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For old coordinates $(x,y)$ the new coordinates $(u,v)$ are related like this:

$x = u\cos(\theta) - v\sin(\theta)$

$y = u\sin(\theta) + v\cos(\theta)$

So would it be correct to say that to rotate the axis for a parametric equation defined by $x = f(u)$ and $y = g(u)$ I need to multiply $f(u)$ with the $x$ rotation equation and $g(u)$ with $y$ rotation equation?

I can't find any resources online for this topic.

Avitus
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Cobold
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  • The parametric equations you are referring to are those described in the question, am I right?In this notation, $f=f(u,v,\theta)$ and similarly for $g$. – Avitus Jul 22 '13 at 14:03
  • No, the equations which I have given are for rotating Cartesian equation axis. I'm asking for the parametric equation axis rotation equations. – Cobold Jul 22 '13 at 14:22
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    When you say "parametric equation axis rotation equations", I'm not sure what you mean. – Eric Auld Jul 22 '13 at 14:27
  • I am struggling with "parametric equation (of?) the axis rotation equation": what do you mean with parametric? What you wrote is the coordinate transformation $(x,y)\mapsto (u,v)$, with $u=x\cos\theta+y\sin\theta$ and $v=-x\sin\theta+y\cos\theta$. Do you need a matricial form? – Avitus Jul 22 '13 at 14:28

1 Answers1

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If you have a parametric equation $$\begin{cases}x = f(u) \\ y = g (u)\end{cases}$$ and you want to rotate the image by $\theta$, you can just take $$\begin{cases}x' = x\cos \theta - y \sin \theta = (\cos\theta) f(u) - (\sin\theta)g(u) \\ y' = x \sin\theta + y\cos\theta = (\sin\theta) f(u) + (\cos\theta)g(u). \end{cases}$$

Eric Auld
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