Business Calc Homework w answers_Part_36

Business Calc Homework w answers_Part_36 - 176 Section 4.6...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 176 Section 4.6 4. Use implicit differentiation. d (x sin y) dx (x)(cos y) dy dx (x (sin y)(1) x cos y) 3. (a) Since d (1 dx dy x dx dy dx dy dx dy dx y 2x dy dx dy dx dV dh dV , we have dh dt dt (b) Since dV dt dV dr dV , we have dr dt dt xy) y(1) sin y (c) y sin y x x cos y y sin y x x cos y 5. Use implicit differentiation. d2 x dx dV dt 4. (a) d tan y dx dy sec2 y dx 2x sec2 y 2x cos2 y dV dt dV dt dV dt d2 rh dt dh r2 dt dh r2 dt dP dt dP dt dP dt dP dt d ln (x y) dx dy 1 1 dx xy dy 1 dx dy dx dh . dt dr dt 2 rh . d (RI 2) dt dR d R I2 I2 dt dt dI dR R 2I I2 dt dt dI dR 2RI I2 dt dt d2 (r h) dt dr h(2r) dt dr 2 rh dt (b) If P is constant, we have 6. Use implicit differentiation. r2 2RI d (2x) dx dI dt dR dt 0, or y2 I2 dP dt 0, which means dR dt 2R dI I dt 2P dI . I 3 dt z2 2 2(x 2x 5. y) 2y 8. Using A(0, 4), we create the parametric equations x 0 at and y 4 bt, which determine a line passing through A at t 0. We now determine a and b so that the line passes through B(5, 0) at t 1. Since 5 0 a, we have a 5, and since 0 4 b, we have b 4. Thus, one parametrization for the line segment is x 5t, y 4 4t, 0 t 1. (Other answers are possible.) 10. One possible answer: 2 3 2 d dt x2 1 ds dt 7. Using A( 2, 1) we create the parametric equations x 2 at and y 1 bt, which determine a line passing through A at t 0. We determine a and b so that the line passes through B(4, 3) at t 1. Since 4 2 a, we have a 6, and since 3 1 b, we have b 4. Thus, one parametrization for the line segment is x 2 6t, y 1 4t, 0 t 1. (Other answers are possible.) 9. One possible answer: ds dt t dA dt dA dr dA , we have dr dt dt 2. Since dS dt dS dr dS , we have dr dt dt 6. 2x 2 y y 2 2 dx x dt dy y dt x2 dA dt dA dt dA dt y2 (x 2 y2 dx dt 2y z 2) z dt 1 z z 2 2x dy dt 2z dz dt dz dt z2 d1 ab sin dt 2 1 da db b sin a 2 dt dt 1 da db b sin a sin 2 dt dt dV dt 2 sin ab ab cos d sin dt d dt 1 volt/sec. (b) Since I is decreasing at the rate of dI dt Section 4.6 Exercises 1. Since ds dt 2x d 2 7. (a) Since V is increasing at the rate of 1 volt/sec, 3 2 t ds dt 1 2 dr dt 2r . dr dt 8r . 1 amp/sec. 3 (c) Differentiating both sides of V dV dt 1 amp/sec, 3 I dR dt dI dt R. IR, we have Section 4.6 (d) Note that V IR gives 12 2R, so R 6 ohms. Now substitute the known values into the equation in (c). 1 2 dR dt (b) P 1 3 6 dR 2 dt 3 ohms/sec 2 3 dR dt 3 ohms/sec. Since this 2 At the instant in question, dr dt 0.01 cm/sec, r dA . dt Step 5: 2r dr dt 52 At the instant in question, cm2/sec 2 (50)(0.01) At the instant in question, the area is increasing at the rate cm2/sec. At the instant in question, dw dt 2 cm/sec, 2 cm/sec, l 12 cm, 5 cm. dy dt y = 3 m, and z We want to find 2 m/sec, dz dt 1 m/sec, x 4 m, 2 m. dA dP dD , , and . dt dt dt dV dS ds , , and . dt dt dt Steps 4, 5, and 6: (a) V xyz xy dz dt xz (4)(3)(1) dy dt yz dx dt (4)(2)( 2) dS dt dS dt 2(xy xz dy 2x dt yz) dx y dt x dz dt z dx dt y lw 2[(4)( 2) (3)(1) (3)(1) (2)( 2)] dz dt z dy dt (4)(1) (2)(1) Steps 4, 5, and 6: 2 m3/sec (3)(2)(1) The rate of change of the volume is 2 m3/sec. (b) S Step 3: dw l dt 1 m/sec, dV dt dV dt Step 2: We want to find dx dt Step 3: 9. Step 1: l length of rectangle w width of rectangle A area of rectangle P perimeter of rectangle