math203spring2009sols - 203 Spring 2009 Group Final...

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203 Spring 2009 Group Final Solutions Q1a: unit vectors are: v1 = 8 4, 0, 1 < Sqrt @8 4, 0, 1 < . 8 4, 0, 1 <D : 4 17 , 0, 1 17 > v2 = - 8 4, 0, 1 < Sqrt @8 4, 0, 1 < . 8 4, 0, 1 <D : - 4 17 , 0, - 1 17 > Q1b: substitute and solve to get the point of intersection as (-8, 4, 0): Solve @ 4 t + 2 H 4 L + 3 H 2 + t L ã 0 D 88 t Ø - 2 << 8 4 t, 4, 2 + t < ê . t Ø - 2 8 - 8, 4, 0 < Q1c: Not perpendicular as the normal vector to the plane {1,2,3} is not parallel to the direction of the line {4,0,1} Q2a: Find the gradient at the point and its length: f = x 2 - Cos @ x y D + Exp @ y z D y z + x 2 - Cos @ x y D v = 8 D @ f, x D , D @ f, y D , D @ f, z D< ê . 8 x Ø 3, y Ø 0, z Ø 1 < 8 6, 1, 0 < a unit vector in the direction of greatest increase: v Sqrt @ v.v D : 6 37 , 1 37 , 0 > the rate of increase is the length of the gradient: Sqrt @ v.v D 37 Q2b: take the dot product to get the desired directional derivative:
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v . 8 - 5, 12, 0 < 8 - 5, 12, 0 < . 8 - 5, 12, 0 < ê . 8 x Ø 3, y Ø 0, z Ø 1 < - 18 13 Q2c: gradient vector gives the normal to tangent plane: 6 H x - 3 L + 1 H y - 0 L + 0 H z - 1 L ã 0 Q3: the region is given by: Plot @8 8 x, 8, x^3 < , 8 x, 0, 2 <D 0.5 1.0 1.5 2.0 5 10 15 To find its mass, integrate to get: 0 8 y ê 8 y 3 I 3 x 2 M x y 30 Q4: Set the first derivatives equal to zero and solve to get:
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This note was uploaded on 01/29/2012 for the course MATH 203 taught by Professor Snell during the Spring '11 term at City College of San Francisco.

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math203spring2009sols - 203 Spring 2009 Group Final...

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