- #1
popsquare
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- 0
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?
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?