ps5sol - 18.02 Fall 2008 Problem Set 5, Part B Solutions 1....

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Unformatted text preview: 18.02 Fall 2008 Problem Set 5, Part B Solutions 1. In general, dg ds vextendsingle vextendsingle vextendsingle vextendsingle u = g u is largest when u = dir g and g (dir g ) = g g | g | = | g | 2 | g | = | g | . Similarly, the directional derivative is smallest when u = dir g and its value is | g | . Therefore the answers to (a) and (b) are as follows. a) Since f = ( 1 / 2 , 1 ) , the maximum directional derivative is |( 1 / 2 , 1 )| = 5 / 2; the minimum is 5 / 2. b) The maximum is achieved in the direction u = dir ( 1 / 2 , 1 ) = ( 1 , 2 ) / 5. The minimum is achieved in the opposite direction, u = dir ( 1 / 2 , 1 ) = ( 1 , 2 ) / 5. c) We need u ( 1 / 2 , 1 ) = 0. In other words, u = dir ( 2 , 1 ) = ( 2 , 1 ) 5 d) The maximum of df/ds is approx. 1 . 12 ( . 02) achieved when u = ( cos 1 , sin 1 ) , 1 243 ( 5 ). u points parallel to (and the same direction as) the gradient vector and perpendicular to the contour line....
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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ps5sol - 18.02 Fall 2008 Problem Set 5, Part B Solutions 1....

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