0
$\begingroup$

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!

  • 0
    I've changed [tag:algebra] tag to [tag:algebra-precalculus], since we don't use algebra tag anymore, see [meta](http://meta.math.stackexchange.com/questions/473/the-use-of-the-algebra-tag/3081#3081) for details.2012-11-18
  • 0
    Regarding the notation $x'$ versus $x'(\theta)$: Sometimes when it is clarified the variable may simply be omitted.2012-11-18

2 Answers 2

1

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

  • 0
    @Lukas Arvidsson You are welcome!2012-11-15
3

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

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

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

  • 0
    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...2012-11-15
  • 0
    The result is the same but the factor $2$ is missing in $-r(\theta) r'(\theta) \sin \theta \cos \theta$ and $r(\theta) r'(\theta) \sin \theta \cos \theta$.2012-11-15
  • 0
    you're right, I added the 2.2012-11-16