Winter 2008 - Cioaba's Class - Final Exam (Version 2)

# Winter 2008- - Math 20C Final(Lecture D Winter 2008 Duration 3 hours Please close your books and turn off your phones You can use one page of

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Unformatted text preview: Math 20C - Final (Lecture D, Winter 2008) Duration: 3 hours Please close your books and turn off your phones. You can use one page of handwritten notes. To get full credit, you should support your answers. 1. Consider the points A (2 , 4 , 5) and B (4 , 2 , 1). a) (2 points) Calculate the length of the segment AB and the area of the triangle OAB , where O is the origin. Solution. | AB | = radicalbig (4 − 2) 2 + (2 − 4) 2 + (1 − 5) 2 = √ 4 + 4 + 16 = √ 24 = 2 √ 6 . vector OA × vector OB = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vector i vector j vector k 2 4 5 4 2 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = − 6 vector i + 18 vector j − 12 vector k. Area ( OAB ) = | vector OA × vector OB | 2 = √ 6 2 + 18 2 + 12 2 2 = √ 36 + 324 + 144 2 = √ 126 . b) (3 points) Find an equation of the set of all points equidistant from the points A and B . Describe the set. Solution. Let P ( x, y, z ) be a point equidistant to A and B . It means that | PA | = | PB | which implies radicalbig ( x − 2) 2 + ( y − 4) 2 + ( z − 5) 2 = radicalbig ( x − 4) 2 + ( y − 2) 2 + ( z − 1) 2 . Squaring and expanding the brackets, we get that x 2 − 4 x + 4 + y 2 − 8 y + 16 + z 2 − 10 z + 25 = x 2 − 8 x + 16 + y 2 − 4 y + 4 + z 2 − 2 z + 1 Canceling the x 2 , y 2 , z 2 we get that − 4 x − 8 y − 10 z + 45 = − 8 x − 4 y − 2 z + 21 which implies − 4 x + 4 y + 8 z = 24 Thus, − x + y + 2 z = 6. This is a plane. 2. Let f ( x, y ) = ye x − 2 y . a) (2 points) Calculate the first order partial derivatives f x and f y . Solution. f x = ye x − 2 y f y = 1 e x − 2 y − 2 ye x − 2 y = (1 − 2 y ) e x − 2 y b) (3 points) Calculate the second order partial derivatives f xx , f xy , f yx and f yy . Solution. f xx = ye x − 2 y . f yy = − 2 e x − 2 y − 2(1 − 2 y ) e x − 2 y = ( − 4 + 4 y ) e x − 2 y . f xy = f yx = 1 e x − 2 y − 2 ye x − 2 y = (1 − 2 y ) e x − 2 y . 3.a) (2 points) Find parametric equations for the line through (3 , , 4) that is perpendicular to the plane x + 2 y − 3 z = 2....
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## This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.

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Winter 2008- - Math 20C Final(Lecture D Winter 2008 Duration 3 hours Please close your books and turn off your phones You can use one page of

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