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

r = 6 sin ⁡ θ − cos ⁡ θ r=\frac{6}{\sin\theta-\cos\theta} r = s i n θ − c o s θ 6

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 six. 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 six. Now, both of these terms have an R in them. And remember that our goal here is going to be to solve for R. So I'm going to 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 six. 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 six 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 = 6 y-x=6 y − x = 6

r = 6 sin ⁡ θ − cos ⁡ θ r=\frac{6}{\sin\theta-\cos\theta} r = s i n θ − c o s θ 6

6 r = sin ⁡ θ − cos ⁡ θ 6r=\sin\theta-\cos\theta 6 r = sin θ − cos θ

r = sin ⁡ θ − cos ⁡ θ 6 r=\frac{\sin\theta-\cos\theta}{6} r = 6 s i n θ − c o s θ

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

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Struggling with Calculus?

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Multiple Choice

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

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

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

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

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

Verified step by step guidance

Step 1: Recall the relationship between Cartesian coordinates (x, y) and polar coordinates (r, θ). In polar coordinates, x = r * cos(θ) and y = r * sin(θ).

Step 2: Substitute x = r * cos(θ) and y = r * sin(θ) into the given Cartesian equation y - x = 6. This transforms the equation into r * sin(θ) - r * cos(θ) = 6.

Step 3: Factor out r from the left-hand side of the equation. This gives r * (sin(θ) - cos(θ)) = 6.

Step 4: Solve for r by dividing both sides of the equation by (sin(θ) - cos(θ)). This results in r = 6 / (sin(θ) - cos(θ)).

Step 5: The polar form of the equation is now expressed as r = 6 / (sin(θ) - cos(θ)).

5:32m

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