I tried to find the coordinates parallel to the initial line by setting the derivative of s*Sin(theta) to zero but it was unsolveable. Is there another way, using curvature or something? Must... get... brain... food... ******** A Problem Worthy of Attack Proves It''s Worth by Fighting Back

Sorry, but I have no idea what you''re talking about...

solution that jumps to my mind is projecting the vector down onto the initial line as you call it.

What initial line?

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Original post by walkingcarcass
Must... get... brain... food...

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Try oatmeal.

i think he means the initial line is the euclidean x+ axis (since this is how it''s drawn in most books).

quote:

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Original post by walkingcarcass
...coordinates parallel to the initial line...

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What the hell are you talking about? Coordinates parallel? A coordinate can''t be parallel, only two or more lines can be parallel. Initial line? What line?

And where the hell did s*sin(theta) come from?

Polar coordinates, as the title suggests.

Your''re right, it is meaningless, I mistyped and didn''t proof-read, sorry for wasting your time.

I MEANT to ask how to find points on a polar curve whose tangents are parallel to the original line, other than setting the derivative of r*sin(theta) to zero (ie the turning point).

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A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

Well, still there is the problem of what the original line is, and what r*sin(theta) means. anyway...

The best way of finding the slope of a polar curve r = f(theta)is to treat it as a parametric curve, by setting x = f(theta) * cos(theta) and y = f(theta) * sin(theta), and then taking the parametric derivative, finding in terms of theta. The formula for a parametric derivative is dy/dx = (dy/dt)/(dx/dt). In this case, t becomes theta, so you will have dy/dx in terms of theta.

Take the initial line to mean the X axis.

The derivative of r*Sin(theta) will be zero at the value of theta for which the y coordinate (in cartesian interpretation) is stationary, ie at a maximum. Is there another way of detecting stationary points on polar graphs?

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A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

r*sin(theta) is 0 when theta = n*PI -> on the x axis. What do you mean, by "stationary points"?

Sorry for asking again; I could probably figure out what you mean if I try hard, but you have to learn to communicate effectively what you mean. If you have a term that you are not sure about, try looking it on MathWorld.

Cédric