chs3-9review

chs3-9review - Cumulative Review Solutions Chapters 39 2....

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Cumulative Review Solutions 247 5. d. s = } e e t t + e e t t } } d d s t } = = = } ( e t + 4 e t ) 2 } f. s = (1n t + e t ) t } d d s t } = 3 } 1 t } + e t 4 t + 3 ln t + e t 4 =1+ te t + ln t + e t 6. c. w = ! 3 x + } 1 x } § } d d w x } = } 1 2 } 1 3 x + } 1 x } 2 } 1 2 } 1 3– } x 1 2 } 2 = } 1 2 } 1 } 3 x 2 x +1 } 2 } 1 2 } 1 } 3 x 2 x 2 1 } 2 = } 1 2 } h. 1n ( x 2 y )=2 y } x 1 2 y } 3 x 2 } d d y x } +2 xy 4 =2 } d d y x } } 1 y } } d d y x } + } 2 x } =2 } d d y x } } d d y x } 1 2– } 1 y } 2 = } 2 x } } d d y x } = } 2 x } 1 } 2 y y –1 } 2 = } 2 x 2 y y x } (3 x 2 –1) } 1 2 } } x } 3 2 } e 2 t +2+ e –2 t –( e 2 t –2+ e –2 t ) } ( e t + e t ) 2 [ e t + e t ][ e t + e t ]–( e t e t )( e t e t ) } ( e t + e t ) 2 Cumulative Review Solutions Chapters 3–9 2. c. lim x 2 } x x 3 2 8 } =lim x 2 } ( x –2)( x x 2 + 2 2 x +4) } =lim x 2 x 2 +2 x +4 =12 f. lim x 0 } ( Ï 2+ Ï 2 w x w x Ï 2 w ) } 3 } ( ( Ï Ï 2 2 + + x w x w + + Ï Ï 2 w 2 w ) ) } =lim x 0 } x Ï 2 w ( 2 Ï + 2 x + x w 2 + Ï 2 w ) } =lim x 0 } Ï 2 w ( Ï 2+ 1 x w + Ï 2 w ) } = } 1 4 } 4. b. y = } 2 x 2 x } } d d y x } =lim h 0 3 } 2 ( x ( + x h + ) 2 h ) } } 2 x 2 x } 4 =lim h 0 =lim h 0 =lim h 0 = } x 2 x 4 4 x } = } x x 3 4 } h [– x 2 –4 x +2 x 2 –2 h + xh ] } h ( x 2 )( x + h ) 2 2 x x 3 x 2 h –2 x 2 + x 3 –4 xh +2 x 2 h –2 h 2 + xh 2 } h ( x 2 )( x + h ) 2 2 x 2 x 2 ( x + h )–(2– x )( x 2 +2 xh + h 2 ) } hx 2 ( x + h ) 2
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8. c. s =3 ! } 2 2 + –3 3 t t } § s =3 1 } 2 2 + –3 3 t t } 2 } 1 2 } } d d s t } =3 3 } 1 2 } 1 } 2 2 + –3 3 t t } 2 } 1 2 } 43 4 = } 3 2 } 31 } 2 2 + –3 3 t t } 2 } 1 2 } 43 4 = f. x 3 +3 x 2 y + y 3 = c 3 3 x 2 +6 xy +3 x 2 } d d y x } +3 y 2 } d d y x } =0 } d d y x } (3 x 2 +3 y 2 )=–3 x 2 –6 xy } d d y x } = } –3 3 ( ( x x 2 2 + + 2 y x 2 ) y ) } = } –( x x 2 2 + + 2 y x 2 y ) } 9. } d d s t } = e t 2 + te t 2 (2 t ) At x = π , } d d s t } = e π 2 + π e π 2 (2 π ) = e π 2 (1+2 π 2 ). At t = π , s = π e π 2 or the point is ( π , π e π 2 ). The equation is y π e π 2 = e π 2 (1+2 π 2 )( x π ). 10. y = e kx , y' = ke kx , y'' = k 2 e kx , y''' = k 3 e kx a. y'' –3 y' +2 y =0 k 2 e kx –3 ke kx +2 e kx =0 Since e kx 0, k 2 –3 k +2=0 ( k – 2)( k –1)=0 k =2 or k =1 . b. y''' y'' –4 y' +4 y =0 e kx [ k 3 k 2 –4 k +4]= 0 or k 3 k 2 –4 k +4 =0 k 2 ( k –1)+4( k –1)=0 ( k – 1)( k – 2)( k +2)=0 k =1 ,2 or–2 18 } (2+3 t ) } 1 2 } (2–3 t ) } 3 2 } 12+12 t } (2–3 t ) 2 3(2–3 t )–(–3)(2 + 3 t ) } (2–3 t ) 2 248 Cumulative Review Solutions 12. y 2 = e 2 x +2 y e When y =2 , therefore 4 = e 2 x +4– e e 2 x = e 2 x =1 x = } 1 2 } y 2 = e 2 x +2 y e 2 y } d d y x
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This note was uploaded on 02/10/2010 for the course MATHS MCV-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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chs3-9review - Cumulative Review Solutions Chapters 39 2....

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