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Sec 14.1-sm.dvi copy

# Sec 14.1-sm.dvi copy - DIFFERENTIATION IN 15 SE VE RA L VA...

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15 DIFFERENTIATION IN SEVERAL VARIABLES 15.1 Functions of Two or More Variables (ET Section 14.1) Preliminary Questions 1. What is the difference between a horizontal trace and a level curve? How are they related? SOLUTION A horizontal trace at height c consists of all points ( x , y , c ) such that f ( x , y ) = c . A level curve is the curve f ( x , y ) = c in the xy -plane. The horizontal trace is in the z = c plane. The two curves are related in the sense that the level curve is the projection of the horizontal trace on the xy -plane. The two curves have the same shape but they are located in parallel planes. 2. Describe the trace of f ( x , y ) = x 2 - sin ( x 3 y ) in the xz -plane. SOLUTION The intersection of the graph of f ( x , y ) = x 2 - sin ( x 3 y ) with the xz -plane is obtained by setting y = 0 in the equation z = x 2 - sin ( x 3 y ) . We get the equation z = x 2 - sin 0 = x 2 . This is the parabola z = x 2 in the xz -plane. 3. Is it possible for two different level curves of a function to intersect? Explain. SOLUTION Two different level curves of f ( x , y ) are the curves in the xy -plane defined by equations f ( x , y ) = c 1 and f ( x , y ) = c 2 for c 1 = c 2 . If the curves intersect at a point ( x 0 , y 0 ) , then f ( x 0 , y 0 ) = c 1 and f ( x 0 , y 0 ) = c 2 , which implies that c 1 = c 2 . Therefore, two different level curves of a function do not intersect. 4. Describe the contour map of f ( x , y ) = x with contour interval 1. SOLUTION The level curves of the function f ( x , y ) = x are the vertical lines x = c . Therefore, the contour map of f with contour interval 1 consists of vertical lines so that every two adjacent lines are distanced one unit from another. 5. How will the contour maps of f ( x , y ) = x and g ( x , y ) = 2 x with contour interval 1 look different? SOLUTION The level curves of f ( x , y ) = x are the vertical lines x = c , and the level curves of g ( x , y ) = 2 x are the vertical lines 2 x = c or x = c 2 . Therefore, the contour map of f ( x , y ) = x with contour interval 1 consists of vertical lines with distance one unit between adjacent lines, whereas in the contour map of g ( x , y ) = 2 x (with contour interval 1) the distance between two adjacent vertical lines is 1 2 . Exercises In Exercises 1–4, evaluate the function at the specified points. 1. f ( x , y ) = x + yx 3 , ( 2 , 2 ) , ( - 1 , 4 ) , ( 6 , 1 2 ) SOLUTION We substitute the values for x and y in f ( x , y ) and compute the values of f at the given points. This gives f ( 2 , 2 ) = 2 + 2 · 2 3 = 18 f ( - 1 , 4 ) = - 1 + 4 · ( - 1 ) 3 = - 5 f 6 , 1 2 = 6 + 1 2 · 6 3 = 114 2. g ( x , y ) = y x 2 + y 2 , ( 1 , 3 ) , ( 3 , - 2 ) SOLUTION We substitute ( x , y ) = ( 1 , 3 ) and ( x , y ) = ( 3 , - 2 ) in the function to obtain g ( 1 , 3 ) = 3 1 2 + 3 2 = 3 10 ; g ( 3 , - 2 ) = - 2 3 2 + ( - 2 ) 2 = - 2 13 3. h ( x , y , z ) = xyz - 2 , ( 3 , 8 , 2 ) , ( 3 , - 2 , - 6 ) SOLUTION Substituting ( x , y , z ) = ( 3 , 8 , 2 ) and ( x , y , z ) = ( 3 , - 2 , - 6 ) in the function, we obtain h ( 3 , 8 , 2 ) = 3 · 8 · 2 - 2 = 3 · 8 · 1 4 = 6 h ( 3 , - 2 , - 6 ) = 3 · ( - 2 ) · ( - 6 ) - 2 = - 6 · 1 36 = - 1 6

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604 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 4. Q ( y , z ) = y 2 + y sin z , ( y , z ) = ( 2 , π 2 ), ( - 2 , π 6 ) SOLUTION We have Q 2 , π 2 = 2 2 + 2 sin π 2 = 4 + 2 · 1 = 6 Q - 2 , π 6 = ( - 2 ) 2 - 2 sin π 6 = 4 - 2 · 1 2 = 3 In Exercises 5–16, sketch the domain of the function.
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Sec 14.1-sm.dvi copy - DIFFERENTIATION IN 15 SE VE RA L VA...

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