Chapter 3. Derivatives

Transcription

1 Section. Capter Derivatives Section. Eploration Derivative of a Function (pp. ) Reaing te Graps. Te grap in Figure.b represents te rate of cange of te ept of te water in te itc wit respect to time. Since y is measure in inces an is measure in ays, te erivative y woul be measure in inces per ay. Tose are te units tat soul be use along te y-ais in Figure.b.. Te water in te itc is inc eep at te start of te first ay an rising rapily. It continues to rise, at a graually ecreasing rate, until te en of te secon ay, wen it acieves a maimum ept of about inces. During ays,, 5, an 6, te water level goes own, until it reaces a ept of inc at te en of ay 6. During te sevent ay it rises again, almost to a ept of inces.. Te weater appears to ave been wettest at te beginning of ay (wen te water level was rising fastest) an riest at te en of ay (wen te water level was eclining te fastest).. Te igest point on te grap of te erivative sows were te water is rising te fastest, wile te lowest point (most negative) on te grap of te erivative sows were te water is eclining te fastest. 5. Te y-coorinate of point C gives te maimum ept of te water level in te itc over te 7-ay perio, wile te -coorinate of C gives te time uring te 7-ay perio tat te maimum ept occurre. Te erivative of te function canges sign from positive to negative at C, inicating tat tis is wen te water level stops rising an begins falling. 6. Water continues to run own sies of ills an troug unergroun streams long after te rain as stoppe falling. Depening on ow muc ig groun is locate near te itc, water from te first ay s rain coul still be flowing into te itc several ays later. Engineers responsible for floo control of major rivers must take tis into consieration wen tey preict wen floowaters will crest, an at wat levels. Quick Review... ( + ) ( + + ) + ( + ) ( + ) y. Since y for y y <,. y y. 8 ( + )( ) ( + ) ( + ) 8 5. Te verte of te parabola is at (, ). Te slope of te line troug (, ) an anoter point (, + ) on te parabola is ( + ). Since, te slope of te line tangent to te parabola at its verte is. 6. Use te grap of f to fin tat (, ) is te coorinate of te ig point an (, ) is te coorinate of te low point. Terefore, f is increasing on (, ] an [, ). 7. f( ) ( ) ( ) + + f( ) ( + ) + Copyrigt 6 Pearson Eucation, Inc.

2 8. f + f ( ) + + ( ) (see Eercise 7). 9. No, te two one-sie its are ifferent (see Eercise 7).. No, f is iscontinuous at because f( ) oes not eist. Section. Eercises. f f + f ( ) ( ) ( ) + + ( + ) ( + ) ( + ) ( + ).. ( ) ( ) Section. f( + ) f( ) f ( ) [ ( + ) ] [ ( ) ] [ ( + )] ( ) f( + ) f( ) f () f( ) f( ) ( + ) ( + ) ( + ) ( + ). f( + ) f( ) f () [( + ) + ] [ + ] ( + ) ( + ) 5. f( ) f( ) f ( ) ( ) Copyrigt 6 Pearson Eucation, Inc.

3 Section f( ) f( ) f () ( + ) ( + ) ( )( + ) ( + ) f( ) f( ) f () + + ( + ) ( + + ) ( ) ( + + ) ( + ) ( )( + + ) ( )( + + ) + + f( ) f( ) f ( ) ( ) ( + ) ( ( ) + ) ( + ) + f f f ( + ) ( ) ( ) [( + ) ] ( ). y y( + ) y( ) 7( + ) Let f( ) ( ) f ( ) f( + ) f( ) ( + ) + + ( + ). f( + ) f( ) f( ) ( + ) ( 6+ ) 6. Te grap of y f () is ecreasing for < an increasing for >, so its erivative is negative for < an positive for >. (b). Te grap of y f () is always increasing, so its erivative is always. (a) 5. Te grap of y f () oscillates between increasing an ecreasing, so its erivative oscillates between positive an negative. () 6. Te grap of y f () is ecreasing, ten increasing, ten ecreasing, an ten increasing, so its erivative is negative, ten positive, ten negative, an ten positive. (c) Copyrigt 6 Pearson Eucation, Inc.

4 Section (a) Te tangent line as slope 5 an passes troug (, ). y 5( ) + y 5 7 (b) Te normal line as slope y ( ) y an passes troug (, ) y f ( + ) f ( ) [ ( + ) ( + ) + 5] ( + 5) ( + ) y At, ( ), so te tangent line as slope an passes troug (, y()) (, 6). y ( ) 6 y 9. Let f( ). f( + ) f( ) f () ( + ) ( + + ) (a) Te tangent line as slope an passes troug (, ). Its equation is y ( ) +, or y. (b) Te normal line as slope or y +. an passes troug (, ). Its equation is y ( ) +, Copyrigt 6 Pearson Eucation, Inc.

5 6 Section.. Let f( ). f( + ) f( ) f ( ) + ( + ) ( + + ) ( + + ) ( + ) ( + + ) ( + + ) + + (a) Te tangent line as slope an passes troug (, ). Its equation is y +. (b) Te normal line as slope an passes troug (, ). Its equation is y (a) Te amount of ayligt is increasing at te fastest rate wen te slope of te grap is largest. Tis occurs about onefourt of te way troug te year, sometime aroun April. Te rate at tis time is approimately ours ays or 6 our per ay. (b) Yes, te rate of cange is zero wen te tangent to te grap is orizontal. Tis occurs near te beginning of te year an alfway troug te year, aroun January an July. (c) Positive: January troug July Negative: July troug December. Te slope of te given grap is zero at an at, so te erivative grap inclues (, ) an (, ). Te slopes at an at are about 5 an te slope at.5 is about.5, so te erivative grap inclues (, 5), (, 5), an (.5,.5). Connecting te points smootly, we obtain te grap sown.. (a) Using Figure.a, te number of ares is largest after years an smallest after 6 years. Using Figure.b, te erivative is at tese times. (b) Using Figure.a, tere were, ares wen te population was te largest an 6, ares wen te population was te smallest. (c) Te peak for te ares occurs at years an te peak for te lyn occurs at years. Terefore, years elapse between te peaks.. Since te grap of y ln is ecreasing for < < an increasing for >, its erivative is negative for < < an positive for >. Te only one of te given functions wit tis property is y ln. Note also tat y ln is unefine for <, wic furter agrees wit te given grap. (ii) 5. Eac of te functions y sin, y, y, as te property tat y() but te grap as nonzero slope (or unefine slope) at, so none of tese functions can be its own erivative. Te function y is not its own erivative because y() but ( + ) y ( ) + ( + ). Tis leaves only e, wic can plausibly be its own erivative because bot te function value an te slope increase from very small positive values to very large values as we move from left to rigt along te grap. (iv) Copyrigt 6 Pearson Eucation, Inc.

6 Section (a) Te slope from to is ( ). Te slope from to is. Te slope from to is ( ). Te slope from to 6 is ( ). 6 Note tat te erivative is unefine at,, an. (Te function is ifferentiable at an at 6 because tese are enpoints of te omain an te one-sie erivatives eist.) Te grap of te erivative is sown. 8. y Mipoint of Interval () Slope ( y ) (b),, 7. For >, te grap of y f() must lie on a line of slope tat passes troug (, ): y. Ten y( ) ( ), so for <, te grap of y f() must lie on a line of slope tat passes troug (, ): y ( + ) + or y +. +, < Tus f(), y A grap of te erivative ata is sown. 5 [,] by [,8] (a) Te erivative represents te spee of te skier. (b) Since te istances are given in feet an te times are given in secons, te units are feet per secon. Copyrigt 6 Pearson Eucation, Inc.

