HW6 - 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 2(62 4(2 p 4 y 2 y dy =(62 4 R

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Unformatted text preview: 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- 2 (62 . 4)(2 p 4- y 2 )(- y ) dy = (62 . 4) R- 2 (4- y 2 ) 1 / 2 (- 2 y ) dy = (62 . 4) 2 3 (4- y 2 ) 3 / 2- 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 ( x ) = 3 x 2 ⇒ f (1) = 3 , f (- 1) = 3; (1) g ( x ) = 1 3 x- 2 / 3 ⇒ g (1) = 1 3 , g (- 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...
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This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).

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