Mathematics Bridging course for Applied Sciences WS 2016/2017

Transcription

1 Mthemtics Bridging course for Applied Sciences WS 016/017 1

2 Writing with deciml power nd eponents Writing with deciml power nd eponents it is possible to write very smll nd very big numbers in compct wy. They build the bse for scientific writing (SCI).

3 nme deciml nottion eponenttion (scientific nottion) quintillion (trillion) =10 18 qudrillion =10 15 trillion (billion) =10 1 billion (Millird) =10 9 million =10*10*10*10*10*10=10 6 hundred thousnd =10*10*10*10*10=10 5 ten thousnd =10*10*10*10= 10 4 thousnd 1000 =10*10*10= 10 hundred 100 =10*10=10 ten 10 =10 1 one 1 =10 0 tenth 0,1 =1/10=10 1 hundredth 0,01 =1/100=10 thousndth 0,001 =1/1000=10 Ten thousndth 0,0001 =1/10000=10 4

4 Scientific nottion presenttion of numbers in the wy: A 10 n with 1 A < 10 nd n integer number e.g.: 0, = 6,54* =,50010*10 5 0,0800 =,800 *10 useful Link:

5 Eercise `scientific nottion 1.) 580=.) 0,007=.) 87,9654= 4.) 818,5000= 5.) 1,85*10 = 6.) 91,64*10 6 = 5

6 Units /Prefies SI units (french: Système interntionl d unités) Interntionl system for physicl quntities The SI system eists of 7 bse units All other physicl units derive from these bse units derived units

7 SI units mesurement unit symbol length metre m mss kilogrm kg time second s temperture kelvin K mount of substnce mole mol electric current mpere A luminous intensity cndel cd 7

8 frequency hertz force newton pressure pscl energy joule power wtt voltge volt s Hz 1 s kg m N s m kg m N P s kg m m N J s kg m s J W A s kg m A W V Derived SI units, e.g.

9 Prefies re dded to unit nmes to produce multiple nd sub multiples of the originl unit. All multiples re integer powers of ten to void lot of positions fter deciml point e.g. 7000m = 7*10 m = 7 km unit is m (metre) k (stnding for kilo) is the prefi nd replces the fctor 1000 respectively 10

10 SI Prefies fctor prefi symbol/bbrevition pet P 10 1 ter T 10 9 gig G 10 6 meg M 10 kilo k 10 hekto h 10 1 dek d 10 1 dezi d 10 centi c 10 milli m 10 6 micro µ 10 9 nno n (10 10 Angström Å) 10 1 pico P femto f 10

11 Eercise `use of prefies Write the following vlues with SI prefi nd vice vers: e.g. 4,85* 10 9 g = 4,85 ng oder,58 mg=,58*10 g 1.),16*10 m = 4.) 4, cl =.) 5,98*10 9 s = 5.),50 ng =.) 58,89*10 g = 6.) 5µmol =

12 Conversion of units emple km = cm considertions: km => prefi kilo = 10 cm => prefi centi = 10 difference of the two prefies: 5 powers of ten (from to ) conversion from higher prefi to lower prefi: number hs to be multiplied by the fctor (difference of power of ten) 159 km = 159 *10 5 cm

13 Conversion of units emple 4 nm = cm considertions : nm => prefi nno = 10 9 cm => prefi centi = 10 difference of the prefies: 7 powers of ten (from 9 to ) conversion from lower prefi to higher prefi: number hs to be divided by the fctor (difference of power of ten) 4 nm = 4/10 7 cm = 4*10 7 cm

14 Eercise `conversion of units Convert the following quntities into the stted units. 1. 7m (dm). 6 km (m). 5 dm (cm) 4. 5 dm (µm) 5. nm (cm) cm (dm) mm (cm) 8. 5 µm (mm) 9. 5m (µm) dm (mm) m (mm) mm (dm) 1. 6 µm (nm) m (km) Useful links/interctive tsks:

15 Eercise `conversion of units Convert the following quntities into the stted units t (kg) (note: the tonne (t) is no SI unit but is gently used for msses). mg (kg). 5 ng (mg) µg (g) 5. 6 ms (s) 6. 5 µs (s) 7. 5 L (nl) 8. 4 Gbit (Mbit) mol (µmol) 10. 5,5 mmol (mol) 15