7 8 Section.. (a) (c) Te grap appears to be approimately linear an passes troug (, ) an (9.5, 6.), so te slope is Te equation of te erivative is approimately D 6.65t. (c) Since te elevation y is given in feet an te istance own river is given in miles, te units of te graiant are feet per mile. () Since te elevation y is given in feet an te istance ownriver is given in miles, te units of te erivative y are feet per mile. (b) Mipoint of Interval () Slope ( y ) (e) Look for te steepest part of te curve. Tis is were te elevation is ropping most rapily, an terefore te most likely location for significant rapis. (f) Look for te lowest point on te grap. Tis is were te elevation is ropping most rapily, an terefore te most likely location for significant rapis.. We sow tat te rigt-an erivative at oes not eist. f( + ) f( ) ( + ) Does not eist We sow tat te rigt-an erivative at oes not eist. f( + ) f( ) ( + ) ( ) A grap of te erivative ata is sown. [, ] by [.5,.5] Te cosine function coul be te erivative of te sine function. Te values of cosine are positive were sine is increasing, zero were sine as orizontal tangents, an negative were sine is ecreasing. Copyrigt 6 Pearson Eucation, Inc.

8 Section. 9. f f ( + ) ( ) Tus, te rigt-an erivative at oes not eist. 5. Two parabolas are parallel if tey ave te same erivative at every value of. Tis means tat teir tangent lines are parallel at eac value of. Two suc parabolas are given by y an y +. Tey are grape below. Te parabolas are everywere equiistant, as long as te istance between tem is always measure along a vertical line. 6. True. f () + 7. False. Let f(). Te left-an erivative at is an te rigt-an erivative at is. f () oes not eist. 8. C; f f f ( + ) ( ) ( ) [ ( + )] [ ( )] A;. B;. C;. (a) f( + ) f( ) f () [ ( + ) ] [ ( ) ] 6 ( 6 ) ( 6 ) 6 f( + ) f() f() f() ( ) ( ) f( + ) f() f() f() ( ) ( ) f ( + ( ) ) f ( f ) ( + ) + ( + ) + (b) f( + ) f( ) f ( ) ( + ) Copyrigt 6 Pearson Eucation, Inc.

9 Section. (c) () f ( ) ( ) f ( ) + + (e) Yes, te one-sie its eist an are te same, so f ( ). (f) (g) f( + ) f( ) ( + ) ( + ) ( + ) f( + ) f( ) ( + ) Te rigt-an erivative oes not eist. () It oes not eist because te rigt-an erivative oes not eist.. (e) Te y-intercept of te erivative is b a.. Since te function must be continuous at, we ave ( + k) f( ), so + k, + or k. Tis gives f( ),, >. Now we confirm tat f() is ifferentiable at. f( + ) f( ) ( + ) ( ) + + ( + + ) f( + ) f() [ ( + ) ] () ] ( + ) Since te rigt-an erivative equals te leftan erivative at, te erivative eists (an is equal to ) wen k. 5. (a) P Alternate meto: (b) Using te answer to part (a), te probability is about (c) Let P represent te answer to part (b), P.8. Ten te probability tat tree people all ave ifferent birtays is P. Aing a fourt person, te probability tat all ave ifferent 6 birtays is ( P), so te 65 probability of a sare birtay is 6 ( P) () No; clearly February 9 is a muc less likely birt ate. Furtermore, census ata o not support te assumption tat te oter 65 birt ates are equally likely. However, tis simplifying assumption may still give us some insigt into tis problem even if te calculate probabilities aren t completely accurate. 6. Te erivative looks like a quaratic wit roots at an 6. All suc quaratics ave te form a ( )( 6) a ( 8+ ). Since f (), ten a, an f ( ) ( 8+ ). Using te meto of Eample, a function tat as f as its erivative is +. Aing a constant to tis Copyrigt 6 Pearson Eucation, Inc.

10 Section. trial version of f oes not cange f, so since f(), te function is f( ) + +. Section. Differentiability (pp. 7) Eploration Zooming in to See Differentiability. Zooming in on te grap of f at te point (, ) always prouces a grap eactly like te one sown below, provie tat a square winow is use. Te corner sows no sign of straigtening out.. Zooming in on te grap of g at te point (, ) begins to reveal a smoot turning point. Tis grap sows te result of tree zooms, eac by a factor of orizontally an vertically, starting wit te winow. [, ] by [. 6,. 6]. Eploration Looking at te Symmetric Difference Quotient Analytically... f( + ) f( ) (. ).. f ( ) Te ifference quotient is. away from f (). f( + ) f( ) (. ) ( 9. 99). Te symmetric ifference quotient eactly equals f (). f( + ) f( ) (. ).. f ( ) Te ifference quotient is. away from f (). f( + ) f( ) (. ) ( 9. 99)... Te symmetric ifference quotient is. away from f (). Quick Review.. Yes. On our graper, te grap became orizontal after 8 zooms. Results can vary on ifferent macines.. As we zoom in on te graps of f an g togeter, te ifferentiable function graually straigtens out to resemble its tangent line, wile te nonifferentiable function stubbornly retains its same sape.. No (Te f () term in te numerator is incorrect.). Yes. Yes 5. No (Te enominator for tis epression soul be ). 6. All reals 7. [, ) 8. [, ) 9. Te equation is equivalent to y. + (. + 5), so te slope is.. Copyrigt 6 Pearson Eucation, Inc.

11 Section.. f( +. ) f(. ). 5( +. ) 5(. ). 5 (. ). 5 Section. Eercises. Left-an erivative: f( + ) f( ) Rigt-an erivative: f( + ) f( ) Since, te function is not ifferentiable at te point P.. Left-an erivative: f( + ) f( ) Rigt-an erivative: f( + ) f( ) ( + ) Since, te function is not ifferentiable at te point P.. Left-an erivative: f( + ) f( ) + ( + )( + + ) ( + + ) ( + ) ( + + ) + + Rigt-an erivative: f( + ) f( ) [ ( + ) ] Since, te function is not ifferentiable at te point P.. Left-an erivative: f( + ) f( ) ( + ) Rigt-an erivative: f( + ) f( ) ( + ) + ( + ) + ( + ) + + Since, te function is not ifferentiable at te point P. 5. (a) All points in [, ] (b) None (c) None 6. (a) All points in [, ] (b) None (c) None 7. (a) All points in [, ] ecept (b) None (c) 8. (a) All points in [, ] ecept,, (b) (c), Copyrigt 6 Pearson Eucation, Inc.

12 9. (a) All points in [, ] ecept (b) (c) None. (a) All points in [, ] ecept, (b), (c) None. Since tan tan y( ), te problem is a iscontinuity.. 5 / f( + ) f( ) 5 / 5 / f( + ) f( ) / Te problem is a cusp.. Note tat y , +, >. f( + ) f( ) f( + ) f( ) ( + ) Te problem is a corner.. Section. ( ) f( + ) f( ) / Te problem is a vertical tangent. 5, 5. Note tat y, > f( + ) f( ) ( 5 ) ( ) 5 5 f( + ) f( ) ( ) ( ) Te problem is a corner. 6. f( + ) f( ) / f( + ) f( ) / Te problem is a cusp. Copyrigt 6 Pearson Eucation, Inc.

