Polar Coordinates and Polar Graphs

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October 2nd 2007, 01:17 PM
googoogaga

Polar Coordinates and Polar Graphs

Hello, I am having a dificult time working these problesms. I made an attempt to work them out unfortunately my book does not show the answers. So I have no clue if I am following the right steps. Could anyone please correct my answers if they are wrong and please explain in details how you found the answers? Thank you!

1) Plot the points in polar coordinates and find the corresponding rectangular coordinates for the points.

a) (-2, 7pi/4 ) answer: (-1.41, -1.41)
b) (0, -7pi/6 ) answer: ( 0, 0)
c) (-3, -1.57 ) answer: ( 3.39, 0.42)

2) The rectangular coordinates are given. Find two sets of polar coordinates for the given points for
0 is less than or equal to Theta which is less than 2 pi.

a) (0, -5) answer: (5, -pi/4) ; (-5, -5pi/4)
b) (4, -2) answer: (4.47, -26.56 degrees) ; (-4.47, 180 degrees)
c) (3, (-3)^1/2 ) answer: ( 3.46, -pi/ 6) ; -3.46, 7pi/6)

( I change my radicals in decimals I hope you don't mind. It makes it easier to type)
October 2nd 2007, 02:10 PM
Jhevon

Quote:

Originally Posted by googoogaga

Hello, I am having a dificult time working these problesms. I made an attempt to work them out unfortunately my book does not show the answers. So I have no clue if I am following the right steps. Could anyone please correct my answers if they are wrong and please explain in details how you found the answers? Thank you!

1) Plot the points in polar coordinates and find the corresponding rectangular coordinates for the points.

a) (-2, 7pi/4 ) answer: (-1.41, -1.41)

this is incorrect. $\sin \left( \frac {7 \pi}4 \right)$ is negative. if you didn't catch that, you should have realized that the original point is in the second quadrant while your new point is in the third

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b) (0, -7pi/6 ) answer: ( 0, 0)

correct

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c) (-3, -1.57 ) answer: ( 3.39, 0.42)

your answer is way off here. what formulas did you use? did you take the angle to be in degrees or radians?

Quote:

2) The rectangular coordinates are given. Find two sets of polar coordinates for the given points for
0 is less than or equal to Theta which is less than 2 pi.

a) (0, -5) answer: (5, -pi/4) ; (-5, -5pi/4)

no. the point is one the y-axis. $- \frac {\pi}4$ does not get you there

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b) (4, -2) answer: (4.47, -26.56 degrees) ; (-4.47, 180 degrees)

why did you do this one in degrees?

anyway, your first coordinates are ok (just change them to radians), your second is not. (-4.47, 180 deg) lies on the x-axis, the point you're considering does not. if you change 4.47 to -4.47 then you should ADD 180 deg to -26.56 deg (but of course, do the angles in radians)

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c) (3, (-3)^1/2 ) answer: ( 3.46, -pi/ 6) ; -3.46, 7pi/6)

obviously you made a typo here. (-3)^(1/2) is not a real number. did you mean -(3)^(1/2) ?
October 2nd 2007, 02:48 PM
Soroban

Hello, googoogaga!

Are you making sketches?

Quote:

1) Plot the points in polar coordinates
and find the corresponding rectangular coordinates for the points.

$a)\;\left(-2,\,\frac{7\pi}{4}\right)$

The angle: . $\theta \:=\:\frac{7\pi}{4} \:=\:315^o$ is in Quadrant 4.

$r = -2$ puts the point in Quadrant 2.

The rectangular coordinates are: . $\left(-\sqrt{2},\:\sqrt{2}\right)$ .

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$b)\;\left(0,\,-\frac{7\pi}{6}\right)$ . . answer: $(0,\,0)$ . . . . Right!

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$c)\;(-3,\,-1.57)$

That angle is in radians . . . and is approximately $\frac{\pi}{2}$

We have: . $\left(-3,\,-\frac{\pi}{2}\right)$

$\theta = -\frac{\pi}{2}$ places us on the negative y-axis.

$r = -3$ puts us on the positive y-axis.

The rectangular coordinates are: . $(0,\:3)$

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2) The rectangular coordinates are given.
Find two sets of polar coordinates for the given points for $0 \leq \theta \leq 2\pi$

$a)\;(0,\,-5)$

You have the wrong angles.
And you forgot that they want positive angles.

Answers: . $\left(5,\,\frac{3\pi}{2}\right),\:\left(-5,\,\frac{\pi}{2}\right)$

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$b)\;(4,\,-2)$

This point is in Quadrant 4.

$r \:=\:\sqrt{4^2 + (-2)^2} \:=\:\sqrt{20} \:=\:\pm2\sqrt{5}$

The angle is: . $\theta \:=\:\tan^{-1}\left(\frac{-2}{4}\right) \:=\:-26.56^o$

The polar coordinates are: . $\left(2\sqrt{5},\:333.43^o\right),\;\left(-2\sqrt{5},\:153.43^o\right)$

But I believe they want the angles in radians.

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$c)\;\left(3,\:-\sqrt{3}\right)$

Your answers are correct ... except for the negative angle.

Answers: . $\left(2\sqrt{3},\,\frac{11\pi}{6}\right),\:\left(-2\sqrt{3},\:\frac{5\pi}{6}\right)$
October 2nd 2007, 03:11 PM
googoogaga

Mr. Soroban, What equation did you use for 2a?
October 2nd 2007, 04:44 PM
Jhevon

Quote:

Originally Posted by googoogaga

Mr. Soroban, What equation did you use for 2a?

we cannot find the angle in the usual way here, since $\tan^{-1} \left( \frac yx \right)$ is not defined here. we just have to realize that since we are on the y-axis, the angle must be $\frac {\pi}2 + n \pi$ for $n \in \mathbb{Z}$