240final2sols

# 240final2sols - SOLUTIONS 1 Put a = P Q =(1 1 1 b = R Q =(2...

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SOLUTIONS 1. Put a = P Q =( 1 , 1 , 1), b = R Q =( 2 , 2 , 1). (a) n = a · b =( 3 , 3 , 0). Plane 3( x +3)+3( y 1) = 0 (b) A = | a · b | / 2=3 2 / 2 (c) c = T Q =(1 , 0 , 2), Volume V =( a · b ) · c =3 2. (a) r 0 ( t )=(1 , 2 t, 3 t 2 ), T (1) = r 0 (1) / | r 0 (1) | =(1 / 14 , 2 / 14 , 3 / 14) (b) ( x 1) + 2( y 1) + 3( z 1) = 0 (c) r 0 ( t )=(0 , 2 , 6 t ), κ (1) = | r 0 (1) × r 0 (1) | / | r 0 (1) | 3 =2 19 / 14 3 / 2 3. Z π/ 2 0 Z π/ 2 y sin x x dxdy = Z π/ 2 0 Z x 0 sin x x dydx = Z π/ 2 0 sin x x [ y ] x 0 dx = Z π/ 2 0 sin xdx = [cos x ] π/ 2 0 =1 4. (a) f x =3 x 2 3 = 0 solutions x =1 , 1, f y =2 y 6 = 0, solution y =3 . Crit ica lpo ints ( x, y )=( 1 , 3) , (1 , 3) (b) D = f xx f yy ( f xy ) 2 =12 x First critical point: D ( 1 , 3) = 12, (-1,3) is a saddle. Second critical point: D (1 , 3) = 12, f xx (1 ,

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240final2sols - SOLUTIONS 1 Put a = P Q =(1 1 1 b = R Q =(2...

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