P1_soln

P1_soln - Math 1920 Prelim 1 Solutions Tuesday Feb 17th...

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Unformatted text preview: Math 1920 Prelim 1 - Solutions Tuesday, Feb 17th, 2009 1. [15pts] Consider the vectors ~v = i + 2 j + a k and ~w = i + 2 j + k . (a) Find all values of the parameter a (if any) such that ~v is perpendicular to ~w . A: ~v · ~u = < 1 , 2 , a > · < 1 , 2 , 1 > = 1 + 4 + a = 0 → a =- 5 (b) Find all values of the parameter a (if any) such that the area of a parallelogram determined by ~v and ~w is equal to √ 20 . A: ~v × ~u = < 1 , 2 , a > × < 1 , 2 , 1 > = 2(1- a ) i + ( a- 1) j || ~v × ~u || = p 4(1- a ) 2 + ( a- 1) 2 = | a- 1 | √ 5 = √ 20 → a =- 1 , 3 (c) Find the projection of ~w onto the plane 2 x + 1 y + 2 z = 4 A: Note first that ~w = ( proj ~n ~w ) + ( ~w- proj ~n ~w ) we are interested in ( ~w- proj ~n ~w ). proj plane ~w = ~w- proj ~n ~w = ~w- ~w · ~n | ~n | 2 ~n = < 1 , 2 , 1 >- < 1 , 2 , 1 > · < 2 , 1 , 2 > 9 < 2 , 1 , 2 > =- 1 3 , 4 3 ,- 1 3 2. [10pts] Let ~ r ( t ) be the vector function defined by ~ r ( t ) = i + 1 2 t 2 j + 1 3 t 3 k . Compute the arc length parameter s ( t ) of ~ r ( t ) starting from t=0. A: s ( t ) = Z t ds = Z t √ T 2 + T 4 dT = Z t T √ 1 + T 2 dT where we make the substitution u = 1 + T 2 and du = 2 T dT . s ( t ) = Z 1+ t 2 1 1 2 √ u du = Z 1+ t 2 1 1 2 2 3 u 3 / 2 1+ t 2 1 = 1 3 ( 1 + T 2 ) 3 / 2 t = 1 3 ( 1 + t 2 ) 3 / 2- 1 3 3. [10pts] Find ∂ 4 f ∂ 2 x∂y∂z , where f ( w, x, y, z ) = arctan( x z ) + z y e xy + y ln ( w x ) z + sin( y ) A: ∂ 4 f ∂ 2 x∂y∂z = ∂ 4 ∂ 2 x∂y∂z arctan( x z ) + z y e xy + y ln ( w x ) z + sin( y ) = ∂ 4 ∂ 2 x∂y∂z [tan- 1 ( x z )] + ∂ 4 ∂ 2 x∂y∂z z y e xy + ∂ 4 ∂ 2 x∂y∂z [ y ln ( w x ) z ] + ∂ 4 ∂ 2 x∂y∂z [sin(...
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This note was uploaded on 12/25/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

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P1_soln - Math 1920 Prelim 1 Solutions Tuesday Feb 17th...

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