.  How would you do that?

anonymous · Answer

I think the easier equation to do it for would be the y equation. Would you end up with $$\sqrt{y}-2=t$$

anonymous · Answer

Do you have any restrictions on $t$? I don't think you can assume the positive root of $y$ just because it works nicely that way.

anonymous · Answer

My problem has no restrictions on t. It just says to find the rectangular version of the parametric curve.

anonymous · Answer

I'm trying to think if your method faces an obstacle at any point... so far, it looks fine, but I would have suggested eliminating $t+2$ rather than $t$ alone.
$$\begin{cases}x=(t+3)^3\y=(t+2)^2\end{cases}~~\Rightarrow~~\begin{cases}x=((t+2)+1)^3\y=(t+2)^2\end{cases}~~\Rightarrow~~\begin{cases}x=(t+2)^3+3(t+2)^2+3(t+2)+1\\sqrt{y}=t+2\end{cases}$$
then substitute from there.

If you're confident with your solution, keep it.

anonymous · Answer

The cut-off part is the rest of the expansion of the binomial.

anonymous · Answer

Okay. That is very helpful! If I use the y equation and substitute it back in for x I believe I would get this $$x=(\sqrt{y}-2+3)^{3}$$ How could this be expanded so it is a y = form? Is this why it is easier to use the x equation?

anonymous · Answer

You should be able to easily isolate the $y$: $$x=\left(\sqrt y-2+3\right)^3~~\Rightarrow~~x^{1/3}=\sqrt{y}+1~~\Rightarrow~~y=\left(x^{1/3}-1\right)^2$$ The thing is, I'm not sure if the parametric equations and this rectangular equation give the same function: parametric: http://www.wolframalpha.com/input/?i=x%3D%28t%2B3%29%5E3%2C+y%3D%28t%2B2%29%5E2 rectangular: http://www.wolframalpha.com/input/?i=Plot%5B%28x%5E%281%2F3%29-1%29%5E2%2C%7Bx%2C-.2%2C5%7D%5D

anonymous · Answer

Okay. I didn't think to move the 3 over make it a 1/3 on the x. It doesn't look like they exactly put out the same equation. Do you have any suggestions?

anonymous · Answer

Also is the x1/3 -1 to the second or third power? If it is too the third. Why is this?