# vb - MATH 241H EXAM 2 Solutions 1(a lim(x,y(2,1 f(x y =...

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MATH 241H - EXAM 2 Solutions - March 12, 2007 1. (a) lim ( x,y ) (2 , 1) f ( x, y ) = 1 / 5. (b) lim ( x,y ) (0 , 0) f ( x, y ) = lim r 0 r sin 3 θ = 0. (c) f ( x, 0) = 0 for all x so f x (0 , 0) = 0. (d) f (0 , y ) = y for all y so f y (0 , 0) = 1. 2. P = kT/V , so dP/dt = ( ∂P/∂T )( dT/dt ) + ( ∂P/∂V )( dV/dt ) = ( k/V )( - 2) + ( - kT/V 2 )(3) = k ( - 2 / 200 - 120 / (200) 2 ) = - 13 k/ 1000. 3. (a) f x = 2 y 3 z 2 / p xy 3 z 2 + 2 = 1 , f y = 6 xy 2 z 2 / p xy 3 z 2 + 2 = 6 , f z = 4 xy 3 z/ p xy 3 z 2 + 2 = - 4, so direction is i + 6 j - 4 k , the gradient. (b) u = (1 / 6) i - (1 / 6) j + (2 / 6) k , so D u f (2 , 1 , - 1) = (1 / 6) - (6 / 6) - (8 / 6) = - 13 / 6. (c) ( x - 2) + 6( y - 1) - 4( z + 1) = 0. 4. f x = 2 x - 4 = 0 so x = 2; f y = 3 y 2 + 6 y = 3 y ( y + 2) = 0 so y = 0 , - 2. Thus critical points are (2 , 0) and (2 , - 2). f xx = 2 , f yy = 6 y + 6 , f xy = 0 so D = 12 y + 12. At (2 , 0) , D = 36 > 0 and f xx = 2 > 0 so (2 , 0) is a relative min. At (2 , - 2) , D = - 12 < 0 so (2 , - 2) is a saddlepoint. 5. f x = 2 x + 1 = 0 if x = - 1 / 2. f y = 2 y = 0 if y = 0. Thus ( - 1 / 2 , 0) is a critical point inside the region. On the boundary, 2
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