Engineering Calculus Notes 305

Engineering Calculus Notes 305 - 3.5. SURFACES AND THEIR...

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Unformatted text preview: 3.5. SURFACES AND THEIR TANGENT PLANES 293 − with →(0) = P . Then the velocity vector p ′ (0) lies in the plane tangent to p → the surface L(f, c) at − (0). p Now on one hand, by the Chain Rule (3.3.6) we know that d dt t=0 → − → [f (− (t))] = ∇ f (P ) · p ′ (0) ; p → → on the other hand, since − (t) lies in the level set L(f, c), f (− (t)) = c for p p all t, and in particular, d dt t=0 → [f (− (t))] = 0. p It follows that → − ∇ f (P ) · p ′ (0) = 0 for every vector tangent to L(f, c) at P ; in other words,10 Remark 3.5.4. If P is a regular point of f (x, y, z ), then the tangent plane to the level set L(f, c) through P is the plane through P perpendicular to → − the gradient vector ∇ f (P ) of f at P . If we write this out in terms of coordinates, we ﬁnd that a point (x, y, z ) = (x0 + △x, y0 + △y, z0 + △z ) lies on the plane tangent at (x0 , y0 , z0 ) to the surface f (x, y, z ) = c = f (x0 , y0 , z0 ) if and only if ∂f (x0 , y0 , z0 ) △x + ∂x ∂f (x0 , y0 , z0 ) △y + ∂y ∂f (x0 , y0 , z0 ) △z = 0, ∂z in other words, if d(x0 ,y0 ,zso) f (x − x0 , y − y0 , z − z0 ) = 0. Yet a third way to express this is to add c = f (x0 , y0 , z0 ) to both sides, noting that the left side then becomes the linearization of f at P : Tx0 ,y0 ,z0 f (x, y, z ) = f (x0 , y0 , z0 ) . We summarize all of this in 10 Strictly speaking, we have only shown that every tangent vector is perpendicular to → − → − ∇ f ; we need to also show that every vector which is perpendicular to ∇ f is the velocity vector of some curve in L(f, c) as it goes through P . See Exercise 14. ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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