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Unformatted text preview: Solutions Padmini February 24, 2008 Note: These solutions have been made in a span of two hours. There may be egregious computational errors, but I hope you get how to solve the problems. 1. vector PQ = < 3 , 1 , 7 > and vector PR = < , 5 , 5 > . And hence normal vector is vector PQ × vector PR = < 40 , 15 , 15 > , hence equation of the plane is 40( x 1) 15( y 4) + 15( z 6) = 0 or 8 x + 3 y 3 z = 2. 2. Since we have sin(2 t ), period is π and hence limits for t would be ≤ t ≤ π . r ′ ( t ) = < 4cos(2 t ) , 4sin(2 t ) , 3 > and hence,  r ′ ( t )  = 5 and arc length would be integraltext π 5 dt = 5 π 3. Direction vector of the desired line is < 2 , 3 , 1 > and hence, equation would be x = 2 t + 2 , y = 3 t 7 , z = t + 5. 4. Volume is given by the determinant: vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 0 1 2 1 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 5. f ( x, y ) = z = sin( x 2 + y ) hence, f x = 2 x cos( x 2 + y ) and f x (1 , 1) = 2, f y = cos( x 2 + y ) and f y (1 , 1) = 1, and hence, equation of the plane is z 0 = 2( x 1) + ( y + 1) or 2 x + y z = 1.= 1....
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This note was uploaded on 08/10/2008 for the course MATH 251 taught by Professor Beck during the Spring '08 term at Rutgers.
 Spring '08
 Beck

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