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Unformatted text preview: Math 100 Homework 4 fall 2010 1. Let f ( x, y ) = 2000 . 02 x 2 . 04 y 2 , p = ( 20 , 5 , 1991) (a) The directional derivative of f along (1 , 0) (the unit vector pointing to the east) at p is D (1 , 0) f ( p ) = ∂f ∂x ( 20 , 5) = . 04( 20) > . The climber would ascend by going east. (b) The directional derivative of f along 1 √ 2 ( 1 , 1) (the unit vector point ing to northwest) at p is 1 √ 2 ( 1 , 1) ·∇ f ( p ) = 1 √ 2 ( 1 , 1) · ( . 04( 20) , . 08(5)) = 1 . 2 √ 2 < . The climber would decend by going northwest. (c) If the climber begins a level path by going towards the direction v = ( v 1 , v 2 ) (so that v is a unit vector), then the directional derivative of f along v at p is 0. Hence 0 = v · ∇ f ( p ) = ( v 1 , v 2 ) · ( . 04( 20) , . 08(5)) = 0 . 4(2 v 1 v 2 ) . Therefore 2 v 1 = v 2 , and recall that v is a unit vector, we have v = 1 √ 5 (1 , 2) or 1 √ 5 (1 , 2) . Consequently, the climber begins a level path by going along the direction with compass bearing (east is regarded as 0 ◦ ) tan 1 2 or π + tan 1 2. 2. Let F ( x, y, z ) = xf ( x/y ) z . For each point ( a, b, af ( a/b )) on the level surface F ( x, y, z ) = 0, we have n = ∇ F ( a, b, af ( a/b )) = ( f ( a/b ) + af ( a/b ) /b, a 2 f ( a/b ) /b 2 , 1) ....
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 Spring '11
 ching
 Calculus, Derivative

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