Convert each equation to its polar form.y−x=6y-x=6y−x=6

r=6sin⁡θ−cos⁡θr=\frac{6}{\sin\theta-\cos\theta}r=sinθ−cosθ6

Convert each equation to its polar form. y − x = 6 y-x=6 y − x = 6

In this problem, we're asked to convert the equation into its polar form, and the equation that we're given is y minus x is equal to 6. Now remember that we can convert equations from their rectangular form into their polar form by simply using these formulas here. So let's go ahead and do that. Now here I have a y and an x in my equation, so that means I'm going to use these two formulas that x is equal to r cosine of theta and y is equal to r sine of theta. So let's go ahead and sub those into our equation, giving me that r sine theta minus r cosine theta is equal to 6. Now both of these terms have an r in them. And remember that our goal here is gonna be to solve for r. So I'm gonna go ahead and pull those r's out, factoring it out, giving me that r times the sine of theta minus the cosine of theta is equal to 6. Now from here to fully isolate r I'm just going to divide both sides by sine theta minus cosine theta letting it cancel out on that left side and leaving me on the right side with this fraction here. Now having it in its final form with r isolated, this is equal to 6 over sine theta minus cosine theta, and we're completely done here having solved for r. So this is my equation in its polar form. Thanks for watching, and let me know if you have questions.

Convert each equation to its polar form.y−x=6y-x=6y−x=6

r=6sin⁡θ−cos⁡θr=\frac{6}{\sin\theta-\cos\theta}r=sinθ−cosθ6

6r=sin⁡θ−cos⁡θ6r=\sin\theta-\cos\theta6r=sinθ−cosθ

r=sin⁡θ−cos⁡θ6r=\frac{\sin\theta-\cos\theta}{6}r=6sinθ−cosθ

r=6(sin⁡θ−cos⁡θ)r=6\left(\sin\theta-\cos\theta\right)r=6(sinθ−cosθ)

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15. Polar Equations

Convert Equations Between Polar and Rectangular Forms

Precalculus

15. Polar Equations

Convert Equations Between Polar and Rectangular Forms

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객관식

Convert each equation to its polar form.
$y-x=6$

$6r=$\sin\[\theta$-$\cos\]\theta$$

$r=$\frac{\sin\theta-\cos\theta}{6}$$

$r=6$\left$($\sin\]\theta$-$\cos\[\theta\]\right$)$

$r=$\frac{6}{\sin\theta-\cos\theta}$$

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검증된 단계별 안내

Start by understanding the relationship between Cartesian coordinates (x, y) and polar coordinates (r, θ). The conversion formulas are: x = r * cos(θ) and y = r * sin(θ).

Given the equation y - x = 6, substitute the polar coordinate expressions for x and y: r * sin(θ) - r * cos(θ) = 6.

Factor out r from the left side of the equation: r * (sin(θ) - cos(θ)) = 6.

To solve for r, divide both sides of the equation by (sin(θ) - cos(θ)): r = 6 / (sin(θ) - cos(θ)).

This is the polar form of the given equation, where r is expressed in terms of θ.

6:5m

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