Question #5b70d

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Question #5b70d

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Answer 1 · 2017-03-07T21:22:59

$r = \pm 1$ $r = pm 1$

Explanation:

This is the unit circle about the Origin

Take the parameterisation in polar coordinates:

$x = r cos θ, y = r sin θ$ $x = r cos theta, y = r sin theta$

Observe that:

$x^{2} + y^{2} = r^{2} ({cos}^{2} θ + {sin}^{2} θ) = r^{2}$ $x^2 + y^2 = r^2(cos^2 theta + sin^2 theta) = r^2$ ,

Then note that you actually have:

$r^{2} = 1, r = \pm 1$ $r^2 = 1, r = pm 1$

I'd go with the $\pm$ $pm$ bit too. I've seen it argued that $r$ $r$ should only ever be positive, though I can't remember the detail and it makes no sense to me (albeit I say that as a user of maths as opposed to mathematician).

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Answer 2 · 2017-03-07T21:26:49

$r^{2} = 1$ $r^2=1$

Explanation:

To convert from $cartesian to polar form$ $color(blue)"cartesian to polar form"$

$color(red)(bar(ul(|color(white)(2/2)color(black)(x=rcostheta;y=rsintheta)color(white)(2/2)|)))$

$x^2+y^2=1$

$rArr(rcostheta)^2+(rsintheta)^2=1$

$rArrr^2cos^2theta+r^2sin^2theta=1$

$rArrr^2(cos^2theta+sin^2theta)=1$

$rArrr^2=1" since "cos^2theta+sin^2theta=1$

This is the equation of a circle, centred at the origin with radius 1

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Question #5b70d

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