Convert the point to polar coordinates.(1,1)(1,1)(1,1)

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

Convert the point to polar coordinates. ( 1 , 1 ) (1,1) ( 1 , 1 )

In this problem, what we're asked to convert the point given in rectangular coordinates into polar coordinates, and the point that we're given here is 11. So let's start with step 1 and go ahead and plot our point. Now plotting our point, I end up right here in quadrant 1. Now moving on to step 2, calculating our r value, this gives me the square root of 1 squared plus 1 squared, which will end up being the square root of 2 for that r value. Now moving on to finding theta, here we look at the location of our point. Now here it's located in quadrant 1, which tells me that in order to find theta, all I need to do is take the inverse tangent of y over x. Now Now putting my y and x values in there I end up getting that theta is equal to the inverse tangent of 1 over 1, which we know is just 1. So this is the inverse tangent of 1. And if I type this into my calculator or I use my unit circle, I will see that this is equal to pi over 4 for that theta value. Now I have my point in polar coordinates at root 2pi over 4, and we're done here with our final answer. Thanks for watching and I'll see you in the next one.

Convert the point to polar coordinates. ( 1 , 1 ) (1,1) ( 1 , 1 )

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

( 1 , π 4 ) (1,\frac{\pi}{4}) ( 1 , 4 π )

( 2 , 5 π 4 ) (\sqrt{2},\frac{5\pi}{4}) ( 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 5 π )

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

9. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

Trigonometry

9. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

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Multiple Choice

Convert the point to polar coordinates.
$(1,1)$

$(1,\frac{\pi}{4})$

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

$(\sqrt{2},\frac{5\pi}{4})$

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

Verified step by step guidance

Start by understanding that polar coordinates are represented as (r, θ), where r is the radius or distance from the origin, and θ is the angle from the positive x-axis.

Given the Cartesian coordinates (1, 1), calculate the radius r using the formula r = \sqrt{x^2 + y^2}. Substitute x = 1 and y = 1 into the formula.

Next, determine the angle θ using the formula θ = \arctan\left(\frac{y}{x}\right). Substitute x = 1 and y = 1 into the formula.

Since both x and y are positive, the point lies in the first quadrant, and the angle θ should be between 0 and \frac{\pi}{2}.

Combine the calculated values of r and θ to express the point in polar coordinates as (r, θ).

6:17m

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