Transform the rectangular equation into polar form: y^2 = x^2 (2+x / 2-x)??

Question

Ive been messing around with this one for a while and cant seem to get it into the appropriate form.

Answer 1

I think you mean

y^2= x^2 (2+x)/(2-x)
add x^2 on both sides to get y^2+x^2 on LHS

y^2 + x^2 = x^2((2+x)/(2-x) + 1) = 4x^2/(2-x)
put y = r sin t and x = r cos t

y^2 + x^2 = r^2 so
r^2 = 4 x^2/(2-x)
r^2= 4 r^2 cos^2 t/( 2 - r cos t)
or 1 = 4 cos ^2 t/(2- r cos t)
or 2- r cos t = 4 cos ^2 t
r = (2/cos t - 4 cos t)

Answer 2

nicely, one factor is that in many cases you place the type earlier the letter, for this reason 2x + 2y yet another factor that could have long previous incorrect is which you have misinterpret the question. If the '2' grew to become into after the (x+y), (x+y) could actually be squared (the type is smaller and fairly raised next to the bracket). if it is so, then do in basic terms (x+y)(x+y) that's x^2 + 2xy + y^2 in case you probably did no longer understand, in many cases notation '^' ability, to the ability of; so x^2 ability x squared desire this helps!

Answer 3

Transform the rectangular equation into polar form:
y^2 = x^2 (2+x / 2-x).

I assume you mean:

y² = x²(2 + x)/(2 - x)

Remember the identities.

x = rcosθ
y = rsinθ
x² + y² = r²
________

y² = x²(2 + x)/(2 - x)

(2 - x)y² = x²(2 + x)
2y² - xy² = 2x² + x³
0 = 2x² - 2y² + xy² + x³
2x² - 2y² + xy² + x³ = 0

Plug in the identities.

2r²cos²θ - 2r²sin²θ + r³cosθsin²θ + r³cos³θ = 0

Divide by r².

2cos²θ - 2sin²θ + rcosθsin²θ + rcos³θ = 0
rcosθ(sin²θ + cos²θ) = 2(sin²θ - cos²θ)
rcosθ = -2(cos²θ - sin²θ)
rcosθ = -2(cos 2θ)

r = -2(cos 2θ)/cosθ

Answer 4

Multiply both sides by (2 - x) to get something like:

2y^2 - xy^2 = 4x^2 - x^4

Use x=rcos(@) and y = rsin(@) to make it 100% polar. From there, trig identities might help you make it neater and more compact.

Answer 5

First convert x and y to polar
x= r*cos(theta)
y=r*sin(theta)
substitute to get
r^2*sin^2(theta)
=r^2*cos^2(theta)*[2+r*cos(theta)]/[2-r*cos(theta)]
so, for r not 0 (note that y=0, x=0 is a point in the equation, so the origin [r=0] is in the set of solutions)
sin^2(theta)
=cos^2(theta)*[2+r*cos(theta)]/[2-r*cos(theta)]
=cos^2(theta)*[(4-2+rcos(theta))]/[2-r*cos(theta)]
=cos^2(theta)*([4/(2-r*cos(theta))]-1)
so
sin^2(theta)+cos^2(theta)
=cos^2(theta)*[4/(2-r*cos(theta))]
1 = 4*cos^2(theta)/(2-r*cos(theta))
2-r*cos(theta)=4*cos^2(theta)
2*sec(theta) - r = 4*cos(theta)
2sec(theta) - 4*cos(theta) = r