Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
92 LECTURE 12. THE CHAIN RULE Revised December 6, 2006. Math 250, Fall 2006
Lecture 13 Tangent Planes Since the gradient vector is perpendicular to a surface in three dimensions, we can find the tangent plane by constructing the locus of all points perpendicular to the gradient vector. In fact, we will define the tangent plane in terms of the gradient vector. Definition 13.1 The tangent plane to a surface f(x, y, z) = k at a point P 0 = (x 0 , y 0 , z 0 ) is the plane perpendicular to the gradient vector at p 0 . To get a formula for the tangent plane, let P = (x, y, z) run over all points in the plane. Then any vector of the form P − P 0 = (x − x 0 , y − y 0 , z − z 0 ) = (x − x 0 )i + (y − y 0 )j + (z − z 0 )k Since ∇f(x 0 , y 0 , z 0 ) is normal to the tangent plane, the critical equation defining the tangent plane is (P − P 0 ) · ∇f(x 0 , y 0 , z 0 ) = 0 (13.1) Example 13.1 Find the equation of the plane tangent to the surface at the point P 0 = (1, 2, 3). 3x 2 + y 2 − z 2 = −20 Solution. First, we rewrite the equation of the surface into the form f(x, y, z) = 0, as f(x, y, z) = 3x 2 + y 2 − z 2 + 20 = 0 The general form of the gradient vector at any point on the surface is then ∇f = ∂f ∂x i + ∂f ∂y j + ∂f ∂z At the point P 0 = (1, 2, 3) the gradient is k = 6xi + 2yj − 2zk ∇f(P 0 ) = 6(1)i + 2(2)j − 2(3)k = 6i + 4j − 6k 93
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92 LECTURE 12. THE CHAIN RULE<br />
Revised December 6, 2006. Math 250, Fall 2006
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