Martens_WA3 - 3.1 14 f x = x 2 3 f x = lim(3 x ∆x 2 −(3...

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Unformatted text preview: 3.1 14 f ( x ) = x + 2 3 f ' ( x ) = lim (3 ( x + ∆x ) + 2) − (3 x + 2) ∆x →0 ∆x (3x + 3∆x + 2) − (3x + 2) f ' ( x ) = lim ∆x →0 ∆x 3∆x f ' ( x ) = lim ∆x →0 ∆x f ' ( x ) = lim 3 = 3 ∆x →0 22 f ( x) = 1 x2 1 f '( x) = lim ( x + ∆x ) 2 − 1 x2 ∆x →0 ∆x x 2 − ( x + ∆x ) 2 f '( x) = lim ( x + ∆x ) ∆x 2 x2 ∆x →0 f '( x) = lim f '( x) = lim f '( x) = lim f '( x) = ∆x →0 ∆x ( x + ∆x ) x 2 2 x 2 − ( x + ∆x ) 2 −(2 x∆x + ∆x 2 ) ∆x ( x + ∆x ) x 2 2 ∆x →0 −( 2 x + ∆x) ∆x →0 ( x + ∆x ) 2 2 x2 −2 x −2 =3 x4 x −(2 x + 0) ( x + 0) x 2 = 26(a) f ( x) = 4 x 4 4 − ( x + ∆x) x ∆x f '( x ) = lim ∆x →0 4 x − 4 ( x + ∆x ) ( x + ∆x) x f '( x ) = lim ∆x →0 ∆x f '( x ) = lim 4 x − 4 ( x + ∆x) ∆x ( x + ∆x) x 4 x − 4 ( x + ∆x) ∆x ( x + ∆x) x )( x + ( x + ∆x) x + ( x + ∆x) ) ∆x →0 f '( x ) = lim ( ∆x →0 f '( x ) = lim f '( x ) = lim 4 x − 4( x + ∆x) ∆x →0 [ ∆x ( x + ∆x ) x ][ x + ( x + ∆x )] 4 x − 4 x + 4∆x ∆x →0 [ ∆x ( x + ∆x ) x ][ x + ( x + ∆x )] ∆x →0 f '( x ) = lim f '( x ) = lim f '( x ) = −4 ( x + ∆x) x [ x + ( x + ∆x )] −4 ( x + 0) x [ x + ( x + 0)] ∆x →0 −4 −2 = x2 x x x 32(a) f ( x) = 1 x +1 1 1 − f '( x) = lim ∆x + x + 1 x + 1 ∆x → 0 ∆x ( x + 1) − (∆x + x + 1) (∆x + x + 1)( x + 1) f '( x) = lim ∆x → 0 ∆x f '( x) = lim f f f f ( x + 1) − ( ∆x + x + 1) ∆x → 0 ∆x ( ∆x + x + 1)( x + 1) ∆x '( x) = lim ∆x → 0 ∆x ( ∆x + x + 1)( x + 1) 1 '( x) = lim ∆x → 0 ( ∆x + x + 1)( x + 1) 1 '( x) = (0 + x + 1)( x + 1) 1 '( x) = ( x + 1) 2 Find the slope of the tangent a t (0,1) 1 (0 + 1) 2 1 m = =1 1 m= The equation of the line is y = -x+1. 3.2 12 y = t 2 + 2t − 3 y ' = 2t + 2 22 y = 5 + sin x y ' = cos x 24 3 y = e x + 2 cos x 4 3 y ' = e x − 2sin x 4 32 f (t ) = 3 − f '(t ) = 33 , ( ,3) 5t 5 3 5t 2 3 3 5 f '( ) = = 5 5( 3 ) 2 3 5 42 2 x 2 − 3x + 1 f ( x) = x x(4 x − 3) − (2 x 2 − 3x + 1) f '( x) = x2 (4 x 2 − 3 x) − (2 x 2 − 3 x + 1) f '( x) = x2 2 x2 −1 f '( x) = x2 48 f (t ) = t 2/3 − t1/3 + 4 2 1 