13 Section f( a+ ) f( a ) f a f a ( + ) ( ) f a f a ( + ) ( ) f a f a ( + ) ( ) f a f a ( + ) ( ) f( a+ ) f( a ) f a f a ( + ) ( ) (. ) (. ) ( (. ) (. ) )., yes it is ifferentiable. (. ) (. ) ( (. 999) (. 999)., yes it is ifferentiable. (. ) + (. ) (. 999) (. 99)., yes it is ifferentiable. (. ) (. ) ((. ) (. )) , yes it is ifferentiable. (. 999) (. 999) ((. ) (. )). 8., yes it is ifferentiable. (. ) (. ) ((. 999) (. 999)). 8., yes it is ifferentiable. / / (. ) (. )., no it is not ifferentiable. (CUSP). f( a+ ) f( a ) , no it is not ifferentiable. (CORNER) f a f a ( + ) ( ) f( a+ ) f( a ) 5 / 5 / (. ) (. )., no it is not ifferentiable. (CUSP) 5 / 5 / (. ) (. )., no it is not ifferentiable. (CUSP) [, ] by [.5,.5] y sin Copyrigt 6 Pearson Eucation, Inc.

14 8. 9. y Section. 5 ( + k) ( ) k k [ ( + k) 6+ 5] 5 k k k k k k k Te function as a vertical tangent at. It is ifferentiable for all reals ecept.. y abs ( ) or [, ] by [, ] y tan Note: Due to te way NDER is efine, te grap of y NDER () actually as two asymptotes for eac asymptote of y tan. Te asymptotes of y NDER () occur at + k ±., were k is an integer. A goo winow for viewing tis beavior is [.566,.576] by [, ].. Fin te zeros of te enominator. 5 ( + )( 5) or 5 Te function is a rational function, so it is ifferentiable for all in its omain: all reals ecept, 5.. Te function is ifferentiable ecept possibly were 6, tat is, at. We ceck for ifferentiability at, using k instea of te usual, in orer to avoi confusion wit te function ().. Note tat te sine function is o, so sin, < P ( ) sin ( ) sin,. Te grap of P() as a corner at. Te function is ifferentiable for all reals ecept.. Since te cosine function is even, Q ( ) cos( ) cos. Te function is ifferentiable for all reals. 5. Te function is piecewise-efine in terms of polynomials, so it is ifferentiable everywere ecept possibly at an at. Ceck : g( + ) g( ) ( + ) + ( + ) g( + ) g( ) ( + ) Te function is ifferentiable at. Ceck : Since g() ( ) an g ( ) ( + ) ( ) + 7, te function is not continuous (an ence not ifferentiable) at. Te function is ifferentiable for all reals ecept. Copyrigt 6 Pearson Eucation, Inc.

15 6 Section., < 6. Note tat C ( ), so it is, Ceck : C( + ) C( ) Te function is ifferentiable for all reals. 7. Te function f() oes not ave te intermeiate value property. Coose some a in (, ) an b in (, ). Ten f(a) an f(b), but f oes not take on any value between an. Terefore, by te Intermeiate Value Teorem for Derivatives, f cannot be te erivative of any function on [, ]. 8. (a) is not in teir omains, or, tey are bot iscontinuous at. (b) For NDER :,,, For : NDER, (c) It returns an incorrect response because even toug tese functions are not efine at, tey are efine at ±.. Te responses iffer from eac oter because is even (wic automatically makes NDER, 9. (a) an is o. ( ) ( ) f f ( ) a( ) + b( ) a+ b Te relationsip is a + b. (b) Since te function nees to be continuous, we may assume tat a + b an f(). f( + ) f( ) ( ( + )) ( ) f( + ) f( ) + a( + ) + b( + ) + a+ a+ a + b+ b + a + a + b + ( a + b ) + ( a+ a+ b) + a+ b Terefore, a + b. Substituting a for b gives a + ( a), so a. Ten b a ( ) 5. Te values are a an b 5.. True. See Teorem.. False. Te function f() is continuous at but is not ifferentiable at.. B f( a+ ) f( a ). A; NDER( f, a) Te symmetric ifference quotient gets larger as gets smaller, so f () is unefine. ( + ) + ( ( ) + ). B; ( + ) + ( + ) 5. C; (a) Copyrigt 6 Pearson Eucation, Inc.

16 Section. 7 (b) You can use Trace to elp see tat te value of Y is for every < an is for every. It appears to be te grap, < of f( ),. (c) () You can use Trace to elp see tat te value of Y is for every < an is for every. It appears to be te grap, < of f( ),. 7. (a) (tat is, for in any open interval containing ). () No, because te one-sie its (as in part (c)) o not eist. (e) ( ) g( + ) g( ) sin sin As note in part (a), te it of tis as approaces zero is, so g (). Section. Rules for Differentiation (pp. 8 8) Quick Review... ( )( + ) (b) See eercise 6. (c) () NDER(Y,.)., NDER(Y, ).9995, NDER(Y,.). Te numerical erivative gets it rigt at. an., but gets it wrong at were te function is iscontinuous an nonifferentiable ( + )( + ) ( + ) + 8. (a) Note tat sin( / ) for all ecept, so sin by te Sanwic Teorem. Terefore, f is continuous at. (b) f( + ) f( ) sin sin (c) Te it oes not eist because sin oscillates between an an infinite number of times arbitrarily close to 6 6 At. 7, 5 5. At. 9, 5 9,. After rouning, we ave: 6 At, At, 5 9,. 8. (a) f ( ) 7 (b) f () 7 Copyrigt 6 Pearson Eucation, Inc.

17 8 Section. (c) f( + ) 7 () f( ) f( a) 7 7 a a a a a 9. Tese are all constant functions, so te grap of eac function is a orizontal line an te erivative of eac function is.. (a) (b) Section. Eercises.. f( + ) f( ) f ( ) + + f( + ) f( ) f ( ) + ( + ) ( + ) ( + ) ( + ) y ( ) + ( ) + y ( ) y. ( ) + ( ) +. y ( ) + ( ) + ( ) y ( ) y () ( ) + ( ) ( ) + + y ( + + ) + Horizontal tangents at y ( + + ) 8+ Horizontal tangents at,. y ( ) +,. + 8 Horizontal tangents at, ±. y ( 6 ) Horizontal tangents at,. y 5 ( 5 ) 5 5 Horizontal tangents at,,. y ( ) + Horizontal tangents at, , 8 Copyrigt 6 Pearson Eucation, Inc.

18 . (a) (b). (a) (b) y [( + )( + )] ( + ) ( + ) + ( + ) ( + ) ( + )( ) + ( + )( ) y [( + )( + )] ( ) y ( + ) ( + ) ( ) ( ) ( + ) + y ( + ) Tis is equivalent to te answer in part (a) Section. 9 y + 5 ( )( ) ( + 5)( ) ( ) 9 ( ) + 5 y ( + 5 ) ( )( + + ) y ( ) + + y ( + )( ) ( )( ) ( + ) ( + ) ( + + )( + + ) 7 5 ( ) ( + )( + ) 5 ( ) y ( )( ) ( ) ( ) + ( ) Copyrigt 6 Pearson Eucation, Inc.

