15.6%20Prel%20Q%20and%20Ex%201%20-%207

15.6%20Prel%20Q%20and%20Ex%201%20-%207 - 702 C H A P T E R...

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702 CHAPTER 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: ydy = tan xdx Z = Z tan 1 2 y 2 =− ln | cos x |+ k y 2 2ln | cos x k ln ³ cos 2 x ´ + k y q ln ( cos 2 x ) + k Since y ( 0 ) = 1 > 0, the appropriate sign is the positive sign. That is, y = q ln ( cos 2 x ) + k (1) We ±nd the constant k by substituting x = 0, y = 1 and solve for k .Thisgives 1 = q ln ( cos 2 0 ) + k = ln 1 + k = k Hence, k = 1 Substituting in (2) gives the following solution: y = q 1 ln ( cos 2 x ) (2) The following ±gure 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? (a) The primary derivatives of f are f x and f y . (b) The independent variables are u and v ,onwhich x and y depend. In Questions 2–4, suppose that f ( u ,v) = ue v ,whereu = rs and v = r + s. 2. The composite function f ( u is equal to: (a) rse r + s (b) re s (c) rs
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SECTION 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 by the corre- sponding functions u = rs and v = r + s .Thisgives 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 at ( r , s ) = ( 1 , 1 ) ? We compute u = and v = r + s at the point ( r , s ) = ( 1 , 1 ) : u ( 1 , 1 ) = 1 · 1 = 1 ; 1 , 1 ) = 1 + 1 = 2 Substituting in f ( u = ue v ,weget f ( u ¯ ¯ ¯ ¯ ( 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 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 (b) f x (c) f z z (d) f By the Chain Rule, the derivative f is f = f x x + f y y + f z z Therefore (c) is the only correct answer.
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.6%20Prel%20Q%20and%20Ex%201%20-%207 - 702 C H A P T E R...

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