s4 - ) = p 1 − x 2 − y 2 (an ellipsoid) (b) f ( x, y )...

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Solution to Set 4 1. (a) The square root is real only if the argument is non-negative, namely 1 x 2 y 2 0 x 2 + y 2 1. The domain is a closed disk (boundary included) centered at (0 , 0) with radius 1. (b) The logarithmic function is deFned only for a positive argument, namely x + y> 0. The domain is an open half plane above the line x + y = 0 (135 with the x-axis). 2. (a) c = 0 case: The level curve is deFned by the equation f ( x, y )= p 1 x 2 y 2 =0 x 2 + y 2 =1. It is a circle centered at (0 , 0) with radius 1. c =1 / 2 case: The level curve satisFes x 2 + y 2 =1 / 2 and is a circle centered at (0 , 0) with radius 1 / 2. (b) f ( x, y )= x 2 y 2 ; c = 1 , 0 , 1 c = 1 case: The level curve is the hyperbola y 2 x 2 =1. c = 0 case: The level curve is composed of the two lines x = ± y . c = 1 case: The level curve is the hyperbola x 2 y 2 =1. 3. Sketch the graph of f (a) f ( x, y
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Unformatted text preview: ) = p 1 − x 2 − y 2 (an ellipsoid) (b) f ( x, y ) = p x 2 + y 2 (a cone) (c) f ( x, y ) = x 2 + y 2 + 1 (a paraboloid) (d) f ( x, y ) = x 2 − y 2 + 1 (a hyperbolic paraboloid) 4. (a) lim ( x,y ) → (1 , 0) x 2 − xy + 1 x 2 + y 2 = 2 (b) lim ( x,y ) → ( ln 2 , 0) e 2 x + y 2 = 4 5. Let x = r cos θ , y = r sin θ , then lim ( x,y ) → (0 , 0) xy x 2 + y 2 = lim r → r 2 sin θ cos θ r 2 = sin θ cos θ and is dependent on θ (not a unique constant). Therefore, the limit does not exist. 6. Let x = r cos θ , y = r sin θ , then lim ( x,y ) → (0 , 0) ( x 2 + y 2 ) sin 1 x 2 + y 2 = lim r → r 2 sin 1 r 2 = 0 = f (0 , 0) Therefore, the function is continuous at (0 , 0). 1...
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