Question

If the original coordinate axes are rotated 45° to obtain the x' and y' axes, what is the value of x in terms of x' and y'?

Answer 1

@jdoe0001

Answer 2

This seems fun

Answer 3

a is set up like this \[x=\frac{ \sqrt{2x'} }{ 2}-\frac{ 2y' }{ 2 }\]

Answer 4

you can use a simple rotation matrix
\[ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\]

Answer 5

Hmm interesting if that is A, I remember doing these guys in Calc 2 and the formula *we can derive it if you want would be

\[\large x = \hat xcos\theta - \hat y sin(theta)\]

Answer 6

D is set up like this \[x=\frac{ 1 }{ 2}x'+\frac{ \sqrt{3} }{ 2}y'\]

Answer 7

...

Answer 8

http://www.wolframalpha.com/input/?i=rotation+matrix&a=*C.rotation+matrix-_*Calculator.dflt-&f2=45%C2%B0&f=RotationCalculator.alpha_45%C2%B0&a=*FP.RotationCalculator.dir-_plus&f4=%7Bx%2C+y%7D&f=RotationCalculator.point_%7Bx%2C+y%7D&a=*FVarOpt.1-_***RotationCalculator.point--.***RotationCalculator.axis---.*--