f '(t ) = t −1/3 − t −2/3 3 3 2 1 f '(t ) = 1/3 − 2/3 3t 3t 56 1 h(t ) = sin t + et 2 1 h '(t ) = cos t + et 2 At point (π , 1π e) 2 1 h '(π ) = cos π + eπ 2 1 h '(π ) = −1 + eπ 2 Tangent line: 1 1 y − eπ = (−1 + eπ )(t − π ) 2 2 1 1 y = (−1 + eπ )(t − π ) + eπ 2 2 94 s (t ) = −16t 2 − 22t + 220 s '(t ) = −32t − 22 s '(3) = −32(3) − 22 = −118 ft / sec Velocity after falling 108 feet. 0 = −16t 2 − 22t + 108 = 2 s '(2) = −32(2) − 22 = −86 3.3 14 f ( x ) = ( x 2 − 2 x + 1)( x 3 − 1) f '( x ) = ( x 2 − 2 x + 1)(3x 2 ) + ( x 3 − 1)(2 x − 2) f '( x ) = 3 x 2 ( x − 1) 2 + 2( x − 1) 2 ( x 2 + x + 1) f '( x ) = ( x − 1) 2 (5 x 2 + 2 x + 2) f '(1) = (1 − 1) 2 (5(1) 2 + 2(1) + 2) = 0 18 sin x x ( x)(cos x) − (sin x) f '( x) = x2 f ( x) = π 3 1 ( )( ) − ( ) π 2 f '( ) = 6 2 2 π 6 36 π 3(π 3) − 18 f '( ) = 6 π2 28 x3 + 3x + 2 x2 + 1 ( x 2 + 1)(3x 2 + 3) − ( x3 + 3 x + 2)(2 x) f '( x) = ( x 2 + 1) 2 f ( x) = f '( x) = f '( x) = f '( x) = ( x 2 + 1)(3x 2 + 3) − ( x3 + 3 x + 2)(2 x) ( x 2 + 1) 2 (3 x 4 + 3 x 2 + 3 x 2 + 3) − (2 x 4 + 6 x 2 + 4 x ) ( x 2 + 1) 2 x4 − 4 x + 3 ( x 2 + 1) 2 52 y = x cos x + sin x y ' = − x sin x + cos x + cos x y ' = − x sin x + 2 cos x 104 h(t ) = et sin t h '(t ) = et cos t + et sin t h '(t ) = et (− sin t + cos t ) + et (cos t + sin t ) h '(t ) = 2et cos t 3.4 10 y = (2 x 3 + 1) 2 y ' = 2(2 x 3 + 1)6 x 2 y ' = 12 x 2 (2 x3 + 1) 16 g ( x) = 5 − 3x g ( x) = (5 − 3x )1/2 1 (5 − 3 x) −1/2 (−3) 2 −3 g '( x) = 2 5 − 3x g '( x) = 22 s (t ) = (t 2 + 3t − 1) −1 s '(t ) = −1(t 2 + 3t − 1) −2 (2t + 3) s '(t ) = −(2t + 3) (t + 3t − 1) −2 2 60 g (t ) = 5cos3 π t g '(t ) = 15(cos π t ) 2 (− sin π t )(π ) 66 y = e− x 2 dy = −2 xe − x dx 102 2 1 x−2 f ( x) = ( x − 2) −1 f ( x) = f '( x ) = ( x − 2) −2 −1 f '( x ) = ( x − 2) 2 f ''( x) = 2( x − 2) −3 f ''( x) = 2 ( x − 2)3 3.5 6 x 2 y + y 2 x = −3 x 2 y '+ y 2 + 2 yx + 2 yxy ' = 0 x 2 y '+ 2 yxy ' = − y 2 − 2 yx y '( x 2 + 2 yx) = − y 2 − 2 yx y' = 20 − y 2 − 2 yx x 2 + 2 yx ln( xy ) + 5 x = 30 ln x + ln y + 5 x = 30 1 1 dy + +5 = 0 x y dx 1 dy 1 = − −5 