- #1

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Given a surface S defined as the end point of the vector:

[tex]

\mathbf{r}(u,v) = u\mathbf{i} + v\mathbf{j} + f(u,v)\mathbf{k}

[/tex]

and any curve on the surface represented by

[tex]

\mathbf{r}(\lambda) = \mathbf{r}(u(\lambda),v(\lambda))

[/tex]

and it mentions the tangent to the curve [tex]\mathbf{r}(\lambda)[/tex] is given by

[tex]

\frac{d\mathbf{r}}{d\lambda} = \frac{\partial \mathbf{r}}{\partial u} \frac{du}{d\lambda} + \frac{\partial \mathbf{r}}{\partial v} \frac{dv}{d\lambda}

[/tex]

It then goes on to focus on a specific case dealing with the curves u = constant and v = constant. It says:

The curves u = constant and v = constant passing through any point P on the surface S are said to be called coordinate curves.

I follow the above however it then states...(for u and v being coordinate curves)

If the surface is smooth, then the vectors [tex]\partial \mathbf{r} / \partial u[/tex] and [tex]\partial \mathbf{r} / \partial v[/tex] are linearly independant.

It gives no explanation as to why the two vectors are linearly independant... any ideas as to how to prove this?

The only thing i could come up with was that since the surface is smooth then any curves on it must be also be smooth and thus differentiable at all points and therefore [tex]d\mathbf{r}/ d\lambda[/tex] must be a non zero vector since the curve [tex]\mathbf{r}(\lambda)[/tex] must have a defined tangent.

The equation:

[tex]

\frac{\partial \mathbf{r}}{\partial u} \frac{du}{d\lambda} + \frac{\partial \mathbf{r}}{\partial v} \frac{dv}{d\lambda}= \mathbf{0}

[/tex]

cannot be true for any non trivial combination of the vectors [tex]\partial \mathbf{r} / \partial u[/tex] and [tex]\partial \mathbf{r} / \partial v[/tex] therefore they are linearly independant.

Seems a bit of a fuzzy proof though...