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10 - 220 C CHAPTER3 DIFFERENTIATION RULES(b If a is the...

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Unformatted text preview: 220 C CHAPTER3 DIFFERENTIATION RULES (b) If a is the angle between the tangent line and the av— axis then tana— — slope of the line: sinh7 %, so a = tan (sinh 2—70) m 0.343 rad~ N 19.66". Thus. the angle between the line and the pole IS 6 = 900 — a m 70.34". 50. We differentiate the function twice, then substitute into the differential equation: y = E cosh $135 => 09 T dy T p996 pg p927 dzy m w _:_ h(__ )_:»h__ __ (pg)pg:p9 pg dm pg sin T T 8111 T —> (1932 cosh T T T cosh T . 2 We evaluate the two sides separately: LHS— — —:—2-— — pT_g cosh p—gf x RHS— — tip—9 1 + (:__>2y>_p_gT1 I1 —I— sinh2—— p g coshp 91E by the identity proved 1n Example 1(a) 51. (a) y : Asinhmm + Bcoshmz => y' : mAcoshmsc + mBsinhmac => y" : m2A sinh mm + mZB cosh mm = m2(A sinh ms: + B cosh mm) = (b) From part (a), a solution of y” : 9y is y($) = Asinh 3m + B cosh 336. So —4 : y(0) = AsinhO + BcoshO = B, so B : '4. Now y'(m) = 3Acosh 3x — 1251nh3ac => 6=y'(0)=3A : A:2.soy:2sinh3x—4cosh3w. sinhw . emfle’x . 1—6—2“c 1—0 1 52.1‘ :1 ———=1 __—:__:_ $3130 e1 mi“; 2am $13.30 2 2 2 53. The tangent to y : cosha: has slope 1 when 3/ = sinhm = 1 :> as = sinh—1 1 = ln(1 + J5) by Equation 3. Since sinhx : 1 and y = coshx = V 1 + sinh2 m, we have coshsc :2 fl. The point is (ln(1 + J?) , x5). 54. coshac = cosh[1n(sec 9 + 1221116)] 2 % [emsec 9H“ 0) + 6— ”(sec 9““ 9)] —1 0+t 0+——i-—— #lsec6+tan0+————S—[email protected]—" _ 2 sec an seCG +tan6 _ 2 (sec6+tan€)(sec9 — tan0) — —;[sec 6+tan0+ 92%;}:2706] : %(sec9+tan6+sec0—tan6) :sec0 55. If aez + be” 2 a cosh(m + B) [or a sinh(:c + m], then aeI + be"c : %(ez+6 :I: ei$’:) : %(ee ex ea :I: (fie—3) = (geek:I :I: (ge’fik—m. Comparing coefficients of e” and e we have a — —efl (l) and b— — 21:30‘6‘6 (2). We need to find a and B. Dividing equation (1) by equation (2) gives us 5; = :l:e2fl => (*) 2B- ln(:l:% ) ¢ [3: 121n(:l:%). SOIVing equations (1) and (2) forefi givesusefi:ggandefizzlz%J so——=:I:§ag => a2=:I:4ab => a:2\/:I:ab. (*) Ifa b > 0, we use the + sign and obtain a cosh function, whereas if g < 0, we use the — sign and obtain a sinh function. In summary. if a and b have the same sign, we have ae“c + be’3E : 2v ab cosh (cc + % ln 96) , whereas, if a and b have the opposite sign, then aem + be” : 2V —ab sinh (x + % ln(#%)). 10 Related Rates dV dV dav 2 dm _ 3 . _.__ .— _. 1‘V’$ :> dt dccdt 35” dt dA dA dr dr dA d7" 2 2 ___.____ ———=2 ——:2 30 1 =60 S 2. (a) A — 7M" —> dt dr dt 271'?" dt (b) dt 71'?“ dt 7r( m)( m/s) 71' m/ SECTION 3.10 RELATED RATES E 221 3.y2x3+2:1: => %:j—Z%:(3x2+2)(5)25(3m2+2).Whenx=2.