Business Calc Homework w answers_Part_37

Business Calc Homework w answers_Part_37 - Section 4.6 20....

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Unformatted text preview: Section 4.6 20. Step 1: r radius of spherical balloon S surface area of spherical balloon V volume of spherical balloon (b) 16 cos l 1 d dt Step 2: At the instant in question, dV dt 100 ft3/min and r 1 d dt dr dS We want to find the values of and . dt dt 6 dl l 2 dt 62 l 5 ft. Step 3: 1 0.6 1 6 ( 2) 102 2 The rate of change of angle 3 radian/sec 20 3 radian/sec. 20 is Steps 4, 5, and 6: (a) V dV dt dr dt 4 (5)2 100 dr dt 22. Step 1: x distance from origin to bicycle y height of balloon (distance from origin to balloon) s distance from balloon to bicycle 43 r 3 4 r2 dr dt Step 2: 1 ft/min dS dt dS dt dS dt 4 r2 8r dy dx is a constant 1 ft/sec and is a constant dt dt We know that The radius is increasing at the rate of 1 ft/min. (b) S 17 ft/sec. Three seconds before the instant in question, the values of x and y are x 0 ft and y 65 ft. Therefore, at the instant in question x 51 ft and y 68 ft. Step 3: dr dt 8 (5)(1) We want to find the value of 40 ft2/min ds at the instant in question. dt Step 4: x2 The surface area is increasing at the rate of s 40 ft2/min. y2 Step 5: ds dt 21. Step 1: l length of rope x horizontal distance from boat to dock angle between the rope and a vertical line dl At the instant in question, dt 2 ft/sec and l ds dt 10 ft. 2 dx dt dx dt 2 51 dc dt l2 102 dx dt x2 y dy dt y2 11 ft/sec 682 6x 2 12x [3(2)2 15x) 15) 12(2) dx dt 15](0.1) 0.3 dr dt dp dt dl 36 dt 10 dy 2y dt (68)(1) d3 (x dt (3x 2 dx d and . dt dt 36 l x2 (51)(17) Step 4, 5, and 6: l2 dx 2x 2 dt y The distance between the balloon and the bicycle is increasing at the rate of 11 ft/sec. 23. (a) Step 3: We want to find the values of x 1 Step 6: Step 2: (a) x 181 ( 2) 2.5 ft/sec 36 The boat is approaching the dock at the rate of (b) dx d (9x) 9 9(0.1) 0.9 dt dt dr dc 0.9 0.3 0.6 dt dt dc dt d3 x dt 3x 2 2.5 ft/sec. 45 x 45 dx x 2 dt 6x 2 12x 3(1.5)2 12(1.5) 45 (0.05) 1.52 1.5625 dr dt dp dt d (70x) dt dr dc dt dt 70 dx dt 3.5 70(0.05) ( 1.5625) 3.5 5.0625 182 Section 4.6 24. (a) Note that the level of the coffee in the cone is not needed until part (b). Step 1: V1 volume of coffee in pot y depth of coffee in pot Step 2: dV1 dt Step 2: 10 in3/min At the instant in question, Q Step 3: We want to find the value of dy dt dD dt . dt y dy dt dy dt Q D Step 5: Step 6: 9 0 mL/min . Step 4: 9 10 41 mL/L, 2 dy . dt We want to find the value of 9y Step 5: dV1 233 mL/min, D dQ 2 (mL/L)/min, and dt Step 3: Step 4: V1 25. Step 1: Q rate of CO2 exhalation (mL/min) D difference between CO2 concentration in blood pumped to the lungs and CO2 concentration in blood returning from the lungs (mL/L) y cardiac output dy dt 10 9 dy dt 0.354 in./min D dQ dt Q dD dt D2 The level in the pot is increasing at the rate of Step 6: approximately 0.354 in./min. dy dt (b) Step 1: V2 volume of coffee in filter r radius of surface of coffee in filter h depth of coffee in filter (41)(0) 466 1681 0.277 L/min2 The cardiac output is increasing at the rate of approximately 0.277 L/min2. 