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File0058 - 10 11 12 13 Chapter 3 Differentiation(a = 1...

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Unformatted text preview: 10. 11. 12. 13. Chapter 3 Differentiation (a) % = 1 volt/sec (b) % z — § amp/sec dV _ (.11 d dR dV dI dR _ 1 (1V V d1 (C) a—R(a)+1(a—1)->T—i(nr Re) a T(E_Td_r) (d) ‘13}:— : % [ — Q (~ 3] : G) (3) : gohms/sec, R is increasing _ 2 d? A. 2 dR d1 m)P—RI=>E_J a+2121a _ 2 _gl_P_2d_l3 g Q _2R1fl_ 2(i)dl__2l9g (b)P—RI :0“dr“1dr+2RIdr—’ dt_ 12 dt_ 12 Fat _ / _"‘_ 2 21/2 d‘_. d (a) 3— X2_‘yZ—(X +y) :> 3%“ X2X+y2fi _ 2, 2 21/2 d‘,_ d d @s—¢fi+y—o+y) sn- g¢§+xuwfi , 2.. 2 2 d~_ d d d d d ,_ d (c) S—\/)l2——y2 => 3 —x +y : ZSfi—2x§+2ya¥ —> 25‘0'2x d’: : 2ya¥ -> (T): ~¥a% X _ 2 2 2 2 _ 2 2 2 :12 V d_x 9; (1; (a) s—x/x +y +z =>s-x+y +z =>ZSdl—Zxah+2yd‘+22dI (.1; _ x 22 ' «Lv Z d_2 :> (11— ;;x7+y2+z'2 d! + :;x24ry2JrzZ (11+ x2+y2+27 dl - vie _ g _ 92 2 g (b) From part (a) With d‘ — 0 :> dl — 7+2+y2+22 d‘ + 7——=-=——x2+y2+z2 d: (c) Frompart(a)with$=0 2) O:2x%§+2y%¥+22% => %+X§X+§%=O x dt (a) A:%absin0 => %?=%abcos0% (b) A=%absin0 => %=%abcos€%+%bsin0%fi (c) A=%absin0 :> %=éabCO§0i—€+%bsin0%+%asin9% GivenA:7rr2,%:0.01cm/sec,andr:50cm. Since‘L—‘i‘ =27rrg,then‘ii—T =50=27r(50)(fi) =7rcm2/min. Givend—g :—2¢m/sec,fl =2cm/sec,€=12cmandw=50m. d! (11 (a) A : 13w :> ‘11—? = e 93 + w g—f => ‘11—? 2 12(2) + 5(—2) : 14 cm2/sec, increasing (b) P : 26 + 2w => Q = 2 9! + 2 d—W = 2(—2) + 2(2) : 0 cmlsec, constant dt d! dt _ w d_w [ d! (c) D : x/w'2 +62 : (w2 +€2)1/2 => id? = %(w2 +62)1/2(2w ‘3—‘2’ +23 3—?) :> % 2 “12:6; 2 (”(32:23:52) 2 — % cm/sec. decreasing (a) V = xyz => ‘3‘: = yZ ‘3’: + xz ‘3: + xy ‘3; - ‘3’ (4,3,2) - (3)(2)(1) + (4)(2)(—2) + (4)(3)(l) = 2 WIS/sec (b) s :2xy+2xz+2yz => § =(2y+22 3—: +(2x+2z) 9% +(2x+2y) g—f :> dd—f (4,3,2) : (10x1) + (12)(—2) + (14)(1) = 0m2/sec _ 2 2 2 ,_ 2 2 21/2 il_z _ x g EX 2 g (c) €— i/x +y +z —(x +y +2) :> m — x2+y7+zz dl+——L—\/x9+y2+z2 Lll+ x2+y2+zz d1 : 3—? (4,3,2) : (7%) (1)+ (732—9) (—2) + (722—9) (1) 2 O m/sec Given: % = S stec, the ladder is 13 ft long, and x z 12, y = 5 at the instant of time (21) Since x2 + y2 4169 gr = § 3;; = (152) (5) — —12 ft/sec, the ladder is sliding down the wall (b) The area of the triangle formed by the ladder and walls is A : %xy 2 fil—g‘ = (g) (x 9g + y fj—f) . The area is changing at § [12(-12)+ 5(5)] = 7 % = 759.5 ftZ/sec. —i -- Q 2.95 d0- 1.dx__l __ (C) cos6713 :> Smam—l3 (it “> dl‘ 13sin0 dl (5)(5)— lrad/sec ...
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