‘:—Z=5(14)=70. 4.m2+y2225 => 2x%+2y%=0 :> x%:—y% => $=~§%. Wheny24,x2+42225 z»'xzi3.For%=6,%=—fi(6)=¢s. 5-z2=962+y2 2 22%:2x%+2y:—f 2 §I2(x:—:+yj—ztj>.Whenx2Sandy=12. 22=52+122 :> 222169 :> z=i13.For%=2and%=3,§=fm(5-2+12.3)=if—:. 6.3.12x/1-t—a:3 2> %:dd—::—:2%(1+$3)71/2(3m2)::22%Z:.With:§—4whenx22and y23,wehave42:—E:%:—: => 2—:22cm/s. 7. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi / h passes directly over a radar station. If we let it be time (in hours) and m be the horizontal distance traveled by the plane (in mi). then we are given that dm/dt 2 500 mi/h. (b) Unknown: the rate at which the distance from the plane to the station is (c) x increasing when it is 2 mi from the station. If we let y be the distance from 1'7 y the plane to the station, then we want to find dy/dt when y 2 2 mi. (d) By the Pythagorean Theorem, y2 2 x2 + 1 2> 2y(dy/dt) 2 2m(dm/dt). d . (e 2—3: 2 5% 2 E(500). Since y2 2 x2 + 1, when 3/ 2 2. a: 2 \/§ so CT? 2 [email protected](500) 2 250x/gz 433 mi/h. 31 y 8. (a) Given: the rate of decrease of the surface area is 1 cm2/ min. If we let (c ) t be time (in minutes) and S be the surface area (in cm2), then we are given that dS/dt 2 —1 cmz/s. (b) Unknown: the rate of decrease of the diameter when the diameter is 10 cm. If we let :1: be the diameter, then we want to find dm/dt when ac 2 10 cm. (d) If the radius is 1' and the diameter x = 21", then r = gas and S : 47rr2 : 47r(%:c)2 : 7rx2 :> E _ Edi _ 2 dz dt _ dzc dt _ mdt‘ dS dx dx 1 dx 1 _1:_:2 _ _:__, = _:_2 (e) dt 7rw dt 2 dt 27:10 When at 10. dt 207r' So the rate of decrease . 1 . 18 W cm/min. 9. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s) and :c be the distance from the pole to the man (in ft), then we are given that dsc/dt 2 5 ft/s. (b) Unknown: the rate at which the tip of his shadow is moving when he is (c) 40 ft from the pole. If we let y be the distance from the man to the tip of 15 his shadow (in ft), then we want to find dita + y) when a: 2 40 ft. 222 C CHAPTERB DIFFERENTIATION RULES 15 (d) By Sim1lar tr1angles ~6— : CE :4— y :> 15y : 6x + 6y => 9y = 60: => y 2 gm. . d d 5 d (e) The Up of the shadow moves at a rate of EQ + y) — dt (3: I gm) — gd—fl: — 2(5) — % ft/s. 10. (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km/h, and ship B is sailing north at 25 km / h. If we let it be time (in hours), a: be the distance traveled by ship A (in km), and y be the distance traveled by ship B (in km), then we are given that dcc/dt = 35 km/h and dy/dt : 25 km/h. (b) Unknown: the rate at which the distance between the ships is changing at (c) 4:00 PM. If we let z be the distance between the ships. then we want to find dz/dt when t = 4 h. x l50—x dz dy 2 2 2 (d) z i (150 — :13) +y :> 22 dt — 2(150 x)( (it) +2y— dt (e) At 4:00 p.11. w 2 4(35) : 140 and y _ 4(25) _ 100 _> z _ ./(150 140)2 + 1002 2 «10,100. So dz 1 dm dy —10(35) + 100(25) 215 _:_ -150 ———%21.4km h. dt z I“ )d t+y dt] ‘/—1o,100' m1 / din dy . 