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Coordinate curves problem

  1. Jun 6, 2008 #1
    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 [tex]\mathbf{r}(\lambda)[/tex] 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 [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:

    \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 [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...
  2. jcsd
  3. Jun 6, 2008 #2
    Proof seems legit, if awkwardly phrased.
  4. Jun 7, 2008 #3


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    Homework Helper

    Hi Ceres629! :smile:

    Since you're given:
    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 [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}\,=\,0[/tex] if ∂r/∂u and ∂r/∂v are parallel;
    and λ can be chosen to be arc-length, so that dr/dλ ≠ 0. :smile:
  5. Jun 20, 2008 #4
    thanks tiny tim, your answer also makes sense.
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