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# HW6 - Math 191 6.7 17 When the water reaches the top of the...

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Math 191 HW 6 Solutions 6.7: 17 When the water reaches the top of the tank, the force on the movable side is R 0 - 2 (62 . 4)(2 p 4 - y 2 )( - y ) dy = (62 . 4) R 0 - 2 (4 - y 2 ) 1 / 2 ( - 2 y ) dy = (62 . 4) 2 3 (4 - y 2 ) 3 / 2 0 - 2 = 332 . 8ft · lb. The force compressing the spring is F = 100 x , so when the tank is full we have 332 . 8 = 100 x x 3 . 33ft. Therefore the movable end does not reach the required 5ft to allow drainage the tank will overflow. 7.1: 14 Step 1: y = x 2 x = - y , since x 0 Step 2: y = - x = f - 1 ( x ) 7.1: 16 Step 1: y = x 2 - 2 x + 1 y = ( x - 1) 2 y = x - 1, since x 1 x = 1 + y Step 2: y = 1 + x = f - 1 ( x ) 7.1: 29 a. f ( g ( x )) = ( 3 x ) 3 = x, g ( f ( x )) = 3 x 3 = x b. f 0 ( x ) = 3 x 2 f 0 (1) = 3 , f 0 ( - 1) = 3; (1) g 0 ( x ) = 1 3 x - 2 / 3 g 0 (1) = 1 3 , g 0 ( - 1) = 1 3 (2) c. The line y = 0 is tangent to f ( x ) = x 3 at (0 , 0); the line x = 0 is tangent to g ( x ) = 3 x at (0 , 0) 7.1: 32 df dx = 2 x - 4 df - 1 dx x = f (5) = 1 df dx x =5 = 1 6 7.1: 34 dg - 1 dx x =0 = dg - 1 dx x = f (0) = 1 dg dx x =0 = 1 2 7.2: 2 a. ln 1 125 = ln 1 - 3 ln 5 = - 3 ln 5 b. ln 9 . 8 = ln 40 5 = ln 7 2 - ln 5 = 2 ln 7 - ln 5 c. ln 7 7 = ln 7 3 / 2 = 3 2 ln 7 d. ln 1225 = ln 35 2 = 2 ln 35 = 2 ln 5 + 2 ln 7 e. ln 0 . 056 = ln 7 125 = ln 7 - ln 5 3 = ln 7 - 3 ln 5 f. ln 35+ln 1 2 ln 25 = ln 5+ln 7 - ln 7 2 ln 5 = 1 2 7.2: 22 y = x ln x 1+ln x y 0 = (1+ln x )(ln x + x · 1 x ) - ( x ln x )( 1 x ) (1+ln x ) 2 = (1+ln x ) 2 - ln x (1+ln x ) 2 = 1 - ln x (1+ln x ) 2 7.2: 38 R 0 - 1 3 3 x - 2 dx = [ln | 3 x - 2 | ] 0 - 1 = ln 2
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