2 2 2 11. y We are g1ven thatE — 60 mi/h and— dt :25 m1/h. z :2: +3; => dz dsc dy dz da: dy dz 1 xda; d_y> — 4 — — — 2» — - + z x 2zdt 2mdt+2ydt E> zdt mdt+ydt dt z dt ydt After 2 hours, a: = 2 (60) : 120 and y = 2 (25) = 50 => 2 : \/1202 + 502 : 130, dz 1 d3: dy 120(60) + 50(25) . _ z _ ._——————— = 65 m1 h. 5° dt z< dt+ydt>2130 / dw y a _2 _ E 12. We are given that E — —.1 6 m/s. By similar triangles 12 35 => # a: y dy_ 24 d1: 24 d__y_ _24(1.6) _ ___._ — _ — —— : #06 , A => dt— $2 dt- _:c2 (1.6). When a:— 8 d—t— 64 m/s '2 1 so the shadow 1s dec1easing at a rate of 0.6 m/s. 6:7: :4ft/s sandcé? :5ft/s.z2=(m+y)2+5002 => dz — 2(92 + y) (d_;:+:11t> . 15 minutes after the woman stans, we have We are given that 22—— dt ac : (4 ft/s)(20 min)(60 s/min) : 4800 ft and y = 5- 15-60 = 4500 i z— _ \/(4800 + 4500)2 + 500?— [6 740 000 so dz :c + y due 4800 + 4500 4 f 837 ~ —- —_——— +5 _ ~8.99fts. dt 2 < + dt) \/86J40,000( ) x/8674 / SECTION 3.10 RELATED RATES :l 223 . do: 14. We are g1ven that d—t = 24 ft/s. d (a) 3122(90—ac)2+902 : 2yj—y:2(907x)(—d—:), 313;»13 When m— — 45 y— — V452 + 902: 45 \/5, so _yd_ 90—x _d_x>_ _24 27% And y ( dt 45435 ) x/E’ so the distance from second base 13 decreasing at a rate of %~ ~ 10 7 ft/s (b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer—and we do. dz dac dz 45 24 2: 2 2 _ _ :4 _= 24 —m10.7ft . z :10 + 90 :> 22 dt 2x dt When a: 45 z 5J5, so dt 4—5¢5( )2 fl /s 15. A = ébh, where b is the base and h is the altitude. We are given that %— — 1 cm/min and (:34:— — 2 cm2/min dA_ b— dh db dt dt dt Using the Product Rule we have— — + h— —) When h — 10 and A — 100, we have 100=§b(10) => $12210 :> :220,so2:;<201+10%) :> 4220+1og : d _ —b : fl = —1.6cm/min. d d _y ~1m/s f‘1ndd—fwhen11328my2 2562 +1 3 9. 5 3 III 16. dy d3: d2: ydy y _ _ dt_ dt _> dt_xE__E'Whenm_8’y_‘/fiso dx x/6_5 x/(E — : 4—. Thus, the boat approaches the dock at T m 1.01 m/s. 17. We are given that ‘51—: 2 35 km/h and ill—f : 25 km/h. 22 = (a: + 3;)2 +1002 d i, 22 _ : 2(m + y)<— + d—i). At 4:00 P.M.. m = 4(35) : 140 and y 2 4(25) 2 100 3 z : ./(140 +100)2 +1002 = «67,600 = 260, so g_m+y dy _140+100 720 (d—td +— “T —3~55,4km/h. dt 2 (35 + 25) : 18. Let D denote the distance from the origin (0. 0) to the point on the curve y = fl D:\/(m—0)2+(y—0)2=x/x2+(\/:E>2=vw2+w => dB 1 2 71/2 dz 2x+1 d2: dar: —=-m+m 2x+1—=K_wt_: — dt 2( ) ( )dt 2 (132+;ndt. lhdt 3whenx— 4 dB 9 27 —=“3 2—m302 t 2 20” 4 5 cm/S 224 3 CHAPTER3 DIFFERENTIATION RULES dV 19. 