How do you find the points of intersection of $θ = \frac{π}{4}, r = 2$ $theta=pi/4, r=2$ ?

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How do you find the points of intersection of $θ = \frac{π}{4}, r = 2$ $theta=pi/4, r=2$ ?

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Answer 1 · 2018-02-21T15:54:06

Their point of intersection is at $x = \sqrt{2}$ $x=sqrt2$ and $y = \sqrt{2}$ $y=sqrt2$

Explanation:

$θ = \frac{π}{4}$ $theta = pi/4$ is a ray emerging from the origin with an angle of $\frac{π}{4}$ $pi/4$ from the positive $x$ $x$ -axis. (cyan)

$r = 2$ $r=2$ is a circle of radius 2 centered at the origin. (magenta)

Drawn with Graphmatica

The point of intersection in polar coordinates is (unsurprisingly)

$(r, θ) = (2, \frac{π}{2})$ $(r,theta) = (2,pi/2)$

To get the cartesian coordinates, apply the following

$x = r cos θ = 2 cos (\frac{π}{2}) = 2 (\frac{1}{\sqrt{2}}) = \sqrt{2}$ $x = rcos theta = 2 cos(pi/2) = 2 (1/sqrt2) = sqrt2$
$y = r sin θ = 2 sin (\frac{π}{2}) = 2 (\frac{1}{\sqrt{2}}) = \sqrt{2}$ $y = rsin theta = 2 sin(pi/2) = 2 (1/sqrt2) = sqrt2$

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How do you find the points of intersection of $θ = \frac{π}{4}, r = 2$ $theta=pi/4, r=2$ ?

1 Answer

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How do you find the points of intersection of $θ = \frac{π}{4}, r = 2$ $theta=pi/4, r=2$ ?