mathematics –bridging course for applied sciences · 2015-08-06 · mathematics –bridging...
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Writing with decimal power and exponents
Writing with decimal power and exponents it is possible to write very small and very big numbers in a compact way. They build the base for scientific writing (SCI).
name decimal notation exponentation (scientific notation)
quintillion (trillion) 1 000 000 000 000 000 000 =1018
quadrillion 1 000 000 000 000 000 =1015
trillion (billion) 1 000 000 000 000 =1012
billion (Milliard) 1 000 000 000 =109
million 1 000 000 =10*10*10*10*10*10=106
hundred thousand 1000000 =10*10*10*10*10=105
ten thousand 10000 =10*10*10*10= 104
thousand 1000 =10*10*10= 103
hundred 100 =10*10=102
ten 10 =101
one 1 =100
tenth 0,1 =1/10=10-1
hundredth 0,01 =1/100=10-2
thousandth 0,001 =1/1000=10-3
Ten thousandth 0,0001 =1/10000=10-4
Scientific notation
presentation of numbers in the way:
A 10n
with 1 ≤ A < 10 and n integer number
e.g.: 0,000654 = 6,54*10-4
350010 = 3,50010*105
0,02800 = 2,800 *10-2
useful Link: http://science.widener.edu/svb/tutorial/scinot.html
Exercise `scientific notation´
1.) 25802=
2.) 0,0027=
3.) 87,9654=
5
4.) 818,5000=
5.) 12,85*102=
6.) 913,64*10-6=
Units /Prefixes
SI-units (french: Système international d’unités)
�International system for physical quantities
�The SI-system exists of 7 base units
�All other physical units derive from these base units � derived units
SI-units
7
measurement unit symbol
length metre m
mass kilogram kg
time second s
temperature kelvin K
amount of substance mole mol
electric current ampere A
luminous intensity candela cd
frequency hertz
force newton
pressure pascal
energy joule
power watt
voltage volt
sHz
1=
2s
kgmN
⋅=
22sm
kg
m
NPa
⋅==
s
kgmmNJ
⋅=⋅=
2
3
2
s
kgm
s
JW
⋅==
As
kgm
A
WV
⋅
⋅==
3
2
Derived SI-units, e.g.
Prefixes are added to unit names to produce multiple and sub-multiples of the original unit. All multiples are integer powers of ten to avoid a lot of positions after decimal point
e.g. 7000m = 7*103 m = 7 km
� unit is m (metre)
� k (standing for kilo) is the prefix and replaces the factor 1000 respectively 103
10
SI-Prefixesfactor prefix symbol/abbreviation
1015 peta P
1012 tera T
109 giga G
106 mega M
103 kilo k
102 hekto h
101 deka da
10-1 dezi d
10-2 centi c
10-3 milli m
10-6 micro µ
10-9 nano n
(10-10 Angström Å)
10-12 pico P
10-15 femto f
Exercise `use of prefixesWrite the following values with SI-prefix and vice versa:
e.g. 4,85* 10-9 g = 4,85 ng oder 2,58 mg= 2,58*10-3 g
1.) 3,16*10-3 m = 4.) 34,2 cL =
2.) 5,98*109 s = 5.) 2,50 ng =
3.) 58,89*103 g = 6.) 5µmol =
Conversion of units
example 1 159 km = ________cm
considerations:
km => prefix kilo = 103
cm => prefix centi = 10-2
conversion from a higher prefix to a lower prefix: number has to be multiplied by the factor (difference of power of ten)
� 159 km = 159 *105 cm
difference of thetwo prefixes:5 powers of ten(from 3 to -2)
Conversion of units
example 2 4 nm = ________cm
considerations :
nm => prefix nano = 10-9
cm => prefix centi = 10-2
conversion from lower prefix to a higher prefix: number has to be divided by the factor (difference of power of ten)
� 4 nm = 4/107 cm = 4*10-7cm
difference of the prefixes:7 powers of ten(from -9 to -2)
Exercise `conversion of units´
1. 7m (dm)
2. 6 km (m)
3. 5 dm (cm)
4. 5 dm (µm)
5. 32 nm (cm)
6. 560 cm (dm)
7. 940 mm (cm)
8. 5 µm (mm)
9. 5m (µm)
10. 72 dm (mm)
11. 37 m (mm)
12. 6300 mm (dm)
13. 6 µm (nm)
14. 700 m (km)
14
Convert the following quantities into the stated units.
