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Unformatted text preview: 702 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 72. Find the curve y = g ( x ) passing through ( , 1 ) that crosses each level curve of f ( x , y ) = y sin x at a right angle. If you have a computer algebra system, graph y = g ( x ) together with the level curves of f . SOLUTION Using f x = y cos x , f y = sin x , and y ( ) = 1, we get d y d x = tan x y ⇒ y ( ) = 1 We solve the differential equation using separation of variables: y d y = tan x d x . y d y = . tan x d x 1 2 y 2 =  ln  cos x  + k y 2 =  2 ln  cos x  + k =  ln . cos 2 x . + k y = ± . ln ( cos 2 x ) + k Since y ( ) = 1 > 0, the appropriate sign is the positive sign. That is, y = . ln ( cos 2 x ) + k (1) We find the constant k by substituting x = 0, y = 1 and solve for k . This gives 1 = . ln ( cos 2 ) + k = √ ln 1 + k = √ k Hence, k = 1 Substituting in (2) gives the following solution: y = . 1 ln ( cos 2 x ) (2) The following figure shows the graph of the curve (3) together with some level curves of f . x y y sin x = c c = 0.15 y = √ 1ln (cos 2 x ) 15.6 The Chain Rule (ET Section 14.6) Preliminary Questions 1. Consider a function f ( x , y ) where x = u v and y = u /v . (a) What are the primary derivatives of f ? (b) What are the independent variables? SOLUTION (a) The primary derivatives of f are ∂ f ∂ x and ∂ f ∂ y . (b) The independent variables are u and v , on which x and y depend. In Questions 2–4, suppose that f ( u , v) = ue v , where u = r s and v = r + s. 2. The composite function f ( u , v) is equal to: (a) r se r + s (b) r e s (c) r se r s S E C T I O N 15.6 The Chain Rule (ET Section 14.6) 703 SOLUTION The composite function f ( u , v) is obtained by replacing u and v in the formula for f ( u , v) by the corre sponding functions u = r s and v = r + s . This gives f ( u ( r , s ), v( r , s ) ) = u ( r , s ) e v( r , s ) = r se r + s Answer (a) is the correct answer. 3. What is the value of f ( u , v) at ( r , s ) = ( 1 , 1 ) ? SOLUTION We compute u = r s and v = r + s at the point ( r , s ) = ( 1 , 1 ) : u ( 1 , 1 ) = 1 · 1 = 1 ; v( 1 , 1 ) = 1 + 1 = 2 Substituting in f ( u , v) = ue v , we get f ( u , v) . . . . ( r , s ) = ( 1 , 1 ) = 1 · e 2 = e 2 . 4. According to the Chain Rule, ∂ f ∂ r is equal to (choose correct answer): (a) ∂ f ∂ x ∂ x ∂ r + ∂ f ∂ x ∂ x ∂ s (b) ∂ f ∂ x ∂ x ∂ r + ∂ f ∂ y ∂ y ∂ r (c) ∂ f ∂ r ∂ r ∂ x + ∂ f ∂ s ∂ s ∂ x SOLUTION For a function f ( x , y ) where x = x ( r , s ) and y = y ( r , s ) , the Chain Rule states that the partial derivative ∂ f ∂ r is as given in (b). That is, ∂ f ∂ x ∂ x ∂ r + ∂ f ∂ y ∂ y ∂ r 5. Suppose that x , y , z are functions of the independent variables u , v, w . Given a function f ( x , y , z ) , which of the following terms appear in the Chain Rule expression for ∂ f ∂w ?...
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This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher

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