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1 Chapter 5 Review 95 (c) f( ) f ( 7) ( 7) ( 6 7) Chapter 5 Review Eercises (pp. 60 6). y y ( ) + ( )( ) + ( ) The first derivative has a zero at. 6 Critical point value: y 9 Endpoint values: y y 0 The global maimum value is 6 at 9, and the global minimum value is at.. Since y is a cubic function with a positive leading coefficient, we have lim y and lim y. There are no global etrema.. / / y ( )( e )( ) + ( e )( ) / e + / e ( )( + ) Intervals < < < 0 0 < < > Sign of y + + Behavior of y Decreasing Increasing Decreasing Increasing Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

2 96 Chapter 5 Review d / y [ e ( + )] d / / ( e )( + ) + ( + )( e )( ) / ( e )( + + ) / e ( + ) / e [( 05. ) +. 75] The second derivative is always positive (where defined), so the function is concave up for all 0. (a) [, 0 ) and[, ) (b) (, ] and (0, ] (c) (, 0 ) and ( 0, ) (d) None (e) Local (and absolute) minima at (, e) and (, e) (f) None. Note that the domain of the function is [, ]. ( y ( ) + )( ) + ( ) Intervals < < < < < < Sign of y + Behavior of y Decreasing Increasing Decreasing ( )( ) ( ) ( ) y ( 6) ( ) Note that the values ± 6 are not zeros of y because they fall outside of the domain. Intervals < < 0 0< < Sign of y + Behavior of y up Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

3 Chapter 5 Review (a), (b), and, (c) (, 0) (d) (0, ) (e) Local maima: (, 0), (, ) Local minima: (, 0), (, ) Note that the etrema at ± are also absolute etrema. (f) (0, 0) y Using grapher techniques, the zero of y is Intervals < < Sign of y + Behavior of y Increasing Decreasing y ( + 6 ) The second derivative is always negative so the function is concave for all. (a) Approimately (, 05. ] (b) Approimately [ 05., ) (c) None (d) (, ) (e) Local (and absolute) maimum at ( 05., 5. ) (f) None y e Intervals < < Sign of y + Behavior of y Decreasing Increasing y e The second derivative is always positive, so the function is concave up for all. (a) [, ) (b) (, ] (c) (, ) (d) None (e) Local (and absolute) minimum at (, 0) (f) None 7. Note that the domain is (, ). / y ( ) 5/ y ( ) ( ) 5/ ( ) Intervals < < 0 0< < Sign of y + Behavior of y Decreasing Increasing 5 ( ) 5/ / 5/ ( ) ( ) ( )( ) ( ) ( ) y ( ) / ( ) [ + 5 ] 5/ ( ) + 9/ ( ) The second derivative is always positive, so the function is concave up on its domain (, ). (a) [0, ) (b) (, 0] (c) (, ) (d) None (e) Local minimum at (0, ) (f) None Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

4 9 Chapter 5 Review. ( )( ) ( )( ) ( + ) y ( ) ( ) Intervals < / / < < < Sign of y + Behavior of y Increasing Decreasing Decreasing ( ) ( 6 ) ( + )( )( )( ) y ( ) ( )( 6 ) ( + )( 6 ) ( ) 6 ( + ) ( ) Intervals < / / < < 0 0< < < Sign of y + + Behavior of y up (a) (, + / ] (, 0.79] (b) [ + /, ) [ 0.79, ) and (, ) (c) (, + / ) (,.60) and (, ) (d) ( + /, ) (.60, ) (e) Local minimum at / /, ( 0. 79, 0. 59) up (f) / /, (. 60, 0. 0) 9. Note that the domain is [, ]. y Since y is negative on (, ) and y is continuous, y is decreasing on its domain [, ]. d / y [ ( ) ] d / ( ) ( ) / ( ) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

