Convert each equation to its polar form.x2+(y−2)2=4x^2+(y-2)^2=4x2+(y−2)2=4

r=4sin⁡θr=4\sin\thetar=4sinθ

Convert each equation to its polar form. x 2 + ( y − 2 ) 2 = 4 x^2+(y-2)^2=4 x 2 + ( y − 2 ) 2 = 4

In this problem, we're asked to convert the equation into its polar form, and the equation that we're given is x squared plus y minus two squared is equal to four. So let's go ahead and dive in. Now remember that we want to replace x, y, and x squared plus y squared using these formulas. Now here you might be tempted to try to replace this using that r squared value, but since we have y minus two being squared and not just y, we can't do that. So from here, we actually have to replace our squared x and y values using our cosine theta and our sine theta. So if I replace x with our cosine theta, y with our sine theta, that gives me our cosine theta, and that whole term is squared, plus r sine theta minus two, and all of that is squared. Then on that right side, I still just have a four. Now we're gonna have to do a bit more algebra than we have for our other practice problems, but that's fine. Make sure that you take your time here because it can be easy to just miss a step in your algebra. Now I want to go ahead and get all of these values squared. Remember that if you have an r cosine theta being squared, remember to square that r. So squaring that whole term will give me r squared times the cosine squared of theta. Now here, since I have r sine of theta minus two, this ends up giving me more than just one term. So if I go ahead and square these, this will give me r squared times the sine squared of theta minus four r sine theta and then a plus four. Then all of this is equal to four. Now from here, let's see if we can do anything here to cancel some stuff out. Now I see that I have this four and also this four on the right. So if I were to subtract four from both sides, those fours are going to cancel. Now I have r squared times the cosine squared of theta plus r squared times the sine squared of theta minus four r sine theta is equal to zero. Now where do I go from here? Because this looks like a a lot is going on and we still wanna try to solve for r. Now remember that you want to be constantly scanning for and recognizing if there are trig identities. Now here one thing that I notice is that I have a cosine squared theta and a sine squared theta. So let's go ahead and factor out our r squared term from these. This gives me r squared times cosine squared of theta plus the sine squared of theta, and then still having that term on the end minus four r sine theta, and all of that is equal to zero. Now based on my Pythagorean Shirk identities, I know that this is just equal to one. So this entire term is just r squared because r squared times one is just r squared. Now I have r squared minus four r sine beta. So this is looking much more simple. So let's go ahead and do a bit more algebra. Now I'm gonna go ahead and add four r sine beta to both sides. It will cancel on that left side, leaving me with r squared, and that's equal to four r sine theta. Now, since I have an r and an r squared, one of those will cancel out. And my final equation that I'm left with here is R is equal to four times the sine of theta. And this is my equation in its polar form. Thanks for watching and let me know if you have any questions.

Convert each equation to its polar form.x2+(y−2)2=$$4x^2$+(y-2)^2=4x2+(y$$−2)2=4

r=4sin⁡θr=4\sin\thetar=4sinθ

r2=4sin⁡θ$$r^2=4$$\sin\thetar2=4sinθ

r=2sin⁡θr=2\sin\thetar=2sinθ

r=cos⁡θ−sin⁡θr=\cos\theta-\sin\thetar=cosθ−sinθ

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Difficoltà con Calculus?

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Convert each equation to its polar form.
$x^2+(y-2)^2=4$

$r^2=4$\sin$$\theta$$

$r=4$\sin$$\theta$$

$r=2$\sin$$\theta$$

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

Guida verificata passo dopo passo

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

Step 2: Substitute x = r * cos(θ) and y = r * sin(θ) into the given equation x² + (y - 2)² = 4. This transforms the equation into polar form.

Step 3: Expand (y - 2)² using the substitution y = r * sin(θ). This results in (r * sin(θ) - 2)².

Step 4: Simplify the expanded equation and combine terms to isolate r². This involves squaring the terms and combining like terms.

Step 5: Factorize or simplify the resulting equation further to express r explicitly in terms of θ, completing the conversion to polar form.