New coordinates from the rotation of an axis

[rajeshmarndi]

Homework Statement

There is a point P(x,y) and now I rotate the x-y axis, say by θ degree. What will be the coordinates of P from this new axis.

I have google but found formula for new coordinates when the points is rotated by θ degree. So I tried my own. So is there other simplified formula for the above situation.

Homework Equations

The Attempt at a Solution

Plz see the attached figure.
AC-AD is the new axis and (x',y') are the new coordinates of point P.

y'=PC
cos θ= PC/PB = y'/PB
y'= PB cos θ

PB= y- BE
tan θ= BE/x , BE = x tanθ
PB = y - xtanθ
y'= (y- xtanθ) cosθ
= ycosθ - xsinθ --------------eq(1)x'= AB + BC
sinθ=BE/AB
AB=BE/sinθ, tanθ=BE/x
BE=xtanθ
AB=xtanθ/sinθ = x/cosθ

tanθ = BC/y'
BC= y'tanθ
= (ycosθ-xsinθ)tanθ , ( y' from eq(1) )
= ysinθ - x sinθtanθ
x' = AB + BC
= x/cosθ + ysinθ - x sinθtanθ

So the new coordinates are
x'= x/cosθ + ysinθ - x sinθtanθ
y'= ycosθ - xsinθ

 

[Orodruin]

x'= x/cosθ + ysinθ - x sinθtanθ

Can you think of a way of rewriting x(1/cosθ - sinθ tanθ) to a form which is slightly more pleasant to look at? Otherwise this formula is correct.

 

[rajeshmarndi]

Can you think of a way of rewriting x(1/cosθ - sinθ tanθ) to a form which is slightly more pleasant to look at? Otherwise this formula is correct.

Thanks.
x'= x/cosθ + ysinθ - x sinθtanθ
x'= x/cosθ - xsin²θ/cosθ + ysinθ
x'= x(1-sin²θ)/cosθ) + ysinθ
x'= x(cos²θ/cosθ) + ysinθ
x'= xcosθ + ysinθ

Now it is pleasant to look at.

This looks very much similar to, when a points is rotated by θ. The new coordinates are,

x' = x cos θ - y sin θ
y' = y cos θ + x sin θ

Only difference is the -/+ sign. Seems like the difference is because as we increase the θ(when a points is rotated), x' gets shorter. Thats just a guess with a first look.

 

[Orodruin]

Yes, the sign depends on the active-vs-passive transformation, i.e., if you rotate the points or the coordinate system.

 

[Ray Vickson]

Thanks.
x'= x/cosθ + ysinθ - x sinθtanθ
x'= x/cosθ - xsin²θ/cosθ + ysinθ
x'= x(1-sin²θ)/cosθ) + ysinθ
x'= x(cos²θ/cosθ) + ysinθ
x'= xcosθ + ysinθ

Now it is pleasant to look at.

This looks very much similar to, when a points is rotated by θ. The new coordinates are,

x' = x cos θ - y sin θ
y' = y cos θ + x sin θ

Only difference is the -/+ sign. Seems like the difference is because as we increase the θ(when a points is rotated), x' gets shorter. Thats just a guess with a first look.

If you fix the point and rotate the coordinate system, the new coordinates $(x',y')$ are given by
$$x' = \cos(\theta) x + \sin(\theta) y \\ y' = -\sin(\theta) x + \cos(\theta) y$$
If you fix the coordinate system and rotate the point, the new coordinates $(x',y')$ are given by
$$x' = \cos(\theta) x - \sin(\theta) y\\ y' = \sin(\theta) x + \cos(\theta) y$$

 

[ND3]

ycosθ-xsinθ=(ysinθ+xcosθ)^2

 

[ND3]

ycosθ-xsinθ=(ysinθ+xcosθ)^2

Plug in the angle you are rotating by for new equation of parabola (counterclockwise is positive angle)

 

[SammyS]

Plug in the angle you are rotating by for new equation of parabola (counterclockwise is positive angle)

Hello, @ND3 .

This thread is 3 & 1/2 years old.

Many of us have posted to a similarly old thread at one time or another.