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I need help with the following trigonometric development:

$ x = r(\theta)\cos\theta$

$ y = r(\theta)\sin\theta$

this gives:

$ x' = r'(\theta)\cos\theta - r(\theta)\sin\theta$

$ y' = r'(\theta)\sin\theta + r(\theta)\cos\theta$

My problem is that I cannot understand this development:

$(x')^2 + (y')^2 = r(\theta)^2 + r'(\theta)^2$

Can someone please explain to me how the last development is made and how you do / see that it is valid.

I am also a bit puzzled about why $x'$ and $y'$ is not written $x'(\theta)$ & $y'(\theta)$.

Thank you!

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I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Nov 18 '12 at 20:34
Regarding the notation $x'$ versus $x'(\theta)$: Sometimes when it is clarified the variable may simply be omitted. – AD. Nov 18 '12 at 20:48
up vote 1 down vote accepted

Just expand $$\begin{eqnarray*} \left( x^{\prime }\right) ^{2} &=&\left( r^{\prime }\left( \theta \right) \cos \theta -r\left( \theta \right) \sin \theta \right) ^{2} \\ &=&\left( r^{\prime }\left( \theta \right) \right) ^{2}\cos ^{2}\theta -2r\left( \theta \right) r^{\prime }\left( \theta \right) \sin \theta \cos \theta +\left( r\left( \theta \right) \right) ^{2}\sin ^{2}\theta \\ \left( y^{\prime }\right) ^{2} &=&\left( r^{\prime }\left( \theta \right) \sin \theta +r\left( \theta \right) \cos \theta \right) ^{2} \\ &=&\left( r^{\prime }\left( \theta \right) \right) ^{2}\sin ^{2}\theta +2r\left( \theta \right) r^{\prime }\left( \theta \right) \sin \theta \cos \theta +\left( r\left( \theta \right) \right) ^{2}\cos ^{2}\theta, \end{eqnarray*}$$

sum $(x^{\prime})^2+(y^{\prime })^2$ and use the identity $\sin ^{2}\theta +\cos ^{2}\theta =1$.

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Thank you for the answer! – Lukas Arvidsson Nov 15 '12 at 15:06
@Lukas Arvidsson You are welcome! – Américo Tavares Nov 15 '12 at 15:07

you need to use the fact that $\sin^2(\theta) + \cos^2(\theta) = 1$:

$$ x' = r(\theta) \cos \theta - r(\theta) \sin \theta \Rightarrow x'^2 = r'(\theta)^2 \cos^2 \theta - 2 r(\theta) r'(\theta) \sin \theta \cos \theta + r(\theta)^2 \sin^2 \theta $$ $$ y' = r(\theta) \sin \theta + r(\theta) \cos \theta \Rightarrow y'^2 = r'(\theta)^2 \sin^2 \theta + 2 r(\theta) r'(\theta) \sin \theta \cos \theta + r(\theta)^2 \cos^2 \theta $$

$$ \Rightarrow x'^2 + y'^2 = r(\theta)^2 (\sin^2 \theta + \cos^2 \theta) - 2 r(\theta) r'(\theta) \sin \theta \cos \theta + 2 r(\theta) r'(\theta) \sin \theta \cos \theta + r'(\theta)^2 (\sin^2 \theta + \cos^2 \theta) = \left(r(\theta)^2 + r'(\theta)^2\right) (\sin^2 \theta + \cos^2 \theta) = / using~\sin^2 \theta+\cos^2 \theta = 1 / = r(\theta)^2 + r'(\theta)^2 $$

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Thank you! Very helpful! I think that one thing that is making these kinds of equations hard for me is that my handwriting is not the best. Always something that gets missing as I develop the equations... – Lukas Arvidsson Nov 15 '12 at 15:17
The result is the same but the factor $2$ is missing in $-r(\theta) r'(\theta) \sin \theta \cos \theta$ and $r(\theta) r'(\theta) \sin \theta \cos \theta$. – Américo Tavares Nov 15 '12 at 18:47
you're right, I added the 2. – Joakim Nohlgård Nov 16 '12 at 7:45

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