Why Does Rotating Coordinate Axes Affect Calculations?

[bobsmith76]

Homework Statement

17. xy = 2

The Attempt at a Solution

Do you see that step where they do the following:

√2/2 - √2/2 = my answer is 0

and they multiply that to

√2/2 + √2/2 = my answer is √2

So to me the answer is 0 * √2 = 0, but the book shows that that calculation = 2, then they set x over 2 and y over 2, don't understand why.

I don't understand what the book is doing. I understand everything else except that one part.

 

[tiny-tim]

hi bobsmith76!

So to me the answer is 0 * √2 = 0, not 2

the answer to what?

 

[bobsmith76]

see revision

 

[tiny-tim]

Do you see that step where they do the following:

√2/2 - √2/2 = my answer is 0

and they multiply that to

√2/2 + √2/2 = my answer is √2

do you mean the LHS of the equation at the beginning of the last line?

then you're ignoring the x' and y'

that LHS multiplies out to 1/2 x'² - 1/2 y'², which you can see is the LHS of the next equation

(and the RHS of both equations stays as 2)

 

[bobsmith76]

are you saying they square √2/2 - √2/2? if they do that is still zero. I don't know what you mean by I'm ignoring x'. What x' am i ignoring.

 

[tiny-tim]

are you saying they square √2/2 - √2/2?

they don't have √2/2 - √2/2 !

are we looking at the same question? ​

all i'm seeing is something similar with x' and y'

 

[HallsofIvy]

They are NOT dealing with $\sqrt{2}/2- \sqrt{2}/2$, they are dealing with $(\sqrt{2}/2)x'- (\sqrt{2}/2)y'$.

 

[bobsmith76]

ok, thanks. I misread what I was reading I thought it was

(√2/2x' - √2/2x')(√2/2y' + √2/2y')