ch05 - Chapter 5 Applications of Derivatives Review of...

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Review of Prerequisite Skills 5. a. 3( x – 2) + 2( x – 1) – 6 = 0 3 x – 6+2 x – 2 – 6 = 0 5 x =14 x = } 1 5 4 } e. } 6 t } + } 2 t } =4 12 + t 2 =8 t t 2 – 8 t +12=0 ( t – 6)( t – 2) = 0 t =2 or t =6 f. x 3 +2 x 2 – 3 x =0 x ( x 2 x – 3) = 0 x ( x + 3)( x – 1) = 0 x or x = –3 or x =1 g. x 3 – 8 x 2 +16 x x ( x 2 – 8 x + 16) = 0 x ( x – 4) 2 x or x h. 4 t 3 +12 t 2 t – 3=0 4 t 2 ( t + 3) – 1( t +3)=0 ( t +3)(4 t 2 – 1) = 0 ( t +3)(2 t – 1)(2 t +1)=0 t = –3 or t = } 1 2 } or t =– } 1 2 } i. 4 t 4 – 13 t 2 +9=0 (4 t 2 – 9)( t 2 – 1) = 0 t } 9 4 } or t =±1 6. a. 3 x – 2 > 7 3 x >9 x >3 b. x ( x – 3) > 0 x <0 or x + + 0 3 c. x 2 +4 x >0 x ( x – 4) < 0 0< x <4 Exercise 5.1 2. d. 3 xy 2 + y 3 3 y 2 +3 x 2 y } d d y x } y 2 } d d y x } } d d y x } (2 xy + y 2 )=– y 2 } d d y x } = } 2 xy y + 2 y 2 } f. 9 x 2 – 16 y 2 = –144 18 x – 32 y } d d y x } } d d y x } = } 1 9 6 x y } g. } 1 x 6 2 } + } 1 3 3 } y 2 } 1 2 6 x } + } 1 6 3 } y } d d y x } 26 x +96 y } d d y x } } d d y x } } 1 4 3 8 x y } h. 3 x 2 xy 3 =9 6 x y 3 x 3 y 2 } d d y x } 6 xy 2 } d d y x } =–3 x – 2 y 3 } d d y x } = } –3 x 6 x y 2 2 y 3 } + 0 4 Chapter 5: Applications of Derivatives 101 Chapter 5 • Applications of Derivatives
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j. x 3 + y 3 =6 xy 3 x 2 +3 y 2 } d d y x } y + } d d y x } (6 x ) (3 y 2 – 6 x ) } d d y x } =–3 x 2 +6 y } d d y x } = } 3 3 y x 2 2 + 6 6 x y } = } y x 2 2 + 2 2 x y } k. x 3 y 3 = 144 3 x 2 y 3 y 2 } d d y x } x 3 =0 } d d y x } =– } x x 2 3 y y 3 2 } } y x } m. xy 3 x 3 y =2 y 3 y 2 } d d y x } x 3 3 x 2 y + } d d y x } x 3 4 (3 y 2 x 3 ) } d d y x } =3 x 2 y y 3 } d d y x } = } 3 3 x y 2 2 y x y 3 3 } n. Ï x w + Ï y w =5 x } 1 2 } + y } 1 2 } } 1 2 } x } 1 2 } + } 1 2 } y } 1 2 } } d d y x } } d d y x } = } Ï Ï y w x w } o. ( x + y ) 2 = x 2 + y 2 2( x + y ) 3 1+ } d d y x } 4 x +2 y } d d y x } } d d y x } [ x + y y ]= x x y } d d y x } = } x y } x } 1 2 } } y } 1 2 } 3. a. x 2 + y 2 =13 2 x y } d d y x } At (2, –3), 2(2) + 2(–3) } d d y x } } d d y x } = } 2 3 } . The equation of the tangent at (2, –3) is y = } 2 3 } x + b . At (2, –3), –3 = } 2 3 } (2) + b –9=4+3 b –13=3 b } 1 3 3 } = b. Therefore, the equation of the tangent to x 2 + y 2 = 13 is y = } 2 3 } 3 } 1 3 3 } . c. } 2 x 5 2 } } 3 y 6 2 } =–1 } 2 2 5 x } } 3 2 6 y } } d d y x } 36 x – 25 y } d d y x } At (5 Ï 3 w , –12), 180 Ï 3 w + 300 } d d y x } } d d y x } } 3 Ï 5 3 w } . 102 Chapter 5: Applications of Derivatives
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The equation of the tangent is y = mx + b . At (5 Ï 3 w , –12) and with m = – }} 3 Ï 5 3 w } , –12=– 3 Ï 5 3 w } ( 5 Ï 3 w ) + b –12 = –9+ b –3 = b Therefore, the equation of the tangent is y =– } 3 Ï 5 3 w } x – 3. 4. x + y 2 =1 The line x +2 y =0 has slope of – } 1 2 } . 1+2 y } d d y x } Since the tangent line is parallel to x y , then } d d y x } } 1 2 } . y 1 } 1 2 } 2 1 – y y Substituting, x +1=1 x Therefore, the tangent line to the curve x + y 2 is parallel to the line x y a t (0, 1). 5. a. 5 x 2 – 6 xy +5 y 2 =16 10 x 1 6 y + } d d y x } (6 x ) 2 +10 y } d d y x } (1) At (1, –1), 10 – 1 –6+6 } d d y x } 2 – 10 } d d y x } 16 – 16 } d d y x } } d d y x } . b. When the tangent line is horizontal, } d d y x } . Substituting, 10 x – (6 y +0)+0=0 . y = } 5 3 } x at the point ( x 1 , y 1 ) of tangency: substitute y 1 = } 5 3 } x 1 into 5 x 1 2 – 6 x 1 y 1 y 1 2 = 16.
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ch05 - Chapter 5 Applications of Derivatives Review of...

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