16 Eercise `conversion of units Convert the following quntities into the stted units. 1. 0,5 m (dm ). 5 L (dm³). 1 ml (dm³) 4. 1 m³ (mm³) 5. 6 cm³ (mm³) 6. 0,5 mm³ (cm³) 7. 0,6 L (cm³) 8. 1 mm³ (L) emples: 1 m = 1m*1m = 1*10 cm*1*10 cm = 1*10 4 cm 1 m = 1m*1m*1m = (1*10 cm)*(1*10 cm)*(1*10 cm) = 1*10 6 cm 1 m = 1000 L 16

17 Eercise `conversion of units Convert the following quntities into the stted units. 1. min (s). 5 h (min). 5 min (h) 4. 1 d (s) 5. 1,5 (s) 6.,5 min (s) 7. min 50s (s) 8. 7,15 K ( C) 9. 5 C (K) emples: in min: = *65d/*4h/d*60min/h= (*65*4*60) min=105100min 0 C = (0+7,15) K 17

18 1. Write the vlue in scientific nottion (using the writing of the power of ten with one pre deciml plce). Write the vlue in suitble unit with prefi. Emple: 0,00876 m step 1 8,76 10 m step 8,76 mm 18

19 1. 0,0048 g. 0,00478 mm. 0,0098 L 6. 0,006 m 4. 0,049 kg 7. 0,00145 s 5. 0,004 µm 8. 0, cm 9. 0, dm² 10. 0,00846 m² 19

20 Eercise `clculting with units Convert into the stted units nd clculte m 1 mm [ cm ]. 5 cm 51 µm [ mm ]. 4 cm 16 nm [ mm ] 4. 1g 15 kg [ kg ] 5. 5 mg µg [ mg ] 0

21 Eercise `clculting with units Convert into the stted units nd clculte g 78mg [ g ] µg 674 ng [ mg ] 8. 1mL 5dm [ dm ] 9. 6L cm [ L] 1

22 Eercise `clculting with unit Convert into the stted units nd clculte min 60 s [min] s,5h [min] 1. 4d 50 min [ h] s 4 s 16 min [ h]

23 Eercise `clculting with units Convert into the stted units nd clculte. 1. V 5µm 6mm cm [ cm³]. V 4dm16cm5µm [ m³]. V 4mm 16dm [ L]

24 Eercise `clculting with units.) An nt is moving with 9 km/d. Wht is the velocity in km/h respectively cm/min? b.) 5 mg/100ml = mg/l c.) 50 mmol/l = mol/l d.) 490 mg/l = g/ml 4

25 5 Eercise `derived units N s kg m s kg km 50 9 e) J s kg m g m 5min 500 ) (5 f) N s kg m h g s mm 1 4 ) (4 g)

26 Eercise `derived units h) 8,8t kg 0,km (min) m s P i) ng 0,nm 0,µs kg m s P 6

27 Summtion nottion (cpitl sigm nottion) Stopping point upper limit of summtion summtion sign Inde of summtion (continous inde) function of inderepresenting ech successive terme in the rw Strting point lower limit of summtion In tht emple: the inde k gets vlues from 1 (strting point) to 5 (stopping point): 1,,,4,5. The inde is lwys incremented by 1

28 Summtion nottion (cpitl sigm nottion) Emple: In tht emple the inde k cn only ccept vlues from 1 (strting point) to 4 (stopping point) in entired steps: 1,, nd 4, tht will be dded.

29 Emple summtion nottion: In tht emple the inde i cn only ccept vlues from 1 (strting point) to (stopping point) in entired steps: 1, 0, 1, nd, tht will be inserted in the function for i nd thn will be dded.

30 Emple summtion nottion: In tht emple the inde i cn only ccept vlues from 1 (strting point) to 7 (stopping point) in entired steps: 1,,, 4, 5, 6 nd 7, tht will be inserted in the function for i nd thn will be dded.

