How do you express the polar equation $r^{2} = 16 cos θ$ $r^2 = 16cos theta$ in cartesian form?

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Answer 1 · 2017-10-23T21:02:29

${(x^{2} + y^{2})}^{\frac{3}{2}} - 16 x = 0$ $(x^2+y^2)^(3/2) - 16x = 0$

To convert from polar to rectangular coordinates we can use:

$x = r cos θ$ $x = r cos theta$

$y = r sin θ$ $y = r sin theta$

and the consequence:

$r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2+y^2)$

So, given:

$r^{2} = 16 cos θ$ $r^2 = 16cos theta$

we can multiply both sides by $r$ $r$ to get:

$r^{3} = 16 r cos θ$ $r^3 = 16 r cos theta$

Then use some of our formulae to rewrite as:

${(x^{2} + y^{2})}^{\frac{3}{2}} = 16 x$ $(x^2+y^2)^(3/2) = 16x$

Subtract $16 x$ $16x$ from both sides to get the form:

${(x^{2} + y^{2})}^{\frac{3}{2}} - 16 x = 0$ $(x^2+y^2)^(3/2) - 16x = 0$

graph{(x^2+y^2)^(3/2) - 16x=0 [-10, 10, -5, 5]}

How do you express the polar equation r^2 = 16cos thetar2=16cosθr^2 = 16cos theta in cartesian form?