Question

r=2-6 cos(θ) convert this polar coordinate equation into an equation in rectangular coordinates(x,y)


given (-8,8) convert this point into polar coordinates with 6 >= 0 and 0<=θ<= 2pi

given x^2+ (y-5)^2=25 cover this rectangular coordinate equation in to an equation in polar coordinates (r,θ)

Answer 1

cos(theta)=x/r
sin(theta)=y/r
r^2=x^2+y^2
-----------------------------
so we have r=2-6cos(theta)
sqrt{x^2+y^2}=2-6*x/r

Answer 2

but r=sqrt{x^2+y^2}

sqrt{x^2+y^2}=2-6*x/(sqrt{x^2+y^2}

Answer 3

if you want you can multiply both sides of that equation by sqrt{x^2+y^2}

Answer 4

for the second one we have (x,y)=(rcostheta,rsintheta)
rcostheta=x rsintheta=y
costheta=x/r sintheta=y/r

draw a traingle in the second quadrant
you should get the hypethuese or r=8sqrt{2}

and you should get theta=3pi/4

Answer 5

for the last one
just use rcostheta=x and rsintheta=y
r^2cos^2theta+(rsintheta-5)^2=25
now multiply (rsintheta-5)(rsintheta-5) and you should be able to use cos^2x+sin^2x=1
eventually