31 Eercises summ Eercise `summtion nottion : Write in eplicit/elborted wy

32 Eercise `summtion nottion : Write in eplicit /elborted wy One more detiled video eplntory video To how to hndle summtion nottions (in english):

33 Frctions

34 Epnding + Reducing Epnding mens multiplying the numertor nd denomintor of frction by the sme non zero number b b c c Reducing mens dividing the numertor nd denomintor of frction by the sme non zero number b b : : c c 4

35 Addition: Subtrction: Multipliction: Division: d b b c d d c b d b b c d d c b d b c d c b c b d c d b d c b : 5

36 Eercise `frctions 1 d b b c d d c b d b b c d d c b 5 4 ) b) c) 5 1 ) b) c) 6

37 d b c d c b Eercise `frctions

38 c b d c d b d c b : 5 : : : : : : Eercise `frctions

39 Eercise `frctions

40 Percentge clcultions 40

41 Percentge % mens: divide number by 100 Emple: % = /100 = 0, Converting in percentge: multiply number with 100 Emple: 0,144 = 0,144*100 = 14,4% 41

42 Eercise `epress s percentge , , ,9 4

43 Eercise : convert the percentges into frctions ,5% 5. 16,09%. 6,% %. 44,4% 7. 0,5% 4. 78% 4

44 X% of Y (percentge of ) Emple: 5% of 100 5% 100 0,

45 Eercise ` percentge of 1. 1% of 4. % of % of ,88% of ,% of

46 Systems of liner equtions Aim: the simplest method for solving system of liner equtions is to repetedly eliminte vribles. methods ( possibilities) : by Equlizing by Substitution by Addition 46

47 System of liner equtions with two vribles Generel formul: (1) 1 b 1 y c 1 () b y c 47

48 Equlizing Emple (1) ( ) y 8 y y 5 8 y 48

49 Eercise `system of liner euqtions solve by the method of `equlizing` y 8.) y 17 4 y b.) y 49

50 Substitution () (1) y y y y y y y y y Emple 50

51 Eercise `system of liner euqtions solve by the method of `substitution` 6 11y 4.) 6 5 y b.) 15 4y 90 4y 6 51

52 Addition Emple y y y y y y 5

53 Eercise `system of liner euqtions solve by the method of `ddition` 1 1 y 6.) 4 4 y y 7 b.) 8 y 115 5

54 Binominl theorem ( b) b b ( b) b b b ( b) *( b) 54

55 Pscl`s tringle 55

56 Eercise Pscl`s tringle Determine the following equtions with Pscl`s tringle. 56

57 Eercise `binominl theorem clculte: 17 ² ³ 768 ² ² 6144 ³ ² ) (5 6) (5 4. ) (5. 6)³ (5. 5)² ( m zm m z m z z y 57

58 Eponents n m n m 1) n m n m ) m n n m ) ( ) n n 1 ) : 0, (³) 1 4 4

59 5) b n b n n ) ( b) n n b n ( 4b)³ 4³ ³ b³ 64³ b³ 7) )

60 Eercise `eponents ( n) m n m 4 ) 4 ) d 0 4 ) b b b 4 ) b e 4 4 ) b c ( n) m n m 4 ) 4 c) 4 0 ) u u b 1 ) d 60

61 Roots 1) n m m n ) n n ) n n

62 n n n 4) b b ) n n b n b ) m n m 1 n 1 mn mn ) n 1 1 n ,5 6

63 Eercise `roots )( ) b)( ) c)( 1 ) 4 d) 0 5 e) f ) g) 16 h) i) j) k)

64 Logrithmic functions 64

65 Logrithmic functions The eqution b one rel number. eits of ect This is the logrithm of the number b in respect to the bse log ( b) short : log ( b) b log ( b) b ³ 8 65

66 Specil logrithms common logrithm: (logrithm of with respect to bse 10) log 10 ( r) lg( r) Binry logrithm: (logrithm of with respect to bse ) log ( r) lb( r) nturl logrithm: (logrithm of with respect to bse e) (mthemticl constnt e =,718, clled Euler's number ) log e ( r) ln( r) 66

67 Clcultion of logrithm log ( b) log log ( b) ( ) lg(56) log 4 (56) lg(4) 67

68 Eercises `logrithm 1 1 log ) (100) log ) (1) log ) 7 log ) 5 1 log ) (8) log ) f e d c b 68 0,001 log g) 10

69 Logrithmic rules y y log log ) ( log ) 1 v u v u log log log ) log(15) ) log(5 log() 5) log( log(0,65) 8 5 log log(8) 5) log( 69

70 k k log log ) ) ( log 1 log ) 4 b n b n ) ( log 4 ) ( log 4 )) ( log ) ( (log 1 ) ( log 1 ) ( log log 1 y y y y 70