Useful links/interactive tasks:http://exercises.murov.info/ex1-2.htmhttp://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i17/bk8_17i2.htm
1. 5 t (kg) (note: the tonne (t) is no
SI-unit- but is gently used for masses)
2. 3 mg (kg)
3. 5 ng (mg)
4. 16 µg (g)
5. 36 ms (s)
6. 5 µs (s)
7. 25 L (nL)
8. 4 Gbit (Mbit)
9. 17 mol (µmol)
10. 25,5 mmol (mol)
15
Exercise `conversion of units´
Convert the following quantities into the stated units.
16
1. 0,5 m2 (dm2)
2. 5 L (dm³)
3. 21 mL (dm³)
4. 1 m³ (mm³)
5. 36 cm³ (mm³)
6. 0,5 mm³ (cm³)
7. 0,6 L (cm³)
8. 12 mm³ (L)
examples: • 1 m2 = 1m*1m = 1*102 cm*1*102 cm = 1*104 cm2
• 1 m3 = 1m*1m*1m = (1*102 cm)*(1*102 cm)*(1*102 cm) = 1*106 cm3
• 1 m3 = 1000 L
Exercise `conversion of units´
Convert the following quantities into the stated units.
17
1. 3 min (s)
2. 5 h (min)
3. 35 min (h)
4. 1 d (s)
5. 1,5 a (s)
6. 2,25 min (s)
7. 2 min 50s (s)
8. 273,15 K (°C)
9. 25 °C (K)
examples: • 2 a in min: 2a= 2a*365d/a*24h/d*60min/h= (2*365*24*60) min=1051200min • 20 C = (20+273,15) K
Exercise `conversion of units´
Convert the following quantities into the stated units.
1. Write the value in scientific notation (using the writing of the power of ten with one pre decimal place)
2. Write the value in a suitable unit with a prefix.
18
Example:
mm
m
m
76,8
1076,8
00876,03
=
⋅= −step 1
step 2
µm
kg
L
mm
g
0024,0.5
0349,0.4
0098,0.3
00478,0.2
0048,0.1
19
²008463,0.10
²000574,0.9
000789,0.8
00145,0.7
0036,0.6
m
dm
cm
s
m
20
Exercise `calculating with units
][2325.5
][151.4
][1634.3
][5125.2
][125.1
mgµgmg
kgkgg
mmnmcm
mmµmcm
cmmmm
+
+
+
−
+
⋅
Convert into the stated units and calculate.
21
Exercise `calculating with units
][236.9
][51.8
][367490.7
][78456.6
3
33
LcmL
dmdmmL
mgngµg
gmgg
−
+
+
+
⋅
Convert into the stated units and calculate.
22
Exercise `calculating with unit
][min162415.13
][min504.12
[min]5,345.11
[min]60min35.10
hss
hd
hs
s
++
+
+
+⋅
Convert into the stated units and calculate.
23
][164.3
³][51634.2
³][2265.1
2LdmmmV
mµmcmdmV
cmcmmmµmV
⋅=
⋅⋅=
⋅⋅=
Exercise `calculating with units
Convert into the stated units and calculate.
24
a.) An ant is moving with 9 km/d. What is the velocity in km/h respectively cm/min?
b.) 25 mg/100mL = ________ mg/L
c.) 250 mmol/L = ________ mol/L
d.) 490 mg/L = ________ g/mL
Exercise `calculating with units
25
Exercise `derived units
Ns
kgm
s
kgkm________________
50
2922
=⋅
=⋅
e)
Js
kgmgm______________
min5
500)25(2
2
2
2
=⋅
=⋅
f)
Ns
kgm
h
gs
mm
______________1
4)43(
2=
⋅=
⋅g)
26
Exercise `derived units
Pasm
kg
km
t______________
min)2(2,0
8,2822
=⋅
=⋅h)
Pasm
kg
µsnm
ng______________
2,02,0
222
=⋅
=⋅
i)
Summation notation (capital-sigma notation) ∑
In that example: the index k gets values from 1 (starting point) to 5 (stopping point): 1,2,3,4,5. The index is always incremented by 1
summation sign
Stopping pointupper limit of summation
function of index-representing each successive terme in the raw
Starting pointlower limit of summation
Index of summation(continous index)
Summation notation (capital-sigma notation) ∑
Example:
In that example the index k can only acceptvalues from 1 (starting point) to 4 (stoppingpoint) in entired steps:1,2,3 and 4, that will be added.
