Convert the point to polar coordinates.(−1,−3)(-1,-\sqrt{3})(−1,−3)

( 2 , 4 π 3 ) (2,\frac{4\pi}{3}) ( 2 , 3 4 π )

Convert the point to polar coordinates. ( − 1 , − 3 ) (-1,-\sqrt{3}) ( − 1 , − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> )

Here we're asked to convert the point given in rectangular coordinates into polar coordinates, and the point that we're given is the negative one, negative square root of three. So let's start with step number one and plot our point. Now it may not be easy to locate the square root of three on your graph, but you can always type this into your calculator and you would get that this is equal to about negative 1.73. So with that in mind, graphing this point at negative one, negative 1.73, we end up right about here in quadrant three. Now we can move on to step number two in calculating r. Now in calculating r, I get that r is equal to that x value negative one squared plus that y value negative root three squared. Now in actually squaring these values, this ends up giving me the square root of one plus three, which is really just the square root of four, which we know is just two. So my r value here is two and we can go ahead and move on to step number three in finding theta. Now looking at the location of our point here, I see that it's located in quadrant three. Now because it's located in quadrant three, that tells me that I need to take the inverse tangent of y over x and then add pi to it. So let's go ahead and do that here. Now here theta will be equal to the inverse tangent of my y value negative root three over negative one. And remember, we're going to add pi after we take that inverse tangent. Now taking the inverse tangent here ends up giving me the inverse tangent of the square root of three, still adding that pi to the end, and the inverse tangent of the square root of three is simply pi over three. So pi over three plus pi, making sure to add that factor to get our point located in the right quadrant. This ends up giving me four pi over three for my angle theta. Now I have my point in polar coordinates two four pi over three and this is my final answer. Let me know if you have any questions. Thanks for watching.

Convert the point to polar coordinates. ( − 1 , − 3 ) (-1,-\sqrt{3}) ( − 1 , − 3 <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"> )

( 2 , 4 π 3 ) (2,\frac{4\pi}{3}) ( 2 , 3 4 π )

( 2 , π 3 ) (2,\frac{\pi}{3}) ( 2 , 3 π )

( 2 , 7 π 6 ) (2,\frac{7\pi}{6}) ( 2 , 6 7 π )

( − 2 , 4 π 3 ) (-2,\frac{4\pi}{3}) ( − 2 , 3 4 π )

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Polar Coordinates

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Multiple Choice

Convert the point to polar coordinates.
$(-1,-\sqrt{3})$

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

$(2,\frac{7\pi}{6})$

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

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

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 calculated as r = √(x² + y²), and the angle θ is calculated as θ = arctan(y/x).

Step 2: Compute the radius r. Substitute x = -1 and y = -√3 into the formula r = √(x² + y²). This will give the magnitude of the point's distance from the origin.

Step 3: Compute the angle θ. Use the formula θ = arctan(y/x) with x = -1 and y = -√3. Since both x and y are negative, the point lies in the third quadrant, and you need to adjust θ accordingly to reflect its position in the polar coordinate system.

Step 4: Express θ in terms of radians. Ensure the angle is represented in the standard polar form, which typically uses radians. Adjust the angle to be within the range [0, 2π] if necessary.

Step 5: Combine the radius r and angle θ to write the polar coordinates in the form (r, θ). Ensure the final representation reflects the correct quadrant and is simplified if possible.

5:32m

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