2 If C : the rate at which water is pumped in, then E = C’ — 10,000. Where I V = émflh is the volume at time t. By similar triangles. g = % => 6 _ l _ 1 1 2 _ 7r 3 dV 7r dh i T—3h => V—§W(§h) h—fih => EZEhza. dh When h : 200 cm. dt— : 20 cm/min, so C — 10,000 = §(200)2(20) => C = 10,000 + fl‘gfl‘lw z 289,253 cm3/min. 3 . . . 3 b 20. By Similar triangles, 1 : hi so b : 3h. The trough has volume !!l W db ‘7 l V : ébhuo) : 5(3h)h :15h2 => 12 = E : 3mg 2 4 dh 2 dh —:—.Wh 22—: :_ ‘. dt 5h en h 2 dt 5 ‘% 5 ft/min 21. The figure is labeled in meters. The area A of a trapezoid is %(base1 + baseg)(height). and the volume V of the lO—meter—long trough is 10A. Thus, the volume of the trapezoid with height h is V : (10); [0.3 + (0.3 + 211)] h. By similar triangles. % 2 9035:) : i, so dV dV dh dh 2a h => V 5(06+h)h 3h+5h Now dt dh dt 0 (3+ )dt dh 0.2 dh 0.2 0.2 . 1 , 10 . a — 3+10h‘ Whenh — 0.3, a — 3+ 10(03) — —6- m/min — 30 m/min or 3 cm/min. 22. _l 34 ’l The figure is drawn without the top 3 feet. 2' V 1 b 12 h(20 10(b+12)h df ’ 'l t‘ 1e 2 — : an , rom simi ar man s, i V-V 2< + > > g I4—6—>l<—-12—>I<——-16——>I a: 6 y 16 8 8h 11h _:— —:—-:-— : Z 2 —=12 —-.Th, h andh 6 ,sob m+12+y h+1 +3 + 3 US 11h 110h2 dV 220 dh : —— : ,:—-: 4 — ——.Wh h=5, V 10(24-I— 3 >h 240h-I— 3 andsoO8 dt (2 0+ 3 h) dt en dh 0.8 3 . —— : _—————— : —— m 0.00132 ft n. dt 240 + 5(220/3) 2275 /”“ 2 3 . dV _ 3 . _ l 2 _ i E _ 1’: 23. We are given that E — 30 ft /min. V — 371'?” h — 37r(2 h — 12 dV dV dh 77h2 dh dh 120 __ : ___ = —— — = —. Wh Z) dt dhdt T 30 4dt :> dt whz 6“ 120 dh h —— 10 fl. a? — 102W : 3 m 0.38 ft/min. 571' SECTION 3.10 RELATED RATES 3 225 24. We are given dx/dt 2 8 ft/s. COM 2 ——1:0 2> as 2 100cot6 2> y '00 d2: 2 d0 d0 sin2¢9 _:_ _ —:— -8.When 2200, . dt 100 CSC 9 dt :’ dt 100 y x 2 100 1 d0 (1/2) 1 . ' :_:_ _:— ‘82——rads.Theanleis ““9 200 2 :5 dt 100 50 / g - 1 decreasmg at a rate of 5 rad / s. 25. A2§bh,butb25mandsin62 2> h24sin0,so m: d0 A 2 %(5)(4 sin 0) 2 10 sin 6. We are given a 2 0.06 rad/s. so 5 % 2 %% 2 (100080)(0.06) 2 0.6C050. When (9 2 g % 2 0.6(cos g) 2 (06)(%) 2 0.3 mZ/s. 26. We are given dQ/dt 2 2°/min 2 % rad/min. By the Law of Cosines, 12 X m2 :122 +152 — 2(12)(15) C050 = 369 — 360cos6 2 dw d0 dx 180 sing d0 —:3 ‘ — —:——.Wh 626°, 15 2x dt 60 $1119 dt 2) dt w dt en 0 x = #369 — 360 cos 60° 2 M189 2 3 Vi. so dz_1805in60°1 m/é f7” E‘ 3V2? 90‘3x/2—1: 21 m 0.396 m/min. 27. Differentiating both sides of PV 2 C’ with respect to t and using the Product Rule gives us P (iii: + V15; 2 0 dV V dP dP fl 600 t E 2 _Fd_t' When V 2 600. P 2 150 and (1—15 2 20, so we have 2 ~fi(20) 2 —80. Thus. the volume is decreasing at a rate of 80 cm3/ min. 28.1w”: 0 s p. live-4 % + w g _ 0 2 g _ ijjw jg; _ $3.13 When V 2 400, P 2 80 and Cell—1: 2 —10, so we have % 2 —1:?