5 Chapter 5 Review 99 Intervals < < 0 0< < Sign of y + Behavior of y up (a) None (b) [, ] (c) (, 0) (d) (0, ) (e) Local (and absolute) maimum at (, π ); local (and absolute) minimum at (, 0) π (f) 0, 0. Note that the denominator of y is always positive because it is equivalent to ( + ( + + )( ) ( )( + ) y ( + + ) + ( + + ) Intervals < < < < Sign of y + Behavior of y Decreasing Increasing Decreasing ) +. ( + + ) ( ) ( + )( )( + + )( + ) y ( + + ) ( + + )( ) ( + )( + ) ( + + ) ( + + ) Using graphing techniques, the zeros of (and hence of y ) are at 5., 0706., and. 90. Intervals (,.5) (.5, 0.706) ( 0.706,.90) (.90, ) Sign of y + + Behavior of y (a), (b) (, and, ) up up Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

6 00 Chapter 5 Review (c) Approimately (.5, 0.706) and (.90, ) (d) Approimately (,.5) and ( 0.706,.90) (e) Local maimum at local minimum at,, (. 7, 0. ); ( 7., 06. ) (f) (.5, 0.57), ( 0.706, 0.), and (.90, 0.6) d. For > 0, y ln d d For < 0: y ln( ) ( ) d Thus y for all in the domain. Intervals (, 0) (0, ) Sign of y + Behavior of y Decreasing Increasing y. The second derivative is always negative, so the function is concave on each open interval of its domain. (a) (0, ] (b) [, 0) (c) None (d) (, 0) and (0, ) (e) Local (and absolute) maima at (, ln ) and (, ln ) (f) None. y cos sin Using graphing techniques, the zeros of y in the domain 0 π are 0.76, 0.99, π 57.,., and.965,., π, 5.59 Intervals 0 < < < < < < π π < <.. < <.965 Sign of y Behavior of y Increasing Decreasing Increasing Decreasing Increasing Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

7 Intervals.965 < <.. < < π π < < 5.59 Chapter 5 Review < < π Sign of y + + Behavior of y y 9sin 6cos Decreasing Increasing Decreasing Increasing Using graphing techniques, the zeros of y in the domain 0 π are 0. 5,. 66,. 76, 600., 5.,., 5. and Intervals 0 < < < < < < < < < <.5 Sign of y + + Behavior of y up up Intervals.5 < <.. < < < < < < π Sign of y + + Behavior of y up π π up (a) Approimately [0, 0.76], 099.,, [., 965. ],.,, and [ 559., π ] (b) Approimately [0.76, 0.99], π π,., [ 965.,. ], and, 559. (c) Approimately (0.5,.66), (.76,.600), (,5,.), and (5., 6.000) (d) Approimately (0, 0.5), (.66,.76), (.600,.5), (., 5.), and( 6000., π ) π (e) Local maima at ( ,. ),, 0 and (.965,.66), π,, and ( π, ); local minima at (,),( , 05. ),(., 0.5), (.,.06), and (5.59,.06) π Note that the local etrema at.,, and are also absolute etrema. (f) (0.5, 0.7), (.66, 0.67), (.76, 0.67), (.600, 0.7), (.5, 0.9), (., 0.0), (5., 0.0), and (6.000, 0.9) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

8 0 Chapter 5 Review. y < 0 e,, > 0 Intervals < 0 Sign of y y 0 < < < + Behavior of y Decreasing Increasing Decreasing y e, < 0 6, > 0 Intervals < 0 0< Sign of y + Behavior of y up (a) 0, (b) (, 0] and, (c) (, 0) (d) (, 0 ) (e) Local maimum at 6, ( 55.,. 079) (f) None. Note that there is no point of inflection at 0because the derivative is undefined and no tangent line eists at this point.. y Using graphing techniques, the zeros of y are and. 69. Intervals < < < < Sign of y + Behavior of y Decreasing Increasing Decreasing y Using graphing techniques, the zero of y is Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

