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Sec 14.6-sm.dvi

# Sec 14.6-sm.dvi - 702 C H A P T E R 15 D I F F E R E N T I...

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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 ( 0 , 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 ( 0 ) = 1, we get dy dx = tan x y y ( 0 ) = 1 We solve the differential equation using separation of variables: y dy = tan x dx y dy = tan x dx 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 ( 0 ) = 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 0 ) + 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 0 y y sin x = c c = 0.15 y = 1-ln (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 = rs and v = r + s. 2. The composite function f ( u , v) is equal to: (a) rse r + s (b) re s (c) rse rs

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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 = rs and v = r + s . This gives f ( u ( r , s ), v( r , s ) ) = u ( r , s ) e v( r , s ) = rse 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 = rs 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 ? (a) f v x v (b) f w w x (c) f z z w (d) f v v w SOLUTION By the Chain Rule, the derivative f w is f w = f x x w + f y y w + f z z w Therefore (c) is the only correct answer. 6. With notation as in the previous question, does x v appear in the Chain Rule expression for f u ?
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