Why does this formula yield a clockwise rotation for curves defined by a cartesian equation, and a counterclockwise rotation in parametric form

algebra-precalculusanalytic geometrygeometryrotations

Here is the reasoning I've attemped in order to espablish a formula for the counterclockwise rotation of a curve.

Let $M=(x,y)$ be any point on the plane, associated with a cartesian coordinate system $<0, \vec { i}, \vec{j}>$.

The position vector of ths point is $\vec {OM} = \vec A+ \vec B= x\vec { i} + y \vec{j} $

Rotating point $M$ counterclockwise by a positve angle $\alpha$ amounts to rotate counterclockwise the components of its postion vector by the same angle.

So , with $M'$ the image of $M$ under ths rotation, we will have :

$\vec {OM'} = \vec {A'}+\vec {B'}$ with $\vec {A'}$, the image of $\vec {A}$ and $\vec {B'}$ , the image of vector $\vec {B}$.

Now ,

$\vec {A'}= x cos (\alpha) \vec i + x sin (\alpha) \vec j$.

$\vec {B'}= y cos (\alpha + \pi /2) \vec i + x sin (\alpha+ \pi/2) \vec j $

$= y ( – sin (\alpha) )\vec i + x cos(\alpha) \vec j$

$= -y sin (\alpha) )\vec i + x cos(\alpha)\vec j$.

Therefore ,

$\vec{OM'} = \vec {A'} + \vec {B'} = \Large{[} x cos (\alpha) – y sin (\alpha)\Large{]} \vec i + \Large{[}x sin( \alpha) + y cos(\alpha)\Large{]} \vec j$,

implying that

$M' = (x cos (\alpha) – y sin (\alpha) , x sin( \alpha) + y cos(\alpha))$.

My problem is that :

(1) when I apply this reasoning to a parametric curve, it seems to work , that is, if I have a curve $C= < f(t), g(t)> $ and ask for

$$C' = <f(t) cos (\alpha) – g(t) sin (\alpha) , f(t) sin( \alpha) + g(t) cos(\alpha)>$$

what I get is that $C'$ rotates counterclockwise, as desired.

Example with $\alpha = 2 $ rd: https://www.desmos.com/calculator/daaouk87yw

(2) but if I substitute the coordinates of a generic $M'$ point in a cartesian " $x-y$" equation, the curve rotates clockwise.

Example : https://www.desmos.com/calculator/vm5pbtxzge

I'm sure some confusion prevents me from understanding what is happening, but I can't manage to see which one.

Best Answer

We'll compare the two operations you describe. Instead of reasoning with coordinates, let's use directly the point $M$, and the operation $R$ applied to $M$. $R$ can be a rotation, or any other bijective function.

  • In the first case, you directly change the parametric equation $M = F(t)$ that expresses $M$ according to some parameter $t$. This becomes $M' = R \circ F (t)$.
  • In the second case, you transform the usage of $M$ coordinates in an equation they verify. If the equation is $E(M) = 0$, it becomes $E(R(M''))=0$.

$R$ being bijective, $E(R(M''))=0$ is equivalent to $M=R(M'')$.
While in the first case $M'=R(M)$.
We can see the role of $R$ is reversed; hence what you remarked: if you express $M''=S(M)$, we see that $S=R^{-1}$.

This is very intuitive, in fact. Let's say it with a scaling operation; compare:

  • "I'll give you twice the amount $M$".
  • "The amount I'll give you is such that, when multiplied by two, it is the same as $M$".

Another similar situation happens when transforming a vector, versus transforming the vector base: that's covariance and contravariance.

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