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chs5-7review

# chs5-7review - Cumulative Review Solutions Chapters 57 1 c...

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Cumulative Review Solutions 181 Chapters 5–7 1. c. x 2 + 16 y 2 = 5 x + 4 y We differentiate both sides of the equation with respect to x : d d x ( x 2 + 16 y 2 ) = d d x (5 x + 4 y ) 2 x + 32 y d d y x = 5 + 4 d d y x (32 y – 4) d d y x = 5 – 2 x d d y x = 3 5 2 y 2 x 4 . d. 2 x 2 xy + 2 y = 5 We differentiate both sides of the equation with respect to x : d d x (2 x 2 xy + 2 y ) = d d x (5) 4 x y + x d d y x + 2 d d y x = 0 (2 – x ) d d y x = y – 4 x d d y x = y 2 4 x x . f. ( 2 x + 3 y ) 2 = 10 We differentiate both sides of the equation with respect to x : d d x (2 x + 3 y ) 2 = d d x (10) 2(2 x + 3 y ) 2 + 3 d d y x = 0 4(2 x + 3 y ) + 6(2 x + 3 y ) d d y x = 0 d d y x = – 2 3 . 2. c. xy 2 + x 2 y = 2 at (1, 1) The slope of the tangent line at any point on the curve is given by d d y x . We differentiate both sides of the equation with respect to x : (1) y 2 + x 2 y d d y x + (2 x ) y + x 2 d d y x = 0. At (1, 1), 1 + 2 d d y x + 2 + d d y x = 0 d d y x = –1. An equation of the tangent line at (1, 1) is y – 1 = –( x – 1) or x + y – 2 = 0 d. y 2 = 3 7 x x 2 2 + – 4 9 at (1, 2) The slope of the tangent line at any point on the curve is given by d d y x . We differentiate both sides of the equation with respect to x : 2 y d d y x = . At (1, 2), 4 d d y x = (6)(3) – 3 ( 2 12)(14) d d y x = – 2 6 5 . An equation of the tangent line at (1, 2) is y – 2 = – 2 6 5 ( x – 1) or 25 x + 6 y – 37 = 0 3. d. f ( x ) = x 4 x 1 4 f' ( x ) = 4 x 3 + x 4 5 f'' ( x ) = 12 x 2 2 x 0 6 (6 x )(7 x 2 – 4) – (3 x 2 + 9)(14 x ) (7 x 2 – 4) 2 Cumulative Review Solutions

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4. b. y = ( x 2 + 4) (1 – 3 x 3 ) d d y x = (2 x )(1 – 3 x 3 ) + ( x 2 + 4)(–9 x 2 ) = 2 x – 6 x 4 – 9 x 4 – 36 x 2 = 2 x – 15 x 4 – 36 x 2 d dx 2 y 2 = 2 – 60 x 3 – 72 x 5. s ( t ) = 3 t 3 – 40.5 t 2 + 162 t for 0 t 8 a. The position of the object at any time t in the interval is s ( t ) = 3 t 3 – 40.5 t 2 + 162 t . The velocity of the object at any time t in the interval is v ( t ) = s' ( t ) = 9 t 2 – 81 t + 162. The acceleration of the object at any time t in the interval is a ( t ) = v' ( t ) = 18 t – 81.
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