If you multiply both sides of r=cos(θ) by r you get
r2=rcos(θ) which is the same as x2+y2=x or
x2−x+1/4+y2=1/4
(x−1/2)2+y2=1/4
a circle with center at (1/2, 0) and radius 1/2 so it is tangent to the y-axis.
While you can have r negative in the equation, that is interpreted in polar coordinates as the radius going the other way. With θ=π, r=−1 which gives the point (1, 0) on the positive x-axis. Note that cos(θ) goes from 0 to 0 as θ goes from 0 to π and again as θ goes from π to 2π.
As θ goes from 0 to 2π, the point goes around the circle twice.
Show that r = cos theta is a graph of a circle with diameter 1 from 0,0 to 0,1
How to do this algebraically?
I know r = sqrt(x^2 + y^2) so what should i do?
4 answers
a circle of radius .5 with center at(0, .5) is:
x^2 + ( y - .5)^2 = .25
x^2 + y^2 - y + .25 = .25
x^2 + y^2 - y = 0
now
let theta = T
x = r cos T
y = r sin T
then for our circle
r^2 cos^2T + r^2sin^2T = r sin T
or
r^2 = r sin T
r = sin T
so I think you have a typo
x^2 + ( y - .5)^2 = .25
x^2 + y^2 - y + .25 = .25
x^2 + y^2 - y = 0
now
let theta = T
x = r cos T
y = r sin T
then for our circle
r^2 cos^2T + r^2sin^2T = r sin T
or
r^2 = r sin T
r = sin T
so I think you have a typo
Tam also has circle from 0,0 to 1,0
not from 0,0 to 0,1
not from 0,0 to 0,1
thanks xD
1 answer