I would like to ask how to differentiate a curve lying on a regular parametrised surface. I came across several questions that involve a curve lying on a surface. Differentiating the parametrised curve was necessary to prove the statements. My teacher gave us the following theorem (or definition?) as a 'clue' but I am struggling to understand why this is true.

$\vec r$' = $\vec x $$_u$$\cdot$u'+$\vec x$$_v$$\cdot$v'

Let $\vec x$(u, v) = (f$_1$(u,v), f$_2$(u,v), f$_3$(u,v)) be a regular parametrised surface
and $\vec r$(t) = $\vec x$(u(t), (v(t)) be a curve lying on the surface

$\frac{d\vec r}{ds}$ = ($\frac{df_1(u, v)}{du}$$\cdot$$\frac{du}{dt}$ + $\frac{df_1(u, v)}{dv}$$\cdot$$\frac{dv}{dt}$, ..., ...) = $\vec x $$_u$$\cdot$u'+$\vec x$$_v$$\cdot$v'

I understand the chain rule, but I don't quite understand why the two derivatives of vector are added together. Is it possible that this is because the following is (might be?) true:

$\vec r$(u, v) = $\alpha$$\vec r$$_u$ + $\beta$$\vec r$$_v$
where they are linearly independent?

I am new to this and am still struggling with vector differentiation. I know this question may sound stupid but try not to laugh... Your help will be greatly appreciated. :)

  • $\begingroup$ It might help to read the multivariate case of the chain rule math.libretexts.org/Bookshelves/Calculus/… $\endgroup$ – irchans Aug 16 at 6:31
  • $\begingroup$ Thank you so much. You have been very helpful. @irchans I edited my post because this question should fall under the category of multivariable-calculus, rather differential-geometry. $\endgroup$ – EF160 Aug 16 at 6:57

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