§0) (~10) 2 @ Thus, the volume is increasing at a rate of £70 2 36 cm3/min. 29. With R1 2 80 and R2 21001—1% 2 Bil + Rig 2 i + % 2 £{% 2 % 50R 2 %. Differentiating % 2 Bil + Rig with respect to t. we have 2%? 2 —Rif% — Rig—(2% 2> g = R2<Ri§% + Ri§%>' When 31 = 80and R2 = 100. dB _ 4002 1 1 107 dt _ 92 [8—02(03) + 1002 (02)] — m ~ 0.132 9/8 226 E CHAPTER3 DIFFERENTIATION RULES dB 30. We want to find — when L : 18 using B = 0.007W2/3 and W = 0.12L2‘53. dt @_dBdeL_ 2 ,1/3 153 20—15 dt _ dW dL dt _ (0007' EW )(0'12‘2‘53'L (10,000,000) 5 _ 2 2,53 *1/3 . _ _ [0007- §(0.12- 18 ) ](0.12 . 2.53 ~ 181 53) (W) m 1.045 x 10 8 g/yr . d 31. We are given that (1—: : 2 ft/s. sin0 : % => :0 = 103in0 => dsc d9 7r 7r d6 dt cos dt hen 6 4, 2 10 cos 4 dt => 49 2 Q x dt 10(1/fi) 5 II due 32. P Using Q for the origin, we are given a = —2 ft / s and need to find di: when as = —5. Using the Pythagorean Theorem twice, we have x dsc y dy d 2 122d y_ x y + j.Nowwhenm=#5.39:t/(—5)2+122+ :t/2+122=13—I—t/y2+122 4:) dt _ f “gt/$2 + 122 dt My? + 122 = 26, and y : V262 - 122 = V532. So when :1: = —5. dy : (—5)(26) ( 2) ~ ———10— m —0.87 ft/s. So cart B is moving towards Q at about 0.87 ft/s. dt «532(13) J133 33. (a) By the Pythagorean Theorem. 40002 + y2 = 82. Differentiating with respect (5 y ~ (Ly _ all ”Ly _ to t, we obtain 2y dt — 23 dt' We know that dt 7 600 ft/s, so when 4000 y : 3000 ft, 6 = \/ 40002 + 30002 : V25,000,000 = 5000 ft and de y dy 3000 1800 ___._, 6 ————360ft. dt 0 t 5000( 00) 5 /S y d d y 2 d9 1 dy d6 cos2 6 dy (b) Here ta“ 6 4000 $ dt (tan 9) dt (4000) :> see dt 4000 t 2) dt 4000 dt _ dy , _ , _ 4000 4000 _ 4 When y — 3000 ft. E — 600 ft/s. l — 5000 and c056 — —_€ ~ 5000 # 5 do , <4/5>2 E dt _ 4000 (600) — 0.096 rad/s. 34. 35. 36. 37. SECTION 3.10 RELATED RATES 227 We are given that g? = 4(271') : 87r rad/min. m = 3tan¢9 => dt 3 d—m:3se020d—6.Whenx—1tan0—§sosec0—1+(%):lgoand dt dt dz ,r . P i = 3(—9—0)(87r)— ~ % av, 83.8 km/min. We are given that (511—: : 300 km / h. By the Law of Cosines, y2 = x2 + 12 , 2(1)(x) cos 120° 2 $2 + 1 * 254—5) : x2 + x + 1, so d_y d“? dx dy 2x + 1 d_:1: After 1 minute, dt 2’02 5+5 3 a: 2g dt $:3—0600:5km =>y=\/52+5+1:\/3—1km => @szOO): w~296km/h. dt 2\/?H WE 2y— dy dt 22 2262+y2 —2mycos45° :m2+y2 — flmy => We are given that %— fl 3 mi/h and— — —2 mi/h. By the Law of Cosines, x 22 d: 22: d: + 2y d7: fix (1?: fly div. After 15 minutes [2 2 <§>2+<§J2we>e> :» “3‘6” 22%and h]. bird _§ _2_ wehavew—4andy—4— dz 2 E _ \/13—6\/§ [2( a: 2.125 mi/h. Let the distance between the runner and the friend be Z. Then by the Law € A of Cosines, 200—4 (2 : 2002 + 1002 — 2 . 200 . 100 . 0050 = 50,000 — 40,000 c080 (i). 