71 Further rules 1 1) log log ) log 1 0 log 1 6 log(1) log(6) log(6) 71

72 Eercise `logrithm Simplify the following terms: ) log log 10 5 b) log log c) log n f ) log 7

73 Eercise `logrithm 1. log y rs. lg 6³ 7z. * lg( ) 5 * lg( y) * lg( z) 4. 1 * lg( ² y²) * lg( ) 7

74 Clcultion of emple 5 14 log(5 ) log(14) log(5) log(14) log(14) log(5) ( 1,64) 74

75 Determine the solution e d c 75 log 1 56 ) log ) b

76 e function/ln function 76

77 Clcultion of emple ln() 6 ln(); 6 ln() 6 ² ln() 6 ² ln() ² 0,5 0 1 ² 0,5 ² 0,5 e e 77

78 Eercise `e function Clculte: ) e 10 b) ep() ep() c) ln() 16 d) 4 ln( ) ln( 4) 78

79 Determine 1. ³ ln( ³) 0. ( e ) ( e ) 0. ( e 1) (ln( ) 1) 0 4. e ² 1 79

80 Trigonometric functions rdin b 80

81 Importnt ngles in rdin nd degree rdin degree 0 0 π/6 0 π/4 45 π/ 60 π/ 90 π 180 π 60 81

82 Conversion of ngles from degree to rdin: 180 from rdin to degree: 180 clcultor: deg = degree; rd = rdin Sense of rottion: clockwise rotting ngles re negtiv, counterclockwise rotting ngles re positiv 8

83 Eercise `converting ngles 1. 4 =. 16 =. 17 = 4. 0,94 = 5. 1,81 = 6. 5,97 = 8

84 side b: vis à vis point B respectively the ngle β. Side b is opposite the ngle β `oppsite to the ngle β. Side b is close to the ngle α `djcent to the ngle α Trigonometric functions for right ngled tringles side : vis à vis point A respectively the ngle α. Side is opposite the ngle α `oppsite to the ngle α. Side is close to the ngle β `djcent to the ngle β The side opposite to the right ngle (90 ) is clled hypotenuse, here: c = hypotenuse 84

85 Definitions in right ngled tringles: Sine (sin) of n ngle= opposite of n ngle hypotenuse cosine (cos) of n ngle = djcent of n ngle hypotenuse tngent (tn) of n ngle = opposite of n ngle djcent of n ngle cotngent (cot) of n ngle = djcent of n ngle opposite of n ngle

86 Specil ngles sin cos tn , ,

87 Determintion of ngles Angles re determined by the inverse trigonometric functions: 87

88 Determintions in right ngled tringles emplel: = 7,6 cm; c = 15,5 cm; γ = 90 88

89 Eercise `trigonometric functions Clculte the missing vlues! 1. b =,4cm; c =,cm; γ = 90. = 5,cm; α = 66,5 ; γ = 90. c = 1,5cm; β = 7, ; γ = b = 1,6cm; α =, ; γ = 90 89

90 Function of sine nd cosine function of sine nd cosine sin() bzw.cos() 1,5 1 0, ,5-1 -1,5 ngle in degree sin cos 90

91 Importnt correltions: sin ) cos( Additions theorem: ) sin( ) sin( ) cos( ) cos( ) cos( ) cos( ) sin( ) cos( ) sin( ) sin( y y y y y y cos ) sin( ) cot( 1 ) cos( ) sin( ) tn( 91

92 Solving equtions Qudrtic equtions Root equtions Frction equtions 9

93 Solving qudrtic equtions q p p q p 1; ) ( 0 qudrtic eqution (p-q-formul) 7 1; 91) ( ² 1 1, 9

94 About qudrtic equtions qudrtic equtions hve t most solutions The solving formul: gives: one solution if: two solutions if : no solution if: q p p ) ( 0 ) ( q p 0 ) ( q p 0 ) ( q p 94

