Differentiating the Dot Product's Evil Twin

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In summary, the conversation discusses the use of cosine and sine in dot and cross products. The main question is why the "evil twin" of the dot product is not differentiable at parallel vectors. The answer lies in the presence of a square root function in the formula, which is not differentiable at zero. This is demonstrated through examples of other non-differentiable functions. The regular dot and cross products do not have this issue and can smoothly pass through zero angles.

Trying2Learn

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Hello,

I found this link very useful:
https://www.quora.com/Why-is-cosine-used-in-dot-products-and-sine-used-in-cross-products

I understand all of Anders Kaseorg's discussion except for ONE PONT.

At the very end, he writes: "[the evil twin of the dot product] is not differentiable at parallel vectors."

Could someone explain why? (I can see the issue with the evil twin of the cross, but not the dot)

Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

The actual function being used is something like ##\sqrt{|x|}##. If you graph that function on the interval ##[-1,1]## you'll see that it's continuous but not differentiable at ##x=0##. It bounces off the x-axis there.
Also look at the graph of ##|\sin x|## on ##[-1,1]## to see another relevant non-differentiable bounce.

By contrast, the formulas for the usual dot and the usual cross have no square roots, or anything else that can upset differentiability, such as the denominators in the evil twin cross. They pass zero angles (parallel for dot, perp for cross) smoothly.

andrewkirk said:
Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

OH! I see!

THANK YOU!