Example summation notation:
In that example the index i can only acceptvalues from -1 (starting point) to 3 (stoppingpoint) in entired steps:-1, 0, 1, 2 and 3, that will be inserted in thefunction for i and than will be added.
Example summation notation:
In that example the index i can only accept valuesfrom 1 (starting point) to 7 (stopping point) in entired steps:1, 2, 3, 4, 5, 6 and 7, that will be inserted in thefunction for i and than will be added.
One more detailed video explanatory videoTo how to handle summation notations (in english): http://www.youtube.com/watch?v=hEPk36Yncxg
Exercise `summation notation :Write in explicit /elaborated way
Expanding + Reducing
• Expanding means multiplying the numerator and denominator of a fraction by the same non-zero number
• Reducing means dividing the numerator and denominator of a fraction by the same non zero number
cb
ca
b
a
⋅
⋅=
cb
ca
b
a
:
:=
34http://www.wyzant.com/help/math/elementary_math/fractions/expanding_and_reducing_fractions
• Addition:
• Subtraction:
• Multiplication:
• Division:
db
bcda
d
c
b
a
⋅
⋅+⋅=+
db
bcda
d
c
b
a
⋅
⋅−⋅=−
db
ca
d
c
b
a
⋅
⋅=⋅
cb
da
c
d
b
a
d
c
b
a
⋅
⋅=⋅=:
35
Exercise `fractions 1´
db
bcda
d
c
b
a
⋅
⋅+⋅=+
db
bcda
d
c
b
a
⋅
⋅−⋅=−
=+5
2
4
3)a =+
7
8
6
5)b
=++3
1
11
5
9
8)c
=−2
5
3
1)a
=−5
2
6
1)b
=−−11
5
9
7
5
8)c
36
db
ca
d
c
b
a
⋅
⋅=⋅
=⋅
=⋅
2
5
6
3.2
9
4
8
7.1
=⋅
=⋅
4
52
8
56.6
4
33
16
3.5
37
=⋅
=⋅
42
4
7
35.4
27
39
14
12.3
Exercise `fractions 2
cb
da
c
d
b
a
d
c
b
a
⋅
⋅=⋅=:
=
=
2
5:
6
3.2
9
4:
8
7.1
=
=
4
52:
8
56.6
4
33:
16
3.5
38
=
=
42
4:
7
35.4
2
19:
14
11.3
Exercise `fractions 3´
=
−
+
=
16
12
8
34
5
3
2
.2
2
36
5
.1
=
+
⋅
=
+
−
28
25
16
35
7
21
19
.4
9
5
27
128
3
11
5
.3
39
Exercise `fractions 4´
Percentage
• % means: divide a number by 100
• Example: 32% = 32/100 = 0,32
• Converting in percentage: multiply a number with 100
• Example: 0,144 = 0,144*100 = 14,4%
41
Exercise :convert the percentages into fractions.
=
=
=
=
%78.4
%4,44.3
%2,63.2
%5,81.1
=
=
=
%35,0.7
%94.24.6
%09,16.5
43
Exercise `…percentage of…´
1. 12% of 243
2. 33% of 148
3. 39% of 3290
4. 0,88% of 5
5. 40,2% of 23910
45
Systems of linear equations
• Aim: the simplest method for solving a system of linear equations is to repeatedly eliminate variables.
• methods (3 possibilities) :
• by ´Equalizing´
• by ´Substitution´
• by ´Addition´
46
Exercise `system of linear euqations –solve by the method of `equalizing`
4933
24.)
17
82.)