4 _ 2 13— 26f )3] _ “m“ :—\/13 6f Mm v OJ + [\D A NIH V [\D I § A film V [\D | E A rob—- Differentiating implicitly with respect to t, we obtain 222—: 2 ~40 000(— sin 9) d—9 .Now if D [S the distance run when the angle is (9 radians, then by the formula for the length of an arc on a circle, 3 = r0, we have D = 1000, so d0 1 dB 7 d3 = —D :> — : —— — b t — 100 dt 100 dt 1—00 To su sti ute into the expression for dt we must know sinfi at the time when 15 = 200, which we find from (at): 2002 = 50,000 — 40,000 c030 c) c050 2 sin0 : 1 — (if 2 ?. Substituting we get 2(200) 2—5: 40,000‘2—1_5(&> 2 ash- :> df/dt : L? m 6.78 m/s. Whether the distance between them is increasing or decreasing depends on the direction in which the runner is running. 228 1: CHAPTER3 DIFFERENTIATION HULES 38. The hour hand of a clock goes around once every 12 hours or, in radians per hour. 3—72' : % rad / h. The minute hand goes around once an hour, or at the rate of 27r rad/ h. So the angle (9 between them (measuring clockwise from the minute hand to the hour hand) is changing at the rate of dO/dt : g 1 277 = —%71 rad/h. Now, to relate 0 to K, we use the Law of Cosines: £2 2 42 +82 — 2-4 ~ 8 ~ 0056 : 80 - 64c0s0(*). Differentiating implicitly with respect to t. we get 26 (cit—f : —64(— sin (9)6572. At 1:00, the angle between the two hands is one—twelfth of the circle, that is, 21—72” : % radians. We use (wk) to find 6 at 1:00: 6 = i /8() — 64 005% = V 80 — 32 x/fi. Substituting, we get 26% = 64 sin % (11:1) 2) dz 64 l 7&3 8 — (2) ( 6 ) — ———E—— m —18.6. So at 1:00, the distance between the tips of the hands is dt 2 80—32%; 3\/80—32\/§ decreasing at a rate of 18.6 mm/h m 0.005 mm/s. 3.11 Linear Approximations and Differentials __’______________———— 1. As in Example 1, T(0) : 185, T(10) = 172. T(20) = 160. and T(10) — T(20) _ 172 1 160 10 1 20 _ —10 T(30) z T(20) + T’(20)(30 — 20) z 160 — 1.2(10) : 148 OF. T’(20) % : #12 OF/min. We would expect the temperature of the turkey to get closer to 75 °F as time increases. Since the temperature decreased 13 °F in the first 10 minutes and 12 0F in the second 10 minutes. we can assume that the slopes of the tangent line are increasing through negative values: 71.3, —1.2, . . . . Hence, the tangent lines are under the curve and 148 °F is an underestimate. From the figure. we estimate the slope of the tangent line at t = 20 to be 1%:‘3—105 : — 3%. Then the linear approximation becomes T(30) % T(20) + T’(20) ~ 10 w 160 7 % (10) : 147% ~ 147.7. w : w : —12.2 kilopascals/km. 2 —1 13(3) m P(2) + P’(2)(3 1 2) m 74.9 — 122(1) : 62.7 kPa. 2. P'(2) m From the figure, we estimate the slope of the tangent line at h : 2 to be $22 = — %. Then the linear approximation becomes 13(3) x 19(2) + P’(2) . 1 m 74.9 — 3—35 m 63.23 kPa. ...
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