review_sol

review_sol - Solutions Padmini February 24, 2008 Note:...

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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 and hence limits for t would be t . r ( t ) = < 4cos(2 t ) ,- 4sin(2 t ) , 3 > and hence, | r ( t ) | = 5 and arc length would be integraltext 5 dt = 5 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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review_sol - Solutions Padmini February 24, 2008 Note:...

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