95 Eercise `qudrtic equtions 7 ² 6 4. ) 4 (15 5) ( 9) (7 ) ( ² ² 1. 95

96 Root equtions emple proof squring is not equivlent conversion > tht mens you hve to proof 96

97 Eercise `root equtions

98 Frction equtions emple 4 5; 0 4 9² ² ² 0 10 ) ( ) ( / HN 98

99 Eercise `frction equtions determine t: t t bt t b b b 6b t 99

100 ppliction

101 density (mteril constnt) density( i ) mss( m volume( V i ) i ) typicl units : g cm³ kg dm ; kg m ; g ml ; kg L 101

102 Fresh snowfll hs density of 0,0 g/cm³.. which weight hs fresh snowfll of 0 cm thickness on flt roof of 0 cm length nd 10 m width? b. If this mount of snow melts. How mny liter of wter re formed? 10

103 A irregulr formed piece of jewellery (trinket) weighs 0,177N in ir, t thin fiber the lifting power in wter is 0,017N. Is the trinket mde of gold? obtin: F=m*g g= 9,81 N/kg; density (gold)= 19, kg/dm³; density (wter)= 0,998 kg/dm³ 10

104 Dilutions 104

105 Dilutions dilute: the concentrtion of solved substnce in solution is reduced. Add wter 8 Pkt/100 ml +100 ml wter 100 ml 100 ml shke 8 Pkt/00 ml = 4 Pkt/100 ml 105

106 Dilution fctor F F initil concentrtion finl concentrtion finl volume initil volume Initil volume + wter = finl volume 106

107 emple: Crete 50 ml of physiologic slt solution 0,9% out of 10% slt solution: initil concentrtion: 10% finl concentrtion : 0,9% finl volume: 50 ml Needed initil volume? Needed volume of wter? 107

108 F initil concentrtion finl concentrtion finl volume initil volume 10% 0, 9% 50mL initil volume initil volume 50mL 0,9% 10% 4,5mL V ( wter) 50mL 4,5mL 45,5mL

109 Eercise `dilutions 1. You hve 10 times concentrted buffer which should be diluted to one times. Crete 500 ml of the buffer. How mny ml buffer nd wter re needed? 109

110 Eercise `dilutuions. For determining the concentrtion of etrcted DNA you dilute 100µL of the DNA solution with 400µL wter. The diluted solution hs concentrtion of 50 mg DNA per ml. Wht is the concentrtion of the initil solution? Clculte the dilution fctor! How mny DNA is isolted if the volume of the initil solution ws 5mL? 110

111 Digrms 111

112 is lbeling including units, y is: is of ordintes (dependent dimension) Single dt sign points or crosses Best fit curve (=regression line, = trend line) informtive digrm title Especilly if more thn two grphs re presented in the digrm legend is necesrry. Ais clssifiction in well spent wy of using the given plce is lbeling including units, is: is of bscisse 11 (independent (preset) dimension)

113 Digrm emple 11

114 Eercise `digrms/liner equtions 1.) Mesurements of the solubility L of slt in wter depending on the temperture T give following dt: i T i [ C] L i [g/100ml] 70,7 88, 104,9 14,7 148,0 176,0.) Drw the corresponding digrm. b.) Determine the eqution of the regression line by the digrm c.) Determine the solubility of the slt by 6,5 C c1.) grphiclly (in /by digrm) c.) clcultive (by the liner eqution) 114

115 Eercise `digrms/liner equtions.) The concentrtion of n pple juice smple is determined. Therefore stndrds (solutions of juices with known concentrtions) re prepred. The smple nd the stndrds re mesured by photometer. Tht mens which concentrtion cuses which colour intensity (etinction) nd in the other wy round which colour intensity (etinction) is which concentrtion? Solution Concentrtion in % Etinction.) Drw the digrm on millimetre pper. b.) Determine the eqution of the regression line. c.) Clculte the concentrtion of the smple. Smple 115

116 Eercise `liner equtions.) Two points of stright line re known (pir of vrites): P(/) nd Q ( 1/ ). Determine the liner eqution. 4.) A lyer of lipids is 100 nm nd grows 5 nm per dy..) Give n eqution to clculte the lipid lyer t ny time. b.) How thick is the lipid lyer fter one week?

117 Arithmetic men Definition of the rithmetic men: ) 5,0,8,6,9,5 1 n ( 1... ) 1 n n i n i1 b.),94;,90;,9;,76;,80;,85;,84;,86;,8;,87 c.) 6m, 7m, 4m, 4m, 5m,,m 4m, 7m, 0m, 5m, 5m, 6m, m 117