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The gradient as a normal vector

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) = <x(t),y(t),z(t)>. P. C.

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The gradient as a normal vector

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  1. The gradient as a normal vector

  2. Consider z=f(x,y) and let F(x,y,z) = f(x,y)-z Let P=(x0,y0,z0) 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) = <x(t),y(t),z(t)>.

  3. P C

  4. 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:

  5. In particular, this is true at t0, so that r(t0)=<x0,y0,z0> P r’(t0) C

  6. 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

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

  8. 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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