Polar Coordinates, I hate them already

Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

Sketch the lines (or half-lines) with polar equations:

e) rcos(theta - pi/6) = 2

This is what I did....

Rcos(theta - pi/6) = 2
cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

= Rcos(theta)*(root3/2) - 0.5Rsin(theta)

Now: Rcos(theta) = x, Rsin(theta) = y

(root3/2)x - 0.5y = 2 [*2]
root3*x - y = 4
y = root3*x - 4

I get a line, which I think is the answer.

However, take a point (x,y) such as (4, 2.92)

theta = arctan(2.92/4) = 0.631
R = (4^2 + 2.92^2)^0.5 = 4.952

So if we put it back into the original equation:

4.952 * Cos ( 0.631 - pi/6) = 2

However, the left hand side = 4.92, which is totally wrong.

If you take theta as 1.67, this works. Can anybody explain why?

Ad

Reply 1

Adhsur

15

Maths in the middle of the summer holidays is music to my ears :tongue:

Reply 2

Jonny W

8

byb3

Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

Sketch the lines (or half-lines) with polar equations:

e) rcos(theta - pi/6) = 2

This is what I did....

Rcos(theta - pi/6) = 2
cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

= Rcos(theta)*(root3/2) - 0.5Rsin(theta)

Actually,

cos(theta - pi/6)
= cos(theta)cos(-pi/6) - sin(theta)sin(-pi/6) (minus, not plus!)
= cos(theta)cos(pi/6) + sin(theta)sin(pi/6)
= (sqrt(3)/2) cos(theta) + (1/2) sin(theta)

Substituting into r cos(theta - pi/6) = 2 we get

(sqrt(3)/2) x + (1/2) y = 2
ie, y = 4 - sqrt(3) x

Jonny W

Reply 3

byb3

OP

8

cheers mate, i see where I went wrong..

Cos(A-B) = CosACosB + SinASinB

But because I used (-B) on the right, it messed things up!

Thanks for clearing that one up.

Polar Coordinates, I hate them already

Related discussions

Latest

Trending

Trending

Articles for you