D length of a diagonal of the rectangle dA dt dA dt dl 2l 2 2 dt 2l w (12)( 2) (5)(2) 122 dw l w dt dt dw 2w 2 dt l w2 14 cm/sec 13 1 Step 2: Step 6: (a) A w2 10. Step 1: x, y, z edge lengths of the box V volume of the box S surface area of the box s diagonal length of the box r2 and w 0 cm/sec (d) The area is increasing, because its derivative is positive. The perimeter is not changing, because its derivative is zero. The diagonal length is decreasing, because its derivative is negative. Step 4: dl dt 2(2) 14 cm/sec. 13 50 cm. Step 3: of 2( 2) The rate of change of the length of the diameter is Step 2: dA dt dw dt dl dD dt dD dt 8. Step 1: r radius of plate A area of plate dA dt 2 l2 (c) D value is positive, R is increasing. A 2w dl 2 dt The rate of change of the perimeter is 0 cm/sec. R is changing at the rate of We want to find 2l dP dt dP dt 177 0 m2/sec The rate of change of the surface area is 0 m2/sec. dl w dt (12)(2) (5)( 2) 14 cm2/sec The rate of change of the area is 14 cm2/sec. 178 Section 4.6 13. Step 1: x distance from wall to base of ladder y height of top of ladder A area of triangle formed by the ladder, wall, and ground angle between the ladder and the ground 10. continued x2 (c) s y2 z2 1 ds dt 2 x 2 y2 2x z2 dx dt 2y dy dt 2z dz dy Step 2: dx x dt dy y dt x 2 y dz z dt 2 z (4)(1) ds dt At the instant in question, x 2 (2)(1) 32 0 22 29 Step 4, 5, and 6: 11. Step 1: s (diagonal) distance from antenna to airplane x horizontal distance from antenna to airplane ds dt Step 4: x 2 49 y 300 mph. dx . dt s2 x2 169 169 122 5. rate of 12 ft/sec. (Note that the downward rate of 49 motion is positive.) 2s s2 49 ds dt s s2 ds 49 dt 10 102 (300) 3000 49 mph 1 xy 2 1 dy x 2 dt (b) A Step 6: dx dt 0 The top of the ladder is sliding down the wall at the s 2 or x 1 2 169 dy 2y dt dy Then 2(12)(5) 2(5) 0 dt dy dy 12 ft/sec or 12 ft/sec dt dt Step 5: dx dt y2 To evaluate, note that, at the instant in question, Step 3: We want to find (a) x 2 dx 2x dt Step 2: At the instant in question, 10 mi and 5 ft/sec. dy dA d , , and . dt dt dt We want to find 0 m/sec The rate of change of the diagonal length is 0 m/sec. s dx dt Step 3: (3)( 2) 42 12 ft and dA dt 420.08 mph y dx dt Using the results from step 2 and from part (a), we 51 The speed of the airplane is about 420.08 mph. have 12. Step 1: h height (or depth) of the water in the trough V volume of water in the trough dA dt 1 [(12)( 12) 2 (5)(5)] 119 ft/sec. 2 The area of the triangle is changing at the rate of 59.5 ft2/sec. Step 2: At the instant in question, dV dt 2.5 ft3/min and h 2 ft. y x (c) tan 2 x d dt Step 3: sec dh We want to find . dt Since tan Step 4: 4 3 cos The width of the top surface of the water is h, so we have V 1 4 (h) h (15), or V 2 3 10h 2 20h y dx dt x2 5 , we have for 0 12 1 12 and so sec2 13 dh dt d dt Step 6: of dh dt 2.5 20(2) dh dt 0.0625 1 ft/min 16 The water level is increasing at the rate of 1 ft/min. 16 2 12 2 13 169 . 