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Find w, the angle ∠rox. That is, the angle from the red vector r to the positive x axis, in terms of u and v.

Also find the derivative of w when neither u nor v are held constant but instead are held to remain proportional. That is, doubling u will double v.

This is seeking to prove or disprove that if u and v are held proportional to each other that w changes with them, proportionally.


We can directly project vector $ \vec or $ on positive $x-$ axis as

$$ \cos \angle rox =(\cos v \cos u) $$

It is in effect the same when we take dot product of vectors

$$ \vec {ox}= (\cos v \cos u,0,0);\; \vec {or} = (\cos v \cos u,\cos v \sin u,\sin v) $$

$$\cos w_1=\cos ∠rox =\dfrac{(\cos v \cos u)^2+0+0}{(\cos v \cos u)\cdot (1) }=\cos v \cos u $$

$$\cos w_2=\cos 2v \cos 2u.$$

If $ v=um,$ then $ (\cos m v \cos u)$ is not proportional to $ (\cos v \cos u) $.

For the double angles

If e.g., $ u=v=\pi/3, \cos w_1=\dfrac14; \cos w_2= \dfrac14 $ which does not even change.

There can be no linearity or proportionality with circular trigonometric ratios products. Can be also confirmed by graphing the two cosines.


I would say, $\,\cos ∠rox = \frac{r\cos v \cos u}{r}=\cos v \cos u$


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