s11wk08su

# s11wk08su - Solution We had the gradient< y x ln x>...

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Math 23 B. Dodson Week 4 Homework: 14.3 [partial derivatives and ] 2nd order deriv 14.4 [tangent plane and] diferentials 14.5 chain rule 14.6 directional derivatives 14.7 max/min 14.8 Lagrange multipliers 15.1 approximating sums For double integrals [Just: Section 15.1: 6] Problem 14.6.8: (a) ±ind the gradient oF f ( x, y ) = y ln x . (b) Evaluate the gradient at P (1 , - 3) . (c) ±ind the rate oF change oF f at P in the direction oF the unit vector vu = < - 4 5 , 3 5 > . Solution: The gradient is < y x , ln x >, which at P (1 , - 3) is < - 3 1 , ln 1 > = < - 3 , 0 > . So the rate oF change is D vu f (1 , - 3) = --→ grad · vu = < - 3 , 0 >< - 4 5 , 3 5 > = 12 5 . We say that f is increasing at a rate oF 2.4 in the direction oF vu.

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2 Problem 14.6.8d: Find the maximum rate of change of f at P and the direction in which it occurs.
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Unformatted text preview: Solution. We had the gradient < y x , ln x >, which at P (1 ,-3) is <-3 1 , ln 1 > = <-3 , > . so the max rate of change is the length of the gradient, 3, which occurs in the direction of-v i = <-1 , >, the direction of the gradient. Week 4 Homework: 14.7 max/min Problem 14.7.28: [omitted] We covered the Text Example 3, showing the use of the second deivative test. Week 8 Homework: 14.8 Lagrange multipliers We covered the text Example 2, which resembles the homework turned-in, using 2-variable functions. For . . . an example in 3-variables . . . [omitted]....
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## This note was uploaded on 11/24/2011 for the course MATH 23 taught by Professor Yukich during the Spring '06 term at Lehigh University .

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