polar to cartesian?

Question

i would like to Convert the Cartesian coordinates (-6, 13) to polar coordinates.


iwould like to Firstly, give an exact answer such that r greater than or equal to 0, 0 is less than or equal to pi is less than 2pi

then i would like to find an equivalent (exact) answer with r coordinate negative, in the same manner as above

lastly i would like to give an equivalent (exact) answer with r coordinate positive, but different to my first answer

so far fro the first question i have got

sqrt 205 23/36 pi

23/36 pi is incorrect though!

could someone please help me i reposted this as no one knew the answer well

Answer 1

r is the distance from the origin. Using Pythagoras,
r = sqrt(6^2 + 13^2)
r = sqrt(205)

As for the angle theta, we know two things:
tan theta = 13/-6, and theta is in the 2nd quadrant (since the coordinate (-6,13) is in the 2nd quadrant).
Take the inverse tangent of (13/-6): The calculator gives -1.138 pi. Add 2pi to that, and you get 0.861pi for your angle.

0.861pi is in the 2nd quadrant, so that's your answer!

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For the second part: It's impossible to have a negative r coordinate. r is the distance from the origin to your (r, theta) point, and distance can't be negative.

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For the third part, I'm not sure what you're asking for. You can, of course, add 2pi to your theta and end up at the same coordinate... is that what you're after?

Answer 2

To get polar coordinates out of cartesian coordinates you have to convert the (x,y) of cartesian coordinates to the (r, h) (I'm writing h since theta won't display properly here) of polar coordinates.

OK, first the r, This is the distance from the origin to the point (x, y), which is sqrt(x^2 + y^2).

Then the h. This is the angle between the positive x-axis and the line from the origin to (x, y). You can calculate sin(h) = y/r and cos(h) = x/r. In the example you gave, it's easiest to do this in two steps. First find the angle h* where sin(h*) = |y|/r and cos(h*) = |x|/r, then since you know your point is in the second quadrant, h = pi - h*. (draw the graph and you'll see it).

As for the second and third parts of your question, I won't stand by these answers because I don't completely understand the questions. But here goes anyway.... As far as I understand polar coordinates, r is always non-negative.... but if you insisted on r being negative, then h = 2*pi - h*. And if you wanted an equivalent answer with r positive, remember that the location (r, h) is equivalent to (r, h+2n*pi) where n is any positive or negative integer.

Answer 3

ok so you have the point (-6, 13)

you want to find its distance from the origin

r² = 6² + 13²
r = 14.318

next you want to find the angle - draw it out its in the 3rd quadrant
make a right triangle and call the angle of interest Ø

sin Ø = 13/6
Ø = arcsin(13/6)

the angle for polar coords will be 180 + Ø

then your coordinate is (r, Ø + 180)
tada


if r is negative... its the same deal but the angle will change to Ø instead of 180 + Ø so the coordinates are

(-r, Ø)

lastly you could also write
(r, Ø - 180)

enjoy

Answer 4

x = rcos(#)
y =rsin(#)

r = sqrt(x^2 +y^2) = sqrt(205)

cos(#) = x/r = -6/sqrt(205)

sin(#) = y/r =13/sqrt(205)

therefore # = 114.78 degrees or 2.003288915 radians