Apply & Practice 3.5 Set 1: P #3-18 (mult. of 3); 19 #21 write explicit #27-33 (mult. of 3) point #39-40 eqn tang line from graph

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1 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson WATCH for the product rule and the chain rule. If the order that our terms are in differ from the answer ke, it is not a problem. For ab, I use ab ab where as the book uses ab ab. The are equivalent. Appl & Practice.5 Set : P 7-7 #-8 (mult. of ); 9 # write eplicit #7- (mult. of ) point #9-40 eqn tang line from graph or OR d d note: 5. alternative, if ou distribute st: sin tan cos tan sec cos tan sec cos tan sec Alwas good to get rid of comple fractions. Multipl numerator and denominator b. S. Stirling Page of 6

2 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson since 6 Note: Implicit is MUCH easier! At,0 : is undefined. Note: Could substitute here! Must distribute to isolate the term. Don t miss the chain rule on this! (Which includes a product rule.) u u S. Stirling Page of 6

3 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson 4 or 4 or S. Stirling Page of 6

4 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson Substitute for. b/c: 6 Appl & Practice.5 Set : P 7-74 #5, 5, 56 nd derive. #60, 64, 9 #49, 50 inverse trig. #6 normal line #9, 95 challenge easier to use after dividing both sides b. st derivative rewrite nd derivative Substitute for. OR Using the Quotient Propert: At, 5, Tangent line: 5 5 Normal line: 0 0 so horizontal line. Normal line has perpendicular slope so undefined (vertical line) S. Stirling Page 4 of 6

5 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson Find the derivative: For a horizontal tangent 0 happens when 44 0 and : Find matching -coordinate , 4 Horizontal tangents at,0 and, 4. For a vertical tangent DNE happens when 0 and : Vertical tangents at 0, 0,,. and Find matching -coordinate. Now need to find an or a, so use substitution into 5 to solve simultaneous equations. Use the simplest relationship to find the matching -coordinates. 4 When, 4. 4 When, 4. S. Stirling Page 5 of 6

6 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson Remember, a is the.6 Derivatives of Inverse Functions Appl & Practice.6 sin when 6 f f cos 6 6 or f f f input for and the output of f. IN OUT 6 a 6 P 79-8 #, 6 deriv. a. #7, 8 show slopes inv. # eqn tang. line #-4 (mult. of ) derive. #45, 5 tangent lines #5, 56 linear appro. ONLY! #59 implicit. #6, 64 writing f ( ) So f 8 f ( ) 4 4 IN OUT f 8 f a 8 f (8) f () 4 4 It sas show that, so ou are doing it the long wa! S. Stirling Page 6 of 6

7 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson 4 4 g( ) = 4 4 = 4 u g( ) u u u u g( ) cos cos cos Multipl num. and den. b to get rid of the comple fraction. arccos 0 arccos arccos S. Stirling Page 7 of 6

8 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson Need to get common denominators, so S. Stirling Page 8 of 6

9 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson 4 f ( ) arccos( ) f( ) 4 and When, cos 6 6 When, 5 cos f 4 4 f sin 6 P 6 f 0 0 P 0 S. Stirling Page 9 of 6

10 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson or Inverse Trig. Functions If u is a differentiable function of. d u d u sin u cos u d u d u d u d u tan u cot u d u d u d u d u sec u csc u d u u d u u S. Stirling Page 0 of 6

11 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson.7 Related Rates Appl & Practice.7 Set P #5, 8, 0,, 4, 5 dr 5. cm da? min A r da dr r Note that the area changes more quickl the larger the radius gets! d cm min change in horizontal direction. d? change in vertical direction. moving downward fast. dr 8. in dv? min 4 V r dv dr 4 r no velocit change in direction moving upward fast (b) Because 4, and dr are constants, sa k. dv dr 4 r dv kr So dv is directl proportional to r. Not a constant rate of change. Start with the relationship and accumulate the constants. Remember, definition of directl proportional? is directl proportional to, if k as goes up b one, increases k times A linear relationship! S. Stirling Page of 6

