# Polar co-ordinates

1. Sep 20, 2011

### imy786

1. The problem statement, all variables and given/known data

A point has coordinates (−5, 3*square root of 7).
What is the polar coordinates of this point?

(in the form a,b)

2. Relevant equations

x= rcos theta
y=rsin theta

3. The attempt at a solution

Using phythagoras thrown

-5=x=rcos theta (eq 1)
3*square root 7=Y=rsin theta (eq 2)

r=-5/cos theta
r=3*square root 7 / sin theta

2. Sep 20, 2011

### Staff: Mentor

I don't see that you have actually used the Pythagoras Theorem. Isn't r2 = x2 + y2?

It might be helpful to draw a right triangle, with base x, altitude y, and hypotenuse r.

3. Sep 20, 2011

### grzz

Try to find the value of r by eliminating $\theta$ using sin$^{2}$$\theta$ + cos$^{2}$$\theta$ = 1.

4. Sep 20, 2011

### imy786

(-5)^2=r^2cos^2 theta (eq 1)
25 = r^2cos^2
r^2 = 25 / cos^2
r^2 = 25 / (1 - sin ^2 theta)
r^2 * (1 - sin ^2 theta) = 25
(1 - sin ^2 theta) = 25 / r^2
1 - (25 / r^2) = sin ^2 theta

(3)^2* ( 7) =Y=r^2sin^2 theta (eq 2)
9*7 = r^2sin^2 theta
64 = r^2 (sin^2 theta)
r^2 = 64 / sin^2 theta
r^2* sin^2 theta = 64
sin^2 theta = 64 /r^2

1 - (25 / r^2) = 64 /r^2
r^2 - (25*r^2 / r^2) = 64
r^2 - (25) = 64
r^2 = 64-25
r^2 = 39
r = =/- (square root of 39)

Polar co-ordinates of the point?

5. Sep 20, 2011

### verty

x = r cos(theta)
y = r sin(theta)

hmm, I know x and y but not r and theta. I have two unknowns. I need to eliminate r or theta. How can I do that?

if I can eliminate r, I can find theta.
if I can eliminate theta, I can find r.

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6. Sep 20, 2011

### Staff: Mentor

You're really going at it the hard way.
9*7 != 64.
Add 25 to both sides. Also, as noted above, 64 isn't the right number.
If you redo your calculations for r, you will have one of the coordinates. As already recommended, draw a right triangle with base -5 (i.e., to the left of the origin), altitude 3*sqrt(7). You can find the hypotenuse as r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

7. Sep 23, 2011

### imy786

(1 , pie/3)

8. Sep 23, 2011

### imy786

r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

r = $\sqrt{(25 + (21}$

r = $\sqrt{(41}$

r = 6.40

But still need polar point in the form (f, g/pie)

9. Sep 23, 2011

### Staff: Mentor

$(3 \sqrt{7})^2 \neq 21$
Your continuing algebra errors will make completing this problem very difficult.
If you haven't drawn a coordinate system showing the point in rectangular coordinates, you should do so. The x-coordinate and y-coordinate of the point determine a right triangle. What quadrant is this triangle in?

Label the base, altitude, and hypotenuse of this triangle with the values of x, y, and r. Then use one of the trig functions to find the angle that the hypotenuse makes with the base. To get the value of theta for your polar point, you will need to take into account which quadrant the triangle is in.

10. Sep 23, 2011

### Staff: Mentor

I meant to mention this before. The name of this Greek letter - $\pi$ - is pi, not pie.

11. Sep 26, 2011

### imy786

(3√7)^2 = 63

√ (25 + 63) =

=√88

= √(4 x 22)

=2√22

= 9.38

x co-ordinate = -5
y co-ordinate = 3√7 = 7.94

cos (theta) = -5 / 3√7
theta = 129 degrees

pi = 180 degrees