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

 

[CellCoree]

Squaring out coordinates involves taking the coordinates of a point and finding the squared values of each coordinate. In this case, the coordinates are represented by r and $, and we are using trigonometric identities to find the squared values.

To understand this step, we need to first understand the trigonometric identity for the square of sine and cosine. This identity states that (sin($))^2 + (cos($))^2 = 1. So, when we square out the coordinates (-(s-2a)sin($) , scos($)), we get:

(-(s-2a)sin($))^2 + (scos($))^2 = (s-2a)^2(sin($))^2 + s^2(cos($))^2

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

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

= s^2[(sin($))^2 + (cos($))^2] - 4as(sin($))^2 + 4a^2(sin($))^2

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

= s^2 - 4as(sin($))^2 + 4a^2(1 - (sin($))^2) [using the trigonometric identity mentioned earlier]

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

= s^2 + 4a^2 - 4as[(sin($))^2 + (cos($))^2] [using the trigonometric identity again]

= s^2 + 4a^2 - 4as(1) = s^2 + 4a^2 - 4as

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

This is how we get from r=(-(s-2a)sin($) , scos($)) to (r^2)=(s^2)+(4(a^2)-4as)sin($). We are simply using the trigonometric identities to simplify the squared values of the coordinates. I hope this helps clarify the step