Re: Gradient of dot product

let f = u1.u2 = u11u21 + u12u22 = f(x,y)

delf = df/dx**i** +df/dy**j**

del acosf = [d(acosf)/dx]**i** + [d(acosf) /dy]**j** = -1/sqrt(1-f^2)[(df/dx)**i** +(df/dy)**j**]

del acosf = -1/sqrt(1-f^2)delf

delf = u21delu11 + u21delu22 + u11delu21 + u12delu22, or

delf = delu1.u2 = u1.delu2 + u2.delu1 + u1x(delxu2) + u2x(delxu1) look it up, or google

Same in 3 dimensions just more to write. If u1 and u2 are orthogonal unit vectors, go back and think about it, oh well, then u1.u2 =1 and delf=0, so probably not.

Didn't feel like bolding del and all the vectors, ie, del = **del** and u1 = **u**1 etc. U11..are scalars, d is partial.

Re: Gradient of dot product

if u1 & u2 are unit vectors, then u1.u2 = cosθ. Then

acos(u1.u2) = θ

but what is del(θ)? formally, in 3d,

**del**(θ) = dθ/dx **i**+ dθ/dy **j** + dθ/dz **k**, where θ is angle between u1 & u2

I have the feeling there’s a further simplification, but I don’t see it.