Write y = x in polar form?

Question

I solved it and got it down to this: x^2+y^2 = 2x^2

But I got a point taken off, I guess because it can be taken down one more step into: θ=pi/4

How do you make that step to that? How do you go from what I got to θ=pi/4 ? Explain please? Thanks!

Answer 1

One can easily simplify: x² + y² = 2x²

to: y² = x² or (x/y)² = 1 or tan²θ = 1.

Hence θ = ± pi/4 (radians).

Answer 2

r = sqrt(x^2 + y^2)
θ = inversetangent(1 / 1) = 45 degrees = pi/4

.

Answer 3

x = r cos t
y = r sin t

r cos t = r sin t

So r=0 is a solution.
Suppose now that r is different than 0.
We can divide by cos t because if cos t would be 0, then sin t would be different than 0 and we get a contradiction in the above equality.

Hence we get
sin t/cos t = tan t = 1.
now t is from 0 to 2 pi.
One solution is t = pi/4. But tangent is periodical of period pi, therefore the other solution is pi+pi/4 = 5pi/4.

So the equation of the curve is t = pi/4 or t = 5pi/4.In one expression:
(t-pi/4)(t-5pi/4) =0.
Note that r=0 is already included in solution since we do not have restriction for r.

y=x is actually the first bisector and t=pi/4 corresponds with the bisector on positive side of the coordinate system and t=5pi/4 corresponds with the bisector on the negative side.

t=-pi/4 is wrong solution, that is clearly for y=-x and not
y=x.

Answer 4

Actually, you should have two solutions, since x^2+y^2 = 2x^2
is symmetrical about x-axis and y-axis.
θ = ±pi/4
--------
Ideas:
y^2 = x^2
y = ±x
y/x = ±1=> rsinθ/(rcosθ) = ±1
tanθ = ±1
θ = arctan(±1) = ±pi/4, since the domain of arctan(x) is [-pi/2, pi/2]

Answer 5

Polar form should involve only r and Θ.
x=r cos(Θ) and y= r sin(Θ).
y=x
r sin(Θ)=r cos(Θ)
sin(Θ)=cos(Θ) is the polar form.
This is true for Θ=∏/4

Answer 6

Replace EVERY x with r cos(θ) and EVERY y with r sin(θ)

That should get you:
r sin(θ) = r cos(θ)

Now divide both sides by r cos(θ)
r sin(θ) / r cos(θ) = 1

sin(θ) / cos(θ) = 1

tan(θ) = 1

Finally solve for θ:

θ = arctan(1)
θ = π/4

Answer 7

y=x
y/x = 1 = tan θ
θ = arctan (1) = pi/4