144 Combining this result with the results from step 2 and from part (a), we have Step 5: dV dt dy dt 169 d 144 dt (12)( 12) (5)(5) , so 122 1 radian/sec. The angle is changing at the rate 1 radian/sec. Section 4.6 14. Step 1: 179 Step 5: Kite s Inge dV dt 12 r dr dt Step 6: 300 ft dV dt x 1 3000 12 (1.900) 19 2500 0.0076 3 0.0239 in /min s x length of kite string horizontal distance from Inge to kite The volume is increasing at the rate of approximately 0.0239 in3/min. Step 2: At the instant in question, dx dt 25 ft/sec and s 500 ft 16. Step 1: Step 3: We want to find h ds . dt Step 4: x 2 r 300 2 s 2 r h V Step 5: 2x dx dt 2s ds dx or x dt dt s ds dt base radius of cone height of cone volume of cone Step 2: Step 6: At the instant in question, since x 2 x s2 3002 5002 ds dt Thus (400)(25) (500) , so 3002 3002 ds dt s 2, we have At the instant in question, h 4 m and dV dt 10 m3/min. Step 3: 400. 20 ft/sec. Inge must let the string out at the rate of 20 ft/sec. We want to find dh dr and . dt dt Step 4: 15. Step 1: Since the height is 3 of the base diameter, we have 8 4 h. 3 12 142 16 h 3 We also have V rh hh . We will 3 33 27 16 h 3 4 use the equations V and r h. 27 3 h 6 in. 3 (2r) or r 8 Steps 5 and 6: r (a) The cylinder shown represents the shape of the hole. r radius of cylinder V volume of cylinder Step 2: At the instant in question, dr dt 0.001 in. 3 min and (since the diameter is 3.800 in.), r Step 3: Step 4: r 2(6) 1.900 in. 10 dh dt 16 h 2 dh 9 dt 16 (4)2 dh 9 dt 45 m/min 128 1125 cm/min 32 The height is changing at the rate of 1125 32 11.19 cm/min. (b) Using the results from Step 4 and part (a), we have dV We want to find . dt V 1 in./min 3000 dV dt dr dt 4 dh 3 dt 4 1125 3 32 375 cm/min. 8 The radius is changing at the rate of 6 r2 375 8 14.92 cm/min. 180 Section 4.6 17. Step 1: Step 6: 45 m r 6 6 6m 144 dy dt h or r h V radius of top surface of water depth of water in reservoir volume of water in reservoir r dV dt At the instant in question, 50 m3/min and h 5 m. Step 3: We want to find 0.01326 m/min 25 6 1.326 cm/min y)2 (13 169 132, y)2 (13 26y dh dr and . dt dt Step 2: At the instant in question, Note that Then V h 6 by similar cones, so r 7.5h. r 45 12 1 rh (7.5h)2h 18.75 h3 3 3 We want to find the value of 18.75 h 3, dV dt 56.25 h 2 dh . dt From part (b), r 7.5h, dr dt dr dt dh dt 80 cm/min. The rate 3 or of change of the radius of the water’s surface is 80 3 18. (a) Step 1: y depth of water in bowl V volume of water in bowl 1 y2 26y 2 (26 2y) dy dt 13 26y y dy y 2 dt 13 8 26(8) 5 288 125 72 1 24 82 5 12 1 24 0.00553 m/min 0.553 cm/min Step 2: No numerical information is given. Step 2: dV dt y 2. 26y 19. Step 1: r radius of spherical droplet S surface area of spherical droplet V volume of spherical droplet 8.49 cm/min. At the instant in question, dr . dt Step 6: dr dt 7.5 1 m/min. 24 Step 5: falling is positive.) (b) Since r dy dt 8 m, Step 4: 56.25 (52) 50 6 m3/min, y Step 3: dh , and so dt dh 8 32 m/min cm/min. dt 225 9 32 The water level is falling by 1.13 cm/min. 9 dh (Since 0, the rate at which the water level is dt Thus dV dt and therefore (from part (a)) Step 5 and 6: (a) Since V y 2. (c) Step 1: y depth of water r radius of water surface V volume of water in bowl Step 4: y dy dt dy dt 1 24 (b) Since r 2 Step 2: (82)] [26 (8) 6 m3/min and Step 3: We want to show that 8 m. Step 3: Step 4: dy We want to find the value of . dt S Step 4: dr is constant. dt Steps 5 and 6: V 3 y 2(39 y) or V Step 5: dV dt (26 y y 2) dy dt 13 y 2 3 y3 4 r 2, V 4 3 dV r, 3 dt kS for some constant k 43 dV dr r , we have 4 r2 . 3 dt dt dV Substituting kS for and S for 4 r 2, we have kS dt dr or k. dt Differentiating V dr dt S, ...
View Full Document

This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

Ask a homework question - tutors are online