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HH_14_7 - x-1 4 y-1 4 x y z z = P 2 x y FIGURE 3 FIGURE 4...

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(11/11/08) Math 10C. Lecture Examples. Section 14.7. Second-order partial derivatives Example 1 What are the second-order derivatives of f ( x , y ) = xy 2 + x 3 y 5 ? Answer: f xx = 6 xy 5 f xy = f yx = 2 y + 15 x 2 y 4 f yy = 2 x + 20 x 3 y 3 Example 2 Find the second-degree Taylor polynomial approximation of f ( x , y ) = 1 - cos x cos y at x = 0 , y = 0 . Answer: P 2 ( x,y ) = 1 2 x 2 + 1 2 y 2 (The graph of f ( x,y ) = 1 - cos x cos y is in Figure 1, and the graph of its Taylor polynomial approximation is the circular paraboloid shown in Figure 2 with the graph of f .) x y z z = 1 - cos x cos y x y z z = P 2 ( x,y ) FIGURE 1 FIGURE 2 Example 3 Figure 3 shows the graph of f ( x , y ) = 5 - x + y - ( x - 1 ) 4 - ( y - 1 ) 4 and Figure 25 shows the graph of f with the graph of its second-degree Taylor polynomial P 2 ( x , y ) at x = 1 , y = 1 . Give a formula for P 2 ( x , y ) . What type of surface is its graph? x y
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Unformatted text preview: ( x-1) 4-( y-1) 4 x y z z = P 2 ( x, y ) FIGURE 3 FIGURE 4 Answer: P 2 ( x, y ) = 5-x + y • The graph of P 2 is a plane. † Lecture notes to accompany Section 14.7 of Calculus by Hughes-Hallett et al. 1 Math 10C. Lecture Examples. (11/11/08) Section 14.7, p. 2 Example 4 Figure 5 shows the graph of f ( x , y ) = y 2-1 12 y 4-x , and Figure 6 shows its graph with the graph of its second-degree Taylor polynomial z = P 2 ( x , y ) at x = , y = . Give a formula for P 2 ( x , y ) . x y z z = y 2-1 12 y 4-x x y z z = P 2 ( x, y ) FIGURE 5 FIGURE 6 Answer: P 2 ( x, y ) = y 2-x...
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