9 Chapter 5 Review 0 Intervals < < Sign of y + Behavior of y up (a) Approimately [ 057., 69. ] (b) Approimately (, 057. ] and [. 69, ) (c) Approimately (, 079. ) (d) Approimately (. 079, ) (e) Local maimum at (. 69, 057. ); local minimum at ( 0. 57, 0. 97) (f) (. 079, 60. ) y 9 9 y / 5 / 5 Intervals < < < < 9 Sign of y + Behavior of y Decreasing Increasing Decreasing 65 / 6 5 / ( + 9) y / 5 Intervals 9 < < < 0 0 < 9 Sign of y + Behavior of y up (a) 0, 9 (b) (, 0] and 9, (c) (d), 9 0 9, and (, 0 ) Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

10 0 Chapter 5 Review 5 / 0 (e) Local maimum at, ( 09., 0. ); local minimum at (0, 0) (f) 0,, / 6. We use a combination of analytic and grapher techniques to solve this problem. Depending on the viewing windows chosen, graphs obtained using NDER may ehibit strange behavior near because, for eample, 5 + NDER(y, ) 5,000,000 while y is actually undefined at. The graph of y shown below. is ( )( + ) ( 5 + )( ) y ( ) ( ) The graph of y is shown below. The zero of y is 0.5. Intervals < < < < Sign of y + Behavior of y Increasing Decreasing Decreasing ( ) ( ) ( )( )( ) y ( ) ( )( ) ( ) ( ) ( 6 + ) ( ) The graph of y is shown on the net page. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

11 Chapter 5 Review 05 The zero of 6 + (and hence of y ) is.70. Intervals < < < < Sign of y + Behavior of y up (a) Approimately (, 0.5] (b) Approimately [0.5, ) and (, ) (c) Approimately (,.70) (d) (, ) and approimately (.70, ) (e) Local maimum at (0.5,.7) (f) (.70,.0) 7. y 6( + )( ) Intervals < < < < Sign of y + + Behavior of y Decreasing Increasing Increasing y 6( + )( )( ) + 6( ) ( ) 6( )[( + ) + ( )] ( ) Intervals < 0 0 < < < Sign of y + + Behavior of y up (a) There are no local maima. up (b) There is a local (and absolute) minimum at. (c) There are points of inflection at 0 and at. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

12 06 Chapter 5 Review. y 6( + )( ) Intervals < < < < Sign of y + + Behavior of y d y 6( ) 6( ) d Intervals Increasing Decreasing Increasing < Sign of y + < Behavior of y up (a) There is a local maimum at. (b) There is a local minimum at. (c) There is a point of inflection at d 5 9. Since e + e, d f( ) e + C.. d 0. Since sec sec tan, f( ) sec + C. d d. Since ln , d f( ) ln C. d / /. Since + +, d / / f( ) + + C.. f( ) cos+ sin+ C f ( π ) + 0+ C C f( ) cos+ sin+ Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

13 / f( ) C f () C 0 C / f( ) vt () s () t 9. t+ 5 st () 9. t + 5t+ C s( 0) 0 C 0 st () 9. t + 5t+ 0 at () v () t vt () t+ C v( 0) 0 C 0 vt () s () t t+ 0 st () 6t + 0t+ C s( 0) 5 C 5 st () 6t + 0t+ 5 f( ) tan f ( ) sec π π π L ( ) f f + π π π tan sec + + π + + π + 9. f( ) + tan f ( ) ( + tan ) (sec ) cos ( + tan ) L ( ) f( 0) + f ( 0)( 0) ( 0) + 0. f( ) e + sin f ( ) e + cos L ( ) f( 0) + f ( 0)( 0) + ( 0) + Chapter 5 Review 07. The global minimum value of occurs at.. (a) The values of y and y are both negative where the graph is decreasing and concave, at T. (b) The value of y is negative and the value of y is positive where the graph is decreasing and concave up, at P.. (a) The function is increasing on the interval (0, ]. (b) The function is decreasing on the interval [, 0). (c) The local etreme values occur only at the endpoints of the domain. A local maimum value of occurs at, and a local maimum value of occurs at.. The th day. f( ) sec f ( ) sectan π π π L ( ) f + f π π π π sec + sec tan π + () π + 5. y y f() Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

