How do you convert (0, -6)(0,−6) to polar form?

2 Answers
Shwetank Mauria
Jun 29, 2016

Polar coordinates are (6,-pi/2)(6,−π2).

Explanation:

When Cartesian coordinates (x,y)(x,y) are converted into polar coordinates (r,theta)(r,θ), we have the relation

x=rcosthetax=rcosθ and y=rsinthetay=rsinθ and hence

r=sqrt(x^2+y^2)r=√x2+y2, costheta=x/rcosθ=xr and sintheta=y/rsinθ=yr.

hence for (0,-6)(0,−6)

r=sqrt(0^2+(-6)^2)=sqrt(0+36)=sqrt36=6r=√02+(−6)2=√0+36=√36=6

and as costheta=0/6=0cosθ=06=0 and sintheta=-6/6=-1sinθ=−66=−1,

we have theta=-pi/2θ=−π2

Hence polar coordinates are (6,-pi/2)(6,−π2).

Ratnaker Mehta
Jun 29, 2016

Polar conversion is (6,-pi/2).(6,−π2).

Explanation:

Cartesian (x,y)(x,y) in polar is (r,theta)(r,θ), where, x=rcostheta, y=rsintheta, theta in(-pi,pi], x^2+y^2=r^2.x=rcosθ,y=rsinθ,θ∈(−π,π],x2+y2=r2.

Clearly, r=6.r=6.

Now x=rcostheta rArr 0=6costheta rArr costheta = 0.x=rcosθ⇒0=6cosθ⇒cosθ=0.
y=rsintheta rArr -6=6sintheta rArr sintheta =-1.y=rsinθ⇒−6=6sinθ⇒sinθ=−1.

We conclude that theta=-pi/2.θ=−π2.

Hence, polar conversion is (6,-pi/2).(6,−π2).

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