Tangent vectors to a surface

**mathieumg** · April 22nd 2010, 01:47 AM

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!

**HallsofIvy** · April 22nd 2010, 05:28 AM

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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