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Homework3Solns

# Homework3Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...

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SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #3 Solutions Peter Malkin Section 7.3 Question 18 The domain of the function ln( x + y ) is the set of points ( x,y ) such that x + y > 0. More succinctly, the domain is the set D = { ( x,y ) : x + y > 0 } . Question 24 The domain of the function g ( x,y ) = 1 x - y is the set D = { ( x,y ) : x negationslash = y } . Question 30 The contour map consists of curves e 1 - x 2 + y 2 = C 1 x 2 + y 2 = ln ( C ), which are hyperbolas. The contour map of this surface matches (d). Section 7.4 Question 28 f ( x,y ) = ln( xy ) = 1 2 ln( xy ) = 1 2 (ln x + ln y ). So, f x ( x,y ) = 1 2 x and f y ( x,y ) = 1 2 y . Thus, f x ( 1 , 1) = 1 2( - 1) = 1 2 and f y ( 1 , 1) = 1 2( - 1) = 1 2 . Question 42 f ( x,y ) = ln( x 2 + y 2 + 1). f x ( x,y ) = 2 x ( x 2 + y 2 + 1) and f y ( x,y ) = 2 y ( x 2 + y 2 + 1) . So, f x ( x,y ) = 0 when x = 0, and f y ( x,y ) = 0 when y = 0. Thus, f x ( x,y ) = 0 and f y ( x,y ) = 0 at the point (0 , 0). Question 48 z = f ( x,y ) = x y . f x ( x,y ) = 1 y and f y ( x,y ) = x y 2 . The slope of the surface in the x -direction at the point (3 , 1 , 3) is f x (3 , 1) = 1 1 = 1, and in the y -direction at the same point (3 , 1 , 3) is f y (3 , 1) = 3 1 2 = 3. 1

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Question 60 The first partial derivatives of the function f ( x,y ) = x x + y are f x ( x,y ) = 1 ( x + y ) x ( x + y ) 2 = y ( x + y ) 2 and f y ( x,y ) = x ( x + y ) 2 . The second partial derivatives are f xx ( x,y ) = 2 y ( x + y ) 3 ,f yy ( x,y ) = 2 x ( x + y ) 3 , and f xy ( x,y ) = f yx ( x,y ) = x y ( x + y ) 3 .
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Homework3Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...

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