(a) Use washer cross sections: a washer has

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1 Section 8 V / (sec tan ) / / / / [ tan sec tan ] / / (sec sec tan + tan ) + tan sec / + A( ) s w (sec tan ), and / V (sec tan ), which b / same method as in part (a) equals A cross section has width w and area w r The volume is 8 A cross section has width w and area s w ( ) This volume is 8 ( ) Since the diameter of the circular base of the solid etends from to, for a diameter of and a radius of, the solid has the same cross sections as the right circular cone The volumes are equal b Cavalieri s Theorem (a) The volume is the same as if the square had moved without twisting: V Ah s h Still s h : the lateral distribution of the square cross sections doesn t affect the volume That s Cavalieri s Volume Theorem The functions intersect at (, 8) (a) Use washer cross sections: a washer has inner radius r, outer radius R, and area A( ) ( R r ) ( ) The volume is 7 ( ) 7 Use clindrical shells: a shell has a radius / 8 and height The volumeis 8 8 / ( ) 8 / / / 7 / The functions intersect at (, ) and (, ) (a) Use clindrical shells: A shell has radius and height The volume is ( )( ) Copright Pearson Education, Inc Publishing as Prentice Hall

2 7 Use clindrical shells: a shell has radius and height The volume is ( )( ) ( + ) + The intersection points are,,,, and(, ) (a) A washer has inner radius radius R, and area r, outer ( R r ) The volume is 8 A shell has radius and height The volume is ( ) / / 8 Section 8 7 Use clindrical shells: a shell has radius and height The volume is, which from part (a) is sin [ cos ] 9 (a) A cross section has radius r and area A( ) r ( ) The volume is ( ) cm cm 8 g cm ( / ) 9 g (a) A cross section has radius r and area r The volume is [ ] (a) dv Vh ( ) Ahdh ( ), so Ah ( ) dh dv dv dh dh Ah ( ), dt dh dt dt dh dv so dt A( h) dt For h, the area is () 8, dh units units so dt 8 sec 8 sec (sin ) 8 (a) For <, f( ) sin For, f() sin sin, so f() sin for The remaining solid is that swept out b the shaded region in revolution Use clindrical shells: a shell has radius and Copright Pearson Education, Inc Publishing as Prentice Hall

3 8 Section 8 height r The volume is r ( ) r r r ( r ) r ( 8) The answer is independent of r Partition the appropriate interval in the ais of revolution and measure the radius r ( ) of the shadow region at these points Then use an approimation such as the trapezoidal rule to b estimate the integral r ( ) a Solve a : This is true at a For revolution about the -ais, a cross section has radius r a and area A ( ) r ( a ) ( a a + ) The volume is a ( a a + ) a a a + a For revolution about the -ais, a clindrical shell has radius and height a The volume is a a ( )( a ) a a Setting the two volumes equal, a a ields a, so a The slant height s of a tin horizontal slice can be written as s + + ( g ( )) So the surface area is approimated b the Riemann 7 8 n sum g( k ) + ( g ( )) The limit of k that is the integral g ( ), and + + g ( ), and + ( ) ( + ) ( ) 8 9 g ( ), and ( + ) / / / + / / + Using NINT, this evaluates to g ( ), and 8 / + 8 / / 8 / 997 Copright Pearson Education, Inc Publishing as Prentice Hall

4 Section f ( ), and + ( ) + evaluates, using NINT, to f ( ), and ( ) + ( ) evaluates, using NINT, to 877 f ( ), and [ ] 8 + f ( ), and / + / / 9 9 True; b definition False; the volume is given b e A; V (ln( )) B; V ( ) 8 D 9 A cross section has radius r c sin and area A ( ) r ( c sin ) ( c csin+ sin ) The volume is ( c csin + sin ) c ccos + sin c c+ c c c+ (a) Solve d c c+ dc c c c This value of c gives a minimum for V av because > dc Then the volume is + Since the derivative with respect to c is not zero anwhere else besides c, the maimum must occur at c or c The volume for c is 9, and ( 8) for c it is 8 c maimizes the volume E; V / ( ( 8 ) ) 9 Copright Pearson Education, Inc Publishing as Prentice Hall

