235-08y-t2-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT 235 Y - CALCULUS II TEST #2. JANUARY 22, 2009. PROBLEMS AND SOLUTIONS 1. Compute the following limits, if they exist. If the limit does not exist explain why. a) (5 marks) 22 44 (,) ( 0 , 0 ) lim 2 xy x y x y + Solution: Let (,) 2 fxy . Then, for t 0 , f ( t , 0 ) = 0 and f ( t , t ) = 1 / 3 . x y = + So, f 0 when ( x , y ) ( 0 , 0 ) along the x -axis but f 1 /3 when ( x , y ) ( 0 , 0 ) along the line y = x . Therefore, 4 0 , 0 ) lim 2 4 x y x y + does not exist. b) (5 marks) 32 0 , 0 ) lim 2 x y + Solution: Recall that x 4 + y 4 2 x 2 y 2 . Then, 4 4 2 2 1 0 2 x x y x y ≤≤ = ++ . But 0 , 0 ) 1 lim 0 2 x = . Therefore 0 , 0 ) lim 0 2 = + . 2. a) (5 marks) Find the value of the constant k , if any, for which the function f ( x , y , z ) = x 3 + y 2 z is a solution of the differential equation y f x x ( x , y , z ) = k x f y z ( x , y , z ) . Solution: Notice that f x ( x , y , z ) = 3 x 2 and f y ( x , y , z ) = 2 y z . Therefore, f x x ( x , y , z ) = 6 x and f y z ( x , y , z ) = 2 y . So, we need 6 x y = 2 k x y . That is: k = 3 . b) (5 marks) Let 2 3 x gxy hxy = where h ( x , y ) is an arbitrary di f erentiable function with h ( 1 , 1 ) = 3 , h x ( 1 , 1 ) = 0 and h y ( 1 , 1 ) = 12 . Compute g ( 1 , 1 ) . Solution: Recall that g ( 1 , 1 ) = ( g x ( 1 , 1 ) , g y ( 1 , 1 ) ) . Now, 2 2 6 (,)3 ((,) ) x x xhxy xh xy gx y = and 2 2 3( , ) y y ) x hx y y = . Then, 2 (6)(1)(3) (1,1) 2 (3) x g == and 2 2 3(1) (12) (1,1) 4 y g = =− . Finally, g ( 1 , 1 ) = ( 2 , – 4 ) .
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Page 2 3. Let f ( x , y ) be a function of two variables where x = 3 s – 3 t + 1 and y = s t – 4 . a) (6 marks) Assume that f x ( 1 , 0 ) = 3 and f y ( 1 , 0 ) = 4 . Compute (2,2) f s . Solution: Recall that (,) f st s = f x ( x , y ) x s ( s , t ) + f y ( x , y ) y s ( s , t ) . Here, x s ( s , t ) = 3 , y s ( s , t ) = t and ( x , y ) = ( 1 , 0 ) when ( s , t ) = ( 2 , 2 ) . Therefore, f s = f x ( 1 , 0 ) x s ( 2 , 2 ) + f y ( 1 , 0 ) y s ( 2 , 2 ) = ( 3 ) ( 3 ) + ( 4 ) ( 2 ) = 17 . b) (9 marks) Additionally to the above, assume that f x x ( 1 , 0 ) = 1 , f x y ( 1 , 0 ) = 2 = f y x ( 1 , 0 ) and f y y ( 1 , 0 ) = 5 . Compute 2 f ∂∂ . Solution: From (a), we have f s = 3 f x ( x , y ) + t f y ( x , y ) .
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235-08y-t2-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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