26. Step 1: y Step 2: At the instant in question, h (233)( 2) (41)2 dV2 dt 10 in3/min and (x , y ) 5 in. Step 3: We want to find dh . dt x Step 4: Note that 3 , so r 6 r h 12 rh 3 Then V2 h . 2 h3 . 12 x y Step 5: dV2 dt h 2 dh 4 dt x-coordinate of particle’s location y-coordinate of particle’s location angle of inclination of line joining the particle to the origin. Step 2: Step 6: At the instant in question, (5)2 dh 10 4 dt dh 8 in./min dt 5 dh Note that 0, so the rate at which the level is dt dx dt Step 3: falling is positive. The level in the cone is falling at the Step 4: 8 rate of 5 Since y 0.509 in./min. 10 m/sec and x We want to find x 0, 3 m. d . dt x 2, we have tan tan 1 x. y x x2 x x and so, for Section 4.6 Step 5: d dt 1 1 28. Step 1: x x-coordinate of particle y y-coordinate of particle D distance from origin to particle dx x 2 dt Step 6: d dt 1 32 1 183 (10) Step 2: 1 radian/sec At the instant in question, x The angle of inclination is increasing at the rate of dx dt 1 radian/sec. 1 m/sec, and dy dt 5 m, y 12 m, 5 m/sec. Step 3: 27. Step 1: We want to find y dD . dt Step 4: x2 D (x , y ) y2 Step 5: θ dD dt x x 1 2 dx 2x dt y2 x2 dy 2y dt dx dt x2 y dy dt y2 Step 6: x y (5)( 1) dD dt x-coordinate of particle’s location y-coordinate of particle’s location angle of inclination of line joining the particle to the origin (12)( 5) 52 5 m/sec 122 The particle’s distance from the origin is changing at the rate of 5 m/sec. 29. Step 1: Step 2: At the instant in question, dx dt 8 m/sec and x Street light 4 m. Step 3: We want to find d . dt 16 ft Step 4: 6 ft Since y x, we have tan and so, for x y x x 0, 1 tan [ ( x) 1/2 x tan 1 ( x) ( x) 1/2 . Shadow 1/2 , x x s s distance from streetlight base to man length of shadow Step 2: Step 5: d dt 1 1 [( x) 1 1 x 1 1/2 2 x(x 1 dx ( x) 3/2( 1) 2 dt dx 1 2( x)3/2 dt 1 2 10 ft. Step 3: We want to find ds . dt By similar triangles, dx 1) dt 16s 6s 6x, or s s s 6 x 16 . This is equivalent to 3 x. 5 Step 5: 1 2 5 ft/sec and x Step 4: Step 6: d dt dx dt At the instant in question, 4( 4 ( 8) 1) 2 radian/sec 5 ds dt 3 dx 5 dt The angle of inclination is increasing at the rate of Step 6: 2 radian/sec. 5 ds dt 3 ( 5) 5 3ft/sec The shadow length is changing at the rate of 3 ft/sec. 184 Section 4.6 Step 2: 30. Step 1: s distance ball has fallen x distance from bottom of pole to shadow At the instant in question, dV dt Step 2: At the instant in question, s ds dt 1 32 2 16 12 2 4 ft and Step 4: dx . dt 43 r and S 3 We have V Step 4: x 30 By similar triangles, 50 s x 1500 50x 1 1500s x . This is equivalent to 50 sx, or sx 1500. We will use 2 ds dS dt 1500(4) 2(16) 1500 ft/sec. 31. Step 1: x position of car (x 0 when car is right in front of you) camera angle. (We assume is negative until the car passes in front of you, and then positive.) At the first instant in question, x 1 (264) 2 A half second later, x 0 ft and dx dt 264 ft/sec. 2 1/3 dV 3V 4 dt 4 4000 (10)3 . 