19 Section.. y ( + )( + ) ( )( ) ( + )( + ) ( + + )( ) ( + ) ( 5+ 6) ( + 5 6) ( + ) 6 ( + ). (a) At, ( uv ) u () v () + v () u () ()() 5 + ( )( ) (b) At, (c) At, u v() u () u() v () v [()] v ( )( ) ( 5)( ) ( ) 7 v u() v () v() u () u [ u( )] ()() 5 ( )( ) () () At, ( 7v u) 7v () u () 7 ( ) ( ). (a) At, ( uv ) u ( ) v ( ) + v ( ) u ( ) ()() + ()( ) (c) At, v u( ) v ( ) v( ) u ( ) u [ u( )] ()() ()( ) () 9 () Use te result from part (a) for ( uv ). At, ( u v+ uv) u ( ) v ( ) + ( uv) ( ) ( ) + ( ) 5. y ( ) + 5 y () () + 5 Te slope is. (iii) 6. Te given equation is equivalent to y + 6, so te slope is. (iii) 7. + y ( ) ( + ) () y () () () + y() () y ( ) + + (b) At, u v( ) u ( ) u( ) v ( ) v [()] v ()( ) ( )( ) () Copyrigt 6 Pearson Eucation, Inc.

20 Section y ( ) ( + ) ( ) y ( ) ( ) ( ) + y( ) ( ) y ( + ) + y + 5 y ( 8+ ) 8 8 y y + / / + / / / / ( + ) ( ) / ( + ) / / / [( + ) ( )] / ( + ) / / ( + ) ( + ) y + / y y + + y + 6 y + 6 y y + + y + y y y 6 5 y + y + y + y y + y + + y y 6 y 6 iv 5 y 5 y ( ) y ( ) ( ) 9 Te tangent line as slope 9, so te perpenicular line as slope an passes 9 troug (, ). y 9 ( ) + 9 y y ( ) + Te slope is wen +, at ±. Te tangent at as slope an passes troug (, ), so its equation is y ( + ), or y +. Te tangent at as slope an passes troug (, ), so its equation is y ( ) +, or y. Te smallest slope occurs wen + is minimize, so te smallest slope is an occurs at. Copyrigt 6 Pearson Eucation, Inc.

21 Section y ( ) 6 6 6( ) 6( + )( ) Te tangent is parallel to te -ais wen y, at an at. Since y( ) 7 an y(), te two points were tis occurs are (, 7) an (, ). y ( ) y ( ) Te tangent line as slope an passes troug (, 8), so its equation is y ( + ) 8, or y + 6. Te -intercept is an te y-intercept is 6. ( + )( ) ( ) + y ( ) ( + ) ( + ) At te origin: y () Te tangent is y. At (, ): y (). Te tangent is y. ( + )( ) 8( ) 6 y ( ) ( + ) ( + ) y ( ) Te tangent as slope an passes troug (, ). Its equation is y ( ) +, or y +.. (a) Let f( ). ( ) f ( ) f( + ) f( ) ( + ) ( ) (b) Note tat u u() is a function of. u( + ) [ u( )] ( u) u ( + ) u ( ) u ( + ) u ( ) u Copyrigt 6 Pearson Eucation, Inc.

22 Section ( c f( )) c f( ) + f( ) ( c) c f( ) + c f( ) ( ) f f( ) f ( ) f ( ) [ f( )] [ f( )] P nrt an V V V nb V ( V nb) ( nrt) ( nrt) ( V nb) V V ( an V ) ( V nb) V nrt + an V ( V nb) nrt an + ( V nb) V s ( 9. t ) 98. t t t s (. 98t) 98. t t R C M M M M C M M M CM M 9. If te raius of a circle is cange by a very small amount r, te cange in te area can be tougt of as a very tin strip wit lengt given by te circumference, r, an wit r. Terefore, te cange in te area can be approimate by ( r)( r), wic means tat te cange in te area ivie by te cange in te raius is approimately r. 5. If te raius of a spere is cange by a very small amount r, te cange in te volume can be tougt of as a very tin layer wit an area given by te surface area, r, an a tickness given by r. Terefore, te cange in te volume can be approimate by ( r )( r), wic means tat te cange in te volume ivie by te cange in te raius is approimately r. 5. Let t() be te number of trees an y() be te yiel per tree years from now. Ten t() 56, y(), t (), an y ().. 5 Te rate of increase of prouction is ( ty ) t () y () + y ()() t ( 56)(. 5) + ( )( ) 9 busels or annual prouction per year. Copyrigt 6 Pearson Eucation, Inc.

23 Section. 5. Let m() be te number of members an c() be te pavilion cost years from now. Ten m() 65, c() 5, m () 6, an c (). Te rate of cange of eac member s sare is c m() c () c() m () m [ m( )] ( 65)( ) ( 5)( 6) ( 65). ollars per year. Eac member s sare of te cost is ecreasing by approimately cents per year. 5. False; is a constant, so is also a constant an ence ( ). 5. True; f ( ) is never zero, so tere are no orizontal tangents. 55. B; ( uv) u v + v u ( )( ) + ( )( ) 56. D; f( ) f ( ) + f ( ) 57. E; + ( ) ( + ) ( ) ( ) (b) It was rejecte because it is incomparably smaller tan te oter terms: v u an u v. u v (c) ( uv) v + u. Tis is equivalent to te prouct rule given in te tet. () Because is infinitely small, an tis coul be tougt of as iviing by zero. (e) u u+ u u v v + v v ( u+ u)( v) ( u)( v+ v) ( v+ v)( v) uv + vu uv uv v + vv vu uv v Quick Quiz Sections... D. A; Slope of normal: m Slope of tangent: m m Terefore f ().. C; y + ( + ) ( ) ( + ) ( + ) 58. B; f ( ) ( ) + ( + ) [( ) + ( + )] f ( ) only wen Tere is one orizontal tangent at. 59. (a) It is insignificant in te iting case an can be treate as zero (an remove from te epression).. (a) y ( ) 8 ( ), ± Copyrigt 6 Pearson Eucation, Inc.

24 (b) f () ()( ) y () () y y m( ) + y y ( ) y + Section. 5. Wen te trace cursor is moving to te rigt te particle is moving to te rigt, an wen te cursor is moving to te left te particle is moving to te left. Again we fin te particle reverses irection at about t.6 an t.6. (c) m m y ( ). Wen te trace cursor is moving upwar te particle is moving to te rigt, an wen te cursor is moving ownwar te particle is moving to te left. Again we fin te same values of t for wen te particle reverses irection. Section. Velocity an Oter Rates of Cange (pp. 9 ) Eploration Growt Rings on a Tree. Figure. is a better moel, as it sows rings of equal area as oppose to rings of equal wit. It is not likely tat a tree coul sustain increase growt year after year, altoug cate conitions o prouce some years of greater growt tan oters.. Rings of equal area suggest tat te tree as approimately te same amount of woo to its girt eac year. Wit access to approimately te same raw materials from wic to make te woo eac year, tis is ow most trees actually grow.. We can represent te velocity by graping te parametric equations () t () t t t+ 5, y() t ( part ), 5() t () t t t+ 5, y5() t t ( part ), 6() t t, y6() t () t t t+ 5 ( part ). Since cange in area is constant, so also is cange in area. If we enote tis latter constant by k, we ave k r, wic means tat r cange in raius varies inversely as te cange in te raius. In oter wors, te cange in raius must get smaller wen r gets bigger, an vice-versa. Eploration Moeling Horizontal Motion. Te particle reverses irection at about t.6 an t.6. For (, y ) an ( 5, y 5), te particle is moving to te rigt wen te -coorinate of te grap (velocity) is positive, moving to te left wen te -coorinate of te grap (velocity) is negative, an is stoppe wen te -coorinate of te grap (velocity) is. For Copyrigt 6 Pearson Eucation, Inc.

