Proving the Linear Independence of Coordinate Curves on a Smooth Surface

[Ceres629]

I'm stuck on a problem on vector calculus.

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

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

and any curve on the surface represented by

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

and it mentions the tangent to the curve $$\mathbf{r}(\lambda)$$ is given by

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

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 $$\partial \mathbf{r} / \partial u$$ and $$\partial \mathbf{r} / \partial v$$ are linearly independent.

It gives no explanation as to why the two vectors are linearly independent... 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 $$d\mathbf{r}/ d\lambda$$ must be a non zero vector since the curve $$\mathbf{r}(\lambda)$$ must have a defined tangent.

The equation:

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

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

Seems a bit of a fuzzy proof though...

 

[maze]

Proof seems legit, if awkwardly phrased.

 

[tiny-tim]

Hi Ceres629!

Since you're given:

and it mentions the tangent to the curve $$\mathbf{r}(\lambda)$$ is given by

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

I'd be more inclined to say that if the surface is smooth, then du/dλ and dv/dλ can take any values, and in partiuclar can therefore always be chosen to have (du/dλ)/(dv/dλ) = -|∂r/∂v|/|∂r/∂u|, so satisfying $$\frac{d\mathbf{r}}{d\lambda} = \frac{\partial \mathbf{r}}{\partial u} \frac{du}{d\lambda} + \frac{\partial \mathbf{r}}{\partial v} \frac{dv}{d\lambda}\,=\,0$$ if ∂r/∂u and ∂r/∂v are parallel;
and λ can be chosen to be arc-length, so that dr/dλ ≠ 0.

 

[Ceres629]

thanks tiny tim, your answer also makes sense.