5 7 Section 8 (c) The volume gets large without limit This makes sense, since the curve is sweeping out space in an everincreasing radius C d C 7 (a) Using d, and A ields the following areas (in square inches, rounded to the nearest tenth):,,,,, 8, 7, 9, 7, 7, 9,, If C () is the circumference as a function of, then the area of a cross section is C( ) ( C ( )) A ( ), and the volume is ( C( )) A C [ + ( ) + ] 7 in (c) ( ) ( ( )) 7 Hemisphere cross sectional area: R h A Right circular clinder with cone removed cross sectional area: R h A Since A A, the two volumes are equal b Cavalieri s theorem Thus volume of hemisphere volume of clinder volume of cone R R R 7 Use washer cross sections: a washer has inner radius area ( R r ) b a b a + ( ) The volume is a a b a b a a a a b ab r b a, outer radius R b+ a +, and Copright Pearson Education, Inc Publishing as Prentice Hall

6 Section (a) Put the bottom of the bowl at (, a) The area of a horizontal cross section is a ( a ) The volume for height h is h a a ( a ) a h ( a h) h a a For h, and the area of a cross section is ( ) The rate of rise is dh dv ( ) m/sec dt A dt 7 (a) A cross section has radius r a and area A ( ) r a ( a ) The volume is a ( a ) a a a a a a a + a a A cross section has radius r h and area A ( ) r h r + h h The volume is h r + h h h r + h h r h Quick Quiz Sections 8 8 C; A D (sin ( )) 7 (a) The two graphs intersect where e, which a calculator shows to be 7 Store this value as A The area of R is A( e ) Volume A ( e ) ( ) + + A e (c) Volume Section 8 Lengths of Curves (pp ) Quick Review ( + ), which, since, equals + or + +, equals or which, since, + (tan ) (sec ), which since <, equals sec ( + ) + which, since >, equals Copright Pearson Education, Inc Publishing as Prentice Hall

7 7 Section 8 + cos cos, which since, equals cos (a) cos, so Length + cos f( ) has a corner at 7 d / ( ) is undefined at f( ) has a cusp there [, ] b [, ] (c) Length 8 d 8 ( ) + ( + ) / is undefined for f() has a vertical tangent there (a) / ( ), so / Length + / 9 + has a corner at d ( sin ) cos + is undefined for / (sin ) k, where k is an integer f() has vertical tangents at these values of Section 8 Eercises (a), so Length + ( ) + [, ] b [, ] (c) Length (a) (c) Length 7 + +, so + + ( + ) +, and + Then +, but NINT ma fail using over the entire interval because is undefined at ( + ) So, use Then + and Length + ( + ) (a) sec, so Length / + sec [, 7] b [, ] (c) Length 99 (a) cos + sin cos sin, so Length + sin (c) Length 7 Copright Pearson Education, Inc Publishing as Prentice Hall

8 Section 8 7 (c) Length 98 7 (a) tan, so / Length + tan tan ln (sec ) (a), b [, ] (c) Length 9 t t dt t, p b [, ] (c) Length 9 (c) Length 8 (a) sec, so / Length sec / sec tan, ln (cos ), so ln (cos ), < / ( + ) ( ) +, so the length is ( + ) + (c) Length 98 9 (a) sec tan, so / Length + sec tan /, so thelength is / Copright Pearson Education, Inc Publishing as Prentice Hall

9 7 Section 8, so the length is + +, so the length is + + 8, so the length is (a) + + ( + ) ( + ) ( + ) so the length is + ( + ) ( ) + ( + ) + ( ) + ( + ) ( ) + sec, so the length is / / + (sec ) sec / / / [tan ] /, so the length is + ( ) 7 corresponds to as here, so take Then + C, and, since (, ) lies on the curve, C So Two; we know the value of the function at one value of, and we know the square of the derivative We can also let, which gives the solution Copright Pearson Education, Inc Publishing as Prentice Hall

10 (a) corresponds to here, so take as Then + C and, since (, ) lies on the curve, C So Section 8 7 The area is times the length of the arch sin, so the length is + sin, which evaluates, using NINT, to 7 8 Multipl that b, then b $7 to obtain the cost (rounded to the nearest dollar): $8, Two; we know the value of the function at one value of, and we know the square of the derivative We can also let, which gives the solution + cos, so the length is / / + cos cos / [sin ] / / / ( ), so the length is / / 8 +( ) / / 8 / / 8 / / 8 / 8 Find the length of the curve sin for cos, so the length is + cos, which evaluates, using NINT, to 7 inches / / f ( ) +, but NINT fails on + ( f ( )) because of the undefined slope at So, instead solve for in terms of using the quadratic formula / / ( ) +, and / ± + Using the positive + Then, 8 values, ( ) ( ) +, and 8 + / / + + ( ) ( + ) ( 8 8) f ( ) ( + ) 8, ( + ) 8 so the length is + / ( ) + which evaluates, using NINT, to 89 7 There is a corner at : Break the function into two smooth segments:, and +, <, < +, < The length is Copright Pearson Education, Inc Publishing as Prentice Hall