25 6 Section. ( 6, y 6), te particle is moving to te rigt wen te y-coorinate of te grap (velocity) is positive, moving to te left wen te y-coorinate of te grap (velocity) is negative, an is stoppe wen te y-coorinate of te grap (velocity) is. Eploration Seeing Motion on a Graping Calculator 6. f( ) ( + ) 6( )( 7) or 7 f( ) 8 at an at 7.. Let tmin an tma.. Since te rock acieves a maimum eigt of feet, set yma to be sligtly greater tan, for eample yma.. Te graper procees wit constant increments of t (time), so piels appear on te screen at regular time intervals. Wen te rock is moving more slowly, te piels appear closer togeter. Wen te rock is moving faster, te piels appear farter apart. We observe faster motion wen te piels are farter apart. Quick Review.. Te coefficient of is negative, so te parabola opens ownwar.. Te y-intercept is f() 56.. Te -intercepts occur wen f() ( + 6) 6( )( 8) or 8 Te -intercepts are an 8. Since f( ) 6( + 6) 6( + 5 9) 6( 5) +, te range is (, ]. y y 5 at 8 y 8. > + 6 > > 6 < 5 y wen 5. > < 9. Note tat f ( ) + 6. f( + ) f( ) f ( ) () f ( ) + 6 f ( ) At 7 (an, in fact, at any oter value of ), y. Section. Eercises. (a) V() s s 5. Since f + te verte is at (5, ). ( ) 6( + 6) 6( + 5 9) 6( 5), (b) (c) V s s V s () s V s 5 () 75 s 5 Copyrigt 6 Pearson Eucation, Inc.

26 (). (a) (b) (c) (). (a) (b) (c) in in. C C r r C C A r C AC ( ) A C C C A C A C in in. s C 6 C 6 or square inces per inc s s s s A b s As () s A s s A s ( ) s A 5 s ( ) s. (a) s r r s s r r s Section. 7 Use Pytagorean Teorem on lower rigt triangle: s + s ( r) s r s r A s r A() r r A (b) r r (c) () A r r A 5. (a) s(ft) 6. (a) r r 8 () 8 () in or square inces per inc in. 5 t(sec) (b) s () 8, s (.5), s (.5) 5 () in in. or square inces per inc (b) s () 6, s (.5), s (.5) 7. (a) We estimate te slopes at several points as follows, ten connect te points to create a smoot curve. Copyrigt 6 Pearson Eucation, Inc.

27 8 Section. t (ays) 5 Slope (flies/ ay) increasing, for < t < 6. Te acceleration is negative wen v is ecreasing, for t < an for 6 < t < 7. Te acceleration is zero wen v is constant, for < t < an for 7 < t 9. (c) Te particle moves at its greatest spee wen v is maimize, at t an for < t <. 8. Horizontal ais: Days Vertical ais: Flies per ay (b) Fastest: Aroun te 5t ay Slowest: Day 5 or ay Qt () ( t) ( 9 6t+ t ) 8,, t+ t Q () t, + t Te rate of cange of te amount of water in te tank after minutes is Q ( ) 8 gallons per minute. Note tat Q ( ) <, so te rate at wic te water is running out is positive. Te water is running out at te rate of 8 gallons per minute. Te average rate for te first minutes is Q( ) Q( ) 8, 8,, gal/min. Te water is flowing out at an average rate of, gallons per minute over te first min. 9. (a) Te particle moves forwar wen v >, for t < an for 5 < t < 7. Te particle moves backwar wen v <, for < t < 5. Te particle spees up wen v is negative an ecreasing, for < t <, an wen v is positive an increasing, for 5 < t < 6. Te particle slows own wen v is positive an ecreasing, for t < an for 6 < t < 7, an wen v is negative an increasing, for < t < 5. () Te particle stans still for more tan an instant wen v stays at zero, for 7 < t 9.. (a) Te particle is moving left wen te grap of s as negative slope, for < t < an for 5 < t 6. Te particle is moving rigt wen te grap of s as positive slope, for t <. Te particle is staning still wen te grap of s is orizontal, for < t < an for < t < 5. (b) For t < : v cm/sec Spee v cm/sec For < t < : v cm/sec Spee v cm/sec For < t < : v cm/sec Spee v cm/sec ( ) For < t < 5: v cm/sec 5 Spee v cm/sec ( ) For 5 < t 6: v cm/sec 6 5 Spee v cm/sec Velocity grap: v (b) Note tat te acceleration a is t unefine at t, t, an t 6. Te acceleration is positive wen v is Copyrigt 6 Pearson Eucation, Inc.

28 Section. 9 Spee grap:. (a) It takes 5 secons. (b) Average spee F t furlongs/sec.. (a) Te boy reverses irection wen v canges sign, at t an at t 7. (b) Te boy is moving at a constant spee, v m/sec, between t an t 6. (c) Te spee grap is obtaine by reflecting te negative portion of te velocity grap, < t < 7, about te -ais. Spee(m/sec) 6 8 t(sec) () For t < : a m/sec For < t < : a m/sec ( ) For < t < 6: a m/sec 6 ( ) For 6 < t < 8: a m/sec 8 6 For 8 < t : a 5. m/sec 8 Acceleration (m/sec ) (, ) (6, ) (8, ) (, ) (6, ) 6 8 t(sec) (8,.5) (, ) (, ) (,.5) (c) Using a symmetric ifference quotient, te orse s spee is approimately F t furlongs/sec. () Te orse is running te fastest uring te last furlong (between 9t an t furlong markers). Tis furlong takes only secons to run, wic is te least amount of time for a furlong. (e) Te orse accelerates te fastest uring te first furlong (between markers an ). s. (a) Velocity: vt () t ( t. 8t ) t. 6 t m/sec v Acceleration: at () t (. 6t) t 6. m/sec (b) Te rock reaces its igest point wen vt (). 6t, at t 5. It took 5 secons. (c) Te maimum eigt was s(5) 8 meters. () st () ( 8 ) t. 8t 9. 8t t+ 9 Copyrigt 6 Pearson Eucation, Inc.

