How do you convert $r^{2} θ = 1$ $r^2 theta=1$ into cartesian form?

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How do you convert $r^{2} θ = 1$ $r^2 theta=1$ into cartesian form?

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Answer 1 · 2016-10-01T05:37:25

$y = x tan (\frac{1}{x^{2} + y^{2}})$ $y=x tan(1/(x^2+y^2))$ , representing a spiral from the origin that turns forever, to reach the origin..

Explanation:

As $r^{2} > 0$ $r^2>0$ , its reciprocal $θ > 0$ $theta > 0$

Using the conversion formula #r(cos theta, sin theta) = (x, y),

$θ = {tan}^{- 1} (\frac{y}{x}) = \frac{1}{x^{2} + y^{2}} \to y = x tan (\frac{1}{x^{2} + y^{2}})$ $theta=tan^(-1)(y/x)=1/(x^2+y^2) to y = x tan (1/(x^2+y^2))$ .

I want variation of $θ \in (0, \infty)$ $theta in (0, oo)$ .

The graph is a spiral that approaches origin only in the limit,

as $x, y \to \infty$ $x, y to oo$ .. .

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How do you convert $r^{2} θ = 1$ $r^2 theta=1$ into cartesian form?

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How do you convert r^2 theta=1r2θ=1r^2 theta=1 into cartesian form?

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How do you convert $r^{2} θ = 1$ $r^2 theta=1$ into cartesian form?