3 3 1/3 4000 ( 8) 3 2 33. Step 1: p x-coordinate of plane’s position x x-coordinate of car’s position s distance from plane to car (line-of-sight) At the instant in question, dp dt 0, 120 mph, s d at each of the two instants. dt Step 4: (x Step 4: 1 x 132 p)2 32 s2 Step 5: 2(x Step 5: 1 1 dx 132 dt x2 132 dx dt p) dp dt 2s ds dt Step 6: Note that, at the instant in question, Step 6: 1 d 0: dt 132: 5 mi, and dx . dt We want to find We want to find When x 3 dV 4 dt Step 3: Step 3: When x 1/3 3V 4 area is decreasing at the rate of 1.6 cm2/min. p 132 ft and 264 ft/sec. 1 2 3 4 Step 2: Step 2: d dt 3V 2/3 4 16 3 1.6 cm2/min 3 4 1000 dS Since 0, the rate of decrease is positive. The surface dt 1500 ft/sec The shadow is moving at a velocity of tan 4 Step 5: Note that V dt Step 6: dx dt 3V 1/3 , so S 4 Step 6: 1500s dx dt 4 r 2. These equations can be combined by noting that r dS dt . Step 5: dx dt dS . dt We want to find Step 3: 50x 1 (20) = 10 cm. 2 Step 3: 16 ft/sec. We want to find 8 cm3/min and r 8 mL/min 02 132 1 d dt 1 (264) 132 1 1 132 132 2 32. Step 1: r radius of balls plus ice S surface area of ball plus ice V volume of ball plus ice 1 (264) 132 52 x 2 radians/sec 1 radian/sec 2(4 32 dx dt dx 8 dt dx dt 0) 4 mi. 120 2(5)( 160) 120 120 dx dt The car’s speed is 80 mph. 1600 200 80 mph ds dt 160 mph. 185 Section 4.6 Step 6: 34. Step 1: s shadow length sun’s angle of elevation d dt (20)( 2) (10)(1) 102 202 0.1 radian/sec 5.73 degrees/sec Step 2: To the nearest degree, the angle is changing at the rate At the instant in question, s 60 ft and d dt 0.27 /min of 0.0015 radian/min. 6 degrees per second. 36. Step 1: Step 3: A ds . dt We want to find Step 4: c a 80 or s s tan 80 cot 120° Step 5: ds dt O 80 csc d dt 2 a b c Step 6: 80 and 60 Note that, at the moment in question, since tan 0 2 4 and so csc 5 , we have sin distance from O to A distance from O to B distance from A to B At the instant in question, a b 5 nautical miles, da dt 14 knots, and c2 c2 3 nautical miles, a2 a2 db dt 21 knots. Step 3: 2.25 in./min We want to find 7.1 in./min ds Since dt B Step 2: 5 . 4 52 80 (0.0015 ) 4 12 in ft 0.1875 1 ft min ds dt b dc . dt Step 4: Law of Cosines: 0, the rate at which the shadow length is b2 b2 2ab cos 120 ab decreasing is positive. The shadow length is decreasing at Step 5: the rate of approximately 7.1 in./min. 2c 35. Step 1: a distance from origin to A b distance from origin to B angle shown in problem statement 10 m, and b da dt 2 m/sec, a2 db dt dc dt dc 14 dt dc dt 2(7) 1 m/sec, 20 m. Step 3: We want to find d . dt tan 37. a tan 1 b Step 5: d dt b 1 1 a2 b da dt a b2 2b db dt a db dt b da dt b2 (5)2 ab 2(5)(14) 2(3)(21) (3)2 (5)(3) (5)(21) 49 (3)(14) 413 29.5 knots The ships are moving apart at a rate of 29.5 knots. Step 4: a or b da dt Note that, at the instant in question, c At the instant in question, 2a Step 6: Step 2: a dc dt db dt b da dt a db dt a2 b2 dy dt dy dx dx 10(1 x 2) 2(2x) dx dt dt dx Since 3 cm/sec, we have dt dy 60x cm/sec. dt (1 x2)2 (a) dy dt [1 60( 2) ( 2)2]2 (b) dy dt (1 60(0) 0 2)2 (c) dy dt 60(20) (1 202)2 120 52 (1 24 cm/sec 5 0 cm/sec 0.00746 cm/sec dx 20x x 2 ) 2 dt 7 ...
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