Take a simple polar equation like r = θ/2 that graphs out to:
But, how would I achieve a rotation of the light-grey plot in this image (roughly 135 degrees)? Is there a way to easily shift the plot?
2 Answers
$\begingroup$Just put $\theta-135^\circ$ in place of $\theta$. Or if you're working in radians, then the equivalent in radians.
$\endgroup$ 2 $\begingroup$A way to think about this is is that you want to shift all $\theta$ to $\theta'=\theta +\delta$, where $\delta$ is the amount by which you want to rotate. This question has a significance if you want to rotate some equation which is a function of theta. In the case $r=\theta$ that becomes $r=\theta+\delta$.
Of course if our independent variable in our polar equation was a non-identity function of $\theta$ you might be able to use the angle-sum indentities to help you out:
$$ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $$
In case an anyone is trying to programme this in a Cartesian setting like I was trying to do (for a music visualizer) where I wanted my spiral's rotation to be a function of time. $r = \theta(t)$. Normally where solving $r=\theta$ or $\sqrt{x^2+y^2}=tan(\frac{\sin(\theta)}{\cos(\theta)})=tan(\frac{y}{x})$ you can substitute as follows.
$$ \sqrt{x^2+y^2}= tan(\frac{\sin(\theta+t)}{\cos(\theta+t)}) = tan(\frac{\sin\theta \cos t+\cos \theta \sin t}{\cos \theta \cos t - \sin \theta \sin t}) = tan(\frac{y \cos t +x\sin t }{x\cos t - y \sin t}) $$ /
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