12 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson 4. tan 0 d d sec 0 d d 0 sec 0 sec Need, when = 0: 0 tan when 0 4 and sec 4 d Obs 0 0 Now 0 Balloon d d? when = 0. d radians 0 sec. So the balloon s angle of elevation is increasing at a rate of /0 radians per sec. L (a) Speed of boat as it approaches the dock? dl constant, L is more than. d L dl As 0 the speed of the boat increases (approaches ) as it approaches the dock. d.6 Confirm? faster speed (a) dl 4 ft d sec? when L =. L dl d L 0 d L dl and Need, when L = : 5 and 5 Or know 5:: right triangle. 5 d ft sec So boat is traveling at 0.4 ft sec when there is feet of rope out. Boat (b) d 4 ft dl sec? when L =. dl d L Again, when L =, 5 dl ft sec.58 So the winch pulling in the rope at a rate of.58 ft sec feet of rope out. when there is As L 0, dl increases d constant, L is more than. As L 0, dl Confirm? Tr L.5 dl.5 4.9ft.5 sec faster speed. S. Stirling Page of 6

13 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson P s P (a) d 450 mph when 50 mph d 600 mph when 00 mph ds? s need s! s so s 50 s s ds d d ds ds mph 500 The distance between the planes is decreasing at a rate of 750 mph. (b) since d = rt distance time= rate 50 t hour 750 or 0 min to get one of the planes off of its path. dr in min dv? know h r Appl & Practice.7 Set P #,, 4, 6, 6, 7, 5 V r h note: different variables! V r r get in terms of variables: ou have rates for V and r onl! V r dv dr r Now ou have the info! dv ft? 0 when h 5 know d h min dh V r h note: different variables! From given, need in terms of V and h onl! and know r h or r h V h diam= height h V h Now ou can differentiate. 4 dv dh h dh 4 dh 8 ft 405 or dh ft min min S. Stirling Page of 6

14 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson 5 dv ft? 0 when h = 8? min dh V r h But need to write it onl in terms of V and h because no dr in rate list. Use Similar triangles to find r in terms of h: So r h or 5h r 5 V r h get in terms of V and h onl! 5h V h 5 V h now differentiate! 44 dv 5 dh h dh 44 dh 9 ft 0 min The depth of the water is increasing at a rate of 0.86 ft/min r h Similar triangles: So sm r h lg 5.5 a h (a) dv ft? min when h = ft? dh V Bh prism, hprism ft B a h Need in terms of V and h! Tr using Similar triangles: So a.5 h or a h and a h so the base of the = the height! Note: The area of the base (a triangle): B h h Volume of the water! V Bh prism V h h V 6h dv dh h dh dh ft 6 min simplif The water level is rising at a rate of /6 ft/min. (b) All measures in feet, so dh in ft ft 8 min in min dv? when h = ft? from (a) dv dh h dv ft 4 min The water is being pumped into the trough at a rate of ¾ cubic feet per minute. S. Stirling Page 4 of 6

15 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson (a) d 5 ft d sec? (b) d 5 ft ds sec? Need a relationship Using Similar triangles: So 6 or d 0 d 7 d ft 7 7 sec Need a relationship (length of shadow is s). Using Similar triangles, in terms of and s: s and s so s s s 6s 6 4s 6s s 7 ds d 7 ds ft 7 4 sec OR Since ou know d from the previous problem, use s and the Difference Rule If s then ds d d So ds 50 5 ( 5).4 ft 7 7 sec From 7(a) See net page: = 7, = 4, d ft sec, d 7 ft sec Need second derivative (acceleration) 5 0 d d simplif d d d d d d d d Moving at a constant rate so d 0 Need more information, d : Use d ft sec so d ft 0 sec Now substitute: 7 d 4 70 d ft sec = S. Stirling Page 5 of 6

16 Ch 0 Homework Complete Solutions V Part : S. Stirling Calculus: Earl Transcendental Functions, 4e Larson (c) d ft d sec? when = 7 ft? (a) d ft d sec? when = 7 ft? when = 5 ft? when = 4 ft? 5 d d 0 d d or d When = 7 ft, 7:4:5: d When = 5 ft, 5:0:5: d When = 4 ft, 4:7:5: d ft 4 sec ft 0 0 sec ft 7 7 sec tan d d d sec d d d cos Need more information: From (a) = 7, = 4, d ft sec, d 7 ft sec 4 and need cos d 4 = rad sec (b) d ft da sec? when = 7 ft? A da d d Need more information: From (a) = 7, = 4, d ft sec, d 7 ft sec da ft 4 sec S. Stirling Page 6 of 6