29 Section. (e) ± ( ) (. 8)( 9) t 8 (. ). 9, It took about.9 secons to reac alf its maimum eigt. st () t. 8t 8. t( t) t or t Te rock was aloft from t to t, so it was aloft for secons.. On Mars: Velocity s (. 86t ) 7. t t t Solving.7t 6.6, te ownwar velocity reaces 6.6 m/sec after about.6 sec. On Jupiter: Velocity s (. t ). 88t t t Solving.88t 6.6, te ownwar velocity reaces 6.6 m/sec after about.76 sec. 5. Te rock reaces its maimum eigt wen te velocity s () t 9. 8t, at t. 9. Its maimum eigt is about s(. 9) meters. 6. Moon: st () 8t. 6t 6. t( t) t or t It takes secons to return. Eart: st () 8t 6t 6t( 5 t) t or t 5 It takes 5 secons to return. 7. Te following is one way to simulate te problem situation. For te moon: () t ( t < 6) +.( t 6) y () t 8t. 6t t-values: to winow: [, 6] by [,, 7,] For te eart: () t ( t < 6) +.( t 6) y () t 8t 6t t-values : to 5 winow: [, 6 ] by [,, ] 8. (a) 9 ft/sec (b) secons (c) After 8 secons, an its velocity was ft/sec ten () After about secons, an it was falling 9 ft/sec ten (e) About secons (from te rocket s igest point) (f) Te acceleration was greatest just before te engine stoppe. Te acceleration was constant from t to t, wile te rocket was in free fall. 9. (a) Displacement: s() 5 s() m (b) Average velocity m m/sec 5 sec (c) Velocity s () t t At t, velocity s ( ) ( ) 5 m/sec () Acceleration s () t m/sec. (a) (e) Te particle canges irection wen s () t t, so t sec. (f) Since te acceleration is always positive, te position s is at a minimum wen te particle canges irection, at t sec. Its position at tis time is s m. (b) s vt () ( t 7t t 8) t t + + vt () t + t v at () ( t t ) t t + at () 6t+ Copyrigt 6 Pearson Eucation, Inc.

30 . (a) (c) v(t) t + t t.5,.5 () Te particle starts at te point s 8 wen t an moves left until it stops at s.6 wen t.5, ten it moves rigt to te point s. wen t.5 were it stops again, an finally continues left from tere on. s vt () t [( t ) ( t )] t ( t ) ( ) + ( t ) ( t ) ( t )[( t ) + ( t )] ( t )( t ) v (b) at () [( t )( t )] t t at () 6t 6.. Section. vt () s () t t t+ 9 at () v () t 6t Fin wen velocity is zero. t t+ 9 ( t t+ ) ( t )( t ) t or t At t, te acceleration is a () 6 m/sec At t, te acceleration is a () 6 m/sec at () v () t 6t 8t+ Fin wen acceleration is zero. 6t 8t+ 6( t t+ ) 6( t )( t ) t or t At t, te spee is v () m/sec. At t, te spee is v ( ) m/sec.. (a) (c) vt () ( t )( t ) t, () Te particle starts at te point s 6 wen t an move rigt until it stops at s wen t, ten it moves left to te (b) (c) point s.85 wen t were it stops again, an finally continues rigt from tere on. s vt () ( t 6t 8t ) t t + + vt () t t+ 8 ν at () ( t t+ 8) t t at () 6t ν () t t t+ 8 t. 85, (a) y t 6 t t t t 6 + t 6 6 t t t + t + t (b) Te flui level is falling fastest wen y t is te most negative, at t, y wen. t Te flui level is falling y slowest at t, wen. t (c) () Te particle starts at te point s wen t an moves rigt until it stops at s 5.79 wen t.85, ten it moves left to te point s.79 wen t.55 were it stops again, an finally continues rigt from tere on. Copyrigt 6 Pearson Eucation, Inc.

31 Section. y is ecreasing an y t an te magnitue of y t orizontal tangent. is negative over te entire interval; y ecreases more rapily early in te interval, is larger ten. y t is at t, were te grap of y seems to ave a 6. (a) To grap te velocity, we estimate te slopes at several points as follows, ten connect te points to create a smoot curve. t (ours) v (km/our) To grap te acceleration, we estimate te slope of te velocity grap at several points as follows, an ten connect te points to create a smoot curve. t (ours) a (km/our ) (b) s t t t s 6t t Copyrigt 6 Pearson Eucation, Inc.

32 Section. 9. (a) Te graps are very similar. c( ) 7. (a) Average cost, $ per macine 8. (a) (b) c ( ). Marginal cost c ( ) $ 8 per macine (c) Actual cost of st macine is c() c() $79.9, wic is very close to te marginal cost calculate in part (b). [, 5] by [ 5, ] Te values of wic make sense are te wole numbers,. (b) Marginal revenue r ( ) + + ( + )( ) ( )( ) ( + ) ( + ) (c) r () ( 5+ ) 6. Te increase in revenue is approimately $ () Te it is. Tis means tat as gets large, one reaces a point were very little etra revenue can be epecte from selling more esks. [, ] by [, ] (b) Te values of wic make sense are te wole numbers,. (c) [, ] by [.,.] P is most sensitive to canges in wen P ( ) is largest. It is relatively sensitive to canges in between approimately 6 an 6. () Te marginal profit, P (), is greatest at 6.. Since must be an integer, P( 6). 9 tousan ollars or $9. (e) P ( 5)., or $ per package sol P ( ). 65, or $65 per package sol P ( 5). 8, or $8 per package sol P ( 5)., or $ per package sol P ( 75). 6, or $6 per package sol 6 P ( ), or $. per package sol (f) Te it is. Te maimum possible profit is $, montly. (g) Yes; in orer to sell more an more packages, te company migt nee to lower te price to a point were tey won t make any aitional profit.. Since te particle moves along te line y, it will be at te point (5, ) wen t () t 6t + 5t 5. Use a graper to see tat tis occurs wen t.8.. Grap C is position, grap A is velocity, an grap B is acceleration. A is te erivative of C because it is positive, negative, an zero were C is increasing, ecreasing, an as orizontal tangents, respectively. Te relationsip between B an A is similar. Copyrigt 6 Pearson Eucation, Inc.

33 Section.. Grap C is position, grap B is velocity, an grap A is acceleration. B is te erivative of C because it is negative an zero were C is ecreasing an as orizontal tangents, respectively. A is te erivative of B because it is positive, negative, an zero were B is increasing, ecreasing, an as orizontal tangents, respectively.. Note tat ownwar velocity is positive wen McCarty is falling ownwar. His ownwar velocity increases steaily until te paracute opens, an ten ecreases to a constant ownwar velocity. One possible sketc: Downwar velocity. (a) Time V r r r r v Wen r, ( ) 6 cubic r feet of volume per foot of raius. (b) Te increase in te volume is (. ) ( ) 9. cubic feet. 5. Let v be te eit velocity of a particle of lava. Ten st () vt 6t feet, so te velocity is s v t. t Solving s t gives v t. Ten te maimum eigt, in feet, is v v 6 v s v v 6. Solving v 9 gives v ± Te eit 6 velocity was about 8.7 ft/sec. Multiplying by 6 sec mi, we fin tat tis is 58 ft equivalent to about mi/. 6. By estimating te slope of te velocity grap at tat point. 7. Te motion can be simulate in parametric moe using () t t t + t 5 an y () t in [ 6, 8] by [, 5]. (a) It begins at te point ( 5, ) moving in te positive irection. After a little more tan one secon, it as move a bit past (6, ) an it turns back in te negative irection for approimately secons. At te en of tat time, it is near (, ) an it turns back again in te positive irection. After tat, it continues moving in te positive irection inefinitely, speeing up as it goes. (b) Te particle spees up wen its spee is increasing, wic occurs uring te approimate intervals.5 t.67 an t.8. It slows own uring te approimate intervals t.5 an.67 t.8. One way to etermine te enpoints of tese intervals is to use a graper to fin te minimums an maimums for te spee, 6t 6t+ using function t moe in te winow [, 5] by [, ]. (c) Te particle canges irection at t.5 sec an at t.8 sec. () Te particle is at rest instantaneously at t.5 sec an at t.8 sec. (e) Te velocity starts out positive but ecreasing, it becomes negative, ten starts to increase, an becomes positive again an continues to increase. Te spee is ecreasing, reaces at t.5 sec, ten increases until t.7 sec, ecreases until t.8 sec wen it is again, an ten increases after tat. (f) Te particle is at (5, ) wen t t + t 5 5 at t.75 sec, t.66 sec, an at t.9 sec. 8. (a) Solving 6 9t gives t ± 7. It took of a secon. Te average velocity 7 6cm was sec 8 cm/sec. ( 7 ) Copyrigt 6 Pearson Eucation, Inc.

