# How to differentiate a parametrised curve lying on a surface

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'

i.e.,
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

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

• It might help to read the multivariate case of the chain rule math.libretexts.org/Bookshelves/Calculus/… – irchans Aug 16 at 6:31
• 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. – EF160 Aug 16 at 6:57