How do you convert $y = 6 x^{2}$ $y=6x^2$ to polar form?

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How do you convert $y = 6 x^{2}$ $y=6x^2$ to polar form?

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Answer 1 · 2016-07-21T05:04:17

You can use some basic trigonometry identities and some substitution to convert one form into the other.

Explanation:

Given that $r$ $r$ is the distance from the origin -- basically the length of the hypotenuse of a triangle whose legs are of length $x$ $x$ and $y$ $y$ -- and that $θ$ $theta$ is the angle from the x-axis up to hypotenuse, you can draw from basic trigonometry the following relationships:

$x = r \cdot cos (θ)$ $x = r*cos(theta)$
$y = r \cdot sin (θ)$ $y = r*sin(theta)$

Substituting these directly into your existing equation:

$y = 6 x^{2}$ $y=6x^2$
$r \cdot sin (θ) = 6 r^{2} \cdot {(cos (θ))}^{2}$ $r*sin(theta)=6r^2*(cos(theta))^2$
$\frac{6 r^{2}}{r} = \frac{sin (θ)}{{cos (θ)}^{2}}$ $(6r^2)/r=sin(theta)/(cos(theta)^2$
$r = \frac{sin (θ)}{6 \cdot {(cos (θ))}^{2}}$ $r=sin(theta)/(6*(cos(theta))^2)$
$r = \frac{tan (θ)}{6 \cdot cos (θ)}$ $r=tan(theta)/(6*cos(theta))$
$r = \frac{1}{6} \cdot tan (θ) \cdot sec (θ)$ $r=1/6*tan(theta)*sec(theta)$

Pick whichever of those last three forms seems more "simplified" to you.

For more detailed examples of how to do conversions in both directions, including some triangle diagrams that may make the relationships between $x$ $x$ , $y$ $y$ , $r$ $r$ , and $θ$ $theta$ more clear, see https://www.mathsisfun.com/polar-cartesian-coordinates.html.

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How do you convert $y = 6 x^{2}$ $y=6x^2$ to polar form?

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