34 Section. 5 s (b) V 98t t V a 98 t At s 6 cm, t sec (from part (a)) an 7 V cm/sec 7 a 98 cm/sec (c) Once te balls begin falling, eac flas will prouce a ifferent image. Tere are 6 images of te balls falling, so 6 flases 8 flases per secon. secons 7 9. Since profit revenue cost, te Sum an Difference Rule gives ( profit) ( revenue) ( cost), were is te number of units prouce. Tis means tat marginal profit marginal revenue marginal cost. 5. C; vt () 7 t 7 t 7 t 6. Te growt rate is given by b () t () t, t. At t : b ( ), bacteria/our At t 5: b ( 5) bacteria/our At t : b ( ), bacteria/our 7. (a) g ( ) ( ) ( ) ( ) t ( ) ( + ) (b) Te graps of NDER g(), NDER (), an NDER t() are all te same, as sown.. False; it is te absolute value of te velocity.. True. Te acceleration is te first erivative of te velocity wic, in turn, is te secon erivative of te position function.. C;. D; f ( ) + f ( ) ( ) + ( ) V( ) v s. E; ( + 7t t ) t t vt () 7 t< 7< t 7 < t 7 > (c) f() must be of te form f( ) + c, were c is a constant. () Yes; (e) Yes. f( ) f( ) + 8. For t >, te runners spee in feet per secon is v(t) (.5)t t. At a spee of feet per secon, t 7. t, er approac was 7. secons in time. istance. 5t. 5( 7. ) Her approac was feet in istance. Copyrigt 6 Pearson Eucation, Inc.

35 6 Section.5 9. (a) Assume tat f is even. Ten, f( + ) f( ) f ( ) f( ) f( ), an substituting k, f( + k) f( ) k k f( + k) f( ) f ( ) k k So, f is an o function. (b) Assume tat f is o. Ten, f( + ) f( ) f ( ) f( ) + f( ), an substituting k, f( + k) + f( ) k k f( + k) f( ) k k f ( ) So, f is an even function. 5. ( fg) [ f ( g)] f ( g) + g ( f) g f f g + g + f g g + f + fg Section.5 Derivatives of Trigonometric Functions (pp. 9). Wen te grap of sin stops increasing an starts ecreasing, te grap of (sin ) crosses te -ais from above to below.. Te slope of te grap of sin matces te value of (sin ) at tese points. 5. We conjecture tat (sin ) cos. Te graps coincie, supporting our conjecture. 6. Wen te grap of cos is increasing, te grap of (cos ) is positive (above te -ais). Wen te grap of cos is ecreasing, te grap of (cos ) is negative (below te -ais). Wen te grap of cos stops increasing an starts ecreasing, te grap of (cos ) crosses te -ais from above to below. Te slope of te grap of cos matces te value of (cos ) at tese points. We conjecture tat (cos ) sin. Te graps coincie, supporting our conjecture. Eploration Making a Conjecture by Graping te Derivative. Wen te grap of sin is increasing, te grap of (sin ) is positive (above te -ais).. Wen te grap of sin is ecreasing, te grap of (sin ) is negative (below te -ais). Quick Review Copyrigt 6 Pearson Eucation, Inc.

36 . sin. Domain: All reals Range: [, ] k 5. Domain: for o integers k Range: All reals 6. cos a ± sin a ± ( ) ± 7. If tan a, ten a + k for some integer k, so sin a ± Section.5 7 ( sin ) ( ) (sin ) (sin ) ( ) + [ cos + (sin )( )] cos sin ( + tan ) ( ) + (tan ) (tan ) ( ) + + sec + tan ( sec ) sec tan cos 8. cos ( cos ) ( + cos ) ( + cos ) cos ( + cos ) sin ( + cos ) 8. cos + ( + cos ) ( ) ( + cos ) ( + cos ) + cos + sin ( + cos ) 9.. y ( ) 6 y () Te tangent line as slope an passes troug (, ), so its equation is y ( ) +, or y 5. at () v () t 6t t a () Section.5 Eercises. ( + cos ) + ( sin ) + sin. ( sin tan ) cos sec 9. cot cot + ( + cot ) (cot ) (cot ) ( + cot ) ( + cot ) ( + cot )( csc ) (cot )( csc ) ( + cot ) csc ( + cot ) csc sin ( + cot ) sin (sin + cos ). + 5sin + 5cos. ( sec ) (sec ) + sec ( ) sec tan + sec Copyrigt 6 Pearson Eucation, Inc.

37 8 Section.5 cos. sin + ( + sin ) (cos ) (cos ) ( + sin ) ( + sin ) ( + sin )( sin ) (cos )(cos ) ( + sin ) sin sin cos ( + sin ) ( + sin ) ( + sin ) + sin s. vt () ( 5 sin t) t vt () 5cost v at () ( 5 cos t) t at () 5sint Te weigt starts at, goes to 5, an te oscillates between 5 an 5. Te perio of te motion is. Te spee is greatest wen cos t ± ( t k ), zero wen cos k t t, k o. Te acceleration is k greatest wen sin t ± t, k o, zero wen sin t ( t k ). s. vt () ( 7 cos t) t vt () 7sint v at () ( 7 sin t) t at () 7cost Te weigt starts at 7, goes to 7, an ten oscillates between 7 an 7. Te perio of te motion is. Te spee is greatest wen sin k t ± t, k o, zero wen sin t ( t k ). Te acceleration is greatest wen cos t ± ( t k ), zero wen cos k t t, k o. s. (a) vt () ( sin t) t t + vt ( ) cos t, spee cost v at () ( cos) t sint t t (b) v cos, spee a sin (c) Te boy starts at, goes up to 5, goes own to, an ten oscillates between an 5. Te perio of motion is. G s. (a) vt () ( cos) t t t vt () sin t, spee sin t v at () ( sin t) t t at () cos t (b) v sin, spee a cos (c) Te boy starts at, goes up to 5, an ten oscillates between 5 an. Te perio of te motion is. G s 5. (a) vt () ( sint+ cos) t t t vt ( ) cos t sin t, spee cos t sin t v at () ( cost sin t) t t at () sint cost (b) v cos sin v spee a sin cos 5 a Copyrigt 6 Pearson Eucation, Inc.

38 Section.5 9 (c) Te boy starts at, goes to 66.,an ten oscillates between.66 an.66. Te perio of te motion is. s 6. (a) vt () (cost sin t) t t vt () sint cos, t spee sin t cost v at () ( sint cos) t t t at () cost+ sint (b) v sin cos spee a cos sin + (c) Te boy starts at, goes to. 6, an ten oscillates between.6 an.6. Te perio of te motion is. a s jt () t t f() t cost f () t sint f () t cost f () t sint a s jt () t t f() t + cos t f () t sin t f () t cos t f () t sin t a s jt () t t f() t sint cost f () t cost+ sint f () t sint+ cost f () t cost sint. y sin + y (sin + ) cos y( ) sin + y ( ) cos tangent: y ( ) normal: m m y ( ) +. y sec y (sec ) sec tan y sec. y sec tan. tangent: y. ( ) +. y. +.. normal: m. 77 m y. 77( ) y sin y ( sin ) sin + cos y() () sin. 7 y () ()sin + () cos 86. tangent: y 8. 6( ) normal: m. m y. ( ) +. 7 y a s jt () t t f() t + sint f () t cost f () t sint f () t cost Copyrigt 6 Pearson Eucation, Inc.