14 0 Chapter 5 Review 6. (a) We know that f is decreasing on [0, ] and increasing on [, ], the absolute minimum value occurs at and the absolute maimum value occurs at an endpoint. Since f(0) 0, f (), and f(), the absolute minimum value is at and the absolute maimum value is at. (b) The concavity of the graph does not change. There are no points of inflection. (c) 7. (a) f( ) is continuous on [0.5, ] and differentiable on (0. 5, ). (b) f ( ) ( ) + (ln )( ) + ln Using a 0. 5 and b, we solve as follows. f() f(.) 05 f () c 0. 5 ln 0. 5ln ln c 5. ( ) 05. ln 05. ln c 5. ln c 0. ln( 7 ) 0. c e ( 7 ) 5 c e f() b f() a (c) The slope of the line is m b a 0. ln( 7 ) 0. ln 5, and the line passes through (, ln ). Its equation is y 0.(ln 5)( ) + ln, or approimately y (d) The slope of the line is m 0. ln 5, and the line passes through 5 5 (, c f()) c ( e 5, e 5( + 0. ln 5)) (. 579, 07. ). Its equation is y 0. (ln 5)( c) f( c), 5 y 0. ln 5( e 5 + 5) + e 5( + 0. ln 5), 5 y 0.(ln 5) e 5, or approimately y Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5 Review 09. (a) vt () s () t 6t t (b) at () v () t 6 6t (c) The particle starts at position moving in the positive direction, but decelerating. At approimately t 0.5, it reaches position.

15 Chapter 5 Review 09. (a) vt () s () t 6t t (b) at () v () t 6 6t (c) The particle starts at position moving in the positive direction, but decelerating. At approimately t 0.5, it reaches position. and changes direction, beginning to move in the negative direction. After that, it continues to accelerate while moving in the negative direction. 9. (a) L ( ) f( 0) + f ( 0)( 0) + 0( 0) (b) f( 0. ) L( 0. ) (c) Greater than the approimation in (b), since f () is actually positive over the interval (0, 0.) and the estimate is based on the derivative being (a) Since (b) dy ( )( e ) + ( e )( ) d ( ) e, dy ( ) e d. dy [ ( ) ( ) ]( e )( 00. ) 00. e (a) With some rounding, the population in thousands is given by 50, 9 y t + 9. e where t 0 represents 750. (b) [0, 500] by [ 00,000, 700,000] 50, 9 y + 9, 709 thousand or 9,709,000 (c) ( ) 9. e (d) Using the Second Derivative, we find the maimum rate of growth about 9. We find a point of inflection here, which shows the beginning of a decline in the rate of growth. (e) The regression equation predicts a longterm maimum population of 50,9,000.. f( ) cos + f ( ) sin + f( n ) n+ n f ( ) n cos n + n sin n n + The graph of y f () shows that f () 0 has one solution, near Solution: Let t represent time in seconds, where the rocket lifts off at t 0. Since at () v () t 0 m/sec and v () 0 0m/sec, we have vt () 0t, and so v ( 60) 00 m/sec. The speed after minute (60 seconds) will be 00 m/sec.. Let t represent time in seconds, where the rock is blasted upward at t 0. Since at () v () t 7. m/sec and v () 0 9 m/sec, we have vt (). 7t+ 9. Since s () t 7. t+ 9 and s () 0 0, we have st () 6. t + 9t. Solving vt () 0, we find that the rock attains its maimum height at t 5 sec and its height at that time is s ( 5) 6. 5 m. n Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

