Convert the point to polar coordinates.(−2,2)(-2,2)(−2,2)

( 2 2 , 3 π 4 ) (2\sqrt{2},\frac{3\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

Convert the point to polar coordinates. ( − 2 , 2 ) (-2,2) ( − 2 , 2 )

In this problem, we're asked to convert our point given in rectangular coordinates into polar coordinates, and the point that we're given here is negative two, two. So let's get started with our steps and go ahead and plot this point. Now plotting my point here at negative two, two, I end up here in quadrant two. So we can move on to our step two and calculate our r value. Now here r is going to be equal to the square root of that x value negative two squared plus that y value two squared. Now this will end up giving me the square root of four plus four or the square root of eight. Now here we can use our radical rules in order to simplify this further to get that this is equal to two root two for our r value. Now with r calculated, we can go ahead and move on to calculating theta. Now here we want to pay attention to the location of our point as we plotted it in step one, and it's specifically located here in quadrant two. Now whenever I have a point located in quadrant two, I need to go ahead and take the inverse tangent of y over x, but I also need to add pi to it in order to account for what quadrant it's actually located in. So let's go ahead and calculate this value. Now I'm plotting and calculating this. I get that theta is equal to the inverse tangent of my y value two over my x value negative two. Now this simplifies to the inverse tangent of negative one, which is equal to negative pi over four. Now if I stopped here forgetting that I needed to add this pi in, that would get me a point that is located here in quadrant four, which is not where my actual point is. So that's why I need to add that factor of pi. So adding pi to this, making sure that I don't forget that step, if I add pi to that negative pi over four, I end up getting that this is equal to three pi over four for my value of theta. Then my point in polar coordinates is located at two root two, three pi over four, and this is our final answer. Let me know if you have questions.

Convert the point to polar coordinates. ( − 2 , 2 ) (-2,2) ( − 2 , 2 )

( 2 2 , 3 π 4 ) (2\sqrt{2},\frac{3\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

( 2 2 , − π 4 ) (2\sqrt{2},-\frac{\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , − 4 π )

( 2 2 , π 4 ) (2\sqrt{2},\frac{\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 π )

( − 2 2 , 3 π 4 ) (-2\sqrt{2},\frac{3\pi}{4}) ( − 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

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 the point to polar coordinates.
$(-2,2)$

$(2\sqrt{2},\frac{3\pi}{4})$

$(2\sqrt{2},-\frac{\pi}{4})$

$(2\sqrt{2},\frac{\pi}{4})$

$(-2\sqrt{2},\frac{3\pi}{4})$

Verified step by step guidance

Step 1: Recall the formula for converting Cartesian coordinates (x, y) to polar coordinates (r, θ). The radius r is given by r = √(x² + y²), and the angle θ is given by θ = arctan(y/x).

Step 2: Substitute the given Cartesian coordinates (-2, 2) into the formula for r. Compute r = √((-2)² + 2²). This will give the magnitude of the point's distance from the origin.

Step 3: Determine the angle θ using θ = arctan(y/x). Substitute y = 2 and x = -2 into the formula. Since the point is in the second quadrant (x is negative and y is positive), adjust θ accordingly to ensure it lies within the correct quadrant.

Step 4: Express the polar coordinates as (r, θ), where r is the computed radius and θ is the adjusted angle. Ensure θ is expressed in radians, as polar coordinates typically use radians.

Step 5: Verify the solution by checking that the polar coordinates correctly represent the original Cartesian point (-2, 2). You can do this by converting the polar coordinates back to Cartesian form using x = r * cos(θ) and y = r * sin(θ).

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