=−
=−
−−=
−=−
xy
xyb
xy
xya
´Substitution´
6511
5
11
459
15444
11
154
11)2(
154)1(
=⇒
−=
−=⇒
+=
−=⇒
−=++⇒
+=
−=+⇒
=−
−=+
xx
y
yx
y
yy
yx
yx
yx
yx
Example
50
51
Exercise `system of linear euqations –solve by the method of `substitution`
64
90415.)
256
34116.)
=+
=+
+=
=+
yx
yxb
yx
yxa
´Addition´
Example
10
10
303
10
160166039
10037
203
10037
=
=⇒
=
=
=⇒
=−
=+⇒
=−
=+
y
x
y
x
xyx
yx
yx
yx
52
53
Exercise `system of linear euqations –solve by the method of `addition`
115338
27724.)
443
64
1
3
1
.)
=−
=+
=−
=+
yx
yxb
yx
yxa
Exercise `binominal theorem´
calculate:
17²25.7
65536³32768²²6144³51216.6
8172²16.5
)365()365(.4
)35(.3
)³625(.2
)²52(.1
44
4
−
++++
++
+⋅−
−
+
−
x
mzmmzmzz
xx
xx
xy
x
x
57
Exponentsnmnm
aaa+=⋅)1
nm
n
m
aa
a −=)2
nmnmaa
⋅=)()3
n
n
aa
1)4 =−
74334xxxx ==⋅ +
25757 : xxxx == −
2,05
15 1 ==−
58
409622³)2( 12434 === ⋅
n
nn
b
a
b
a=
)5
nnnbaba ⋅=⋅ )()6
1)7 0 =a
aa =1)8
3222
5
5
55xxx
==
140 =
441 =
59
³³64³³³4)³4( babaab ==
Exercise `exponents´
)( nmnmaaa
+=⋅
=⋅ 34 33)a =⋅− 24) aad
=⋅ 04) bbb=⋅ 34) bae
=⋅ 44) bac
)( nm
n
m
aa
a −==
4
3
2
2)a =
−3
4
2
2)c
=4
0
)u
ub =
2
2
1
)x
xd
60
nn
n
b
a
b
a=)5
nmnmm
nm n aaaa ⋅⋅ ===
11
)6
n
nx
x
11
)7−
=
nnn baba ⋅=⋅)4 5125255 333 ==⋅
5253
75
3
75===
2161616 422 === ⋅
5,044
12
1
==−
62
Exercise `roots´
=32 )2)(a =−32 )2)(b =42
1
))(xc =⋅ 520)d
=⋅ 32)3e =⋅ 22)3f =⋅ 3
2
3 22)h
=5
405)i =
4
3
81
27)j =
88
63)k
3 16) =g
63
The equation exits of exactone real number.
This is the logarithm of the number b in respect to the base a
Logarithmic functions
)(log bx a=
bax =
baxb
short
xa =⇔=)(log
:
8³2
3)(log2
==
=
b
b
65
Special logarithms
• common logarithm:
(logarithm of x with respect to base 10)
• Binary logarithm:
(logarithm of x with respect to base 2)
• natural logarithm:
(logarithm of x with respect to base e) (mathematical constant e = 2,718…, called Euler's number )
)lg()(log10 rr =
)()(log2 rlbr =
)ln()(log rre =
66
Exercises `logarithm 1´
( )
=
=
==
=
=
3
1log))100(log)
)1(log)27log)
25
1log))8(log)
310
53
52
fe
dc
ba
68
001,0log)10
g
Logarithmic rules
yxyx aaa loglog)(log)1 +=⋅
vuv
uaaa logloglog)2 −=
)15log()35log()3log()5log( =⋅=+
)625,0log(8
5log)8log()5log( =
=−
69
xkx a
k
a loglog)3 ⋅=
( ) )(log1
log)4 bn
b an
a ⋅=
)(log4)(log 4xx aa ⋅=
( ) ))(log)((log2
1)(log
2
1)(loglog 2
1
yxxyxyxy aaaaa +⋅=⋅==
70
Exercise `logarithm 2´
Simplify the following terms:
2510
log4010
log) +a
nc 10log)
10
3log) af a
72
5010
log50010
log) −b
Exercise `logarithm 3´
=−+
=−+
=
=
)lg(*3²)²lg(*3
1.4
)lg(*3)lg(*5)lg(*3.3
7
³6lg.2
log.1
xyx
zyx
z
x
rs
xya
73
Calculation of xexample
)64,1()5log(
)14log(
)14log()5log(
)14log()5log(
145
≈=⇒
=⋅⇒
=⇒
=
x
x
x
x
74
Determine the solution
42
11
25
753.