16 0 Chapter 5 Review 5. Note that s 00 r and the sector area is given by s A π r π r rs r( 00 r) 50r r. 6. To find the domain of Ar () 50r r, note that r > 0 and 0 < s < π r, which implies 0< 00 r < π r, which gives 50. < r < 50 π +. Since A ( r) 50 r, the critical point occurs at r 5. This value is in the domain and corresponds to the maimum area because A () r, which is negative for all r. The greatest area is attained when r 5 ft and s 50 ft. y 7 (, 7 ) For 0< < 7, the triangle with vertices at (0, 0) and ( ±, 7 ) has an area given by 7 7 A( ) ( )( ). Since + and A 6, A 7 ( )( ) the critical point in the interval ( 0, 7) occurs at and corresponds to the maimum area because A ( ) is negative in this interval. The largest possible area is A () 5 square units. 7. If the dimensions are ft by ft by h ft, then the total amount of steel used is Therefore, + h 0 and so 0 h. The volume is given by 0 + h ft. V( ) h Then V ( ) ( 6 + )( 6 ) and V ( ) 5.. The critical point occurs at 6, and it corresponds to the maimum volume because V ( ) < 0 for > 0. The 0 6 corresponding height is ft. The 6 () base measures 6 ft by 6 ft, and the height is ft.. If the dimensions are ft by ft by h ft, then we have h and so h. Neglecting the quarter-inch thickness of the steel, the area of the steel used is A ( ) + h +. We can minimize the weight of the vat by minimizing this quantity. Now A ( ) ( ) and A ( ) The critical point occurs at and corresponds to the minimum possible area because A ( ) > 0 for > 0. The corresponding height is ft. The base should measure ft by ft, and the height should be ft. h h 9. We have r +, so r. We wish to minimize the cylinder s volume h h V πr h π h πh π for 0 < h <. Since dv πh π π ( + h )( h dh ) and dv π h dh, the critical point occurs at h and it corresponds to the maimum value because dv < 0 for h > 0. The dh corresponding value of r is. The largest possible cylinder has height and radius. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

17 r h 50. Note that, from similar cones, 6, so h r. The volume of the smaller cone is given by π V πr h πr ( r) πr r for 0 < r < 6. Then dv πr πr πr( r), so the critical dr point occurs at r. This critical point corresponds to the maimum volume because dv 0 dr > for a m f( ) f ( ) B B B+ C + C C C B B C B+ C + C C C B C B + C + BC C C ( ) C B B B + + B dv and < 0 for < 6. dr r < The smaller cone has the largest possible value when r ft and h ft. 5. (b, c) Domain: 0 < < 5 Chapter 5 Review The maimum volume is approimately in and it occurs when 96. in. (d) Note that V( ) , so V ( ) Solving V ( ) 0, we have 50 ± ( 50) ( 6)( 75) 6 () 50 ± ± ± These solutions are approimately 96. and 6. 7, so the critical point in the appropriate domain occurs at y 0 5. Lid (, cos 0.) 0 in. Base 5 in. (a) V( ) ( 5 )( 5 ) 5 For 0 π < <, the area of the rectangle is given by A( ) ( )( cos 0. ) 6cos 0.. Then A ( ) 6( 0. sin 0. ) + 6(cos 0. )( ) 6(cos sin 0. ) Solving A ( ) 0 graphically, we find that the critical point occurs at. 6 and the corresponding area is approimately 9.95 square units. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

