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samsolfinbfs10up

# samsolfinbfs10up - MATH 23 Sample Final Exam April 2003...

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Unformatted text preview: MATH 23 Sample Final Exam April, 2003 NAME : B.Dodson Section 110- 313 (Last, First) 1. ( 15 points ) (a) Find a vector perpendicular to the plane through (0 , 2 , 0) , (2 , 3 ,- 1) and (1 , 2 ,- 4) . (b) give an equation for the plane in part (a). See solutions to the sample exam 1 (course documents). 2. ( 10 points ) Find an equation for the tangent line to the curve-→ r ( t ) = ( t 2- 3) i- 2 tj + t 2 k at the point P (1 ,- 4 , 4) .--→ r ( t ) = 2 ti- 2 j + 2 tk, and-→ r ( t ) = < 1 ,- 4 , 4 > when- 2 t =- 4 , so t = 2 (using the second component). So the direction vector of the tangent line is--→ r (2) = 4 i- 2 j + 4 k = < 4 ,- 2 , 4 >, which gives ( t ) = < 1 ,- 4 , 4 > + t < 4 ,- 2 , 4 > . 3. ( 20 points ) The position function for the motion of a particle is given by-→ r ( t ) = ti + 2 tj + t 2 k = < t, 2 t, t 2 > . (a) Find the velocity and acceleration vectors. v = < 1 , 2 , 2 t >, a = < , , 2 > . (b) Find the tangential component a T of acceleration. | v | = √ 1 + 4 + 4 t 2 , and v · a = 4 t so a T = 4 t √ 5 + 4 t 2 . (c) Find the curvature when t = 0 . v (0) = < 1 , 2 , >, a (0) = < , , 2 >, so | v (0) | = √ 5 , and the cross product is < 4 ,- 2 , > = 2 < 2 ,- 1 , > with length 2 √ 5 so κ (0) = 2 √ 5 ( √ 5) 3 = 2 5 . 4. ( 10 points ) Let f ( x, y, z ) = x 2 y + xyz 3 . Find each of the following. (a) ∂f ∂x (b) ∂ 2 f ∂x 2 (c) ∂ 2 f ∂z∂x (1 , 2 , 3) (direct.) 5. ( 15 points ) Let f be the function f ( x, y ) = x 2- 3 y 2 . (a) Find the directional derivative of f at the point (2 , 3) in the direction of the vector a = <- 4 , 3 > =- 4 i + 3 j. (b) Find the maximal rate of change of f at the point...
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samsolfinbfs10up - MATH 23 Sample Final Exam April 2003...

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