107128.
943.
+
−−
−
=⋅
⋅=⋅
⋅=
xx
xx
xx
e
d
c
75
3
2
1 256log) xa =
2log) 2 −=xb
Calculation of xexample
)2ln(26;)2ln(26
)2ln(26²
)2ln(26²
)2ln(3²5,0
202
21
3²5,03²5,0
+=+−=⇒
+−⇒
−=−⇒
=+−⇒
=⇒=− +−+−
xx
x
x
x
eexx
77
Exercise `e-function´
• Calculate:
10) =xea =⋅ )2exp()3exp()b
16)2ln() =xc =⋅+⋅ )43
4ln()
2
21
9
4ln()d
78
• Conversion of angles
� from degree to radian:
� from radian to degree:
• calculator:
deg = degree; rad = radian
• Sense of rotation:
clockwise rotating angles are negativ, counterclockwise rotating angles are positiv
απ
⋅°
=180
x
x⋅°
=π
α180
82
Trigonometric functions forright-angled triangles
84
The side opposite to the right angle (90 ) is called hypotenuse, here: c = hypotenuse
side b:vis-à-vis point B respectivelythe angle β.Side b is opposite the angle β � `oppsite to the angle β.Side b is close to the angle α� `adjacent to the angle α
side a:vis-à-vis point A respectivelythe angle α.Side a is opposite the angle α � `oppsite to the angle α.Side a is close to the angle β� `adjacent to the angle β
Definitions in right-angled triangles:
Sine (sin) of an angle= opposite of an anglehypotenuse
cosine (cos) of an angle = adjacent of an anglehypotenuse
tangent (tan) of an angle = opposite of an angleadjacent of an angle
cotangent (cot) of an angle = adjacent of an angleopposite of an angle
Exercise `trigonometric functions´
89
Calculate the missing values!
1. b = 2,4cm; c = 3,2cm; γ = 90
2. a = 5,2cm; α = 66,5 ; γ = 90
3. c = 21,5cm; β = 72,3 ; γ = 90
4. b = 12,6cm; α = 32,3 ; γ = 90
Function of sine and cosine
�1,5
�1
�0,5
0
0,5
1
1,5
�450 �360 �270 �180 �90 0 90 180 270 360 450
sin(x) bzw.cos(x)
angle in degree
function of sine and cosine
sin
cos
90
Important correlations:
+=
2sin)cos(
πxx
Additions theorem:
)sin()sin()cos()cos()cos(
)cos()sin()cos()sin()sin(
yxyxyx
xyyxyx
⋅⋅=±
⋅±⋅=±
m
−=
2cos)sin(
πxx
)cot(
1
)cos(
)sin()tan(
xx
xx ==
91
Solving quadratic equations
qpp
x
qpxx
−±−=
=++
2
2;1
2
)2
(2
0
quadratic equation (p-q-formula)
7;13
)91(2
6
2
6
0916²
21
2
2,1
=−=
−−
±−=
=−+
xx
x
xx
93
About quadratic equations• quadratic equations have at most 2 solutions
• The solving formula:
gives:
− one solution if:
− two solutions if :
− no solution if:
qpp
−±− 2)2
(2
0)2
( 2 =− qp
0)2
( 2 >− qp
0)2
( 2 <− qp
94
Exercise `quadratic equations´
37²6.4
)415()53()97()27(.3
4421²15.2
03522²3.1
=+
−⋅−=−⋅−
=+
=+−
xx
xxxx
xx
xx
95
Root equations
example
4
95
352
=→
=+→
=+
x
x
x
proof
33
354
=→
=+
squaring is not a equivalent conversion-- > that means you have to proof
96
Fraction equationsexample
4;5
204
²9
2
9
0209²
2²32010
)2(3)2(10
12
310
21
2/1
==⇒
−±=⇒
=+−⇒
−=−−⇒
−⋅=−−⋅⇒
⋅=−
−
xx
x
xx
xxxx
xxxx
HNxx
98
Exercise `fraction equations´
a
t
a
bb
a
t
b
b
a
bt
a
t
ta
a
22.3
6
32
23.2
12
1.1
2−=−
+=+
=−
−
99
determine t:
density (material constant)
101
)(
)()(
i
ii
Vvolume
mmassdensity =ρ
L
kg
mL
g
m
kg
dm
kg
cm
g
unitstypical
;;;³
:
33=
102
Fresh snowfall has a density of 0,20 g/cm³.