18 Chapter 5 Review 5. The cost (in thousands of dollars) is given by C ( ) 0+ 0( 0 y) Then C ( ) 0 ( ) 0 0. Solving C ( ) 0, we have: Choose the positive solution: +. mi 7 6 y. 607 mi 7 5. The length of the track is given by + π r, so we have + π r 00 and therefore 00 π r. Then the area of the rectangle is Ar () r r( 00 π r) 00 00r π r, for 0 < r <. π Therefore, A () r 00 π rand A () r π, 00 so the critical point occurs at r m and π this point corresponds to the maimum rectangle area because A () r < 0 for all r. The corresponding value of is π 00 m. π The rectangle will have the largest possible 00 area when 00 m and r m. π ( 5 )( ) ( 0 )( ) P ( ) k ( 5 ) k 0+ 0 ( 5 ) The solutions of P ( ) 0 are 0 ± ( 0) ( )( 0) 5± 5, so the () solution in the appropriate domain is Check the profit for the critical point and endpoints: Critical point: 76. P( ) 06. k End points: 0 P ( ) k P ( ) k The highest profit is obtained when.76 and y 5.5, which corresponds to 76 grade A tires and 55 grade B tires. 56. (a) The distance between the particles is f(t) π where f() t cost+ cos t+. Then π f () t sint sin t+ Solving f (t) 0 graphically, we obtain t.7, t.0, and so on. Alternatively, f (t) 0 may be solved analytically as follows. 55. Assume the profit is k dollars per hundred grade B tires and k dollars per hundred grade A tires. Then the profit is given by 0 0 P ( ) k+ k 5 ( 0 5) + ( 5 ) k 5 0 k 5 Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

19 Chapter 5 Review π π π π f () t sin t+ sin t+ + π π π π sin t cos cos t sin π π π π + + sin t cos cos t sin π π sin cos t, + π π so the critical points occur when cos t+ 0, or t + kπ. At each of these values, π f() t ± cos ± units, so the maimum distance between the particles is unit. (b) Solving π cost cos t+ graphically, we obtain t.79, t 5.90, and so on. Alternatively, this problem may be solved analytically as follows. π cost cos t+ π π π π cos t+ cos t+ + π π π π π π cos t cos sin t sin cos t cos π π + sin t sin π π sin t + sin 0 π sin t + 0 7π t + kπ 7π The particles collide when t 79. (plus multiples of π if they keep going.) 57. The dimensions will be in. by 0 in. by 6 in., so V( ) ( 0 )( 6 ) for 0 < < 5. Then V () ( )( 0), so the critical point in the correct domain is. This critical point corresponds to the maimum possible volume because V () > 0 for 0 < < and V () < 0 for < < 5. The bo of largest volume has a height of in. and a base measuring 6 in. by in., and its volume is in. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

20 Chapter 5 Review 5. Step : r radius of circle A area of circle Step : dr At the instant in question, m/sec and π r 0 m. Step : da We want to find. Step : A π r Step 5: da dr π r Step 6: da π ( 0) 0 π The area is changing at the rate of 0 m / sec. 59. Step : -coordinate of particle y y-coordinate of particle D distance from origin to particle Step : At the instant in question, 5 m, y m, d dy m/sec, and 5m/sec. Step : dd We want to find. Step : D + y Step 5: dd d dy y + + y d dy + y + y Step 6: dd ()( 5 ) + ( )( 5) 5m/sec 5 + Since dd is negative, the particle is approaching the origin at the positive rate of 5 m/sec. 60. Step : edge of length of cube V volume of cube Step : At the instant in question, dv 00 cm /min and 0 cm. Step : d We want to find. Step : V Step 5: dv d Step 6: d 00 ( 0) d cm/min The edge length is increasing at the rate of cm/min. 6. Step : -coordinate of point y y-coordinate of point D distance from origin to point Step : At the instant in question, and dd units per sec. Step : d We want to find. Step : Since D + y and y, we have D + for 0. Step 5: dd d ( + ) + + d + d + + Step 6: () + d d units per sec Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