39 Section.5 cos( + ) cos. (cos cos sin sin ) cos cos (cos ) sin sin cos sin (cos ) (sin ) cos sin (cos ) (sin ) (cos )( ) (sin )( ) sin 5. (a) (b) tan sin cos (cos ) (sin ) (sin ) (cos ) (cos ) (cos )(cos ) (sin )( sin ) cos cos + sin cos cos sec sec cos (cos ) ( ) ( ) (cos ) (cos ) (cos )( ) ( )( sin ) cos sin cos sec tan 6. (a) (b) cot cos sin (sin ) (cos ) (cos ) (sin ) (sin ) (sin )( sin ) (cos )(cos ) sin (sin + cos ) sin sin csc csc sin (sin ) ( ) ( ) (sin ) (sin ) (sin )( ) ( )(cos ) sin cos sin csc cot 7. sec sec tan wic is at, so te slope of te tangent line is. cos sinwic is at, so te slope of te tangent line is. 8. tan sec, wic is never. cos cot csc, wic is never sin. y( ) cos sin 9. ( ) y sin Te tangent line as slope an passes troug, cos,, so its equation is y +, or Copyrigt 6 Pearson Eucation, Inc.

40 . y + +. Te normal line as slope an passes troug,, so its equation is y +, or y +. y ( ) tan sec y ( ) ( ) sec sec ± cos ± On,, te solutions are ±. Te points on te curve are, an,.. y ( ) ( + cot csc ) csc + csc cot csc + csc cot (a) csc + csc cot + ()( ) Te tangent line as slope an passes troug P,. Its equation is y +, or y + +. y. Section.5 (b) f () csc + csc cot cos + sin sin ( cos ) sin cos at point Q y + cot csc + Te coorinates of Q are,. Te equation of te orizontal line is y. y ( ) ( + csc+ cot ) + ( csccot ) + ( csc ) csc cot csc (a) y csc cot csc ( )( ) ( ) Te tangent line as slope an passes troug P,. Its equation is y +, or y + +. Copyrigt 6 Pearson Eucation, Inc.

41 Section.5 (b) y ( ) csc cot csc cos sin sin ( sin cos + ) cos at point Q y + csc + cot + ( ) + ( ) Te coorinates of Q are,. Te equation of te orizontal line is y.. (a) Velocity: s (t) cos t m/sec Spee: s (t) cos t m/sec Acceleration: s () t sintm/sec Jerk: s (t) cos t m/sec (b) Velocity: cos m/sec Spee: m/sec Acceleration: sin m/sec Jerk: cos m/sec (c) Te boy starts at, goes to an ten oscillates between an. Spee: Greatest wen cos t ± ( or t k ), at te center of te interval of motion. k Zero wen cos t ( or t, k o), at te enpoints of te interval of motion. Acceleration: Greatest (in magnitue) k wen sin t ± (or t, k o) Zero wen sin t ( or t k ) Jerk: Greatest (in magnitue) wen cos t ± ( or t k ). k Zero wen cos t ( or t, k o). (a) Velocity: s (t) cos t sint m/sec Spee: s () t cost sin t m/sec Acceleration: s (t) sin t cos t m/sec Jerk: s (t) cos t + sin t m/sec (b) Velocity: cos sin m/ sec Spee: m/sec Acceleration: sin cos m/ sec Jerk: cos + sin m/sec (c) Te boy starts at, goes to an ten oscillates between ±. Spee: Greatest wen t + k Zero wen t + k Acceleration: Greatest (in magnitue) wen t + k Zero wen t + k Jerk: Greatest (in magnitue) wen t + k Zero wen t + k 5. y csc csc cot y ( csccot ) (csc ) (cot ) (cot ) (csc ) (csc )( csc ) (cot )( csc cot ) csc + csc cot Copyrigt 6 Pearson Eucation, Inc.

42 Section.5 6. y ( θ tan θ) θ θ (tan ) (tan ) ( ) θ + θ θ θ θ θsec θ + tanθ y ( θsec θ + tan θ) θ θ [(sec θ)(sec θ)] + (sec θ) ( θ) + (tan θ) θ θ θ θ (sec θ) (sec θ) + (sec θ) (sec θ) sec θ sec θ θ θ + + θsec θ tanθ + sec θ ( θ tan θ + )(sec θ) or, writing in terms of sines an cosines, + θ tanθ cos θ cosθ + θsinθ cos θ g() g() cos cos(), an g ( ) ( + b ) b. We 7. Continuous: Note tat + + require g ( ) g( ), so b. Te function is continuous if b. Differentiable: For b, te left-an erivative is an te rigt-an erivative is sin (), so te function is not ifferentiable. For oter values of b, te function is iscontinuous at an tere is no leftan erivative. So, tere is no value of b tat will make te function ifferentiable at. 8. Observe te pattern: 5 cos sin cos sin 5 6 cos cos cos cos 6 7 cos sin cos sin 7 8 cos cos cos cos 8 Continuing te pattern, we see tat n cos sin wen n k + for any wole number k. n Since 999 (9) +, cos sin. Copyrigt 6 Pearson Eucation, Inc.

43 Section.5 9. Observe te pattern: sin cos sin cos 5 6 sin sin sin sin 6 7 sin cos sin cos 7 8 sin sin sin sin 8 Continuing te pattern, we see tat n sin cos wen n k + for any n wole number k. Since 75 (8) +, sin cos.. Te line is tangent to te grap of y sin at (, ). Since y () cos (), te line as slope an its equation is y.. (a) Using y, sin (.).. 5. cos [(cos )(cos ) (sin )(sin )] (cos ) (cos ) + (cos ) (cos ) (sin ) (sin ) + (sin ) (sin ) (cos )( sin ) (sin )(cos ) sin cos ( sincos ) sin. True. s () t cos, t s cos >. Te erivative is positive at t. 5. False. Te velocity is negative an te spee is positive at t.. (b) sin(.).977 to igits. So. is only off by.8779 of te compute actual value. sin ( sincos ) (sin cos ) (sin ) (cos ) + (cos ) (sin ) [(sin )( sin ) + (cos )(cos )] (cos sin ) cos 6. A; y sin + cos y ( ) cos sin y( ) sin + cos y ( ) cos sin y ( ) y + 7. B; See 6. m m y ( ) 8. C; y sin y sin + cos y cos + cos sin k sin + cos s 9. C; vt () ( sin t) t t + vt () cost t Copyrigt 6 Pearson Eucation, Inc.

44 Section (a) Te it is 8 because tis is te conversion factor for canging from egrees to raians. (b) Tis it is still. (c) () (e) sin( + ) sin sin sin cos + cos sin sin sin (cos ) + cos sin cos sin sin cos + (sin )( ) + (cos ) 8 cos 8 cos( + ) cos cos cos cos sin sin cos (cos )(cos ) sin sin cos sin cos sin (cos )( ) (sin ) 8 sin 8 sin cos 8 sin 8 8 sin 8 Copyrigt 6 Pearson Eucation, Inc.