11 7 Section ( ) + ( ) + + ( + ), which evaluates, using NINT for each part, to ( ), True, b definition D; sin( ) / + ( sin( )) 8 C; + ( ) + 9 9, but NINT ma fail using over the entire interval because is undefined at So, split the curve into two equal segments b solving + with : The total length is +, / which evaluates, using NINT, to, but NINT ma fail using over the entire interval, because is undefined at So, use, : and the length is + ( ), which evaluates, using NINT, to 7 Horizontal line segments will not work because the limit of the sum k, as the norm of the partition goes to zero, will alwas be the length (b a) of the interval (a, b) No; the curve can be indefinitel long Consider, for eample, the curve sin + for < < False; the function must be differentiable B; A; For track : at ± ± 7, and NINT fails to evaluate + ( ) because of the undefined slope at the limits, so use the track s smmetr, and back awa from the upper limit a little, and find + ( ) 8 Then, approimating the last little stretch at each end b a straight vertical line, add ( ) to get the total length of track as 7 Using a similar strateg, find the length of the right half of track to be 7 Store the unrounded value as A Now enter Y 7and Y t A+ NINT +, t,, t and graph in a [, ] b [, ] window to see the effect of the -coordinate of the lane- starting position on the length of lane (Be patient!) Solve graphicall to find the intersection at 9 99, which leads to starting point coordinates ( 999, 8) Copright Pearson Education, Inc Publishing as Prentice Hall

12 Section (a) The fin is the hpotenuse of a right triangle with leg lengths k and df ( ) k k f k k n lim ( k) + ( f ( k ) k) n k n lim k + ( f ( k )) n k b + ( f ( )) a Yes; an curve of the form ± + c, where c is a constant, has constant slope ±, so that a a + ( ) a Section 8 Applications from Science and Statistics (pp ) Quick Review 8 (a) (a) (a) (a) (a) e [ e ] e e [ e ] e 78 sin [ cos ] 77 ( + ) + n ( + ) + [ n 9 n ] 9 n ( + )sin 7 ( )( ) 7 cos 7 ( ) 7 sin Section 8 Eercises / / e [ e ( + )] J / e sin sin cos 87 J ( ) 9 J sin sin ( e cos + ) [ e + ] sin e J When the bucket is m off the ground, the water weighs F( ) N Then W ( 9 ) [ 9 ] 9 J Copright Pearson Education, Inc Publishing as Prentice Hall

13 78 Section 8 When the bucket is m off the ground, the water weighs F ( ) N Then W ( 9 9 ) [ ] 88 J 7 When the bag is ft off the ground, the sand weighs 8 F ( ) 8 lb 8 Then W ( ) 8 [ ] 9 ft-lb 8 (a) F ks, so 8 k( ) and k lb/in F(), and [ ] in-lb (c) F s, so s 8in 9 (a) F ks, so,7 k(8 ) and k 78 lb/in F() 78 W 78 [ 9 ] 9 7 9in-lb, and W 78 [ 9 ] 7 7 in-lb (a) F ks, so k and k lb/in Then for s 8, F lb [ ] 8 7 in-lb When the end of the rope is m from its starting point, the ( ) m of rope still to go weigh F() ()( ) N The total work is ( )( ) (a) Work 78 J ( p, V ) ( p, V ) F( ) ( p, V ) (, ) (, ) p V p V ( p, V) pa pdv pv ( )( ) 9,, so 9, p and V ( p, V)9, Work dv ( p, V) V 9, [ ] V V V 7,987 in-lb (a) From the equation +, it follows that a thin horizontal rectangle has area 9, where is distance from the top, and pressure The total force is approimatel n ( k) ( 9 k ) k n 8k 9 k k 8 9 [ ( 9 ) ] lb Copright Pearson Education, Inc Publishing as Prentice Hall