y dx x dy y = − − 5y dx x dy y = − − 5y dx x 26 x3 − y 2 = 0 3 x 2 − 2 yy ' = 0 −2 yy ' = −3 x 2 y'= 3x 2 2y At point:[1, 1] 3(1) 2 2(1) 3 1 y' = =1 2 2 y'= 34 y 2 = ln x 1 2 yy ' = y = 2( x 2 − 3 x) x 1 y'= 2 yx A t point [e,1] 1 2(1)e 1 y' = 2e y' = 3.7 2 Find dy dx =2 when x = 3 given dt dt y = 2( x 2 − 3x ) dy dx = 2(2 x − 3) dt dt dy dx = (4 x − 6) dt dt dy = (4(3) − 6)(2) dt dy = (12 − 6)(2) = (6)(2) = 12 dt Find dx dy =5 when x = 1 given dt dt dx 1 dy = dt (4 x − 6) dt dx 1 = (5) dt (4(1) − 6) dx 1 1 5 = (5) = − (5) = − dt (4 − 6) 2 2 6 dx =2 dt y= 1 1 + x2 dy dx −2 x = dt dt (1 + x 2 ) 2 (a) Where x = −2 −2(−2) dy = 2 2 2 dt (1 + (−2) ) dy 4 = 2 dt 25 dy 8 = dt 25 (b) Where x = 0 −2(0) dy = 2 2 2 dt (1 + 0 ) 0 dy = 2 2 = 0 dt (1) (c) Where x = 2 −2(2) dy = 2 2 2 dt (1 + 2 ) −4 dy −4 −8 = 2 2 = 2 = dt 25 25 (5) 18(a) dr =2 dt r =6 4 v = π r3 3 dv 4 dr = 3 π r2 dt 3 dt dv dr = 4π r 2 dt dt dv = 4π 62 2 = π 288 ≈ 904.778684 dt 22 h = 3r dr = dt 1 v = π r 2h 3 1 v = π r 2 (3r ) 3 v = π r2 dv dr = 3π r 2 dt dt (a) r =6 dv = 3π (6) 2 (2) dt dv = π 216 dt (b) r = 24 dv = 3π (24) 2 (2) dt dv = 3π 576(2) = 3π 1152 dt dv = 3456π dt 24 1 V = π r 2h 3 r (10) 1 2 = r 5 h 12 r= 5 h 12 dV 25π 2 dh = h dt 144 dt 2 dV 25π ( h ) dh = dt 144 dt dh 144 dV = dt 25π (h 2 ) dt dh dt dh dt dh dt dh dt 144 (10) 25π (82 ) 1440 = 25π (64) 1440 = π 1600 9 = π 10 = 30 dx = −4 dt x 2 = 122 + n 2 2x dx dn = 2n dt dt dn dx x = dt dt n dn x = −4 dt n x = 13 n = 132 − 122 n = 169 − 144 n = 25 n=5 dn dt dn dx dn dx dn dx x = −4 n (13) = −4 5 52 =− 5 2 = −10 5 32 ds = 240mi / hr dt dy =0 dt y =5 s = 10 x = 8.66 s2 = x2 + y 2 ds dx dy 2s = 2 x + 2 y dt dt dt 2(10mi )(240mi / hr ) = 17.32 20mi dx (240mi / hr ) = 17.32 dt dx 1.155(240mi / hr ) = dt dx = 277.14mi / hr dt dx + 2(5)(0) dt ...
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This note was uploaded on 12/30/2009 for the course MAT 231 taught by Professor Thurber during the Spring '09 term at Thomas Edison State.

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