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popsquare
Dec7-07, 01:53 PM
I have gotten from this equation: y=3x into its polar form by substituting
x= r cos (theta) and y=r sin (theta) to get me here:



r sin (theta) = 3r cos (theta) Substitute and simplify.

r sin (theta) - 3r cos (theta) = 0 Subtract 3r cos (theta) from both sides.

r [sin (theta) - 3 cos (theta)] = 0 Factor out r from the equation.

Now I set both equal to zero like this...

r = 0 or sin (theta) - 3 cos (theta) = 0



I am assuming r = 0 is included in the graph of the second equation. So the only solution to the original problem comes from solving the second equation for Theta --->

sin (theta) - 3 cos (theta) = 0

arctan (3) = 1.2490 radians

Is 1.2490 radians the correct solution to the original question of convert y=3x into polar form?

Office_Shredder
Dec7-07, 02:02 PM
That might be the right answer to the question of what angle is formed between the line y=3x and the x-axis, but just "1.2490 radians" isn't a polar form of anything, it's just a number. An answer like "theta=1.249" might look better, but keep in mind you're missing the half of the line that goes through the third quadrant (as traditionally r is taken to be non-negative)

popsquare
Dec7-07, 02:20 PM
Thanks for your help. Office Shredder. Are either of these possible as polar points

(0,1.249 rad) , (infinity, 1.249 rad)

because I am thinking since r = 0 it cannot possibly be a line like y=3x. Convert the radian angle I found into a polar point that extends infinitely into the I and III Quadrants like the rectangular line equation.

Diffy
Dec7-07, 03:07 PM
Remember that polar coordinates are not the best for graphing straight lines, they are better with curves.

you have, sin (theta) - 3 cos (theta) = 0. from here, you can get
sin (theta) = 3 cos (theta), divide both sides by cos (theta)
tan (theta) = 3

Now think about all the (r, theta) that satisfy this equation. Well no matter what the r is the tan of theta has to be equal to 3. So as long as the angle formed with the origin is 1.249 rad, then the point (r, theta) is valid. The set of all (r, theta) solutions is the same as your graph of y = 3x.

HTH