What is the Cartesian form of $r^{2} - 4 r = sin (θ) - 5 cos (θ)$ $r^2-4r = sin(theta) - 5 cos(theta)$ ?

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What is the Cartesian form of $r^{2} - 4 r = sin (θ) - 5 cos (θ)$ $r^2-4r = sin(theta) - 5 cos(theta)$ ?

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Answer 1 · 2016-03-02T10:32:19

$4 x^{2} + 4 y^{2} - 5 x + y = {(x^{2} + y^{2})}^{\frac{3}{2}}$ $4x^2+4y^2-5x+y=(x^2+y^2)^(3/2)$

Explanation:

To convert a polar coordinate $(r, θ)$ $(r,theta)$ in Cartesian form we use relation $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ , $r cos θ = x$ $rcostheta=x$ , $r sin θ = y$ $rsintheta=y$ and $θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^(-1)(y/x)$ . Using these relations

$r^{2} - 4 r = sin θ - 5 cos θ$ $r^2−4r=sintheta−5costheta$ cann be written as

$r^{3} - 4 r^{2} = r sin θ - 5 r cos θ$ $r^3-4r^2=rsintheta−5rcostheta$ or

${(x^{2} + y^{2})}^{\frac{3}{2}} - 4 (x^{2} + y^{2}) = y - 5 x$ $(x^2+y^2)^(3/2)-4(x^2+y^2)=y-5x$ or

$4 x^{2} + 4 y^{2} - 5 x + y = {(x^{2} + y^{2})}^{\frac{3}{2}}$ $4x^2+4y^2-5x+y=(x^2+y^2)^(3/2)$

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What is the Cartesian form of $r^{2} - 4 r = sin (θ) - 5 cos (θ)$ $r^2-4r = sin(theta) - 5 cos(theta)$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

What is the Cartesian form of r^2-4r = sin(theta) - 5 cos(theta) r2−4r=sin(θ)−5cos(θ)r^2-4r = sin(theta) - 5 cos(theta) ?

1 Answer

Explanation:

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What is the Cartesian form of $r^{2} - 4 r = sin (θ) - 5 cos (θ)$ $r^2-4r = sin(theta) - 5 cos(theta)$ ?