So,

I have f(v) and f(u), and B = f(v) * f(u) * c

When I take the first derivative of B, this requires the product rule between
the u and v functions, and I should get

Bdv = [f'(v) * f(u) + f(v) * f'(u)] * c

But when I take first PARTIAL derivatives, this does NOT require the product
rule, and I should get both

Bdv = f'(v) * f(u) * c     _and_

Bdu = f(v) * f'(u) * c

and these are essentially tangents to the surface in the direction of u or v



When I take second partial derivatives, I should get

Bdvdv = f''(v) * f(u) * c,
Bdvdu = f'(v) * f'(u) * c,
Bdudv = f'(v) * f'(u) * c, and
Bdudu = f(v) * f''(u) * c

the f''s indicate concave up or down along u or v (the change of the slope of
the tangent while traveling in that direction), but I'm a wee bit lost about the
meaning of the mixed partials (which are equivalent, right?) Are they how the
slope of the tangent changes as it gets "slid sideways" along the orthogonal
direction?

And I presumably just plug in u and v and go on my merry way...

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On 2020-09-22 9:50 PM (-4), Bald Eagle wrote:
> [snip]
> 
> But when I take first PARTIAL derivatives, this does NOT require the product
> rule, and I should get both
> 
> Bdv = f'(v) * f(u) * c     _and_
> 
> Bdu = f(v) * f'(u) * c
> 
> and these are essentially tangents to the surface in the direction of u or v
> 
> 
> 
> When I take second partial derivatives, I should get
> 
> Bdvdv = f''(v) * f(u) * c,
> Bdvdu = f'(v) * f'(u) * c,
> Bdudv = f'(v) * f'(u) * c, and
> Bdudu = f(v) * f''(u) * c
> 
> the f''s indicate concave up or down along u or v (the change of the slope of
> the tangent while traveling in that direction), but I'm a wee bit lost about the
> meaning of the mixed partials (which are equivalent, right?) Are they how the
> slope of the tangent changes as it gets "slid sideways" along the orthogonal
> direction?

Get back to me in the spring of 1983, and I'll explain it all.

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