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Math Help - Tangent vectors to a surface

  1. #1
    Junior Member
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    Tangent vectors to a surface

    I hesitated to put this in calculus but oh well.

    We have a surface S : 2x^2 + 3y^2 + x - z = 0 and a point P(1;2;15) belonging to that surface.

    I was asked to find the normal vector, which I did (its (5;12;-1) ). I am also asked to find vectors linearly independant and tangent to that S surface.

    By using partial derivatives I have found

    z - 15 = 5(x - 1)

    z - 15 = 12(y - 2)

    I don't know how to find the vectors from that point. I know that 5 is the slope for the partial derivative in x and 12 is the slope for the partial derivative in y.

    Thanks for the help!
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  2. #2
    MHF Contributor

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    From the equation 2x^2+ 3y^2+ x- z= 0 we get z= 2x^2+ 3y^2+ x and so can write the "position vector" of any point on the surface in terms of x and y only:
    \vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (2x^2+ 3y^2+ x)\vec{k}

    The derivatives of that vector with respect to x and y, evaluated at (1, 2, 15), will be independent tangent vectors.
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