21 Chapter 5 Review 5 h 0 6. (a) Since r, we may write 5r h h or r 5. (b) Step : h depth of water in tank r radius of surface of water V volume of water in tank Step : At the instant in question, dv 5 ft / min and h 6 ft. Step : dh We want to find. Step : V πr h πh 75 Step 5: dv dh π h 5 Step 6: dh 5 π () 6 5 dh ft/min π Since dh is negative, the water level is dropping at the positive rate of 0.76 ft/min. 6. Step : r radius of outer layer of cable on the spool θ clockwise angle turned by spool s length of cable that has been unwound Step : ds At the instant in question, 6 ft/sec and r. ft Step : dθ We want to find. Step : s rθ Step 5: Since r is essentially constant, ds r d θ 6. Step 6: dθ 6. dθ 5 radians/sec The spool is turning at the rate of 5 radians per second. at () v () t g ft/sec Since v() 0 ft/sec, v() t s () t t+. Since s() 0 7 ft, s() t 6t + t 7. The shovelful of dirt reaches its maimum height when v(t) 0, at t sec. Since s(), the shovelful of dirt is still below ground level at this time. There was not enough speed to get the dirt out of the hole. Duck! dv 65. We have V π r h, so π rh and dr dv π rh dr. When the radius changes from a to a + dr, the volume change is approimately dv π ah dr. 66. (a) Let edge of length of cube and S surface area of cube. Then S 6, ds which means and. d ds d We want ds 00. S, which gives d 0. 0( 6 ) or d The edge should be measured with an error of no more than %. (b) Let V volume of cube. Then V, dv which means and dv d. d We have d 00., which means d ( 00. ) 00. V, so dv 0. 0V. The volume calculation will be accurate to within approimately % of the correct volume. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

22 6 Chapter 5 Review 67. Let C circumference, r radius, S surface area, and V volume. dc (a) Since C π r, we have π and so dr dc π dr. Therefore, dc π dr dr 0cm. < 00. The C π r r 0cm calculated radius will be within approimately % of the correct radius. ds (b) Since S π r, we have π r and dr so ds π r dr. Therefore, ds π r dr dr ( 00. ) S π r r The calculated surface area will be within approimately % of the correct surface area. dv (c) Since V π r, we have dr and so dv π r dr. Therefore dv V π r dr π r π r dr 00 (. ) r 0.. The calculated volume will be within approimately % of the correct volume. 6. By similar triangles, we have a a+ 0, 6 h which gives ah 6a + 0 or h 6 + 0a The height of the lamp post is approimately ( ) ft. The estimated error in measuring a was da in. ft. Since dh 0 a, we have da dh 0a da 0( 5 ) ft, 5 so the estimated possible error is ± ft or ± in. 5 5 dy 69. sin cos. Since sin and cos are d both between and, the value of sin cos is never greater than. Therefore, dy for all values of. d Since dy is always negative, the function d decreases on every interval. 70. (a) f has a relative maimum at. This is where f ( ) 0, causing f to go from positive to negative. (b) f has a relative minimum at 0. This is where f ( ) 0, causing f to go from negative to positive. (c) The graph of f is concave up on (, ) and on (, ). These are the intervals on which the derivative of f is increasing. (d) 7. (a) (b) (c) y A π r da π rdr da in. π( ) π sec V π ( ) h π h 6 dv dr dh π π rh r + dh ( ) 6 + dh + dh da π in. π dh in. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

23 Chapter 5 Review 7 7. (a) a 60 V π b, and b a 5 a, π 0a a so V. π Thus dv ( 60a a ) a( 0 a). The da π π relevant domain for a in this problem is (0, 0), so a 0 is the only critical number. The cylinder of maimum volume is formed when a 0 and b 5. (b) The sign graph for the derivative dv a( 0 a) on the interval (0, 0) da π is as follows: By the First Derivative Test, there is a maimum at a 0. Copyright 0 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 4 Applications of Derivatives. Section 4.1 Extreme Values of Functions (pp ) Section Quick Review 4.1

Chapter 4 Applications of Derivatives. Section 4.1 Extreme Values of Functions (pp ) Section Quick Review 4.1 Section. 6 8. Continued (e) vt () t > 0 t > 6 t > 8. (a) d d e u e u du where u d (b) d d d d e + e e e e e e + e e + e (c) y(). (d) m e e y (). 7 y. 7( ) +. y 7. + 0. 68 0. 8 m. 7 y0. 8( ) +. y 0. 8+.