Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

2 Answers 2

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 $\sin ^{2}\theta +\cos ^{2}\theta =1$.

share|improve this answer
    
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 $\sin^2(\theta) + \cos^2(\theta) = 1$:

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

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

share|improve this answer
    
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 Gebart Nov 16 '12 at 7:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.