235-08-ps3-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT 235 Y – CALCULUS II FALL-WINTER 2008-09 ASSIGNMENT #3. DUE ON NOVEMBER 27. PROBLEMS AND SOLUTIONS 1. a) Given the function 2 1 (,) a r c s i n ( 3 s i n( ) ) 24 4 x y fxy ππ =− + . - Find and sketch the domain of f . Solution: Recall that the domain of the function g ( t ) = arcsin t is the closed interval – 1 t 1 . We can simplify the notations in our problem by making () 44 4 xy wx π =+= + y . Notice that for any value of w , 2 1 3sin 1 22 w −≤ 1 . So, for ( x , y ) to be in the domain of f we just need 2 1 13 s i n 2 w −≤ − . Solving this inequality, we obtain 2 sin 2 w and then kw k −≤≤ + , where k . Combining this result with the fact that w 0 we conclude that the domain of f is the set of all ordered pairs of real numbers ( x , y ) such that Z 01 ≤+≤ or 41 , where k = 1 , 2 , 3 , … kx y k + ≤ + - Find the range of f and draw the level curve f ( x , y ) = M , where M denotes the absolute maximum value of the function f . Solution: Recall that the function g ( t ) = arcsin t is continuous and increasing everywhere in its domain. Our previous results show that the expression 2 1 2 w reaches all the values in the interval 1 [1 . Therefore, the function f reaches all the values from arcsin ( – 1 ) = – / 2 to arcsin ( 1 / 2 ) = / 6 . , ] 2 The range of the function f is the interval [ – / 2 , / 6 ] . So, M = / 6 and the condition f ( x , y ) = M is equivalent to sin w = 0 . That is, w = k , where k , k 0 . The level curve f ( x , y ) = M is just Z the set of all ordered pairs of real numbers ( x , y ) such that x yk +≤ , where k = 0 , 1 , 2 , …
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Page 2 b) Evaluate 3 2 1 322 3 42 (,) ( 0 , 0 ) (2 )( 3 2 ) lim 3 x x y xy xx y y x + ⎛⎞ −+− ⎜⎟ + ⎝⎠ , if the limit exists. Solution: Notice that 2 4 3 2 ) 4 1 33 2 x xyy x x y x yx =− ++ y and 34 22 13 2 x y y + += . Making 2 4 3 u 2 x y = + , the given expression simplifies to 3 2 1 1 3 3 3 2 ) (1 4 ) 3 x x y u x x u + + . Now we show that u 0 as ( x , y ) ( 0 , 0 ) . In effect, if ( x , y ) ( 0 , 0 ) then, 2 0 333 x u y <= = + . But 0 , 0 ) lim 0 3 x = , then 2 0 , 0 ) lim 0 3 = + follows from the Squeeze Theorem. The original problem has now been reduced to finding the limit () 1 3 0 lim 1 4 u u u , which is just a particular case of the well know (first year calculus) property 0 lim 1 b abt t t at e . Therefore, 3 2 1 4 3 3 0 , 0 ) 3 2 ) lim 3 x x y x x e + = + .
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Page 3 2. Let 22 arctan arctan if 0 (,) 0i f yx xy x fxy −≠ = 0 y = . a) Is the function f continuous at ( 0 , 0 ) ? Solution: Yes, f is continuous at ( 0 , 0 ) because f ( 0 , 0 ) = 0 = (,) ( 0 , 0 ) lim ( , ) f xy . The first equation follows from the definition of f . So, we just need to show that 0 , 0 ) lim ( , ) = 0 . In effect, if xy 0 , then 0 < arctan arctan y x x y arctan arctan y x x y + . Recall now that for all t , R arctan 2 t π < . Then, arctan arctan y x x y + < () 2 x y + . But , then 0 , 0 ) lim 0 += 0 , 0 ) lim ( , ) = 0 follows also from the Squeeze Theorem. b) Compute f x y x and f x y y , if they exist.
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235-08-ps3-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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