14 Section 8 79 (a) From the equation +, it follows 8 that a thin horizontal rectangle has area, where is distance from the top, and pressure The total force is approimatel n k ( ) k k n k 7 k k lb (a) From the equation 8, it follows that a thin horizontal rectangle has area, where is the distance from the top of the triangle, the pressure is ( ) The total force is approimatel n ( k ) k k n ( 8 k k) k 8 8 ( 8 ) [ 7 ] 9 ( ) 7lb (a) From the equation, it follows that a thin horizontal rectangle has area where is distance from the bottom, and pressure ( ) The total force is approimatel n ( k) ( k ) k n 8 ( k k ) k 8 ( ) (a) Work to raise a thin slice ( )( ) Total work ( ), 97, ft-lb (,97, ft-lb) ( ft-lb/sec) 99 sec min (c) Work to empt half the tank ( ) 7, ft-lb, and 7, 97 sec min (d) The weight per ft of water is a simple multiplicative factor in the answers So divide b and multipl b the appropriate weight-densit For :, 97,, 9, ft-lb and sec min For :, 97,,, ft-lb and 99 sec min Copright Pearson Education, Inc Publishing as Prentice Hall

15 8 Section 8 8 The work needed to raise a thin disk is ( ) ( )( ), where is height up from the bottom The total work is ( )( ) 7, 8, 9 ft-lb 9 Work to pump through the valve is ( ) ( )( + ) for a thin disk and ( )( + ) 9 + 8, 87 ft-lb for the whole tank Work to pump over the rim is ( ) ( )( + ) for a thin disk and ( )( ) ( )( )( ) 98, 8 8 ft-lb for the whole tank Less work is required to fill the tank from the bottom The work is the same as if the straw were initiall an inch long and just touched the surface, and lengthened as the liquid level dropped For a thin disk, the volume is + 7 and the work to raise it is + 7 ( 8 ) The total work is ( 8 ), which using 9 NINT evaluates to 9 in-oz The work is that alrea calculated (to pump the oil to the rim) plus the work needed to raise the entire amount ft higher The latter comes to r h( 7)( ) 7( ) ( 8), 9ft-lb, and the total is,9 +,,8 ft-lb The weight densit is a simple multiplicative factor: Divide b 7 and multipl b,, 8 8 ft-lb 7 The work to raise a thin disk is r ( ) h ( ) ( )( + ) ( )( ) The total work is ( )( ), which evaluates using NINT to 97, ft-b This will come to (97,)($) $88, so es, there s enough mone to hire the firm Pipe radius ft; Work to fill pipe ( ), ft -b 8 Work to fill tank ( ) ( ) 8,, ft-lb Total work 8,, ft-lb, which will take 8,, +,8 sec hr (a) The pressure at depth is, and the area of a thin horizontal strip is The depth of water is ft, so the total force on an end is ( )( ) 9 7 lb On the sides, which are twice as long as the ends, the initial total force is doubled to 97 lb When the tank is upended, the depth is doubled to ft, and the force on a side becomes ( )( ) 88 9 lb, which means that the fluid force doubles 7 in ft, and 77in ft 8 Force on a side pda 8 ( ) lb Copright Pearson Education, Inc Publishing as Prentice Hall

16 Section 8 8 t 7 f () t otherwise t dt t 8 f () t otherwise t dt 9 Since σ, then 8% of the calls are answered in minutes Use f( ) e ( 98) /( ) (a) f( ) ( % ) Take as a convenientl high upper limit: f 7 7 ( ), which means about 7 () people (a) (%), since half of a normal distribution lies below the mean Use NINT to find f( ), where ( ) /( ) f( ) e The result is (%) (c) ft 7 in Pick 8 in as a convenientl high upper limit and with NINT, find 8 7 f( ) The result is (%) (d) if we assume a continuous distribution Integration is a good approimation to the area (which represents the probabilit), since the area is a kind of Riemann sum The proportion of lightbulbs that last between and 8 hours False; three times as much work is required E; ( ) 7 7 J 7 D; W ( ) [ ] J 8 B; F k k W [ ] J 9 E; ( 8 )( ) 9, ft-lb, 78, MG dr 7,, r MG r which for, 78,, 7,, M 97, G 7 evaluates to J (a) The distance goes from m to m The work b an eternal force equals the work done b repulsion in moving the electrons from a -m distance to a -m distance: Work r 9 dr 9 r 8 J Again, find the work done b the fied electrons in pushing the third one wa The total work is the sum of the work b each fied electron The changes in distance are m to m and m to m, respectivel Work 9 9 dr + r r 9 + r r 9 77 J dr True; the force against each vertical side is 8 lbs Copright Pearson Education, Inc Publishing as Prentice Hall

17 8 Section 8 dv dv F m mv, so dt W F( ) dv mv v mv dv v mv mv Since initial velocit, Work Change in kinetic energ mv oz lb 8 m slug, so ft/ sec ft/ sec Work ( ) ft-lb lb lb 97 slug, ft/ sec and 8 ft hr 9 mph 9 mi sec ft/sec, so Work change in kinetic energ 97 ( )( ) 8 ft-lb lb ( ) oz oz oz slug, ( ft/ sec ) so Work 8 ( )( ) ft-lb lb ( ) oz oz oz slug, and ( ft/ sec ) 8 ft hr mph mph mi sec 8 87ft /sec, so Work 8 87 ( ) ft-lb lb ( ) oz oz oz 8slug, ( ft/sec ) so Work ( )( ) ft-lb lb ( ) oz oz oz 7 slug, so ( ft/sec ) Work 7 ( )( ) ft-lb oz lb slug Compression energ 8 of spring ks ( 8) ft-lb, and final height is given b mgh ft-lb, so h ft Quick Quiz Sections 8 and 8 A; D; ( ) + + t t dt t t dt (( ) + ( ) ) C; ( )( ) t t dt t + 9t dt ( ) 8 in (a) n i h h Copright Pearson Education, Inc Publishing as Prentice Hall

18 Chapter 8 Review 8 h dh h bs 7 (c) h h dh bs Chapter 8 Review Eercises (pp ) vt () dt ( t t ) dt t t 7ft 7 c() t dt 7 ( + t ) dt 7 [ t + t ] gal B ( e ) [ e ] ( ) The curves intersect at and There area is [ ( + )] ( ) implies + ( ) p( ) ( ) [ ] g () t E t dt cos dt t t sin, 9 The area is ( + ) + implies The curves intersect at The area is The curves intersect at 8 The area is 8 8 8, or 8 Copright Pearson Education, Inc Publishing as Prentice Hall

19 8 Chapter 8 Review implies, and + implies + The curves intersect at (, ) and (, ) The area is The curves intersect at ±8 The area 8 is ( cos ) 8 8 sin The curves intersect at ± 8 The area 8 is ( sec ) 8 8 ( sec ) 8 tan The area is / / ( sin ) cos + + Solve + cos cos for the -values at the two ends of the region: ±, ie, 7 or Use the smmetr of the area: 7 [( + cos ) ( cos )] 7 ( cos ) [ ] 7 sin 7 The area is ( sin sin ) cos + cos [( cos ) ( + cos )] ( cos ) [ sin ] + 7 Copright Pearson Education, Inc Publishing as Prentice Hall

20 Chapter 8 Review 8 7 Solve to find the intersection + points at and ± / Then use the area s smmetr: the area is / ( ) + In( + ) + ( ) ln / 8 Use the intersect function on a graphing calculator to determine that the curves intersect at The area is, 89 which using NINT evaluates to 7 9 Use the - and -ais smmetries of the area: sin [sin cos ] A cross section has radius r and area 8 ( ) A r 9 8 [ 9 ] v 9 The graphs intersect at (, ) and (, ) (a) Use clindrical shells A shell has radius and height The total volume is ( ) Use clindrical shells A shell has radius and height The total volume is ( ) ( ) / ( ) / 8 (c) Use clindrical shells A shell has radius and height The total volume is ( ) ( ) / ( ) / / (d) Use clindrical shells A shell has radius and height The total volume is ( ) + + (a) Use disks The volume is ( ) k k k dv dk (c) Since V k, k dt dt When k, dk dv ( ), so the dt k dt Copright Pearson Education, Inc Publishing as Prentice Hall

21 8 Chapter 8 Review depth is increasing at the rate of unit per second The football is a solid of revolution about the -ais A cross section has radius and area r The volume is, given the smmetr, / / / 88 7in The width of a cross section is sin, and the area is r sin The volume is sin sin The volume of the piece cut out is ( ) ( ) ( 8) 8 9 ft 7 The curve crosses the -ais at ±, so the length is + ( ), + which using NINT evaluates to The curves intersect at and ± Use the graphs - and -ais smmetr: d ( ), and the total perimeter is + ( ), which using NINT evaluates to Use washer cross sections A washer has inner radius r, outer radius R e, and area ( R r ) ( e ) The volume is ln ln ( e ) e ( ln ) ( ln ) Use clindrical shells Taking the hole to be vertical, a shell has radius and height equals zero when or The maimum is at, the minimum at The distance between them along the curve is + ( ), which using NINT evaluates to 9 The time taken is about 9 9sec If were true, then the curve k sin would have to get vanishingl short as k approaches zero Since in fact the curve s length approaches instead, is false and (a) is true Copright Pearson Education, Inc Publishing as Prentice Hall

22 F ( ), so + ( F ( )) 9 Chapter 8 Review 87 the rim is ( )( ) The total 8 work is ( ) 8 ( ) 8 97 in-lb (a) ( N)( m) J When the end has traveled a distance, the weight of the remaining portion is ( )(8) 8 The total work to lift the rope is ( 8 ) [ ] J (c) + J The weight of the water at elevation (starting from ) is 7 8 ( 8)( 8) The total work is , 8, ft-lb F 8 8 F ks, so k N/m Then s 8 8 Work J To stretch the additional meter, 8 8 Work J The same amount of work is done, but gravit supplies the downhill force, so less work is done b the person The radius of a horizontal cross section is 8, where is distance below the rim The area is ( ), the weight is ( ), and the work to lift it over 7 The width of a thin horizontal strip is ( ), and the force against it is 8( ) The total force is ( ) ( + ) lb 7 8 7in ft, in ft, and 8 in ft For the base, 7 Force 7 8lb 8 For the front and back, Force lb For the sides, Force lb 8, and its 9 A square s height is ( ) area is ( ) The volume is ( ) ( / / + + ) / / Copright Pearson Education, Inc Publishing as Prentice Hall

23 88 Chapter 8 Review Choose cm as a convenientl large upper limit ( 7 ) /( ) e evaluates, using NINT, to (%) Answers will var Find µ, then use the fact that 8% of the class is within σ of µ to find σ, and then choose a convenientl large number b ( µ ) /( σ ) b and calculate e σ Use (a) / f( ) e f( ) evaluates, using NINT, to 87(87%) f( ) 9 ( 9 %) f( ) 997 ( 997 %) Because f() and f( ) The total area gives the probabilit that the variable takes on one of its possible values Since the variable must take on some value, the probabilit must be 7 A shell has radius and height The total volume is ( ) [ ] ϖ A shell has radius and height sin The total volume is ( )(sin ) [ sin cos ] A shell has radius and height The total volume is ( ) [ ] The curves intersect at and A shell has radius and height ( ) + The total volume is ( )( + ) ( + ) + Copright Pearson Education, Inc Publishing as Prentice Hall

24 Chapter 8 Review 89 8 Use the intersect function on a graphing calculator to determine that the curves intersect at ± 89 A shell has radius and height The volume, which is calculated using the right half of the area, is 89 ( ), which using NINT evaluates to (a) ( + )( ) Revolve about the line, using clindrical shells A shell has radius and height total volume is ( ) + + in The dl Since f + ( ) must equal ( f + ( )), + ( f ( )) + f ( ) + f ( ), and f ( ) Then f( ) ln + C, and the requirement to pass through (, ) means that C The function is ln+ f( ) ln + sec, so the area is (tan ) + (sec ), which using NINT evaluates to 8 and, so the area is, + which using NINT evaluates to (a) The two curves intersect at 78 Store this value as A Area A ( + sin sec ) A Volume (( + sin ) (sec ) ) (c) A Volume ( + sin sec ) 9 (a) Average temp t 8 cos dt 87 F Copright Pearson Education, Inc Publishing as Prentice Hall

25 9 Chapter 8 Review t Ft () 8 cos 78 for 89 t 8 79 Store these two values as A and B B t (c) Cost 8 cos 78 dt A The cost was about $ (a) 7 people 9 ( t t+ ) dt 7 dt + 8 dollars 9 dt, 7 ( t t+ ) ( t t+ ) (c) H ( 7) E( 7) L( 7) 8 people H(7) is the number of people in the park at :, and H ( 7) is the rate at which the number of people in the park is changing at : (d) When H () t E() t L() t ; that is, at t 79 Copright Pearson Education, Inc Publishing as Prentice Hall