review_sol - Solutions Padmini Note These solutions have...

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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 = < 0 , - 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 0 t π . r ( t ) = < 4 cos(2 t ) , - 4 sin(2 t ) , 3 > and hence, | r ( t ) | = 5 and arc length would be integraltext π 0 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 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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6. r ( t ) = (1 - cos t ) vector i - sin t vector j and hence | r ( t ) | = radicalbig 2(1 - cos t ) = 2 sin( t/ 2) using the half angle formula. Hence, tangential component =
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