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Thread: Vector Analysis

  1. #1
    Super Member Aryth's Avatar
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    Vector Analysis

    Let $\displaystyle f(x,y,z) = x^2 + 2y^2 + 3z^2$ and let S be the isotimic surface: $\displaystyle f = 1$. Find all the points $\displaystyle (x,y,z)$ on S that have tangent planes with normals $\displaystyle (1,1,1)$.

    I'm not even sure where to begin...
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  2. #2
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    We may use the fact that the normal to any point on the level curve of $\displaystyle f$ is parallel to $\displaystyle \nabla f$ at that point. We know this because

    $\displaystyle \frac{d}{dt}f(\mathbf{r}(t))=\nabla f(\mathbf{r}(t))\cdot \mathbf{r'}(t)=0$

    for a differentiable curve $\displaystyle \mathbf{r}$ on the surface $\displaystyle f(x,y,z)=c$, which means that $\displaystyle \nabla f$ is perpendicular to the curve and thus to the plane itself.

    In our case, we are to find all those points of $\displaystyle f(x,y,z)=1$ such that

    $\displaystyle \nabla f=\left<c,c,c\right>$

    for some constant $\displaystyle c$.
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