# The gradient as a normal vector. Consider z=f(x,y) and let F(x,y,z) = f(x,y)-z Let P=(x 0,y 0,z 0 ) be a point on the surface of F(x,y,z) Let C be any.

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The gradient as a normal vector

Consider z=f(x,y) and let F(x,y,z) = f(x,y)-z Let P=(x 0,y 0,z 0 ) be a point on the surface of F(x,y,z) Let C be any curve on the surface that passes through P and has the vector equation r(t) =.

C P

Since C is on S, any point (x(t),y(t),z(t)) must also satisfy F(x(t),y(t),z(t))=0 The chain rule says that:

C P r’(t 0 ) In particular, this is true at t 0, so that r(t 0 )=

Interpretation The gradient vector at P is perpendicular to the tangent vector to any curve C on S that passes through P This means the tangent plane at P has normal vector

We know how to write the equation of that plane! Remark: The same idea holds true for any F(x,y,z)=k, since

Consequence: Similarly it holds that for functions of two variables that that the gradient is perpendicular to level curves of the form f(x,y) =k (i.e. contour lines!) Examples….

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