a. which weight has fresh snowfall of 30 cm thicknesson a flat roof of 20 cm length and 10 m width?
b. If this amount of snow melts. How many liter ofwater are formed?
103
A irregular formed piece of jewellery (trinket) weighs0,177N in air, at a thin fiber the lifting power in water is
0,017N .
Is the trinket made of gold?
obtain: F=m*g
g= 9,81 N/kg; density (gold)= 19,3 kg/dm³; density (water)= 0,998 kg/dm³
Dilutions
dilute: the concentration of a solved substance in a solution is reduced.
105
100
ml
8 P
kt/1
00 m
L
+100 mLwater
shake10
0 m
l8 Pkt/200 mL=4 Pkt/100 mL
Add water
106
Initial volume + water = final volume
Dilution factor F
volumeinitial
volumefinal
ionconcentratfinal
ionconcentratinitialF
==
example: Create 50 mL of a physiologic salt solution 0,9% out of 10% salt solution:
initial concentration: 10%
final concentration : 0,9%
final volume: 50 mL
Needed initial volume ? Needed volume of water?
107
mLmLmLwaterV
mLmL
volume initial
lumeinitial vo
mL
%,
%
lumeinitial vo
mefinal volu
entrationfinal conc
nncentratioinitial coF
5,455,450)(
5,4%10
%9,050
50
90
10
=−=⇔
=⋅
=⇔
=⇔
==
Exercise `dilutions´
1. You have a 10 times concentrated buffer which should be diluted to one times. Create 500 mL of the buffer. How many mL buffer and water are needed?
109
Exercise `dilutuions´
2. For determining the concentration of extracted DNA you dilute 100µL of the DNA solution with 400µL water. The diluted solution has a concentration of 50 mg DNA per mL. What is the concentration of the initial solution? Calculate the dilution factor! How many DNA is isolated if the volume of the initial solution was 5mL?
110
112
informative diagram title
Especially if more than twographs are presented in thediagram a legend is necesarry.
axis labeling including units,x-axis: axis of abscissae
(independent (preset) dimension)
axis
lab
elin
gin
clu
din
gu
nit
s,y-
axis
: axi
so
fo
rdin
ates
(dep
end
ent
dim
ensi
on
)
Axis classification in a well spent way ofusing the given place
Single data – sign points or crosses
Best-fit-curve(=regression line, = trend line)
Exercise `diagrams/linear equations´
114
1.) Measurements of the solubility L of a salt in water depending on the temperature T give following data:
a.) Draw the corresponding diagram.b.) Determine the equation of the regression line by the diagramc.) Determine the solubility of the salt by 36,5 C
c1.) graphically (in /by diagram)c2.) calculative (by the linear equation)
i 1 2 3 4 5 6Ti [°C] 0 20 40 60 80 100
Li [g/100mL] 70,7 88,3 104,9 124,7 148,0 176,0
Exercise `diagrams/linear equations´
115
2.) The concentration of an apple juice sample is determined. Therefore standards
(solutions of juices with known concentrations) are prepared. The sample and the
standards are measured by a photometer. That means which concentration causes
which „colour intensity“ (extinction) and in the other way around which „colour
intensity“ (extinction) is which concentration?
a.) Draw the diagram on millimetre
paper.
b.) Determine the equation of the
regression line.
c.) Calculate the concentration of
the sample.
Solution Concentration in % Extinction
Sample
Exercise `linear equations´
3.) Two points of a straight line are known (pair of variates): P(2/3) and Q (-1/-3).Determine the linear equation.
4.) A layer of lipids is 100 nm and grows 5 nm per day .a.) Give an equation to calculate the lipid layer at any
time.b.) How thick is the lipid layer after one week?