Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).
The normal vector at a point on a surface is given by
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(1)
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where and are partial derivatives. and (x0, y0) = (5,5)
we have z = (225 - 6x^2 - 2y^2)^(1/2) = f(x,y), giving us
fx = -6x(225 - 6x^2 - 2y^2)^(-1/2), and fx(5,5) = -30(225 - 150 - 25)^(-1/2) = -30(25)^(-1/2) = -6
fy = -2y(225 - 6x^2 - 2y^2)^(-1/2), and fx(5,5) = -10(225 - 150 - 25)^(-1/2) = -10(25)^(-1/2) = -2
Therefore, N = | -6 |
| -2 |
| -1 |