Plot the point (3,π2)(3,\frac{\pi}{2})(3,2π) & find another set of coordinates, (r,θ)(r,θ)(r,θ), for this point, where:(A) r≥0,2π≤θ≤4πr≥0,2π≤θ≤4πr≥0,2π≤θ≤4π,(B) r≥0,−2π≤θ≤0r≥0,-2π≤θ≤0r≥0,−2π≤θ≤0,(C) r≤0,0≤θ≤2πr≤0,0≤θ≤2π r≤0,0≤θ≤2π.

(3,5π2),(3,−3π2),(−3,3π2)(3,\frac{5\pi}{2}),(3,-\frac{3\pi}{2}),(-3,\frac{3\pi}{2})(3,25π),(3,−23π),(−3,23π)

Plot the point ( 3 , π 2 ) (3,\frac{\pi}{2}) ( 3 , 2 π ) & find another set of coordinates, ( r , θ ) (r,θ) ( r , θ ) , for this point, where: (A) r ≥ 0 , 2 π ≤ θ ≤ 4 π r≥0,2π≤θ≤4π r ≥ 0 , 2 π ≤ θ ≤ 4 π , (B) r ≥ 0 , − 2 π ≤ θ ≤ 0 r≥0,-2π≤θ≤0 r ≥ 0 , − 2 π ≤ θ ≤ 0 , (C) r ≤ 0 , 0 ≤ θ ≤ 2 π r≤0,0≤θ≤2π r ≤ 0 , 0 ≤ θ ≤ 2 π .

Hey everyone. In this problem, we're asked to plot the point three, pi over two and then find some other sets of coordinates given this criteria that are all located at that same point. So let's start by first plotting our point at three, pi over two. Remember when plotting in polar coordinates, we want to locate our angle theta first, which in this case is pi over two, so I know that my point will be along this line. Now using that r value of three, I'm going to count out one, two, three to give me my point right here at three high over two. Now let's move on to finding some other sets of coordinates that will be located at the same point. Now looking at a, here we see that we still want our r value to be greater than or equal to zero like it is in our original point. So that tells me that my r value is still going to be three. Then I wanna change my angle theta so that it is in between two pi and four pi, and I can do that by taking my original angle pi over two and adding two pi to it. Now this gives me five pi over two, which is in this interval, so this gives me my other set of coordinates three five pi over two that are still located at that same point. Now let's move on to b. Now here we still want r to be greater than or equal to zero, so again my r value is still going to be three. But here we want our angle theta to be in between negative two pi and zero, so that tells me I should take my original angle and just subtract a factor of two pi. Now this will end up giving me negative three pi over two, and that gives me yet another set of coordinates here. Three negative three pi over two still located at the same point. Now let's look at our final set of criteria here. Here we want r to be negative, less than zero, so I'm gonna change my r value from three to negative three. Now for my value of theta, I still want it to be in between zero and two pi, but that doesn't mean that we can keep our angle the same. It can't just remain being pi over two because if I try to plot that with that negative three r value, it would not end up in the same place. So in order to compensate for that negative that I added to my r, I need to go ahead and add a factor of pi to my angle here. Now this gives me three pi over two and that gives me my final set of coordinates here, negative three, three pi over two, still located at the same exact point. Now you can always double check by plotting all of these points. They will all end up being at the same exact place. Thanks for watching, and let me know if you have questions.

Plot the point (3,π2)(3,\frac{\pi}{2})(3,2π) & find another set of coordinates, (r,θ)(r,θ)(r,θ), for this point, where:(A) r≥0,2π≤θ≤4πr≥0,2π≤θ≤4πr≥0,2π≤θ≤4π,(B) r≥0,−2π≤θ≤0r≥0,-2π≤θ≤0r≥0,−2π≤θ≤0,(C) r≤0,0≤θ≤2πr≤0,0≤θ≤2π r≤0,0≤θ≤2π.

(3,5π2),(3,−3π2),(−3,3π2)(3,\frac{5\pi}{2}),(3,-\frac{3\pi}{2}),(-3,\frac{3\pi}{2})(3,25π),(3,−23π),(−3,23π)

(3,5π2),(−3,−3π2),(−3,3π2)(3,\frac{5\pi}{2}),(-3,-\frac{3\pi}{2}),(-3,\frac{3\pi}{2})(3,25π),(−3,−23π),(−3,23π)

(−3,5π2),(−3,−3π2),(−3,π2)(-3,\frac{5\pi}{2}),(-3,-\frac{3\pi}{2}),(-3,\frac{\pi}{2})(−3,25π),(−3,−23π),(−3,2π)

(3,5π2),(3,−3π2),(−3,π2)(3,\frac{5\pi}{2}),(3,-\frac{3\pi}{2}),(-3,\frac{\pi}{2})(3,25π),(3,−23π),(−3,2π)

16. Parametric Equations & Polar Coordinates

Polar Coordinates

Calculus

16. Parametric Equations & Polar Coordinates

Polar Coordinates

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Multiple Choice

Plot the point $(3,$\frac{\pi}{2}$)$ & find another set of coordinates, $(r,θ)$ , for this point, where:
(A) $r≥0,2π≤θ≤4π$ ,
(B) $r≥0,-2π≤θ≤0$ ,
(C) $r≤0,0≤θ≤2π$ .

$(3,$\frac{5\pi}{2}$),(-3,-$\frac{3\pi}{2}$),(-3,$\frac{3\pi}{2}$)$

$(3,$\frac{5\pi}{2}$),(3,-$\frac{3\pi}{2}$),(-3,$\frac{3\pi}{2}$)$

$(-3,$\frac{5\pi}{2}$),(-3,-$\frac{3\pi}{2}$),(-3,$\frac{\pi}{2}$)$

$(3,$\frac{5\pi}{2}$),(3,-$\frac{3\pi}{2}$),(-3,$\frac{\pi}{2}$)$

Verified step by step guidance

Step 1: Understand the problem. The given point is in polar coordinates (3, π/2), where 'r' is the radial distance and 'θ' is the angle in radians. We need to find alternative representations of this point under different constraints for 'r' and 'θ'.

Step 2: Recall that polar coordinates are periodic. The angle 'θ' can be adjusted by adding or subtracting multiples of 2π to find equivalent points. For example, π/2 + 2π = 5π/2.

Step 3: For constraint (A), where r ≥ 0 and 2π ≤ θ ≤ 4π, adjust the angle 'θ' by adding 2π to π/2. This gives the new coordinates (3, 5π/2).

Step 4: For constraint (B), where r ≥ 0 and -2π ≤ θ ≤ 0, subtract 2π from π/2 to find an equivalent angle. This gives the new coordinates (3, -3π/2).

Step 5: For constraint (C), where r ≤ 0 and 0 ≤ θ ≤ 2π, negate the radial distance 'r' and adjust the angle 'θ' by adding π to π/2. This gives the new coordinates (-3, 3π/2).

6:17m

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