proof involving polar curve :)

tlk2

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Sep 30, 2010
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Hi, I can't figure out how to continue on this problem. Any help with reasons would be appreciated. Thanks! :)

P any point except origin on curve r=f(theta)

? is the angel between the tangent line at P and the radial line OP

show that tan?=r/(dr/d theta ) (hint- ?=c-theta, where c is the angle where the tangent line meets the horizontal line)

I've got:
?=c-theta
tan?=tan(c-theta)=(tanc-tan theta)/(1+tanc tan theta)
Can't figure out where to go from here...
 
Are you asking to prove that the tangent of the angle between the radial line and the tangent line at the point P is given by

tan(α)=rdrdθ\displaystyle tan({\alpha})=\frac{r}{\frac{dr}{d\theta}}?.

The tangent line intersects the polar axis at Q. Then, the slope of the tangent line through Pand Q is

fcosθ+fsinθfsinθ+fcosθ\displaystyle \frac{fcos{\theta}+f'sin{\theta}}{-fsin{\theta}+f'cos{\theta}}...........[1]

This is equal to:

tan(θ+α)=tanθ+tanα1tanθtanα\displaystyle tan({\theta}+{\alpha})=\frac{tan{\theta}+tan{\alpha}}{1-tan{\theta}tan{\alpha}}

=sinθ+cosθtanαcosθsinθtanα\displaystyle =\frac{sin{\theta}+cos{\theta}tan{\alpha}}{cos{\theta}-sin{\theta}tan{\alpha}}...........[2]

equate [1] and [2], cross multiply:

(fcosθ+fsinθ)(cosθsinθtanα)=(sinθ+cosθtanα)(fsinθ+fcosθ)\displaystyle (fcos{\theta}+f'sin{\theta})(cos{\theta}-sin{\theta}tan{\alpha})=(sin{\theta}+cos{\theta}tan{\alpha})(-fsin{\theta}+f'cos{\theta})

see if you can hammer it into shape.
 
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