Squaring out coordinates (with trig)

[UniPhysics90]

I'm following a worked example and really don't get a certain step in it: the step is going between

r=(-(s-2a)sin($) , scos($))

and

(r^2)=(s^2)+(4(a^2)-4as)sin($)

First of all how do you 'square' coordinates? I would have though it stays as coordinates?

Even when i try different ways of multiplying it out i can't get this answer, i always end up with extra terms or wrong coefficients.

any help would be great.

thanks

 

[CompuChip]

Note that r is actually a vector, with two components (r = (r₁, r₂).
So when they write r², they actually mean
r² = r . r = r₁² + r₂².

You will have to use that sin² + cos² = 1. Just watch out when you calculate (a + b)², it is not а² + b².

 

[HallsofIvy]

r²= r.r is not very good notation (although some books use it).
What they mean, and should use, is |r|².

 

[CompuChip]

Yes, welcome in the world of lazy physicists ;)
Why write two bars if you can have a convention to leave them out :P