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

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 3, pi over 2 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 3 pi over 2. Remember when plotting in polar coordinates, we want to locate our angle theta first, which in this case is pi over 2, so I know that my point will be along this line. Now using that r value of 3, I'm going to count out 1, 2, 3 to give me my point right here at 3 high over 2. 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 0 like it is in our original point. So that tells me that my r value is still going to be 3. Then I wanna change my angle theta so that it is in between 2 pi and 4 pi, and I can do that by taking my original angle pi over 2 and adding 2 pi to it. Now this gives me 5 pi over 2, which is in this interval. So this gives me my other set of coordinates, 3, 5, pi over 2, 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 0. So, again, my r value is still going to be 3. But here, we want our angle theta to be in between negative 2 pi and 0. So that tells me I should take my original angle and just subtract a factor of 2 pi. Now this will end up giving me negative 3 pi over 2, and that gives me yet another set of coordinates here. 3 negative 3 pi over 2, 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 0. So I'm gonna change my r value from 3 to negative 3. Now for my value of theta, I still want it to be in between 0 and 2 pi, but that doesn't mean that we can keep our angle the same. It can't just remain being pi over 2 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 3pi over 2 and that gives me my final set of coordinates here, negative 3, 3pi over 2 still located at the same exact point. 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 , 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 .

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

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

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

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

15. Polar Equations

Polar Coordinate System

Precalculus

15. Polar Equations

Polar Coordinate System

Struggling with Precalculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

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

Understand that the point (3, $\frac{\pi}{2}$) is given in polar coordinates, where 3 is the radius (r) and $\frac{\pi}{2}$ is the angle (θ) in radians.

To find another set of coordinates for this point, we need to adjust the angle θ while keeping the radius r the same or changing its sign, depending on the conditions given.

For condition (A), where $r \geq 0$ and $2\pi \leq \theta \leq 4\pi$, add $2\pi$ to the original angle: $\theta = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$. The new coordinates are (3, $\frac{5\pi}{2}$).

For condition (B), where $r \geq 0$ and $-2\pi \leq \theta \leq 0$, subtract $2\pi$ from the original angle: $\theta = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2}$. The new coordinates are (3, $-\frac{3\pi}{2}$).

For condition (C), where $r \leq 0$ and $0 \leq \theta \leq 2\pi$, change the sign of the radius and add $\pi$ to the original angle: $r = -3$ and $\theta = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$. The new coordinates are (-3, $\frac{3\pi}{2}$).

5:32m

Watch next

Master Intro to Polar Coordinates with a bite sized video explanation from Patrick

Start learning

Polar Coordinate System practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice