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+. 9 e (e) y 0 e 0 e e e e or 0 8. (a) > 0 >, < < d (b) f ( ) ln( ) u d d du du ln( u) d u d d ( ) (c) > 0, < < (d) y / / Chapter Applications of Derivatives Section. Etreme Values of Functions (pp. 87 9) Eploration Finding Etreme Values. From the graph we can see that there are three critical points: 0,,. Critical point values: f( ) 0., f( 0) 0, f( ) 0. Endpoint values: f( ) 0., f( ) 0. Thus f has absolute maimum value of 0. at and, absolute minimum value of 0 at 0, and local minimum value of 0. at and.. The graph of f has zeros at and where the graph of f has local etreme values. The graph of f is not defined at 0, another etreme value of the graph of f.. Using the chain rule and d d ( ), we find df i d ( + ). Quick Review. d. f ( ) i ( ) d d / / d. f ( ) 9 ( ) ( 9 ) i ( 9 ) d d / ( 9 ) ( ) / ( 9 ) d sin (ln ). g ( ) sin (ln ) i ln d d. h ( ) e i d e. Graph (c), since this is the only graph that has positive slope at c. 6. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 7. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a. 8. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b. + 9. As, 9 0. Therefore, lim f( ). + + 0. As, 9 0.Therefore, lim f( ). +
6 Section.. (a) d d ( ) f () () (b) d d ( + ) f () (c) Left-hand derivative: lim ( ) ( ) f + h f [( + h) ( + h)] lim h 0 h h 0 h h + 6h + 0h lim h 0 h lim ( h + 6h+ 0) h 0 0 Right-hand derivative: f lim ( + h ) f ( ) lim [( + h ) + ] h 0+ h h 0+ h h lim h 0+ h lim h 0+ Since the left-and right-hand derivatives are not equal, f ( ) is underfined.. (a) The domain is. (See the solution for.(c)). (b) f ( ), <, > Section. Eercises. Minima at (, 0) and (, 0), maimum at (0, ). Local minimum at (, 0), local maimum at (, 0). Maimum at (0, ) Note that there is no minimum since the endpoint (, 0) is ecluded from the graph.. Local maimum at (, 0), local minimum at (, 0), maimum at (, ), minimum at (0, ). Maimum at b, minimum at c ; The Etreme Value Theorem applies because f is continuous on [a, b], so both the maimum and minimum eist. 6. Maimum at c, minimum at b; The Etreme Value Theorem applies because f is continuous on [a, b], so both the maimum and minimum eist. 7. Maimum at c, no minimum; The Etreme Value Theorem does not apply, because the function is not defined on a closed interval. 8. No maimum, no minimum; The Etreme Value Theorem does not apply, because the function is not continuous or defined on a closed interval. 9. Maimum at c, minimum at a; The Etreme Value Theorem does not apply, because the function is not continuous. 0. Maimum at a, minimum at c; The Etreme Value Theorem does not apply since the function is not continuous.. The first derivative f ( ) + has a zero at. Critical point value: f () + ln Endpoint values: f ( 0. ) + ln 0.. 07 f ( ) + ln. 66 Maimum value is + ln at ; minimum value is at ; local maimum at, ln. The first derivative g ( ) e has no zeros, so we need only consider the endpoints. ( ) g( ) e e g( ) e e Maimum value is e at ; minimum value is at. e. The first derivative h ( ) has no zeros, so we need + only consider the endpoints. h( 0) ln 0 h( ) ln Maimum value is ln at ; minimum value is 0 at 0.. The first derivative k ( ) e has a zero at 0. Since the domain has no endpoints, any etreme value must 0 occur at 0. Since k( 0) e and lim k ( ) 0, the maimum value is at 0. ±. The first derivative f ( ) cos +, has zeros at,. Critical point values: f( ) f( ) Endpoint values: 0 f( ) 7 f( ) 0 Maimum value is at ; minimum value is at ;
Section. 6. Continued local minimum at 0, ; 7 local maimum at, 0 6. The first derivative g ( ) sectan has zeros + 6, ( 0. 86,. 089);local minimum at 9, 6 9 ( 0. 86,. 9 ). at 0 and and is undefined at. Since g ( ) secis also undefined at, the critical points occur only at 0 and. Critical point values: 0 g( ) g( ) Since the range of g ( ) is (, ] [, ), these values must be a local minimum and local maimum, respectively. Local minimum at ( 0, ); local maimum at (, ) / 7. The first derivative f ( ) is never zero but is undefined at 0. Critical point value: 0 f( ) 0 Endpoint value: f( ) ( ) / /. Since f( )> 0 for 0, the critical point at 0 is a local minimum, and since f( ) ( ) / for <, the endpoint value at is a global maimum. Maimum value is / at ; minimum value is 0 at 0. / 8. The first derivative f ( ) is never zero but is undefined at 0. Critical point value: 0 f ( ) 0 Endpoint value: f( ). 9 Since f ( ) < 0 for < 0 and f ( ) > 0 for > 0, the critical point is not a local minimum or maimum. The maimum value is at. 9.. To find the eact values, note that y + 8 ( )( + ),which is zero when or. Local maimum at (, 7); local minimum at, 7 Note that y 6+ ( ),which is zero at. The graph shows that the function assumes lower values to the left and higher values to the right of this point, so the function has no local or global etreme values... Minimum value is 0 at and at. Minimum value is at. 0. To confirm that there are no hidden etrema, note that y ( ) ( ) which is zero only at 0 ( ) and is undefined only where y is undefined. There is a local maimum at (0, ). To find the eact values, note that y, which is zero when ±. Local maimum at
6 Section.. 0. The minimum value is at 0. 6. Maimum value is at 0; minimum value is at.. The actual graph of the function has asymptotes at ±, so there are no etrema near these values. (This is an eample of grapher failure.) There is a local minimum at (0, ). 7. Maimum value is at ; minimum value is on the interval [, ]; local maimum at (, 9). Maimum value is at ; minimum value is 0 at and at. 8.. Maimum value is on the interval [, 7]; minimum value is on the interval [, ]. 9. Minimum value is at ; local maimum at (0, 0); local minimum at, Maimum value is on the interval [, ); minimum value is on the interval (, ].. Maimum value is at ; Minimum value is on the interval [, ]. minimum value is at. / / + y () + ( + )
Section. 6. Continued crit. pt. derivative etremum value 6. 0 local ma / 0. 0 0 undefined local min 0 9. crit. pt. derivative etremum value 0 0 minimum 0 0 local ma /. 6 undefined minimum 0 / / 8 8 y ( ) + ( ) crit. pt. derivative etremum value 0 minimum 0 undefined local ma 0 0 minumum 0., y <, > crit. pt. derivative etremum value undefined minimum 7. y i ( ) + ( ) + ( ) crit. pt. derivative etremum value undefined local ma 0, y < 0, > 0 crit. pt. derivative etremum value 0 undefined local min 0 local ma. 0 minimum 0 maimum undefined local min 0 8. y i ( ) + + ( ) +, y < + 6, > crit. pt. derivative etremum value 0 maimum undefined local min 0 maimum
66 Section.. We begin by determining whether f ( )is defined at, where, f( ) + 6 + 8, > Left-hand derivative: lim ( ) ( ) ( + h) ( + h f + h f lim ) + h 0 h h 0 h h h lim h 0 h lim ( h ) h 0 Right-hand derivative: f lim ( + h ) f ( ) h 0 + h lim ( h ) + ( 6 + h) + 8 ( + h) h 0 + h h h h lim h 0 + h lim ( h h) h 0 + Thus, f ( ) + 8, > Note that 0 when, and ± ()() 8 + 8 0 when () ± 8 ± 6. But 0. 8 <,so the only critical points occur at and +.. crit. pt. derivative etremum value 0 local ma. 0 local ma. 079. (a) V( ) 60 + V ( ) 60 0+ ( )( 0) The only critical point in the interval (0, ) is at. The maimum value of V() is at. (b) The largest possible volume of the bo is cubic units, and it occurs when.. (a) P ( ) 00 The only critical point in the interval (0, ) is at 0. The minimum value of P() is 0 at 0. (b) The smallest possible perimeter of the rectangle is 0 units and it occurs at 0, which makes the rectangle a 0 by 0 square.. False. For eample, the maimum could occur at a corner, where f ()would c not eist. 6. False. Consider the graph below. y d 7. E. ( + 6) d 0 f ( ) ( ) ( ) + 6 0 8. E. See Theorem. d 9. B. d ( ) 6 + 6 6 0 ± 0. B.. (a) No, since f ( ) ( ) /, which is undefined at. (b) The derivative is defined and nonzero for all. Also, f( ) 0 and f( ) > 0 for all. (c) No, f () need not have a global maimum because its domain is all real numbers. Any restriction of f to a closed interval of the form [a, b] would have both a maimum value and a minimum value on the interval. (d) The answers are the same as (a) and (b) with replaced by a.,. Note that f( ) + 9 or 0 < 9, < < 0 or. Therefore, + 9, f < or 0 < ( ) < 9, < < 0 or >. (a) No, since the left- and right-hand derivatives at 0 are 9 and 9, respectively.
Section. 67. Continued (b) No, since the left- and right-hand derivatives at are 8 and 8, respectively. (c) No, since the left- and right-hand derivatives at are 8 and 8, respectively. (d) The critical points occur when f ( ) 0 ( at ± ) and when f ( )is undefined (at 0 or ±). The minimum value is 0 at, at 0, and at ; local maima occur at (, 6 ) and (, 6 ).. (a) f ( ) a + b + c is a quadratic, so it can have 0,, or zeros, which would be the critical points of f. Eamples: (d) Assuming that f ()eists, c the one-sided limits in (b) and (c) above must eist and be equal. Since one is nonpositive and one is nonnegative, the only possible common value is 0. (e) There will be an open interval containing c where f ( ) f (c) 0. The difference quotient for the left-hand derivative will have to be negative (or zero), and the difference quotient for the right-hand derivative will have to be positive (or zero). Taking the limit, the lefthand derivative will be nonpositive, and the right-hand derivative will be nonnegative. Therefore, the only possible value for f ()is c 0.. (a) [, ] by [, ] The function f( ) has two critical points at and. [, ] by [, ] The function f( ) has one critical point at 0. [, ] by [, ] The function f( ) + has no critical points. (b) The function can have either two local etreme values or no etreme values. (If there is only one critical point, the cubic function has no etreme values.). (a) By the definition of local maimum value, there is an open interval containing c where f( ) f( c), so f( ) f( c) 0. (b) Because c +, we have ( c) > 0, and the sign of the quotient must be negative (or zero). This means the limit is nonpositive. (c) Because c, we have ( c) < 0, and the sign of the quotient must be positive (or zero). This means the limit is nonnegative. [ 0., 0.6] by [.,.] f (0) 0 is not a local etreme value because in any open interval containing 0, there are infinitely many points where f () and where f( ). (b) One possible answer, on the interval [0, ]: ( )cos, 0 < f( ) 0, This function has no local etreme value at. Note that it is continuous on [0, ]. Section. Mean Value Theorem (pp. 96 0) Quick Review.. 6< 0 < 6 < < < Interval: (, ). 6> 0 > 6 > < or > Intervals: (, ) (, ). Domain: 8 0 8 The domain is [, ].. f is continuous for all in the domain, or, in the interval [, ].
68 Section.. f is differentiable for all in the interior of its domain, or, in the interval (, ). 6. We require 0, so the domain is ±. 7. f is continuous for all in the domain, or, for all ±. 8. f is differentiable for all in the domain, or, for all ±. 9. 7( ) + C 7 + C C 0. () + () + C + C C Section. Eercises. (a) Yes. (b) f ( ) d + + d c + ( ) 0 c.. (a) Yes. (b) f ( ) d d / / / 0 c 0 8 c. 7. (a) No. There is a verticle tangent at 0.. (a) No. There is a corner at.. (a) Yes. d (b) f ( ) sin d ( / ) ( / ) c ( ) c c / 0. 77. 6. (a) Yes. d (b) f ( ) ln( ) d c ln ln c +. ln ln 80 7. (a) No. The function is discontinuous at 8. (a) No. The split function is discontinuous at 9. (a) The secant line passes through (0., f (0.)) (0.,.) and (, f ()) (,.), so its equation is y.. (b) The slope of the secant line is 0, so we need to find c such that f () c 0. c 0 c c f() c f() The tangent line has slope 0 and passes through (, ), so its equation is y. 0. (a) The secant line passes through (, f ()) (, 0) and (, f ()) (, ), so its slope is 0. The equation is y ( ) + 0 or y, or y 0 707 0 707 (b) We need to find c such that f () c. c c c c f c f The tangent line has slope and passes through,. Its equation is y + or y, or y 0. 707 0... Because the trucker s average speed was 79. mph, and by then Mean Value Theorem, the trucker must have been going that speed at least once during the trip.. Let f (t) denote the temperature indicated after t seconds. We assume that f ()is t defined and continuous for 0 t 0. The average rate of change is0. 6 F/sec. Therefore, by the Mean Value Theorem, f () c 0. 6 F/sec for some value of c in [0, 0]. Since the temperature was constant before t 0, we also know that f ( 0) 0 F/min. But f is continuous, so by the Intermediate Value Theorem, the rate of change f ()must t have been 0. F/sec at some moment during the interval.. Because its average speed was approimately 7.667 knots, and by the Mean Value Theorem, it must been going that speed at least once during the trip.
Section. 69. The runner s average speed for the marathon was approimately.909 mph. Therefore, by the Mean Value Theroem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0 mph and the runner s speed is continuous, by the Intermediate Value Theorem, the runner s speed must have been mph at least twice.. (a) f ( ) Since f ( ) > 0 on,, f ( ) 0 at, and f ( ) < 0 on,, we know that f () has a local maimum at. Since f, the local maimum occurs at the point,. (This is also a global maimum.) (b) Since f ( ) > 0 on,, f( ) is increasing on,. (c) Since f ( ) < 0 on,, f ( ) is decreasing on,. 6. (a) g ( ) Since g ( ) < 0on,, g ( ) 0 at, and g ( ) > 0on,, minimum at. we know that g () has a local Since g 9, the local minimum occurs at the 9 point,. (This is also a global minimum.) (b) Since g ( ) > 0 on,, g( ) is increasing on,. (c) Since g ( ) < 0 on,, g( ) is decreasing on,. 7. (a) h ( ) Since h ( )is never zero is undefined only where h() is undefined, there are no critical points. Also, the domain (, 0) ( 0, ) has no endpoints. Therefore, h() has no local etrema. (b) Since h ( )is never positive, h ( ) is not increasing on any interval. (c) Since h ( )< 0 on (, 0) ( 0, ), h() is decreasing on (, 0) and on (0, ). 8. (a) k ( ) Since k ( )is never zero and is undefined only where k ( )is undefined, there are no critical points. Also, the domain (, 0) ( 0, ) has no endpoints. Therefore, k ( )has no local etrema. (b) Since k ( ) > 0 on (, 0), k( ) is increasing on (, 0). (c) Since k ( ) < 0 on ( 0, ), k( ) is decreasing on ( 0, ). 9. (a) f ( ) e Since f ( )is never zero or undefined, and the domain of f( ) has no endpoints, f( ) has no etrema. (b) Since f ( )is always positive, f( ) is increasing on (, ). (c) Since f ( )is never negative, f( ) is not decreasing on any interval.. 0 0 0. (a) f ( ). e Since f ( )is never zero or undefined, and the domain of f( ) has no endpoints, f( ) has no etrema. (b) Since f ( )is never positive, f( ) is not increasing on any interval. (c) Since f ( )is always negative, f( ) is decreasing on (, ).. (a) y + In the domain, ), y is never zero and is undefined only at the endpoint. The function y has a local maimum at (, ). (This is also a global maimum.) (b) Since y is never positive, y is not increasing on any interval. (c) Since y is negative on (, ), y is decreasing on, ).
70 Section.. (a) y 0 ( + )( ).. The function has critical points at, 0,and. Since y < 0 on (, ) and ( 0, ) and y > 0 on (, 0) and (, ), the points at ± are local minima and the point at 0 is a local maimum. Thus, the function has a local maimum at (0, 9) and local minima at (, 6) and (, 6). (These are also global minima.) (b) Since y > 0 on (, 0) and (, ), yis increasing on [, 0] and [, ). (c) Since y > 0 on (, ) and (0, ), y is decreasing on (, ] and [ 0, ]. (a) f ( ) i ( ) + + 8 The local etrema occur at the critical point 8 and at the endpoint. There is a local (and absolute) 8 6 maimum at, or approimately (.67,.08), and a local minimum at (, 0). 8 (b) Since f ( ) > 0 on,, f( ) is decreasing on, 8. (c) Since f ( ) < 0 on 8,, f( ) is decreasing on 8,. The local etrema can occur at the critical points and 0, but the graph shows that no etrema occurs at 0. There is a local (and absolute) minimum at (, 6 ) or approimately (, 76. ). (b) Since g ( ) > 0on the intervals (, 0) and (0, ), and g() is continuous at 0, g() is increasing on [, ). (c) Since g ( ) < 0on the interval (, ), g() is decreasing on (, ].. ( + )( ) ( )( ) (a) h ( ) ( + ) ( + ) ( + )( ) ( + ) The local etrema occur at the critical points, ±. There is a local (and absolute) maimum at, and a local (and absolute) minimum at,. (b) Since h ( ) > 0 on (, ) and (, ), h() is increasing on (, ] and [, ). (c) Since h ( ) < 0 on (, ), h( ) is decreasing on [, ]. 6. ( )( ) ( ) + (a) k ( ) ( ) ( ) Since k () is never zero and is undefined only where k() is undefined, there are no critical points. Since there are no critical points and the domain includes no endpoints, k() has no local etrema. (b) Since k ( )is never positive, k() is not increasing on any interval. (c) Since k ( )is negative wherever it is defined, k() is decreasing on each interval of its domain; on (, ), (, ), and (, ). / / + 8 (a) g ( ) ( ) + ( + 8) /
Section. 7 7. (a) f ( ) + sin Note that > for. and sin for all, so f ( ) > 0 for.. Therefore, all critical points occur in the interval (.,.), as suggested by the graph. Using grapher techniques, there is a local maimum at approimately (.6, 0.06), and a local minimum at approimately (0.9,.69). (b) f () is increasing on the intervals (,.6] and [0.9, ), where the interval endpoints are approimate. (c) f () is decreasing on the interval [.6, 0.9], where the interval endpoints are approimate. 8. / 6. f( ) + C f () / + C + C C / f( ) 7. f( ) ln( + ) + C f ( ) ln( + ) + C 0+ C C f( ) ln( + ) + 8. f( ) + sin+ C f ( 0) 0+ C C f( ) + sin+ 9. Possible answers: (a) (a) g () sin Since g () for all, there are no critical points. Since there are no critical points and the domain has no endpoints, there are no local etrema. (b) Since g () > 0 for all, g() is increasing on (, ). (c) Since g () is never negative, g() is not decreasing on any interval. 9. f( ) +C 0. f( ) + C (b) (c) [, ] by [, ] [, ] by [0,.]. f( ) + + C. f( ) cos+ C. f( ) e + C. f( ) ln ( ) + C. f( ) + C, > 0 f ( ) + C C f( ) +, > 0 [, ] by [0,.] 0. Possible answers: (a) (b)
7 Section. 0. Continued (c) (d). One possible answer: [, ] by [, ]. One possible answer:. (a) Since v ( t). 6, v() t. 6t+ C. But v( 0) 0, so C 0 and v(t).6t. Therefore, v(0).6(0) 8. The rock will be going 8 m/sec. (b) Let s(t) represent position. Since s () t v() t 6. t, s() t 08. t + D. But s(0) 0, so D 0 and st () 08. t. Therefore, s( 0) 0. 8( 0) 70. The rock travels 70 meters in the 0 seconds it takes to hit bottom, so the bottom of the crevasse is 70 meters below the point of release. (c) The velocity is now given by v(t).6t + C, where v(0). (Note that the sign of the initial velocity is the same as the sign used for the acceleration, since both act in a downward direction.) Therefore, v(t).6t +, and st () 08. t + t+ D, where s(0) 0 and so D 0. Using st () 08. t + tand the known crevasse depth of 70 meters, we solve s(t) 70 to obtain the positive solution t 7. 60, and so v(t) v(7.60).6(7.60) + 8.66. The rock will hit bottom after about 7.60 seconds, and it will be going about 8.66 m/sec.. (a) We assume the diving board is located at s 0 and the water at s 0, so that downward velocities are positive. The acceleration due to gravity is 9.8 m/sec, so v () t 98and. v(t) 9.8t + C. Since v(0) 0, we have v(t) 9.8t. Then the position is given by s(t) where s () t v() t 98. t, so s() t 9. t + D. Since s(0) 0, we have st () 9. t. Sloving s(t) 0 gives t 0 00 9, so the positive solution is t 0. 9 7. The velocity at this time is v 0 98 0 7. 7 m/sec. (b) Again v(t) 9.8t + C, but this time v(0) and so v(t) 9.8t. The s () t 98. t, so s(t).9t t + D. Since s(0) 0, we have s(t).9t t. Sloving s(t) 0 gives the positive solution t + 0. 67 sec. 98. The velocity at this time is v + 0 98 + 0. 0 m/sec or 98. 98. about. m/sec.. Because the function is not continuous on [0, ]. The function does not satisfy the hypotheses of the Mean Value Theorem, and so it need not satisfy the conclusion of the Mean Value Theorem. 6. Because the Mean Value Theorem applies to the function y sin on any interval, and y cos is the derivative of sin. So, between any two zeros of sin, its derivative, cos, must be zero at least once. 7. f ( ) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( ) is zero twice between a and b. Then by the Mean Value Theorem, f ( )would have to be zero at least once between the two zeros of f (), but this can t be true since we are given that f ( ) 0 on this interval. Therefore, f ( ) is zero once and only once between a and b. 8. Let f( ) + +. Then f( ) is continuous and differentiable everywhere. f ( ) +, which is never zero between and. Since f ( ) and f ( ), eercise 7 applies, and f ( ) has eactly one zero between and. 9. Let f ( ) + ln ( + ). Then f( ) is continuous and differentiable everywhere on [0, ]. f ( ) +, + which is never zero on [0, ]. Now f (0) 0, so 0 is one solution of the equation. If there were a second solution, f ( ) would be zero twice in [0, ], and by the Mean Value Theorem, f ( )would have to be zero somewhere between the two zeros of f( ).But this can t happen, since f ( )is never zero on [0, ]. Therefore, f( ) 0 has eactly one solution in the interval [0, ].
Section. 7 0. Consider the function k() f() g(). k() is continuous and differentiable on [a, b], and since k(a) f (a) g(a) 0 and k(b) f (b) g(b) 0, by the Mean Value Theorem, there must be a point c in (a, b) where k () c 0. But since k () c f () c g (),this c means that f () c g (),and c c is a point where the graphs of f and g have parallel or identical tangent lines. (c) Regression equation: y 0. 080 + 0. 96. 6+. 779 (, ) by [, ]. False. For eample, the function is increasing on (, ), but f ( 0) 0.. True. In fact, f is the increasing on [a, b] by Corollary to the Mean Value Theorem.. A. f ( ). f( ) f( 0). B. f ( ) 0. 78 980. 96 0 7. 0,negative slope. d. E. ( 0) d. 6. D. is not differentiable at 0. 7. (a) Increasing: [,.] and [., ]; decreasing: [.,.]; local ma:. local min:. (b) Regression equation: y [.,.] by [ 8, 0] (c) Since f ( ), we have f( ) + C. But f (0) 0, so C 0. Then f( ). 8. (a) Toward: 0 < t < and < t < 8; away: < t < (b) A local etremum in this problem is a time/place where Priya changes the direction of her motion. 9. (d) Using the unrounded values from the regression equation, we obtain f () t 0. 9t +. 8t. 6. According to the regression equation, Priya is moving toward the motion detector when f () t < 0 ( 0 < t <. 8 and. 6 < t < 8), and away from the detector when f () t > 0 (. 8 < t <. 6). f( b) f( a) b a b a b a ab f () c, so and c ab. c c ab Thus, c ab. f( b) f( a) b a 60. b+ a b a b a a f c c c b+ a c + b (), so and. 6. By the Mean Value Theorem, sin b sin a (cos c)(b a) for some c between a and b. Taking the absolute value of both sides and using cosc gives the result. 6. Apply the Mean Value Theorem to f on [a, b]. f( b) f( a) Since f( b) < f( a), is negative, and b a hence f ( )must be negative at some point between a and b. 6. Let f ( ) be a monotonic function defined on an interval D. For any two values in D, we may let be the smaller value and let be the larger value, so <. Then either f( ) < f( ) (if f is increasing), or f( ) > f( ) (if f is decreasing), which means f( ) f( ). Therefore, f is one-to-one. Section. Connecting f and f with the Graph of f (pp. 0 8) Eploration Finding f from f. Any function f( ) + C where C is a real number. For eample, let C 0,,. Their graphs are all vertical shifts of each other.. Their behavior is the same as the behavior of the function f of Eample 8.
7 Section. Eploration Finding f from f and f. f has an absolute maimum at 0 and an absolute minimum of at. We are not given enough information to determine f (0).. f has a point of inflection at.. 0. Left end behavior model: 0 Right end behavior model: 7 Horizontal asymptotes: y 0, y 7 Section. Eercises. y Intervals < > Sign of y + Behavior of y Decreasing Increasing Quick Review. Graphical support:. 9< 0 ( + )( ) < 0 Intervals < < < Sign of ( + )( ) Solution set: (, ). > 0 ( + )( ) > 0 < + + Intervals < < < 0 0< < Sign of ( + )( ) Solution set: (, 0) (, ). f: all reals f : all reals, since f ( ) e + e < + +. f: all reals / f : 0, since f ( ). f: ( )() ( )() f :, since f ( ) ( ) ( ) 6. f: all reals / f : 0, since f ( ) 7. Left end behavior model: 0 Right end behavior model: e Horizontal asymptote: y 0 8. Left end behavior model: e Right end behavior model: 0 Horizontal asymptote: y 0 9. Left end behavior model: 0 Right end behavior model: 00 Horizontal asymptote: y 0, y 00 Local (and absolute) minimum at,. y 6 + 6( ) Intervals < 0 0< < < Sign of y + Behavior of y Decreasing Increasing Decreasing Graphical support: Local maimum: (, ); local minimum: (0, ). y 8 8 8( ) ( + ) Intervals < < < 0 0 < < < Sign of y + + Behavior of y Graphical support: Decreasing Increasing Decreasing Increasing Local maimum: (0, ); local (and absolute) minima: (, ) and (, )
Section. 7 / / /. y e ( ) + e e Intervals < 0 0< < < Sign of y + + Behavior of y Increasing Decreasing Increasing Graphical support: 7. y + + 6 6( + ) ( + ) Intervals < < < < Sign of y + + Behavior of y Increasing Decreasing Increasing y + 6( + 7) Intervals < 7 7 < Sign of y + Local minimum: (, e). y 8 8 + ( ) ( 8 ( ) 8 Behavior of y down up Graphical support: Intervals 8 < < < < < < 8 Sign of y + Behavior of y Decreasing Increasing Decreasing Graphical support: 7 (a), 7 (b), Local maima: ( 8, 0) and (, ); local minima: (, ) and ( 8, 0) Note that the local etrema at ± are also absolute etrema. 6., y < 0, > 0 Intervals < 0 > 0 Sign of y + + Behavior of y Increasing Increasing Graphical support: 8. y + Using grapher techniques, the zeros of y are 0., 0.6, and.88. Intervals <0. 0. < < 06. 06. < < 88. 88. < Sign of y + + Behavior of y Increasing Decreasing Increasing Decreasing y + ( ) Intervals < 0 0 < < < Sign of y + Behavior of y down up down Graphical support: Local minimum: (0, ) (a) (, 0.] and [0.6,.88] (b) [ 0., 0.6] and [.88, ) (c) (0, ) (d) (, 0) and (, )
76 Section. 8. Continued (e) Local maima: ( 0.,.) and (.88, 6.); local minimum: (0.6, 0.68) Note that the local maimum at.88 is also an absolute maimum. (f) (0, ) and (, 9) 9. y / Intervals < 0 0< Sign of y + + Behavior of y Increasing Increasing 8 y 9 / Intervals < 0 0< Sign of y + Behavior of y up down Graphical support: Graphical support: (a) ( 0, ) (b) (, 0), <. y, > Intervals < < Sign of y + Behavior of y Increasing Decreasing y 0, <, > Intervals < < Sing of y 0 Behavior of y Linear down Graphical support: (a) (, ) (b) None (c) (, 0) (d) (0, ) (e) None (f) (0, ) / 0. y Intervals < 0 0< Sign of y Behavior of y Decreasing Decreasing y 9 / Intervals < 0 0< Sing of y + Behavior of y down up (a) None (b) (, ). y e y e Since y and y are both positive on the entire domain, y is increasing and concave up on the entire domain. Graphical support: (a) ( 0, ) (b) None
Section. 77. y e y e + e Intervals < > Sign of y + Behavior of y Decreasing Increasing y e + e Intervals < > Sign of y + Behavior of y down up, e. y 9 y 9 0 9 ± Intervals < < < < Sign of y Behavior of y + < < Decreasing Increasing Decreasing y + 0 / / ( 9 ) ( 9 ) y 0 at 0 Intervals < < 0 0 < < Sign of y + Behavior of y up down. y + since y > 0 for all, y is always increasing: d y ( + ) ( + ) ( ) d ( + ) Intervals < 0 0 < Sign of y + Behavior of y up down (0, 0) 6. y ( ) y Intervals < 0 0 < < > Sign of y + + y Intervals < 0 0 < < > Sign of y + Behavior of y down ( 0, 0) and (, 6) / / / 7. y ( ) / / y / up down Intervals < 0 0 < < < Sign of y + Behavior of y Decreasing Decreasing Increasing / 8 / + 8 y + 9 9 / 9 Intervals < < < 0 0 < Sign of y + + Behavior of y up ( 6, ) ( 76,. ) and ( 00, ) / 8. y ( + ) down up / / y ( + ) + y is always increasing, so there are no critical points for y. y 0 / / ( ) ( ) Intervals 0 < < > Sing of y + Behavior of y up down (, ) 9. We use a combination of analytic and grapher techniques to solve this problem. Depending on the viewing window chosen, graphs obtained using NDER may ehibit strange behavior near because, for eample, NDER ( y, ), 000, 000 while y is actually undefined at +. The graph of y is shown below. Behavior of y Increasing Increasing Decreasing
78 Section. 9. Continued ( )( + ) ( + )( ) y ( ) 8 + 8 ( ) The graph of y is shown below. The zeros of y are 0., 0.,and.. Intervals < 0. 0. < <.0.0 < < < <.. < Sign of y Behavior of y + + Decreasing Increasing Decreasing Decreasing Increasing ( ) ( 6 6+ 8) ( 8 + 8)( )( ) y ( ) ( )( 6 6+ 8) ( 8 + 8) ( ) + ( ) ( )( + 7) ( ) The graph of y is shown below. Note that the discriminant of + 7is ( ) ( )( 7), so the only solution of y 0 is. Intervals < < < < Sign of y + + Behavior of y up down (, ) + 0. ( )( ) ( ) y + ( + ) ( + ) up Intervals < < < < Sign of y + Behavior of y Decreasing Increasing Decreasing ( + ) ( ) ( + )( )( + )( ) y ( + ) ( + )( ) ( + ) ( + ) 6 ( ) ( + ) ( + ) Intervals < < < 0 0 < < Sign of y Behavior of y + + down (0, 0),,, and up,. (a) Zero: ±; positive: (, ) and (, ); negative: (, ) (b) Zero: 0; positive: (0, ); negative: (, 0). (a) Zero: 0, ±.; positive: (., 0) and (., ); negative: (,.) and (0,.) (b) Zero: ± 0.7; positive: (, 0.7) and (0.7, ); negative: ( 0.7, 0.7). (a) (, ] and [0, ] (b) [, 0] and [, ) (c) Local maima: and ; local minimum: 0. (a) [, ] (b) (, ] and [, ) (c) Local maimum: ; local minimum:. (a) vt () () t t (b) at () v () t down < up (c) It begins at position moving in a negative direction. It moves to position when t, and then changes direction, moving in a positive direction thereafter. 6. (a) vt () () t t (b) at () v () t (c) In begins at position 6 and moves in the negative direction thereafter. 7. (a) vt () () t t (b) at () v () t 6t
Section. 79 7. Continued (c) It begins at position moving in a negative direction. It moves to position when t, and then changes direction, moving in a positive direction thereafter. 8. (a) vt () () t 6t6t (b) at () v () t 6t (c) It begins at position 0. It starts moving in the positive direction until it reaches position when t, and then it changes direction. It moves in the negative direction thereafter. 9. (a) The velocity is zero when the tangent line is horizontal, at approimately t., t 6 and t 9.8. (b) The acceleration is zero at the inflection points, approimately t, t 8and t. 0. (a) The velocity is zero when the tangent line is horizontal, at approimately t 0., t,and t. (b) The acceleration is zero at the inflection points, approimately t., t., t 8, t,and t.. Some calculators use different logistic regression equations, so answers may vary. 6. 79 (a) y + t 87e 0.. 06 (b) [0, 0] by [ 00, 000] 6. 79 (c) y, 09, 870. (This is + 0. 06( 80). 87e remarkably close to the 000 census number of,8,0.) (d) The second derivative has a zero at about 78, indicating that the population was growing fastest in 898. This corresponds to the inflection point on the regression curve. (e) The regression equation predicts a population limit of about,6,79.. Some calculators use different logistic regression equations, so answers may vary. 89886. 88 (a) y + t 9 e 0.. 8 (b) [0, 9] by [. 0 6,. 0 7 ] (c) The zero of the second derivative is about.6, which puts the fastest growth during 98. This corresponds to the inflection point on the regression curve. (d) The regression curve predicts that cable subscribers will approach a limit of 8,98,86 +,68,0 subscribers (about million).. y + y y 6 y 0at ±. y ( ) > 0and y () < 0, so there is a local minimum at (, ) and a local maimum at (,7).. y 80+ 00 y 80 y 0 y 0 at ± y ( ) < 0and y ( ) > 0, so there is a local maimum at (, 8) and a local minimum at (, 8).. y + y + 6 y 6+ 6 y 0 at and 0. y ( ) < 0, y ( 0) > 0, so there is a local maimum at (, ) and a local minimum at (0, ). 6. y + 60+ 0 y 7 + 60 y 60 0 y 0 at ± and ±. y ( ) < 0, y ( ) > 0 y () < 0, and y ( ) > 0; so there are local maima at (, ) and (, 8), and there are local minima at (, 8) and (, 6). 7. y e y ( + ) e y ( + ) e y 0at. y () > 0, so there is a local minimum at (, / e). 8. y e y ( ) e y ( ) e y 0at y () < 0, so there is a local maimum at (, /e). 9. y ( ) ( ) Intervals < < < < Sign of y + Behavior of y Decreasing Decreasing Increasing
80 Section. 9. Continued y ( ) ( ) + ( )( )( ) ( )[( ) + ( )] ( )( ) Intervals < < < < Sign of y + + Behavior of y up (a) There are no local maima. down (b) There is a local (and absolute) minimum at. (c) There are points of inflection at and at. 0. y ( ) ( )( ) up. y y f () P y f (). y f() Intervals < < < < < < Sign of y + + + Behavior of y Increasing Increasing Decreasing Increasing d y + d [( ) ( 6 8)] ( ) ( 6) + ( 6+ 8)( )( ) ( )[( )( 6) + ( 6+ 8)] ( )( 0+ ) ( )( 0+ ) Note that the zeros of y are and 0 ± 0 ( )( ) 0 ± ± 6. or 7.. The zeros of y can also be found graphically, as shown.. No f must have a horizontal tangent at that point, but f could be increasing (or decreasing), and there would be no local etremum. For eample, if f( ), f ( 0) 0 but there is no local etremum at 0.. No. f ( )could still be positive (or negative) on both sides of c, in which case the concavity of the function would not change at c. For eample, if f ( ), then f ( 0) 0, but f has no inflection point at 0.. One possible answer: y 6. One possible answer: Intervals < < <.6.6 < <.7.7 < Sign of y + + Behavior of y down up down up (a) Local maimum at (b) Local minimum at (c) Points of inflection at, at.6, and at.7.
Section. 8 7. One possible answer: y (, 8) 0 (c) One possible answer: (0, ) (, 0) 0 8. One possible answer: 9. (a) [0, ], [, ], and [., 6] (b) [, ] and [,.] (c) Local maima:, (if f is continuous at ), and 6; local minima: 0,, and. 0. If f is continuous on the interval [0, ]: (a) [0, ] (b) Nowhere (c) Local maimum: ; local minimum: 0. (a) Absolute maimum at (, ); absolute minimum at (, ) (b) None (c) One possible answer: y y f(). (a) Absolute maimum at (0, ); absolute minimum at (, ) and (, ) (b) At (, 0) and (, 0) (d) Since f is even, we know f() f( ). By the continuity of f, since f( )< 0 when < < <, we know that f () 0, and since f ( ) and f ( ) > 0 when < <, we know that f () >. In summary, we know that f() f( ), < f() 0, and < f ( ) 0.. y. y f(). False. For eample, consider f( ) at c 0. 6. True. This is the Second Derivative Test for a local maimum. 7. A. y a + + say a 6 y 6 + 6+ y + 6 y 0 at Interval < / > / Sign of y + Behavior of y up down 8. E.
8 Section. 9. C. y + + 7 y 0 + y 0 60 y 0 at Interval < > Sign of y + Behavior of y down up is an inflection point. 60. A. b 7 6. (a) In eercise, a and b, so, which is a the -value where the point of inflection occurs. The local etrema are at and, which are symmetric about 7. b (b) In eercise 8, a and b 6, so, which is a the -value where the point of inflection occurs. The local etrema are at 0 abd, which are symmetric about. (c) f ( ) a + b + c and f ( ) 6a + b. The point of inflection will occur where b f ( ) 0, which is at. a If there are local etrema, they will occur at the zeros of f ( ). Since f ( )is quadratic, its graph is a parabola and any zeros will be symmetric about the verte which will also be where f ( ) 0. b b ( + ae )( 0) ( c)( abe ) 6. (a) f ( ) b ( + ae ) b abce b ( + ae ) b abce b ( e + a), so the sign of f ( )is the same as the sign of abc. b b b b ( e + a) ( ab ce ) ( abce ) ( e + a)( be (b) f ( ) b ( e + a) b b b b ( e + a)( ab ce ) ( abce )( be ) b ( e + a) b b ab ce ( e a) b ( e + a) a Since a > 0, this changes sign when ln due to the b e b afactor in the numerator, and f( ) has a point of inflection at the location. b ) 6. (a) f ( ) a + b + c + d f ( ) a + 6b + c Since is f ( )quadratic, it must have 0,, or zeros. If f () has 0 or zeros, it will not change sign and the concavity of f () will not change, so there is no point of inflection. If f ( )has zeros, it will change sign twice, and f( ) will have points of inflection. (b) If f has no points of inflection, then f ( )has 0 or zeros, so the discriminant of f ( )is 0. This gives ( 6b) ( a)( c) 0, or b 8ac. If f has points of inflection, then f ( )has zeros and the inequality is reversed, so b > 8ac. In summary, f has points of inflection if and only if b > 8ac. Quick Quiz Sections... (C) f ( ) ( ) ( + ) + ( ) ( + ) 0, 7 9,. (D) f ( ) ( ) + ( ) ( ) 0 f ( ) ( )( 7) 0 7,. (B) 9 0 ±. (a) d ln ( + ) d + 0, Intervals < < < < < < Sign of y + Behavior of y Decreasing Increasing Decreasing f has relative minima at and f has relative maima at ± d 6 (b) ( ) + f d 6 + f ( ) ( + ) ± 0 f has points of inflection at ± (c) The absolute maimum is at and f( ) ln 6+.
Section. 8 Section. Modeling and Optimization (pp. 9 ) Eploration Constructing Cones. The circumference of the base of the cone is the circumference of the circle of radius minus, or 8. 8 Thus, r. Use the Pythagorean Theorem to find h, and the fomula for the volume of a cone to find V.. The epression under the radical must be nonnegative, that 8 is,6 0. Solving this inequality for gives: 0 6.. The circumference of the original circle of radius is 8. Thus, 0 8.. The maimum occurs at about.6. The maimum volume is about V. 80.. Start with dv rh dr + r dh. d d d dr dh Compute and, substitute these values in d d dv dv, set 0, and solve for to obtain d d 8 ( 6) 6.. Then V 8 7 Quick Review.. 80.. y + ( ) Since y 0 for all ( and y > 0 for ), y is increasing on, and there are no local etrema. ( ). y 6 + 6 6( + )( ) y + 6 The critical points occur at or, since y 0 at these points. Since y ( ) 8< 0, the graph has a local maimum at. Since y () 8> 0, the graph has a local minimum at. In summary, there is a local maimum at (, 7) and a local minimum at (, 0). 00. V r h ()() 8 cm. V r h 000 SA rh + r 600 Solving the volume equation for h gives 000. r Substituting into the surface area equation gives 000 + r 600. Solving graphically, we have r r., r. 0, or r 7.. Discarding the negative value and using h 000 to find the corresponding values r of h, the two possibilities for the dimensions of the cylinder are: r 0. cm and h 98. cm, or, r 7. cm and h 66. cm.. Since y sin is an odd function, sin ( α) sin α. 6. Since y cos is an even function, cos ( α) cos α. 7. sin( α) sin cosα cossin α 0cos α ( )sinα sinα 8. cos( α) coscosα sinsin α ( )cosα + 0sinα cosα 9. + y and y + ( ) + ± Since y, the solution are: and y, or, and y. In ordered pair notation, the solutions are (, ) and (, ). 0. + y and y + 9 ( + ) + 9 9 + ( + ) 6 9 + + + 6 6 + 0 ( + ) 0 0 or
8 Section. 0. Continued Since y +, the solutions are: 0 and y, or, and y. In ordered pair notation, the solution are (0, ) and,. Section. Eercises. Represent the numbers by and 0, where 0 0. (a) The sum of the squares is given by f( ) + ( 0 ) 0+ 00. Then f ( ) 0. The critical point and endpoints occur at 0, 0, and 0. Then f (0) 00, f (0) 00, and f (0) 00. The sum of the squares is as large as possible for the numbers 0 and 0, and is as small as possible for the numbers 0 and 0. Graphical support: (b) The sum of one number plus the square root of the other is given by g ( ) + 0. Then g ( ). The critical point occurs when 0 0, so 0 and 79. Testing the endpoints and critical point, we find g( 0) 0 79 8 7., g 0., g( 0) 0. and The sum is as large as possible when the numbers are 79 and summing 79 +, andisassmall as possible when the numbers are 0 and 0 (summing 0 + 0). Graphical support:. Let and y represent the legs of the triangle, and note that 0 < <. Then + y, so y (since y > 0). The area is A y so da + d ( ). The critical point occurs when 0, which means, (since > 0). This value corresponds to the largest possible area, since da d for, > 0 for 0 < < and < da d < 0 < <. When,wehave y and A y Thus, the largest possible area is cm, and the dimensions (legs) are cm by cm. Graphical support:. Let represent the length of the rectangle in inches ( > 0). Then the wih is 6 and the perimeter is P ( ) 6 +. + ( 6) Since P ( ) this critical point occurs at. Since P ( ) < 0for 0< < and P ( ) > 0for >, this critical point corresponds to the minimum perimeter. The smallest possible perimeter is P( ) 6 in., and the rectangle s dimensions are in. by in. Graphical support:.
Section. 8. Let represent the length of the rectangle in meters (0 < < ). Then the wih is and the area is A ( ) ( ). Since A ( ), the critical point occurs at. Since A ( ) > 0 for 0< < and A ( ) < 0 for < <, this critical point corresponds to the maimum area. The rectangle with the largest area measures m by m, so it is a square. Graphical support:. (a) The equation of line AB is y +, so the y-coordinate of P is +. (b) A ( ) ( ) d (c) Since A ( ) ( ), the critical point d occurs at. Since A ( ) > 0 for 0 < < and A ( ) < 0 for < <, this critical point corresponds Graphical support: 7. Let be the side length of the cut-out square (0 < < ). Then the base measures 8 in. by in., and the volume is V( ) ( 8 )( ) 6 + 0. Then V ( ) 9+ 0 ( )( 6). Then the critical point (in 0 < < ) occurs at. Since V ( ) > 0 for 0< < and V ( ) < 0 for < <, the critical point corresponds to the maimum volume. The maimum volume is V 0 90 7 7. in, and the dimensions are in. by in. by in. Graphical support: to the maimum area. The largest possible area is A square unit, and the dimensions of the rectangle are unit by unit. Graphical support: 6. If the upper right verte of the rectangle is located at (, )for0 < <, then the rectangle s dimensions are by and the area is A () ( ). Then A ( ) 6 6( ), so the critical point (for 0 < < ) occurs at. Since A ( ) > 0 for 0< < and A ( ) < 0 for < <, this critical point corresponds to the maimum area. The largest possible area is A( ), and the dimensions are by 8. 8. Note that the values a and b must satisfy a + b 0 and so b 00 a. Then the area is given by A ab a 00 a for 0 < a< 0, and da a a + a da 00 a 00 ( ) + a ( 00 a ) 00 a. The critical point occurs 00 a 00 a da when a 00.Since > 0 for 0 < a < 00 and da da < 0 for 00 < a < 0, this critical point corresponds to da the maimum area. Furthermore, if a 00 then b 00 a 00, so the maimum area occurs when a b.
86 Section. 8. Continued Graphical support: 9. Let be the length in meters of each side that adjoins the river. Then the side parallel to the river measures 800 meters and the area is A ( ) ( 800 ) 800 for 0 < < 00. Therefore, A ( ) 800 and the critical point occurs at 00. Since A ( ) > 0 for 0 < < 00 and A ( ) < 0 for 00 < < 00, the critical point corresponds to the maimum area. The largest possible area is A( 00) 80, 000 m and the dimensions are 00 m (perpendicular to the river) by 00 m (parallel to the river). Graphical support: 0. If the subdividing fence measures meters, then the pea patch measures m by 6 m and the amount of fence needed is f( ) + 6 +. Then f ( ) and the critical point (for > 0) occurs at. Since f ( ) < 0 for 0< < and f ( ) > 0 for >, the critical point corresponds to the minimum total length of fence. The pea patch will measure m by 8 m (with a -m divider), and the total amount of fence needed is f ( ) 7m. Graphical support:. (a) Let be the length in feet of each side of the square base. Then the height is 00 ft and the surface area (not including the open top) is 00 S ( ) +. + 000 Therefore, ( 000) S ( ) 000 and the critical point occurs at 0. Since S ( ) < 0 for 0< < 0 and S ( ) > 0 for > 0, the critical point corresponds to the minimum amount of steel used. The dimensions should be 0 ft by 0 ft by ft, where the height is ft. (b) Assume that the weight is minimized when the total area of the bottom and the four sides is minimized.. (a) Note that y, so y. Then c ( + y) + 0y + 0y + 0 +, 70 dc 0( 7) 0, 70 d The critical point occurs at. Since dc d < 0for dc 0< < and > 0 for >, the critical point d corresponds to the minimum cost. The values of and y are ft and y ft. (b) The material for the tank costs dollars/sq ft and the ecavation charge is 0 dollars for each square foot of the cross-sectional area of one wall of the hole.. Let be the height in inches of the printed area. Then the wih of the printed area is 0 in. and the overall dimensions are + 8 in. by 0 + in. The amount of paper used is A ( ) ( + 8 0 ) +. + 8 + 00 in Then ( 00) A ( ) 00 and the critical point (for > 0) occurs at 0. Since A ( ) < 0for 0< < 0 and A ( ) > 0 for > 0, the critical point corresponds to the minimum amount of paper. Using + 8 and 0 + for 0, the overall dimensions are 8 in. high by 9 in. wide.. (a) st () 6t + 96t+ vt () s () t t+ 96 At t 0, the velocity is v(0) 96 ft/sec. (b) The maimum height occurs when v(t) 0, when t. The maimum height is s() 6 ft and it occurs at t sec.
Section. 87. Continued (c) Note that st () 6t + 96t+ 6( t+ )( t7), so s 0 at t or t 7. Choosing the positive value, of t, the velocity when s 0 is v(7) 8 ft/sec.. We assume that a and b are held constant. Then A( θ) absin θ and A ( θ) abcos θ. The critical point ( for 0 < θ < ) occurs at θ. Since A ( θ) > 0 for 0 < θ < and A ( θ) 0 for < θ <, the critical point corresponds to the maimum area. The angle that maimizes the triangle s area is θ ( or 90 ). 6. Let the can have radius r cm and height h cm. Then 000 r h 000, so h. The area of material used is r 000 da A r + rh r +,so r000r r dr r 000. The critical point occurs at r 000 da r 0 / cm. Since < 0 dr / da / for 0< r < 0 and > 0 for r > 0, the critical dr point corresponds to the least amount of material used and hence the lightest possible can. The dimensions are r 0 / 6. 8cm and h 0 / 6. 8cm. In Eample, because of the top of the can, the best design is less big around and taller. 000 7. Note that r h 000, so h. Then r 000 A 8r + rh 8r +, so r da 6( r ) 6r 000r dr. The critical point r occurs at r cm. Since da < 0 for 0 < r < and dr da > 0 for r >, the critical point corresponds to the least dr amount of aluminium used or wasted and hence the most economical can. The dimensions are r cm and h 0, 8. (a) The base measures 0 in. by in, so the volume formula is ( 0 )( ) V( ) + 7. (b) We require > 0, < 0, and <. Combining these requirements, the domain is the interval (0, ). (c) The maimum volume is approimately 66.0 when.96 in. (d) V ( ) 6 0+ 7 The critical point occurs when V () 0, at 0 ± ( 0) ( 6)( 7) 0 ± 700 ± 7, 6 () 6 that is,.96 or 6.7. We discard the larger value because it is not in the domain. Since V () 0, which is negative when.96, the critical point corresponds to the maimum volume. The maimum 7 volume occurs when 96., which 6 confirms the result in (c). 9. (a) The sides of the suitcase will measure in. by 8 in. and will be in. apart, so the volume formula is V( ) ( )( 8 ) 8 68 + 86. (b) We require > 0, < 8, and <. Combining these requirements, the domain is the interval (0, 9). [0, 9] by [ 00, 600] so the ratio of h to r is 8 to.
88 Section. 9. Continued (c) The maimum volume is approimately 09.9 when.9 in. (d) V () 6 + 86 ( + 6) The critical point is at ± ( ) ( )( 6) ± 7±, that () is, 9. or 06.. We discard the larger value because it is not in the domain. Since V () ( ), which is negative when.9, the critical point corresponds to the maimum volume. The maimum value occurs at 7. 9, which confirms the results in (c). (e) 8 68 + 86 0 8( + 08 0) 0 8( )( )( ) 0 Since is not in the domain, the possible values of are in. or in. (f) The dimensions of the resulting bo are in., ( ) in., and (8 ) in. Each of these measurements must be positive, so that gives the domain of (0, 9) 0. We discard the negative value of because it is not in the domain. Checking the endpoints and critical point, we have f( 0)., f., f and (6).6. Jane should land her boat 087. miles down the shoreline from the point nearest her boat.. If the upper right verte of the rectangle is located at (, cos 0.) for 0 < <, then the rectangle has wih and height cos 0., so the area is A() 8 cos 0.. Then A () 8 (0. sin 0.) + 8(cos 0.)() sin 0. + 8 cos 0.. Solving A () graphically for 0 < <, we find that.7. Evaluating and cos 0. for.7, the dimensions of the rectangle are approimately. (wih) by.6 (height), and the maimum area is approimately 8.98.. Let the radius of the cylinder be r cm, 0 < r < 0. Then the height is 00 r and the volume is Vr () r 00 r cm. Then V () r r ( r) + ( 00 r )( r) 00 r ( ) + r r 00 r 00 r r( 00 r ) 00 r The critical point for 0 < r < 0 occurs at 00 r 0. Since V r () > 0 for 0 < r < 0 and V () r > 0 for 0 < r <0, the critical point corresponds Let be the distance from the point on the shoreline nearest Jane s boat to the point where she lands her boat. Then she needs to row + mi at mph and walk 6 mi at mph. The total amount of time to reach the village is + f( ) 6 + hours (0 6 ). Then f ( ) ( ). + + Solving f ( ) 0, we have: + + ( + ) 6 ± to the maimum volume. The dimensions are r 0 0 86. cm and h. cm, and the volume is 000 8. 0 cm.. Set r ( ) c ( ):. The only positive critical value is, so profit is maimized at a production level of 000 units. Note that ( r c) ( ) ( ) < 0 for all positive, so the Second Derivative Test confirms the Maimum.. Set r ( ) c ( ): /( + ) ( ). We solve this equation grpahically to find that 0. 9. The graph of y r () c() shows a minimum at 0. 9 and a maimum at., so profit is maimized at a production level of about, units.
Section. 89 c ( ). Set c ( ) : 0 + 0 0 + 0. The only positive solution is, so average cost is minimized at a production level of 000 units. Note that d c ( ) 0 d > for all positive, so the Second Derivative Test Confirms the minimum. 6. Set c ( ) c( ) / : e + e e. The only positive solution is ln, so average cost is minimized at a production level of 000 ln, which is about 69 units. d c ( ) Note that e 0 d > for all positive, so the Second Derivative Test confirms the minimum. 7. Revenue: r ( ) 00 ( 0) + 00 Cost: c ( ) 6000 + Profit: p ( ) r ( ) c ( ) + 686000, 0 80 Since p ( ) + 68( 67), the critical point occurs at 67. This value represents the maimum because p ( ), which is negative for all in the domain. The maimum profit occurs if 67 people go on the tour. 8. (a) f ( ) ( e ) + e ( ) e ( ) The critical point occurs at. Since f () > 0 for 0 < and f () < 0 for >, the critical point corresponds to the maimum value of f. The absolute maimum of f occurs at. (b) To find the values of b, use grapher techniques to solve e 0. e 0., e 0. e 0., and so on. To find the values of A, calculate (b a) ae, using the unrounded values of b. (Use the list features of the grapher in order to keep track of the unrounded values for part (d).) a b A 0..7 0. 0..86 0. 0..6 0.6 0..0 0. 0..76 0.8 0.6. 0. 0.7.8 0. 0.8. 0. 0.9. 0.08.0.00 0.00 (c) [0,.] by [ 0., 0.6] (d) Quadratic: A 09. a + 0. a+ 0. [ 0.,.] by [ 0., 0.6] Cubic: A 7. a 78. a + 86. a+ 09. [ 0.,.] by [ 0., 0.6] Quartic: A 9. a + 96. a 687. a + 7. a+ 0. [ 0.,.] by [ 0., 0.6] (e) Quadratic: According to the quadratic regression equation, the maimum area occurs at a 0.0 and is approimately 0.. Cubic: According to the cubic regression equation, the maiumu area occurs at a 0. and is approimately 0..
90 Section. 8. Continued Quartic: point (for h > 0) occurs at h. Since dv dh > 0 for 0 < h < and dv dh < 0 for < h <, the critical point corresponds to the maimum volume. The cone of greatest volume has radius m, height m, and volume m. According to the quartic regression equation the maimum area occurs at a 0.0 and is approimately 0.6. 9. (a) f ( ) is a quadratic polynominal, and as such it can have 0,, or zeros. If it has 0 or zeros, then its sign never changes, so f() has no local etrema. If f ( )has zeros, then its sign changes twice, and f() has local etrema at those points. (b) Possible answers: No local etrema: y ; local etrema: y 0. Let be the length in inches of each edge of the square end, and let y be the length of the bo. Then we require + y 08. Since our goal is to maimize volume, we assume + y 08 and so y 08. The volume is V( ) ( 08 ) 08, where 0 < < 7. Then V 6 ( 8), so the critical point occurs at 8 in. Since V ( ) > 0 for 0< < 8 and V ( ) < 0 for 8< < 7, the critical point corresponds to the maimum volume. The dimensions of the bo with the largest possible volume are 8 in. by 8 in. by 6 in.. Since + y 6, we know that y 8. In part (a), the radius is and the height is 8, and so the volume is given by r h 8 8 ( ) ( ). In part (b), the radius is and the height is 8, and so the volume is given by r h ( 8 ). Thus, each problem requires us to find the value of that maimizes f( ) ( 8 ) in the interval 0 < < 8, so the two problems have the same answer. To solve either problem, note that f( ) 8 and so f ( ) 6 ( ). The critical point occursat. Since f ( ) > 0 for 0 < < and f ( ) < 0 for < < 8, the critical point corresponds to the maimum value of f (). To maimize the volume in either part (a) or (b), let cm and y 6 cm.. Note that h + r and so r h. Then the volume is given by V r h h h h h ( ) for 0< h <, and so dv h ( h ). The critical dh. (a) We require f () to have a critical point at. Since a f ( ) a, we have f ( ) and so our a requirement is that 0. Therefore, a 6. To verify that the critical point corresponds to a local minimum, note that we now have f ( ) 6 and so f ( ) +, so f ( ) 6, which is positive as epected. So, use a 6. (b) We require f () 0. Since f + a, we have f () + a, so our requirement is that + a 0. Therefore, a. To verify that is in fact an inflection point, note that we now have f ( ), which is negative for 0 < < and positive for >. Therefore, the graph of f is concave down in the interval (0, ) and concave up in the interval (, ), So, use a. a. f ( ) a, so the only sign change in a f ( ) occurs at, where the sign changes from negative to positive. This means there is a local minimum at that point, and there are local maima.. (a) Note that f ( ) + a+ b. We require f ( ) 0 and f () 0, which give a + b 0 and 7 + 6a + b 0. Substracting the first equation from the second, we have + 8a 0 and so a. Substituting into the first equation, we have 9 + b 0, so b 9. Therefore, our equation for f( ) is f( ) 9. To verify that we have a local maimum at and a local minimum at, note that f ( ) 69 ( + )( ), which is positive for <, negative for < <, and positive for >. So, use a and b 9. (b) Note that f ( ) + a+ b and f ( ) 6+ a. We require f ( ) 0 and f () 0, which give 8 + 8a + b 0 and 6 + a 0. By the second equation, a, and so the first equation becomes 8 +b 0. Thus b. To verify that we have a local minimum at, and an inflection point at, note that we now have f ( ) 66. Since f changes sign at and is positive at, the desired conditions are satisfied. So, use a and b.
Section. 9 6. Refer to the illustration in the problem statement. Since + y 9, we have 9 y. Then the volume of the cone is given by V r h ( y+ ) ( 9 y )( y+ ) ( y y + 9y+ 7), for < y <. Thus dv ( y 6y+ 9) ( y + y) dy ( y+ )( y ), so the critical point in the interval (, ) is y. Since dv dy > 0 for < y < and dv < 0 for < y <, the critical point does correspond to dy the maimum value, which is V( ) cubic units. 7. (a) Note that w + d, so d w. Then we may write S kwd kw( w ) kwkw for 0 < w <, so ds k kw k( w 8). dw The critical point (for 0 < w < ) occurs at w 8. Since ds dw > 0 for 0< w < and ds < 0 for dw < w <, the critical point corresponds to the maimum strength. The dimensions are in. wide by 6 in. deep. (b) (c) The graph of S ww is shown. The maimum strength shown in the graph occurs at w 6., 9 which agrees with the answer to part (a). The graph of S d d is shown. The maimum strength shown in the graph occurs at d 6 9., 8 which agrees with the answer to part (a), and its value is the same as the maimum value found in part (b), as epected. Changing the value of k changes the maimum strength, but not the dimensions of the strongest beam. The graphs for different values of k look the same ecept that the vertical scale is different. 8. (a) Note that w + d, so d w. Then we / may write S kwd kw( w ), so ds dw kw i w / w + k w / ( ) ( ) ( ) ( ) ( k w )( w + w ) ( k )( w 6) The critical point (for 0 < w < ) occurs at w 6. Since ds > 0 for 0 6 0 6 dw < w < ds and < for dw < w <, the critical point corresponds to the maimum stiffness. The dimensions are 6 in. wide by 6 in. deep. (b) [0, ] by [ 000, 8000] The graph of S w( w ) / is shown. The maimum stiffness shown in the graph occurs at w 6, which agrees with the answer to part (a). (c) [0, ] by [ 000, 8000] The graph of S d d is shown. The maimum stiffness shown in the graph occurs at d 6 0. agrees with the answer to part (a), and its value is the same as the maimum value found in part (b), as epected. Changing the value of k changes the maimum stiffness, but not the dimensions of the stiffest beam. The graphs for different values of k look the same ecept that the vertical scale is different. 9. (a) vt () s () t 0sint The speed at time t is 0 sin t. The maimum speed is 0 cm/sec and it occurs at t t t,,, and t 7 sec. The position at these times is s 0 cm (rest position), and the acceleration at () v () t 0 cost is 0 cm / sec at these times.
9 Section. 9. Continued (b) Since at () 0 cos t, the greatest magnitude of the acceleration occurs at t 0, t, t, t, and t. At these times, the position of the cart is either s 0 cm or s 0cm, and the speed of the cart is 0 cm/sec. 0. Since di sin t+ cos t, the largest magnitude of the current occurs when sin t + cos t 0, or sin t cos t. Squaring both sides gives sin t cos t, and we know that sin t+ cos t, so sin t cos t. Thus the possible values of t are,,,and so on. Eliminating etraneous solutions, the solutions of sin t cos t are t + k for integers k, and at these times i cost+ sin t. The peak current is amps.. The square of the distance is 9 D ( ) ( ), + + 0 + so D ( ) and the critical point occurs at. Since D ( ) < 0for < and D ( ) > 0for >, the critical point corresponds to the minimum distance. The minimum distance is D( ).. Calculus method: The square of the distance from the point (, ) to (, 6 ) is given by D ( ) ( ) + ( 6 ) + + 6 8 + 0 8. Then + D ( ) ( 6) + 6. 8 8 Solving D ( ) 0, we have: 6 8 6 ( 8 ) 9 8 8 ± We discard as an etraneous solution, leaving. Since D ( ) < 0for < < and D ( ) > 0 for < <, the critical point corresponds to the minimum distance. The minimum distance is D( ). Geometry method: The semicircle is centered at the origin and has radius. The distance from the origin to (, ) is + ( ). The shortest distance from the point to the semicircle is the distance along the radius containing the point (, ). That distance is.. No. Since f( ) is a quadratic function and the coefficient of is positive, it has an absolute minimum at the point where f ( ) 0, and the point is,.. (a) Because f() is periodic with period. (b) No. Since f() is continuous on [ 0, ], its absolute minimum occurs at a critical point or endpoint. Find the critical points in [ 0, ]: f ( ) sin sin 0 sin sin cos 0 (sin ) ( + cos ) 0 sin 0or cos 0,, The critical points (and endpoints) are (0, 8), (, 0), and (, 8). Thus, f() has an absolute minimum at (, 0) and it is never negative.. (a) sin t sin t sin t sin tcos t (sin t)( cos t) 0 sin t 0or cost t k, where k is an integer The masses pass each other whenever t is an integer multiple of seconds. (b) The vertical distance between the objects is the absolute value of f( ) sintsin t. Find the critical points in [ 0, ]: f ( ) cost cost 0 ( cos t) cost 0 ( cos tcos t ) 0 ( cos t+ )(cos t ) 0 cost or cost t,, 0, The critical points (and endpoints) are (0, 0),,,,, and (, 0) The distance is greatest when t sec and when t sec. The distance at those times is meters.
Section. 9 6. (a) sin t sin t+ sin t sin tcos + cos tsin sin t sin t+ cos t sin t cos t tant Solving for t, the particles meet at t sec and at t sec. (b) The distance between the particles is the absolute value of f( t) sin t+ sint cost sin t. Find the critical points in [ 0, ]: f () t sint cost 0 sin t cos t tant The solutions are t and t, so the critical 6 6 points are at,,, 6 and 6 and the interval endpoints are at 0,, and,. The particles are farthest apart at t sec and at t sec, and 6 6 the maimum distance between the particles is m. (c) We need to maimize f (), t so we solve f () t 0. f () t cost+ sint 0 sin t cos t This is the same equation we solved in part (a), so the solutions are t sec and t sec. For the function y f ( t), the critical points occur at,, and, and the interval endpoints are at 0,,. and Thus, f () t is maimized at t and t. But these are the instants when the particles pass each other, so the graph of y f() t has corners at these points and d f() t is undefined at these instants. We cannot say that the distance is changing the fastest at any particular instant, but we can say that near t or t the distance is changing faster than at any other time in the interval. 7. The trapezoid has height (cos θ) ft and the trapezoid bases measure ft and ( + sin θ) ft, so the volume is given by V( θ) (cos θ)( + + sin θ)( 0) 0(cos θ)( + sin θ). Find the critical points for 0 θ < : V ( θ) 0(cos θ)(cos θ) + 0( + sin θ)( sin θ) 0 0 cos θ0sinθ 0sin θ 0 0( sin θ) 0sinθ0sin θ 0 0( sin θ+ sinθ 0 0( sin θ )(sin θ + ) 0 sinθ or sinθ θ 6 The critical point is at 6,. Since V ( θ) > 0 for 0 θ < and V ( θ) < 0 for < θ <, the 6 6 critical point corresponds to the maimum possible trough volume. The volume is maimized when θ. 6 8. (a) Sketch segment RS as shown, and let y be the length of segment QR. Note that PB 8., and so QB (. 8 ) 8.( 8.). Also note that triangles QRS and PQB are similar. QR PQ RS QB y 8. 8.( 8.)
9 Section. 8. Continued y (a) 8. 8.( 8.) 8. y 8. L + y 8. L + 8. ( 8. ) + 8. L 8. L 8. (b) Note that >., and let f( ) L. 8. Since y, the approimate domain of f is.0 8.. Then ( 8. )( 6 ) ( )( ) ( 8 ) f ( ) ( 8. ) ( 8. ) For > 0., the critical point occurs at 8 6. 7in., and this corresponds to a minimum value of f( ) because f ( ) < 0 for. 0 < < 6. 7 and f ( ) > 0 for > 6. 7. Therefore, the value of that minimizes L is 6. 7 in. (c) The minimum value of L is (. 6 7). 0 in. (. 6 7) 8. 9. Since R M C M C M M, we have dr CM M f M CM M dm. Let ( ). Then f ( M) C M, and the critical point for f occurs at M C. This value corresponds to a maimum because f M > M < C C ( ) 0 for and f ( M) < 0 for M >. The value of M that maimizes dr is dm M C. 0. The profit is given by P ( ) ( n)( c) a+ b( 00 )( c) b + ( 00 + cb ) + ( a00bc). Then P ( ) b+ ( 00 + c) b b( 00 + c). 00 + c c The critical point occurs at 0 +, and this value corresponds to the maimum profit because c c P ( ) > 0 for < 0+ and P ( ) < 0 for > 0+. A selling price of 0 + c will bring the maimum profit.. True. This is guaranteed by the Etreme Value Theorem (Section.).. False. For eample, consider f( ) at c 0.. D. f( ) ( 60 ) f ( ) ( ) + ( 60)( ) + 0 + 0 ( 0) 0 or 0 60 60 60 0 ( 60 ) 0 ( 0) ( 0) ( 600)( 0), 000. B. Since f ( ) is negative, f( ) is always decreasing, so f ( ).. B. A bh b + h 00 b 00 h A h h 00 00 h h A 00 h A 0 when h 0 b 00 0 0 A ma 0 0 6. E. length 7. height 0 0 ( 0 ) 600 da ( 60 0 ) 60 0 d ( 0 ( ) ) 0. Let P be the foot of the perpendicular from A to the mirror, and Q be the foot of the perpendicular from B to the mirror.
Section. 9 7. Continued Suppose the light strikes the mirror at point R on the way from A to B. Let: a distance from A to P b distance from B to Q c distance from P to Q distance from P to R To minimize the time is to minimize the total distance the light travels going from A to B. The total distance is D ( ) + a + ( c ) + b Then D ( ) ( ) + + a ( c ) + b c + a ( c ) + b Solving D ( ) 0 gives the equation + a c ( c ) + b [ ( c)] which we will refer to as Equation. Squaring both sides, we have: ( c ) + a ( c ) + b c + b c ( ) ( )( + a ) ( c ) + b ( c ) + ( c) a b ( c) a b c c+ a 0 ( a b ) ac+ ac 0 [( a+ b) ac][( ab) ac] ac ac or a + b a b ac Note that the value a b is an etraneous solution because and c have opposite signs for this value. The ac only critical point occurs at a + b. To verify that critical point represents the minimum distance, note that ( + a )( ) ( ) + a D ( ) + a ( c) ( ( c ) + b )( ) ( c) ( c ) + b ( c ) + b + ( a ) [( c ) + b ] + ( c ) ( + a ) [( c ) + b ] a b +, ( + a ) [( c ) + b ] which is always positive. We now know that D ( ) is minimized when Equation is true, or, equivalently, PR QR. This means that the two AR BR right triangles APR and BQR are similar, which in turn implies that the two angles must be equal. 8. dv ka k d ka a The critical point occurs at, which represents a k maimum value because d v k, which is negative for d all. The maimum value of v is ka k ka a k a ka. 9. (a) v cr r cr dv dr 0 cr r cr cr ( r r) 0 The critical point occurs at r 0 r 0. (Note that r 0 is not in the domain of v.) The critical point represents a maimum because d v cr cr c r r 0 6 ( 0 ),which dr is negative in the domain r 0 r r0. (b) We graph v ( 0. r) r, and observe that the maimum indeed occurs at v 0.. h 60. (a) Since A ( q) kmq +, the critical point occurs when km h km, or q. This corresponds to the q h minimum value of Aq ( ) because A ( q) kmq, which is positive for q > 0. (b) The new formula for average weekly cost is ( k+ bq) m Bq ( ) + cm + hq q km + bm + cm + hq q Aq ( ) + bm Since Bq ( ) differs from Aq ( ) by a constant, the minimum value of Bq ( ) will occur at the same q-value as the minimum value of Aq ( ). The most economical quantity is again km. h
96 Section. 6. The profit is given by p( ) r( ) c( ) 6( 6 + ) + 6 9, for 0. Then p ( ) + 9( )( ), so the critical points occur at and. Since p ( ) < 0 for 0 <, p ( ) > 0 for < <, and p ( ) < 0for >, the relative maima occur at the endpoint 0 and at the critical point. Since p( 0) p( ) 0, this means that for 0, the function p( )has its absolute maimum value at the points (0, 0) and (, 0). This result can also be obtained graphically, as shown. 6. The average cost is given by c ( ) a ( ) 0+ 0, 000. Therefore, a ( ) 0 and the critical value is 0, which represents the minimum because a ( ), which is positive for all. The average cost is minimized at a production level of 0 items. 6. (a) According to the graph, y ( 0) 0. (b) According to the graph, y ( L) 0. (c) y( 0) 0, so d 0. Now y ( ) a + b + c, c 0. Therefore, y( ) a + b and so y ( 0) implies that y ( ) a + b. Then y( L) al + bl H and y ( L) al bl 0, so we have two linear equations in the two unknowns a and b. The second al equation gives b. Substituting into the first al equation, we have al + H, or al H, so a H H L. Therefore, b L and the equation for y H is y ( ) L + H L, or y ( ) H. L + L a 6. (a) The base radius of the cone is r and so the a height is h a r a. Therefore, a V( ) r h a a (b) To simplify the calculations, we shall consider the volume as a function of r: volume f () r r a r, where 0 < r < a. d f r dr r a r () ( ) r i i ( r) + ( a r )( r) a r r + r( a r ) a r ( arr ) a r r( a r ) a r a The critical point occurs when r, which a 6 gives r a. Then a a a h a r a. Using a 6 r and h a, we may now find the values of r and h for the given values of a when a : 6 r h, ; when a : 6 r h, ; when a 6: r 6, h ; 8 6 when a 8 : r, h 8 a 6 (c) Since r and h a, the relationship is r h. 6. (a) Let 0 represent the fied value of at point P, so that P has coordinates ( 0, a) and let m f ( 0 ) be the slope of line RT. Then the equation of line RT is y m( 0 ) + a. The y-intercept of this line is m( 0 0) + a am0, and the -intercept is the m0 a solution of m ( 0 ) + a 0, or. Let O m designate the origin. Then (Area of triangle RST).
Section. 97 6. Continued (a) (Area of triangle ORT) i (-intercept of line RT) (y-intercept of line RT) m0 a i ( a m0 ) m m m 0 a m0 a m m m0 a m a m 0 m Substituting for 0, f ( ) for m, and f ( )for a, we f ( ) have A ( ) f ( ). f ( ) (b) The domain is the open interval (0, 0). To graph, let y f ( ) +, 00 y f ( ) NDER ( y ), and y y A( ) y. y The graph of the area function y below. A( ) is shown [0, 0] by [ 00, 000] The vertical asymptotes at 0 and 0 correspond to horizontal or vertical tangent lines, which do not form triangles. (c) Using our epression for the y-intercept of the tangent line, the height of the triangle is a m f( ) f ( )i + 00 00 + 00 + 00 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A() occurs at 866.. Substituting this value into the epression above, the height of the triangle is. This is times the y-coordinate of the center of the ellipse. (d) Part (a) remains unchanged. The domain is (0, C). To graph, note that f ( ) B + B B B C + C C and B B f ( ) C ( ). C C C Therefore, we have f( ) A ( ) f ( ) f ( ) B + B B C C C C B C C B ( BC + B C ) C C C B B + ( BC + B C )( C ) BC C + + B BC C B( C BC C BC( C + C ) BC C BC( C + C ) C ( C )( )( C+ C C ( C+ C ) + C ( ) C A ( ) BC i ( C ) ( C+ C ) BC( C + C ) ( C ) + C C C BC( C + C ) + C C C C + ( C ) + BC( C C ) C C C C ( C ) C BC( C + C ) C C( C ) C C ( C ) BC ( C + C ) ( C C C ) ( C ) To find the critical points for 0 < < C, we solve: C C C C + C C C C 0 ( C ) 0 The minimum value of A() for 0 < < C occurs at the C C critical point, or. The corresponding triangle height is
98 Section. 6. Continued a m f( ) f ( )i B B C C B + + C C B + B B C C C C + C C C BC B + B C C + C B B B + + B This shows that the triangle has minimum area when its height is B. Section. Linerization and Newton s Method (pp. ) Eploration Appreciating Local Linearity. Quick Review.. dy d. dy d.. d cos ( + ) i ( d + ) cos ( + ) ( + )( sin ) ( + cos )( ) ( + ) sin + sin cos ( + ) cos ( + ) sin ( +) 0. 67 0. y ( + 0.000) / + 0.9 The function appears to come to a point. f f a. f ( a) lim ( ) ( ) a a 0 000 lim (. ) / / + + 0. 9 (( 0 + 0. 000) + 0. 9) a 0 lim ( + 0. / 000) 0. 0 a f() is differentiable at 0, and the equation of the tangent line is y.. The graph of the function at that point seems to become the graph of a straight line with repeated zooming.. The graph will eventually look like the tangent line. Eploration Using Newton s Method on Your Calculator See tet page 7.. f ( ) ( )( e ) + ( e )( ) e e f (0) The lines passes through (0, ) and has slope. Its equation is y +. 6. f ( ) ( )( e ) + ( e )( ) e e f ( ) e ( e ) e The lines passes through (, e + ) and has slope e. Its equation is y e( + ) + ( e+ ), or y e+ e+. 7. (a) + 0 (b) e + e + 0 e ( e + ) e + 0. 68 e
Section. 99 8. f ( ) f () () Since f() and f (), the graph of g() passes through (, ) and has slope. Its equation is g() ( ) + (), or g(). f( ) g ( ) 0.7.7.7 0.8.688.8 0.9.87.9.0..069...07...00. 9. f ( ) cos f (. ) cos. Since f (.) sin. and f (.) cos., the tangent line passes through (., sin.) and has slope cos.. Its equation is y (cos. )(. ) + sin., or approimately y 0. 07+ 0. 89 [0, ] by [ 0.,.] 0. For >, f ( ), and so f ( ). Since f( ) and f ( ), the tangent line passes through (, ) and has slope. Its equation is y ( ) +, or y. (b) Since f (.) 8.06 and L(.) 8, the approimation differs from the true value in absolute value by less than 0.. (a) f ( ) ( ) + 9 + 9 We have f( ) and f ( ). L( ) f( ) + f ( )( ( )) ( + ) 9 + (b) Since f(. 9). 90 and L(. 9). 9, the approimation differs from the true value by less than 0.. (a) f ( ) We have f() and f () 0. L ( ) f() + f ()( ) + 0( ) (b) Since f (.).009 and L(.), the approimation differs from the true value by less than 0. (a) f ( ) + We have f( 0) 0 and f ( 0). L ( ) f( 0) + f ( 0)( 0) 0+ (b) Since f ( 0.) 0.09 and L( 0. ) 0. the approimation differs from the true value by less than 0.. (a) f ( ) sec We have f( ) 0 and f ( ). L ( ) f( ) + f ( )( ) 0+ ( ) (b) Since f( + 0. ) 0. 00 and L( + 0. ) 0., the approimation differs from the true value in absolute value by less than 0.. 6. (a) f ( ) Section. Eercises. (a) f ( ) We have f () 7 and f ( ) 0. L( ) f( ) + f ( )( ) 7+ 0( ) 0 We have f( 0) and f ( 0). L ( ) f( 0) + f ( 0)( 0) + ( )( 0) +
00 Section. 6. Continued (b) Since f( 0. ). 706 and L( 0. ). 7080, the approimation differs from the true value in absolute value by less than 0. 7. f ( ) k( + ) k We have f( 0) and f ( 0) k. L ( ) f( 0) + f ( 0)( 0) + k ( 0) + k 00 00 8. (a) (. 00) ( + 0. 00) + ( 00)(. 0 00).; 00. 00. 0. 0 < 0 / (b). 009 ( + 0. 009) + ( 0. 009). 00; 6. 009. 00 9 0 < 0 6 6 9. (a) f( ) ( ) [ + ( )] + 6( ) 6 (b) f( ) [ + ( )] [ + ( )( )] + / (c) f( ) ( + ) + 0. (a) f( ) ( + ) / / + / + + / / (b) f( ) + + / + + / / (c) f( ) + + +. 00 + + 6+ / f ( 00) ( 00) 0. 0 f ( 00) 0 + 0. 0( 000) 0. 0. 7 / f ( 7) ( ) 7 7 f ( 7) + ( / 7)( 6 7) y. 96 7. 000 / f ( 000) ( ) 000 00 y 0 + ( / 00)( 000) y 0 999. 0. 8 / f ( 8) ( ) 8 8 y 9 + ( 80 8) 8 y 9 89. 8. Let f( ) +. Then f ( ) + and f ( ) + n n n. n+ n n f ( ) + n Note that f is cubic and f is always positive, so there is eactly one solution. We choose 0. 0 07. 0. 68606 0. 6896 0. 6878 6 0. 6878 7 Solution: 0. 688. 6. Let f( ) +. Then f ( ) + and f ( ) + n n n n+ n n f ( ) n + n The graph of y f( )shows that f( ) 0 has two solutions..... 696. 6. 607. 669. 60. 669. 60 Solution:. 67,. 60 7. Let f( ) + sin. Then f ( ) cos and f ( ) + sin n n+ n n f ( ) cos n n n n n n n
Section. 0 7. Continued The graph of y f( )shows that f( ) 0 has two solutions 0. 0. 8699. 9698 0. 869. 9669 0. 8669. 96690 0. 8669. 96690 Solutions: 0. 867,. 9669 8. Let f( ). Then f ( ) and f ( ) n n. n+ n n f ( ) n Note that f( ) 0 clearly has two solutions, namely ±. We use Newton s method to find the decimal equivalents... 78. 9798. 8988. 8907. 8907 6 Solutions: ±. 8907 9. (a) Since dy d (b) At the given values,, dy ( ) d. n dy ( i )( 0. 0) 9( 0. 0) 0.. 0. (a) Since dy d, ( + )( ) ( )( ) ( + ) ( + ) dy d. ( + ) (b) At the given values, dy ( ) 0 8 (. ) ( 0. ) [ + ( ) ] 0. 0.. (a) Since dy d ( ) + (ln )( ) ln +, dy ( ln + ) d. (b) At the given values, dy [ ( ) ln( ) + ]( 00. ) 00 (. ) 00.. (a) Since dy d ( ) ( ) + ( )( ) + + ( ) dy d. (b) At the given values, dy ( 0) ( 0). (a) Since dy d sin (b) At the given values, e cos, dy (cos ) e d. sin ( 0. ) 0.. sin dy (cos )( e )( 0. ) ( )( )( 0. ) 0... (a) Since dy d csc cot csc cot, dy csc cot d. (b) At the given values, dy csc cot (. ) 0 0. csc cot 0. 0. (a) y+ y 0 y( + ) y + Since dy ( + )() ( )() d ( + ) ( + ), d dy. ( + ) (b) At the given values, 00. dy ( 0+ ) 00.. 6. (a) y y dy d dy yd dy( + ) ( y) d y dy + d (b) At the given values, and y from the original ( ) equation, dy + ( 0. 0) 0. 07,
0 Section. 7. dy d 8. dy d 9. dy d 0. dy d dy d dy d e + dy ( e + ) d tan d du tan u d + u d u du d dy + 6 d 8 ( 8 + ) d d a a a (ln ) dy ( 8 ln 8+ 8 7 ) d. (a) Δ f f( 0. ) f( 0) 0. 0 0. (b) Since f ( ) +, f ( 0). Therefore, df d 0 (. ) 0.. (c) Δf df 0. 0. 00.. (a) Δ f f(.) f() 0. 0 0. (b) Since f ( ), f ( ). Therefore, df d 0 (. ) 0.. (c) Δf df 0. 0 0. 0 0. (a) Δ f f( 0. ) f( 0. ) (b) Since f ( ), f ( 0. ). Therefore, df d ( 00. ) 0. (c) Δf df +. (a) Δ f f(. 0) f(). 00600 0. 00600 (b) Since f ( ), f ( ). Therefore, df d ( 00. ) 00... Note that dv r, dv r dr. When r changes from dr a to a + dr, the change in volume is approimately a dr. 6. Note that ds 8r, so ds 8rdr. When r changes from dr a to a + dr, the change in surface area is approimately 8adr. 7. Note that dv, so dv d. When changes from d a to a + d, the change in volume is approimately a d. 8. Note that ds, so ds d. When changes from d a to a + d, the change in surface area is approimately a d. 9. Note that dv rh, so dv rh dr. When r changes dr from a to a + dr, the change in volume is approimately 0. Note that ds r, so ds rdh. When h changes from dh a to a + dh, the change in lateral surface area is approimately r dh.. A r da r dr da ( 0)( 0. ) 6. in. ν r dv r dr dv ()(.) 8 0 in. ν s dv s ds dv ( ) ( 0. ) cm. A s da s ds da 0 0 87 ( )(. ). cm. (a) Note that f ( 0) cos 0. L ( ) f( 0) + f ( 0)( 0) + + (b) f( 0. ) L( 0. ). (c) Δf df 0. 00600 0. 0 0. 000600
Section. 0. Continued (c) The actual value is less than.. This is because the derivative is decreasing over the interval [0, 0.], which means that the graph of f () is concave down and lies below its linearization in this interval. da 6. (a) Note that A r and r, so da rdr. dr When r changes from a to a + dr, the change in area is approimately a dr. Substituting for a and 0.0 for dr, the change in area is approimately ( )( 0. 0) 0. 08 0. (b) da A 008. 00. % 7. Let A cross section area, C circumference, and D diameter. Then D C, so dd dc and dd D C C dc. Also, A, so da C C and da dc. When C increases from dc 0 in. to 0 + in. the diameter increases by dd ( ) 0. 666 in. and the area increases by 0 approimately da 0 ( ) in. 8. Let edge length and V volume. Then V, and so dv d. With 0 cm and d 0.0 0. cm, we have V 0 000 cm and dv (0) (0.) 0 cm, so the percentage error in the volume measurement is approimately dv V 0 00. 000 %. 9. Let side length and A area. Then A and da, so da d. We want da 00. A, which d gives d 0. 0, or d 0. 0. The side length should be measured with an error of no more than %. For θ 7 radians, we have dθ < 00. sin cos 00. radian. The angle should be measured with an error of less than 0.0 radian (or approimately 0.7 degrees), which is a percentage error of approimately 0.76%. 0. (a) Note that V r h 0r. D,where D is the interior diameter of the tank. Then dv D, dd so dv DdD. We want dv 00. V,which gives DdD 00. (. D ),or dd 0. 00D.The interior diameter should be measured with an error of no more than 0.%. (b) Now we let D represent the eterior diameter of the tank, and we assume that the paint coverage rate (number of square feet covered per gallon of paint) is known precisely. Then, to determine the amount of paint within %, we need to calculate the lateral surface area S with an error of no more than %. Note that ds S rh 0D, so 0 and ds 0dD. We dd want ds 00. S, which gives 0dD 0. 0( 0D), or dd 00. D. The eterior diameter should be measured with an error of no more than %.. Note that V r h, where h is constant. Then dv rh. dr The percent change is given by dv rhdr dr r 0.% 0.%. V r h r r. Note that dv dh dv h, so dv h dh. We want 00. V, which gives hdh 0. 0( h), or dh 00h.. The height should be measured with an error of no more than %.. Since V r, we have r dv r dr r 6. The volume error in each case is simply r in. Sphere Type True Radius Tape error Radius Error Volume Error Orange in. / 8in. / 6 in. in. Melon in. / 8in. / 6 in. in. Beach Ball 7 in. / 8in. / 6 in.. in.
0 Section. r. Since A r, we have da 8 rdr 8 r 6. The surface area error in each case is simply r in. Sphere Type True Radius Tape Error Radius Error Volume Error Orange in. / 8in. / 6 in. in. Melon in. / 8 in. / 6 in. in. Beach Ball 7 in / 8in. / 6 in.. in.. We have dw dg bg, so dw bg dg. Then dw moon b dg (.) 7. 87. The ratio is dw earth b( ) dg. about 7.87 to. dt 6. (a) Note that T L g,so L g and dg dt L g dg. (b) Note that dt and dg have opposite signs. Thus, if g increases, T decreases and the clock speeds up. (c) L g dg dt dg ( 00) ( 980) 0. 00 dg 0. 976 Since dg 0. 976, g 980 0. 976 979. 0. 7. True. A look at the graph reveals the problem. The graph decreases after toward a horizontal asymptote of 0, so the -intercepts of the tangent lines keep getting bigger without approaching a zero. 8. False. By the product rule, d( uv) udv + vdu. 9. B. f( ) e f ( ) e L ( ) e+ e( ) L ( ) e 60. A. y tan dy (sec ) d (sec ) 0. dy 0. 6. D. f( ) + f ( ) n n + n+ n n () + () ( ) + 8 ( ) 6. A. f( ), 6 f ( 6) ( ) 6 8 66 + ( 66 6) 8 66. 0 The calculator returns.0, or a 0.0% difference. f( ) 0 6. If f ( ) 0, we have. f ( ) f ( ) Therefore, equal to., and all later approimations are also 6. If h, then f ( ) and h h h h h h. If h, then h h f ( ) and h h+ h h h h 6. Note that f ( ) and so / f ( ) n n n+ n n n n f ( ) / n. For n n, we have,, 8, and n 6; n. [ 0, 0] by [, ] 66. (a) i. Qa ( ) f( a) implies that b f ( a ). 0 ii. Since Q ( ) b + b ( a), Q ( a) f ( a) implies that b f ( a).
Section. 0 66. Continued iii. Since Q ( ) b, Q ( a) f ( a) implies that b f a ( ) f a In summary, b f a b f a b ( ) ( ), ( ), 0 and. (b) f( ) ( ) f ( ) ( ) ( ) ( ) f ( ) ( ) ( ) ( ) Since f( 0), f ( 0), and f ( 0), the coefficients are b, b, and b. The quadratic approimation 0 is Q ( ) + +. (c) As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (d) g ( ) g ( ) g ( ) Since g(), g (), and g (), the coefficients are b b,, and b o. The quadratic approimation is Q ( ) ( ) + ( ). As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical (e) h ( ) ( + ) / / h ( ) ( + ) h ( ) ( + ) / Since h( 0), h ( 0), and h ( 0), the coefficients are b b b,, and. 0 8 The quadratic approimation is Q ( ) +. 8 As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (f) The linearization of any differentiable function u() at a is L( ) u( a) + u ( a)( a) b0 + b( a), where b0 and bare the coefficients of the constant and linear terms of the quadratic approimation. Thus, the linearization for f () at 0 is + ; the linearization for g() at is ( ) or ; and the linearization for h() at 0 is +. 67. Finding a zero of sin by Newton s method would use the recursive formula sin( ) n n+ n n n tan, and that cos( ) n is eactly what the calculator would be doing. Any zero of sin would be a multiple of. 68. Just multiply the corresponding derivative formulas by d. (a) Since d () d c 0, d () c 0. d (b) Since d ( cu) c du dcu cdu d, ( ). d (c) Since d ( u+ v) du dv du v du dv d + d, ( + ) + (d) Since (e) Since (f) Since d d ( u i v) udv v du d u v u dv v du d + d, ( i ) +. d u d v v du d v u dv d d u vdu udv v,. v d d u n nu ndu n n, d( u ) nu du. d tan sin / cos 69. lim lim 0 0 sin lim 0 cos sin lim lim 0 cos 0 ()().
06 Section.6 70. ga ( ) c, so if Ea ( ) 0, then ga ( ) f( a) and c f( a). Then E( ) f( ) g( ) f( ) f( a) m( a). Thus, E ( ) f( ) f( a) m. a a f( ) f( a) E ( ) lim f ( a), so lim a a a a f ( a) m. Therefore, if the limit of E ( ) is zero, then m f ( a) and a g () L(). 7. f ( ) + cos + We have f( 0) and f ( 0) L( ) f( 0) + f ( 0)( 0) + The linearization is the sum of the two individual linearizations, which are for sin and + for +. Section.6 Related Rates (pp. 6 ) Eploration Sliding Ladder. Here the -ais represents the ground and the y-ais represents the wall. The curve (, y ) gives the position of the bottom of the ladder (distance from the wall) at any time t in 0 t. The curve (, y ) gives the position of the top of the ladder at any time in 0 t.. 0 t. This is a snapshot at t.. The top of the ladder is moving down the y-ais and the bottom of the ladder is moving to the right on the -ais. The end of the ladder is accelerating. Both aes are hidden from view. 6. dy T 0 ( T) 7. y (). ft/sec. The negative number means the ladder is falling. 8. Since lim y () t, the speed of the top of the ladder t ( / ) is infinite as it hits the ground. Quick Review.6. D ( 7 0) + ( 0 ) 9 + 7. D ( b 0) + ( 0 a) a + b. Use implicit differentiation. d d ( y + y ) ( d d + y ) dy y() y dy dy + + () + d d d dy ( + y ) y d dy y d + y. Use implicit differentiation. d d y d ( sin ) ( y) d ( )(cos y) dy + (sin y)( ) dy y() d d ( + cos y) dy ysiny d dy y sin y d + cos y dy y+ sin y d + cos y. Use implicit differentiation. d d d tan y d sec y dy d dy d sec y dy cos y d 6. Use implicit differentiation. d d ln( + y) ( ) d d dy + + y d dy + ( + y) d dy + y d 7. Using A(, ) we create the parametric equations + at and y + bt, which determine a line passing through A at t 0. We determine a and b so that the line passes through B(, ) at t. Since + a, we have a 6, and since + b, we have b. Thus, one parametrization for the line segment is + 6t, y t, 0 t. (Other answers are possible.)
Section.6 07 8. Using A(0, ), we create the parametric equations 0 + at and y + bt, which determine a line passing through A at t 0. We now determine a and b so that the line passes through B(, 0) at t. Since 0 + a, we have a, and since 0 + b, we have b. Thus, one parametrization for the line segment is t, y + t, 0 t. (Other answers are possible.) 9. One possible answer: t 0. One possible answer: t Section.6 Eercises. Since da. Since ds. (a) Since dv (b) Since dv da dr dr, we have da ds dr dr, we have ds dv dh dh, we have dv dv dr dr, we have dv (c) dv d d r h r ( h ) dv r dh h r dr + ( ) dv r dh + rh dr. (a) dp d ( RI ) dp R d I I dr + dp R I di dr I + dp RI di dr + I r dr, 8 r dr. r dh. rh dr. (b) If P is constant, we have dp 0, which means RI di dr dr R di P di + I 0, or. I I. ds d + y + z ds d ( + y + z ) + y + z ds + y + z ds d + y dy + + y + z d y dy z dz + + z dz 6. da d ab sinθ da da b a db d + + ab i i sinθ i i sinθ i sinθ da da db d bsinθ + asinθ + abcosθ θ 7. (a) Since V is increasing at the rate of volt/sec, dv volt/sec. (b) Since I is decreasing at the rate of di amp/sec, amp/sec. (c) Differentiating both sides of V IR, we have dv I dr R di +. (d) Note that V IR gives R, so R 6 ohms. Now substitute the known values into the equation in (c). dr + 6 dr dr ohms/sec R is changing at the rate of ohms/sec. Since this value is positive, R is increasing. 8. Step : r radius of plate A area of plate Step : At the instant in question, dr 00. cm/sec, r 0cm. Step : We want to find da. Step : A r Step : da r dr Step 6: da 0 0 0 ( )(. ) cm /sec At the instant in question, the area is increasing at the rate of cm / sec.
08 Section.6 9. Step : l length of rectangle w wih of rectangle A area of rectangle P perimeter of rectangle D length of a diagonal of the rectangle Step : At the instant in question, dl dw cm/sec, cm/sec, l cm, and w cm. Step : We want to find da dp, dd, and. Steps,, and 6: (a) A lw da l dw + da w dl ( )( ) + ( )( ) cm /sec The rate of change of the area is cm /sec. (b) P l+ w dp dl dw + dp ( ) + ( ) 0 cm/sec The rate of change of the perimeter is 0 cm/sec. (c) D l + w dd l dl w dw l dl + w dw + l + w l + w dd ( )( ) + ( )( ) cm/sec + The rate of change of the length of the diameter is 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. 0. Step :, y, z edge lengths of the bo V volume of the bo S surface area of the bo s diagonal length of the bo Step : At the instant in question, d dy dz m/sec, m/sec, m/ sec, m, y m, and z m. Step : We want to find dv ds, ds, and. Steps,, and 6: (a) V yz dv y dz + z dy + yz d dv ( )( )( ) + ( )( )( ) + ( )( )( ) m /sec The rate of change of the volume is m /sec. (b) S ( y+ z+ yz) ds dy y d dz z d y dz z dy + + + + + ds [( )( ) + ( )( ) + ( )( ) + ( )( ) + ()() + ()() + ()( )] 0m /sec The rate of change of the surface area is 0 m /sec. (c) s + y + z ds d y dy z dz + + + y + z dy d + y dy + z dz + y + z ds ( )( ) + ( )( ) + ( )( ) 0 0 m/sec + + 9 The rate of change of the diagonal length is 0 m/sec.. Step : r radius of spherical balloon S surface area of spherical balloon V volume of spherical balloon Step : At the instant in question, dv 00 ft /min and r ft. Step : We want to find the values of dr and ds. Steps,, and 6: (a) V r dv r dr dr 00 () dr ft/min The radius is increasing at the rate of ft/min.
Section.6 09. Continued (b) S r ds 8r dr ds 8 ()() ds 0 ft / min The surface area is increasing at the rate of 0 ft /min.. Step : r radius of spherical droplet S surface area of spherical droplet V volume of spherical droplet Step : No numerical information is given. Step : We want to show that dr is constant. Step : dv S r, V r, ks for some constant k Steps and 6: Differentiating V r, we have dv r dr. Substituting ks for dv and S for r, we have ks S dr, or S dr k.. Step : s (diagonal) distance from antenna to airplane horizontal distance from antenna to airplane Step : At the instant in question, ds s 0 mi and 00 mph. Step : We want to find d. Step : + 9 s or s 9 Step : d s 9 Step 6: d 0 0 9 s ds s s 9 000 ( 00) mph 0. 08 mph The speed of the airplane is about 0.08 mph.. Step : s length of kite string horizontal distance from Inge to kite Step : At the instant in question, d ft/sec and s 00 ft Step : We want to find ds. Step : + 00 s Step : d s ds or d s ds Step 6: At the instant in question, since s 00 00 00 00. + 00 s, we have ds ds Thus ( 00)( ) ( 00), so, so ds 0 ft/sec. Inge must let the string out at the rate of 0 ft/sec.. Step : The cylinder shown represents the shape of the hole. r radius of cylinder V volume of cylinder Step : At the instant in question, dr 0. 00 in. in./min min 000 and (since the diameter is.800 in.), r.900 in. Step : We want to find dv. Step : V r () 6 6r Step : dv r dr
0 Section.6. Continued Step 6: dv 9 (. 900) 000 00 0. 0076 0. 09 in /min. The volume is increasing at the rate of approimately 0.09 im /min. 6. Step : r base radius of cone h height of cone V volume of cone Step : At the instant in question, h m and dv Step : We want to find dh and dr. Step : Since the height is 8 h r 8 ( ) or r h. 0 m /min. of the base diameter, we have 6 h We also have V r h h h. We will 7 h use the equations V 6 and r h 7. Step and 6: (a) dv h dh 6 9 6( ) dh 0 9 dh 8 m/ min cm/ min The height is changing at the rate of 9. cm / min. (b) Using the results from Step and part (a), we have dr dh 7 8 cm/min. The radius is changing at the rate of 7 9 8. cm/ min. 7. Step : 6 m h m r r radius of top surface of water h depth of water in reservoir V volume of water in reservoir Step : At the instant in question, dv 0 m /min and h m. Step : We want to find dh and dr. Step : Note that h r 6 by similar cones, so r 7. h. Then V r h ( 7. h) h 8. 7h Steps and 6: (a) Since V h dv 8. 7, 6. h dh. Thus 0 6. ( dh ), and so dh 8 m/min 9 cm/min. The water level is falling by 9. cm/min. (Since dh < 0, the rate at which the water level is falling is positive.) (b) Since r h dr dh 80 7., 7. cm/min. The rate of change of the radius of the water s surface is 80 89. cm/min. 8. (a) Step : y depth of water in bowl V volume of water in bowl Step : At the instant in question, dv 6 m /min and y 8 m. Step : We want to find the value of dy. Step : V y 9 y ( ) or V y y
Section.6 8. Continued Step : dv dy ( 6yy ) Step 6: 6 6() 8 ( 8 ) dy dy 6 dy 0. 06 m/min or. 6 cm/min 6 (b) Since r + ( y), r 69 ( y) 6y y. (c) Step : y depth of water r radius of water surface V volume of water in bowl Step : At the instant in question, dv 6 m /min, and therefore (from part (a)) dy m/min. Step : We want to find the value of dr. Step : From part (b), r 6y y. Step : dr 6y y ( 6 y) dy y 6y y dy y 8 m, Step 6: dr 8 6() 8 8 0. 00 m/min 88 or 0. cm/min 7 9. Step : distance from wall to base of ladder y height of top of ladder A area of triangle formed by the ladder, wall, and ground θ angle between the ladder and the ground Step : At the instant in question, ft and d ft/sec. Step : We want to find dy da dy, d, and θ. Steps,, and 6: (a) + y 69 d + y dy 0 To evaluate, note that, at the instant in question, y 69 69. dy Then ( )() + () 0 dy dy ft/sec or ft/sec The top of the ladder is sliding down the wall at the rate of ft/sec. (Note that the downward rate of motion is positive.) (b) A y da dy y d + Using the results from step and from part (a), we have da + 9 [( )( ) ( )( )] ft /sec. The area of the triangle is changing at the rate of 9. ft /sec. (c) tanθ y sec θ dθ dy y d Since tan θ, we have for 0 < and so θ cosθ sec θ 69. Combining this result with the results from step and from part (a), we have 69 dθ ( )( ) ()(), so dθ radian/sec. The angle is changing at the rate of radian/sec. 0. Step : h height (or depth) of the water in the trough V volume of water in the trough Step : At the instant in question, dv. ft / min and h ft.
Section.6 0. Continued Step : We want to find dh. Step : The wih of the top surface of the water is h, so we have V ( h) h V h ( ), or 0 Step : dv 0h dh Step 6: dh. 0( ) dh 0. 06 ft/min 6 The water level is increasing at the rate of 6 ft/min.. Step : l length of rope horizontal distance from boat to dock θ angle between the rope and a vertical line Step : At the instant in question, dl ft/sec and l 0 ft. Step : We want to find the values of d and d θ. Steps,, and 6: (a) l 6 d l dl l 6 d 0 ( ). ft/sec 0 6 The boat is approaching the dock at the rate of. ft/sec. (b) θ cos 6 l dθ 6 dl t 6 l dθ 06. 6 0 ( ) radian/sec 0 The rate of change of angle θ is 0 radian/sec.. Step : distance from origin to bicycle y height of balloon (distance from origin to balloon) s distance from balloon to bicycle Step : We know that dy is a constant ft/sec and d is a constant 7 ft/sec. Three seconds before the instant in question, the values of and y are 0 ft and y 6 ft. Therefore, at the instant in question ft and y 68 ft. Step : We want to find the value of ds at the instant in question. Step : s + y Step : ds + y d y dy d + y dy + + y Step 6: ds ( )( 7) + ( 68)( ) ft/sec + 68 The distance between the balloon and the bicycle is increasing at the rate of ft/sec. dy dy d. d 0 d 0( + ) ( ) + ( ) Since d cm/sec, we have dy 60 cm/sec. + ( ). (a) dy 60( ) [ + ( ) ] (b) dy 60( 0) ( + 0 ) (c) dy 60( 0) ( + 0 ) dy 0 0 cm/sec dy d d ( ) d cm/sec 0. 0076 cm/sec Since d dy cm/sec, we have (a) dy 86 ( ) 6 cm/sec (b) dy 8 6 ( ) cm/sec (c) dy 8 6 ( ) 88 cm/sec 86 cm/sec.
Section.6. Step : y Step : We want to find d θ, (, y) -coordinate of particle s location y y-coordinate of particle s location θ angle of inclination of line joining the particle to the origin. Step : At the instant in question, d 0 m/sec and m. Step : We want to find d θ. Step : Since y y, we have tan θ and so, for > 0, θ tan. Step : dθ d + Step 6: dθ + ( 0) radian/sec The angle of inclination is increasing at the rate of radian/sec. 6. Step : y (, y) -coordinate of particle s location y y-coordinate of particle s location θ angle of inclination of line joining the particle to the origin Step : At the instant in question, d 8m/secand m. Step : y / Since y, we have tan θ ( ), and so, for < 0, / / θ + tan [ ( ) ] tan ( ). Step : dθ d / + ( ( ) ( ) / [( ) ] d / ( ) d ( ) Step 6: dθ 8 ( ) ( ) radian/sec The angle of inclination is increasing at the rate of radian/sec. 7. Step : r radius of balls plus ice S surface area of ball plus ice V volume of ball plus ice Step : At the instant in question, dv 8mL/min 8cm /minandr ( 0) 0 cm. Step : We want to find ds. Step : We have V r and S r. These equations can be V combined by noting that r / / V, so S Step : ds V / dv V / dv Step 6: 000 Note that V ( 0). ds / 000 8 6 i ( ).6cm /min 000 Since ds < 0, the rate of decrease is positive. The surface area is decreasing at the rate of.6 cm /min.
Section.6 8. Step : -coordinate of particle y y-coordinate of particle D distance from origin to particle Step : At the instant in question, m, y m, d dy m/sec, and m/sec. Step : We want to find dd. Step : D + y Step : dd + y d y dy d + y dy + + y Step 6: dd ()( ) + ( )( ) m/sec + The particle s distance from the origin is changing at the rate of m/sec. 9. Step : Street light 6 ft 6 ft Shadow s distance from streetlight base to man s length of shadow Step : At the instant in question, d ft/secand 0ft. Step : We want to find ds. Step : By similar triangles, s s+. This is equivalent to 6 6 6s 6s+ 6, or s. Step : ds d Step 6: ds ( ) ft/sec The shadow length is changing at the rate of ft/sec. 0. Step : s distance ball has fallen distance from bottom of pole to shadow Step : At the instant in question, s 6 ft and ds 6ft/sec. Step : We want to find d. Step : By similar triangles, 0 0 s. This is equivalent to 0 0 00 0 s, or s 00. We will use 00s. Step : d ds 00s Step 6: d 00( ) ( 6) 00 ft/sec The shadow is moving at a velocity of 00 ft/sec.. Step : position of car ( 0 when car is right in front of you) θ camera angle. (We assumeθ is negative until the car passess in front of you, and then positive.) Step : At the first instant in question, 0 ft and d 6 ft/sec. A half second later, d ( 6) ft and 6 ft/sec. Step : We want to find d θ at each of the two instants. Step : θ tan Step : dθ + i d
Section.6. Continued Step 6: dθ When 0 : + ( 6) radians/sec 0 dθ When : + ( 6) radians/sec. Step : p -coordinate of plane s position -coordinate of car s position s distance from plane to car (line-of-sight) Step : At the instant in question, dp ds p 0, 0mph, s mi, and 60mph. Step : We want to find d. Step : ( p) + s Step : d dp ( p) s ds Step 6: Note that, at the instant in question, mi. d ( 0) 0 ( )( 60) d 8 0 600 d 0 00 d 80 mph The car s speed is 80 mph.. Step : s shadow length θ sun s angle of elevation Step : At the instant in question, dθ s 60 ft and 0. 7 / min 0. 00 radian/min. Step : We want to find ds. Step : 80 tanθ or s 80 cotθ s Step : ds d 80csc θ θ Step 6: Note that, at the moment in question, since tan 80 θ and 0 < θ <, we have sinθ and so 60 cscθ. ds 80 ( 0. 00 ) 0. 87 ft min i in ft. in./min 7. in./min Since ds < 0, the rate at which the shadow length is decreasing is positive. The shadow length is decreasing at the rate of approimately 7. in./min.. Step : a distance from origin to A b distance from origin to B θ angle shown in problem statement Step : At the instant in question, da db m / sec, m/sec, a 0m, and b 0m. Step : We want to find d θ. Step : a a tanθ or θ tan b b Step : dθ a + b b da a db b da a db b a + b Step 6: dθ ( 0)( ) ( 0)( ) 0. radian /sec 0 + 0 7. degrees/sec To the nearest degree, the angle is changing at the rate of 6 degrees per second.
6 Section.6. Step : A a c 0 O b B a distance from O to A b distance from O to B c distance from A to B Step : At the instant in question, a nautical miles, b nautical miles, da db knots, and knots. Step : We want to find dc, Step : Law of Cosines : c a + b abcos0 Step : c a + b + ab c dc a da + b db + a db + b da Step 6: Note that, at the instant in question, c a + b + ab () + () + ()() 9 7 dc () 7 ()( ) ()( ) ()( ) ()( ) + + + dc dc 9. knots The ships are moving apart at a rate of 9. knots. 6. True. Since dc dr, a constant dr dc results in a constant. 7. False. Since da 8. A. v s dv s ds s ( ) s in r dr da, the value of depends on r. 9. E. sa 6s dsa sds sds ds s V s dv s ds s s s s 8in 0. C. d dy y 06. 08. dy dy., but it is negative because y is decreasing. dy... B. v r h sa rh dv r dh dsa hdr dv dsa r dh hdr dh dr h r dr 00 () cm dr. 0 sec. (a) (b) dc dc d ( 6 + ) d ( + ) [() () + ](.) 0 0. dr d d ( 9) 9 9( 0. ) 09. dp dr dc 09. 0. 06. d 6 + d (.) (.) ( 00. ).. 6
Section.6 7. Continued (b) dr d d ( 70) 70 70( 0. 0). dp dr dc. (. 6 ). 06. (a) Note that the level of the coffee in the cone is not needed until part (b). Step : V volume of coffee in pot y depth of coffee in pot Step : dv Step : 0 in / min We want to find the value of dy. Step : V y 9 Step : dv Step 6: dy 9 dy 0 9 dy 0 0. in./min 9 The level in the pot is increasing at the rate of approimately 0. in./min. (b) Step : V volume of coffee in filter r radius of surface of coffee in filter h depth of coffee in filter Step : At the instant in question, dv 0 in / min and h in. Step : We want to find dh. Step : Note that r h 6, so r Then V h. h r h. Step : dv h dh Step 6: ( ) dh 0 dh 8 in./min Note that dh < 0, so the rate at which the level is falling is positive. The level in the come is falling at the 8 rate of 0 09. in./min.. Step : Q rate of CO ehalation (ml/min) D difference between CO concentration in blood pumped to the lungs and CO concentration in blood returning from the lungs (ml/l) y cardiac output Step : At the instant in question, Q ml/min, D ml/l, dd Step : (ml/l)/min, and dq 0 We want to find the value of dy. Step : Q y D Step : dy Step 6: D dq Q dd D ml/ min. dy ( )( 0) ( )( ) 66. ( ) 68 0 77 L/min The cardiac output is increasing at the rate of approimately 0.77 L/min.. (a) The point being plotted would correspond to a point on the edge of the wheel as the wheel turns. (b) One possible answer isθ 6t, where t is in seconds. (An arbitrary constant may be added to this epression, and we have assumed counterclockwise motion.)
8 Section.6. Continued (c) In general, assuming counterclockwise motion: d d sinθ θ (sin θ)( 6) sinθ dy d cos θ θ (cos θ)( 6) cosθ At θ : d sin 6 ( ) 7. 086 ft/sec dy cos 6 ( ) 7. 086 ft/sec At θ : d sin 00. ft/sec dy cos 0 ft/sec At θ : d sin 0 ft/sec dy cos 00. ft/sec 6. (a) One possible answer: 0 cos θ, y 0 + 0 sin θ (b) Since the ferris wheel makes one revolution every 0 sec, we may let θ 0. t and we may write 0cos 0. t, y 0 + 0 sin 0. t. (This assumes that the ferris wheel revolves counterclockwise.) In general: d 0(sin 0. t)( 0. ) 6sin 0. t dy 0(cos 0. t)( 0. ) 6cos 0. t 7. (a) dy At t : d 6sin 0 ft/sec dy 6cos 6( ) 8. 80 ft/sec At t 8 : d 6sin. 6 7. 97 ft/sec dy 6 cos. 6. 8 ft/sec d ( uv ) u dv + v du u( 00. v) + v( 00. u) 009. uv 009. y Since dy 009y.,the rate of growth of total production is 9% per year. (b) dy d ( uv ) u dv + v du u( 00. v) + v( 00. u) 00. uv 00. y The total production is increasing at the rate of % per year. Quick Quiz Sections..6. B. f ( ) n+ n f ( ) f( ) + f ( ) +. B. z + y () + () () + + 0. 6 + z + z dz d + y dy dy dy + dy d dy. A. t () 70 yt () 60t / zt () (( 60t) + 70 ) dz ( t + 600 / 900) ( 700t) dz 700( ) / ( 600( ) + 900) dz 7. 6. (a) f( ) f ( ) ( ) 0 6 + ( 6 ). 0
Chapter Review 9. Continued (b) f ( ) f 6 0 n+ n f ( ), ( ) () 6. () (c) f( ) f 7 7 ( ) ( ) 7 6 + ( 6 7) 7 6.96 d / y [ e ( + )] d / ( e / )( + ) + ( + )( e )( ) / ( e )( + + ) / e ( + ) / e [( 0. ) +. 7] The second derivative is always positive (where defined), so the function is concave up for all 0. Graphical support: Chapter Review (pp. 6 60). y y ( ) + ( )( ) + ( ) The first derivative has a zero at. 6 Critical point value: y 09. 9 Endpoint values: y y 0 The global maimum value is 6 at, and the global 9 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 (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 down
0 Chapter Review. Continued Graphical support: y e The second derivative is always positive, so the function is concave up for all. Graphical support: (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 0. 8. Intervals < 0.8 0.8 < Sign of y + Behavior of y Increasing Decreasing y ( + 6 ) The second derivative is always negative so the function is concave down for all. Graphical support: (a) Approimately (, 0. 8] (a) [, ) (b) (, ] (c) (, ) (d) None (e) Local (and absolute) minimum at (, 0) (f) None 7. Note that the domain is (, ). / y ( ) / y ( ) ( ) / ( ) Intervals < < 0 0< < Sign of y + Behavior of y Decreasing Increasing / / ( ) ( ) ( )( ) ( ) ( ) y / ( ) / ( ) [ + ] / ( ) + 9/ ( ) The second derivative is always positive, so the function is concave up on its domain (, ). Graphical support: (b) Approimately [ 0. 8, ) (c) None (d) (, ) (e) Local (and absolute) maimum at ( 0. 8,. ) (f) None 6. y e Intervals < < Sign of y + Behavior of y Decreasing Increasing (a) [0, ) (b) (, 0] (c) (, ) (d) None (e) Local minimum at (0, ) (f) None
Chapter Review ( )( ) ( )( ) + 8. 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 Graphical support: down down up Graphical support: (a) None (b) [, ] (c) (, 0) (d) (0, ) (e) Local (and absolute) maimum at (, ); local (and absolute) minimum at (, 0) (f) 0, 0. This problem can be solved graphically by using NDER to obtain the graphs shown below. y y y (a) (, / ] (, 0.79] (b) [ /, ) [0.79, ) and (, ) (c) (, / ) (,.60) and (, ) (d) ( /, ) (.60, ) (e) Local minimum at, i ( 0. 79, 0. 9) / / (f) / /, i (. 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 / ( ) ( ) ( ) / Intervals < < 0 0< < Sign of y + Behavior of y up down An alternative approach using a combination of algebraic and graphical techniques follows. 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 ( + + ) ( + + )( ) ( + )( + ) ( + + ) 8 ( + + )
Chapter Review 0. Continued Using graphing techniques, the zeros of 8 (and hence of y ) are at. 8, 0. 706, and. 90. Intervals (,.8) Sign of y Behavior of y (.8, 0.706) (0.706,.90) (.90, ) + + down up down up (a) [, ] (b) (, ] and [, ) (c) Approimately (.8, 0.706) and (.90, ) (d) Approimately (,.8) and (0.706,.90) (e) Local maimum at, (. 7, 0. 8); local minimum at, (. 7, 0. 68) (f) (.8, 0.7), (0.706, 0.8), 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 (e) Local (and absolute) maima at (, ln ) and (, ln ) (f) None. y cossin Using graphing techniques, the zeros of y in the domain 0 are 0. 76, 0. 99, 7.,. 8, and. 96,. 8,,. 9 Intervals 0 < < 0.76 0.76 < < 0.99 0.99 < < < <.8.8 < <.96 Sign of y + + + Behavior of y Increasing Decreasing Increasing Decreasing Increasing Intervals.96 < <.8.8 < < < <.9.9 < Sign of y + + Behavior of y Decreasing Increasing Decreasing Increasing y 9sin 6 cos Using graphing techniques, the zeros of y in the domain 0 are 0.,. 66,. 876,. 600,.,. 8,. and 6. 000. Intervals 0 < < 0. 0. < <.66.66 < <.876.876 < <.600.600 < <. Sign of y + + Behavior of y down up down up down Intervals. < <.8.8 < <.. < < 6.000 6.00 < < Sign of y + + Behavior of y up Graphical support: down up down y The second derivative always negative, so the function is concave down on each open interval of its domain. Graphical support: (a) (0, ] (b) [, 0) (c) None (d) (, 0) and (0, )
Chapter Review. Continued (a) Approimately [0, 0.76], 0. 99,, [. 8,. 96],. 8,, and. 9, (b) Approimately [0.76, 0.99],,. 8, [. 96,. 8],,. 9 and (c) Approimately (0.,.66), (.876,.600), (,,.8), and (., 6.000) (d) Approimately (0, 0.), (.66,.876), (.600,.), (.8,.), and ( 6000, ) (e) Local maima at ( 0. 76,. 66),, 0 and (.96,.66),,, and (, ); local minima at ( 0, ), ( 0. 99, 0. ), (.8, 0.), (.8,.806), and (.9,.806) Note that the local etrema at. 8,, and. 9are also etrema. (f) (0., 0.7), (.66, 0.67), (.876, 0.67), (.600, 0.7), (., 0.9), (.8, 0.0), (., 0.0), and (6.000, 0.9). e y, < 0, > 0 Intervals < 0 0 < < < Sign of + (a) 0, (b) (, 0] and (c) (, 0) (d) ( 0, ), ) 6 (e) Local maimum at, (.,. 079) (f) None. Note that there is no point of inflection at 0 because the derivative is undefined and no tangent line eists at this point.. y + 7 + 0+ Using graphing techniques, the zeros of y are 0. 78 and. 69. Intervals <0. 78 0. 78 < <. 69. 69 < Sign of y + Behavior of y Decreasing Increasing Decreasing y 0 + + 0 Using graphing techniques, the zeros of y is. 079. Intervals <. 079. 079 < Sign of y + Behavior of y up down Graphical support: Behavior of y Decreasing Increasing Decreasing y e, > 0 6, < 0 Intervals < 0 0< Sign of y + Behavior of y up down Graphical support: (a) Approimately [ 0. 78,. 69] (b) Approimately (, 0. 78] and [. 69, ) (c) Approimately (,. 079) (d) Approimately (. 079, ) (e) Local maimum at (. 69, 0. 7); local minimum at ( 0. 78, 0. 97) (f) (. 079,. 60)
Chapter Review 9. y 8 / 9 / 8 y 9 Intervals < 0 0 < < 8 9 8 9 < Sign of y + 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, NDER (y, ),000,000 while y is actually undefined at. The graph of y + is shown below. Behavior of y Decreasing Increasing Decreasing 8 6 / 6 / ( + 9) y 6 / Intervals < 9 < < 0 9 0 < Sign of y + Behavior of y up down down ( )( + 8 ) ( + )( ) y ( ) + 0 6+ ( ) The graph of y is shown below. Graphical support: The zero of y is 0.. (a) 0, 8 9 (b) (, 0] and 8 9, ) (c), 9 (d) 9, 0 and ( 0, ) (e) Local maimum / 8 0 8 at, i ( 0. 889,. 0); 9 9 9 local minimum at (0, 0) (f) / 0, i, 9 9 9 9 0. 667 Intervals < 0. 0. < < < Sign of y + Behavior of y Increasing Decreasing Decreasing ( ) ( 6 + 06) ( + 0 6+ ) ( )( ) y ( ) ( )( 6 + 06) ( + 0 6+ ) ( ) ( 6 + ) ( ) The graph of y is shown below. The zero of 6 + (and hence of y ) is.70. Intervals < < <.70.70 < Sign of y + Behavior of y down up down
Chapter Review 6. Continued (a) Approimately (, 0.] (b) Approimately [0., ) and (, ) (c) Approimately (,.70) (d) (, ) and approimately (.70, ) (e) Local maimum at (0.,.7) (f) (.70,.0) 7. y 6( + )( ) Intervals < < < < Sign of y + + Behavior of y Decreasing Increasing Increasing y 6( + )( )( ) + 6( ) ( ) 6( )[( + ) + ( )] 8 ( ) Intervals < 0 0 < < < Sign of y + + Behavior of y up down up (a) There are no local maima. (b) There is a local (and absolute) minimum at. (c) There are points of inflection at 0 and at. 8. y 6( + )( ) Intervals < < < < Sign of y + + Behavior of y d y 6( ) 6( ) d Increasing Decreasing Increasing Intervals < < Sign of y + Behavior of y down up (a) There is a local maimum at. (b) There is a local maimum at. (c) There is a point of inflection at. 9. Since d e e d + f( ) e + C. 0. Since d sec sec tan, f( ) sec + C. d. Since d ln + +, d + + f( ) ln + + + C.. Since d / / + d +,, / / f( ) + + C.. f( ) cos+ sin+ C f ( ) + 0+ C C f( ) cos+ sin+ /. f( ) + + + + C f () 0 + + + + C 0 C / f( ) + + +. vt () s () t 98. t+ st () 9. t + t+ C s( 0) 0 C 0 st () 9. t + t+ 0 6. at () v () t vt () t+ C v( 0) 0 C 0 vt () s () t t+ 0 st () 6t + 0t+ C s( 0) C st () 6t + 0t+
6 Chapter Review 7. f( ) tan f ( ) sec L ( ) f f + + + tan sec + + + 8. f( ) sec f ( ) sectan L ( ) f f + sec + sec tan + ( ) + 9. f( ) + tan f ( ) ( + tan ) (sec ) cos ( + tan ) (cos + sin ) L ( ) f( 0) + f ( 0)( 0) ( 0) + 0. f( ) e + sin f ( ) e + cos L ( ) f( 0) + f ( 0)( 0) + ( 0) +. The global minimum value of occurs at.. (a) The values of y and y are both negative where the graph is decreasing and concave down, 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. y y f() 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., ] and differentiable on (0., ). (b) f ( ) ( ) + (ln )( ) + ln Using a 0. and b, we solve as follows. f() f(.) 0 f () c 0. ln 0. ln 0. + lnc. ln 0. 0. lnc. ln c 0. ln( 7 ) 0. c e ( 7 ) c e 8. 79 (c) The slope of the line is m f ( b ) f ( a ) 0. ln( 7 ) 0. ln 8, and the line b a passes through (, ln ). Its equation is y 0.(ln 8)( ) + ln, or approimately y. 7. 07.
Chapter Review 7 7. Continued (d) The slope of the line is m 0. ln 8,and the line passes through (, c f()) c ( e 8, e 8( + 0. ln 8)) (. 79, 0. 7). Its equation is y 0. (ln 8)( c) + f( c), y 0 8 e. ln ( 8) + e 8( + 0. ln 8), y 0.(ln 8) e 8, or approimately y. 7. 79. 8. (a) vt () s () t 6tt (b) at () v () t 66t (c) The particle starts at position moving in the positive direction, but decelerating. At approimately t 0.8, it reaches position.8 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 0. 0. (a) Since dy ( )( e ) + ( e )( ) + ( ) e, d dy ( ) e d. (b) dy [ ( ) ( ) ]( e )( 00. ) 00. e 0. 0068 600. 9. (a) With some rounding, y t + 7 7e 0.. 0678 (b) [0, 80] by [0, 600000] 600. 9 (c) y + 7. 7e 0. 0678( 80) + 89, 0, 0, 7 (d) Using the Second Derivative, we find the maimum rate of growth about 88. We find a point of inflection here, which shows the begining of a decline in the rate of growth. 600. 9 (e) y, 6, 000, which is the 0. 0678( ) + 7. 7e approiate maimum population. (f) There are many possible causes. Advances in transportation began drawing the population southward after 90, and Tennessee was well-situated grographically to become a crossroads of river, railroad, and automobile routes. By the year 000 there had been numerous other demographic changes. It should be pointed out that the census years in the data (8090) include the years of the Civil War and Reconstruction, so the regression is based on unusual data.. f( ) cos + f ( ) sin + f( ) n n+ n f ( ) n cos + n n n sin n + The graph of y f () shows that f () 0 has one solution, near. 0. 8688 0. 888 0. 88608 0. 88608 Solution: 0. 886. 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. n
8 Chapter Review. Let t represent time in seconds, where the rock is blasted upward at t 0. Since at () v () t 7m/sec. and v( 0) 9 m/sec, we have vt () 7. t+ 9. Since s () t 7. t+ 9and s( 0) 0, we have st () 86. t + 9t. Solving vt () 0, we find that the rock attains its maimum height at t sec and its height at that time is s( ) 6. m.. Note that s 00 r and the sector area is given by 6. s A r rs r r r r r ( 00 ) 0. To find the domain of Ar () 0rr, note that r > 0 and 0 0< s< r, which gives. < r < 0. Since + A ( r) 0 r, the critical point occurs at r. 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 ft and s 0 ft. y 7 (, 7 ) For 0< < 7, the triangle with vertices at (0, 0) and ( ±, 7 ) has an area given by A ( ) ( )( 7 ) 7. Since A 7 ( )( + ) and A 6,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( ) square units. 7. If the dimensions are ft by ft by h ft, then the total amount of steel used is + h ft. Therefore, 08 + h 08 and so h. The volume is given 08 by V( ) h 70.. Then V ( ) 7 0. 7 0. 7( 6 + )( 6 ) and V ( ).. The critical point occurs at 6, and it corresponds to the maimum volume because V ( ) < 0 for > 0. The corresponding height is 08 6 ft. The 6 () base measures 6 ft by 6 ft, and the height is ft. 8. 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 8 A ( ) + h +. We can minimize the weight of the vat by minimizing this quantity. Now A ( ) 8 ( ) and A ( ) + 6. 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 9. We have r + h, so r. We wish to minimize the cylinder s volume h h V r h h h for 0 < h <. Since dv h ( + h h dh )( ) and dv h, the critical point occurs at h and it dh corresponds to the maimum value because dv < 0 for dh h > 0. The corresponding value of r is. The largest possible cylinder has height and radius. 0. Note that, from similar cones, r h, so h r. 6 The volume of the smaller cone is given by V r h r ( r) r r for 0 < r < 6. Then dv 8r r r( r), so the critical point dr occurs at r. This critical point corresponds to the maimum volume because dv dr > 0 for 0< r < and dv dr < 0 for < r < 6. The smaller cone has the largest possible value when r ft and h ft.
Chapter Review 9. 0 in. Lid Base in. (a) V( ) ( )( ) (b, c) Domain: 0 < < The maimum volume is approimately 66.09 and it occurs when. 96 in. (d) Note that V( ) + 7,. so V ( ) 6 0+ 7. Solving V ( ) 0, we have 0 ± ( 0) ( 6)( 7) 0 ± 700 6 () 0 ± 0 7 ± 7. 6 These solutions are approimately. 96 and 6. 7, so the critical point in the appropriate domain occurs at 7. 6 y 0 p p p (, 8 cos 0.) p For 0 < <, the area of the rectangle is given by A ( ) ( )( 8cos 0. ) 6cos 0.. Then A ( ) 6( 0. sin 0. ) + 6(cos 0. )( ) 6(cos 0. 0. sin 0. ) Solving A ( ) 0 graphically, we find that the critical point occurs at. 868 and the corresponding area is approimately 9.9 square units.. The cost (in thousands of dollars) is given by C ( ) 0+ 0( 0 y) 0+ 600 0. 0 0 Then C ( ) 0 ( ) 0. Solving C ( ) 0, we have: 0 0 9 6 0 0 7 Choose the positive solution: 8 + 8. mi 7 6 y. 607 mi 7. The length of the track is given by + r, so we have + r 00 and therefore 00 r. Then the area of the rectangle is Ar () r r( 00 r) 00 00r r, for 0 < r <. Therefore, A ( r) 00 r and A () r, so the critical point occurs at r 00 m and this point corresponds to the maimum rectangle area because A () r < 0 for all r. The corresponding value of is 00 00 00 m. The rectangle will have the largest possible area when 00 00 mandr m.. 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+ ki ( 0 ) + ( ) k i k i 0 ( )( ) ( 0 )( ) P ( ) k i ( ) 0 + 0 k i ( )
0 Chapter Review. Continued The solutions of P ( ) 0 are 0 ± ( 0) ( )( 0) ±, so the solution in the () appropriate domain is. 76. Check the profit for the critical point and endpoints: Critical point: 76. P( ) 06. k End points: 0 P( ) 8k P( ) 8k The highest profit is obtained when.76 and y., which corresponds to 76 grade A tires and grade B tires. 6. (a) The distance between the particles is f(t) where f( t) cost+ cos t+. Then f () t sint sin t+ Solving f (t) 0 graphically, we obtain t.78, t.0, and so on. Alternatively, f (t) 0 may be solved analytically as follows. f + + () t sin t sin t 8 8 8 + 8 + 8 8 + sin t cos cos t 8 sin 8 sin + cos cos 8 8 + + t t 8 sin 8 sin cos + t, 8 8 so the critical points occcur when cos t+, t k. 0 or + At each of these values, 8 8 (b) Solving cost cos t+ graphically, we obtain t.79, t.890, and so on. Alternatively, this problem may be solved analytically as follows. cost cos t+ cos t + co s t + 8 8 + 8 8 cos t + cos sin + t+ t 8 8 + sin cos cos 8 8 8 8 sin t + 8 sin 8 sin + 0 t 8 sin 8 sin + t 8 0 7 t + k 8 7 The particles collide when t. 79 (plus multiples 8 of if they keep going.) 7. The dimensions will be in. by 0 in. by 6 in., so V() (0 )(6 ) + 60 for 0 < <. Then V () 0 + 60 ( )( 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 < <. The bo of largest volume has a height of in. and a base measuring 6 in. by in., and its volume is in. Graphical support: f() t ± cos ± 0. 76 units, so the maimum 8 distance between the particles is 0.76 units.