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Standar 11
Writing Exponential form to Logarithmic form
First we must learn how to read logarithmic form:
The expression log b y is read as “log of base b of y”
Examples:
log 5 125
log of base 5 of 125
log 6 36
log of base 6 of 36
1
log 3
5
log of base 3 of 1/5
Standard 11
Rewriting Logarithmic Equations
Exponential Form
2 1 
1
2
Logarithmic Form
log 2
1
 1
2
2 4  16
log 2 16  4
5 x  125
log 5 125  x
6 y  36
log 6 36  y
1
x
3 
9
1
log 3  x
9
Standard 11
Simplifying Logarithmic Equations
Logarithmic Form
Exponential Form
Solution
log 4 16
4 x  16
x2
log 3 1
3y  1
y0
1
log 2
8
1
2 
8
z
z  3
log 4 2
4a  2
1
a
4
log 27 3
27  3
b
b
1
6
EXAMPLE 1
Evaluate logarithms
Evaluate (solve) the logarithm.
a. log 4 64
SOLUTION
To help you find the value of log b y, ask yourself what
power of b gives you y.
a.
4 to what power gives 64?
43 = 64, so log 4 64 = 3.
b. log 5 0.2
b. 5 to what power gives 0.2?
5–1 = 0.2, so log 5 0.2 = –1.
EXAMPLE 2
Evaluate logarithms
Evaluate (solve) the logarithm.
c.
log 1/5 125
SOLUTION
To help you find the value of log b y, ask yourself what
power of b gives you y.
–3
c.
1 to what power gives 125? 1 125, so log 125 –3.
=
1/5
5 =
5
d. log 36 6
d. 36 to what power gives 6?
361/2 = 6, so log 36 6 = 1 .
2
EXAMPLE 3
Evaluate common and natural logarithms
Expression Keystrokes
Display
Check
a. log 8
8
0.903089987
100.903
b. ln 0.3
.3
–1.203972804 e –1.204
8
0.3
EXAMPLE 4
Evaluate a logarithmic model
Tornadoes
The wind speed s (in miles
per hour) near the center of a
tornado can be modeled by
s = 93 log d + 65
where d is the distance (in
miles) that the tornado
travels. In 1925, a tornado
traveled 220 miles through
three states. Estimate the
wind speed near the
tornado’s center.
EXAMPLE 4
Evaluate a logarithmic model
SOLUTION
s = 93 log d + 65
= 93 log 220 + 65
93(2.342) + 65
= 282.806
Write function.
Substitute 220 for d.
Use a calculator.
Simplify.
ANSWER
The wind speed near the tornado’s center was about
283 miles per hour.
GUIDED PRACTICE
for Examples 2, 3 and 4
Evaluate (solve) the logarithm.
Use a calculator if necessary.
5. log 2 32
SOLUTION
7. log 12
5
6. log 27 3
SOLUTION
SOLUTION
1.079
8. ln 0.75
1
3
SOLUTION
–0.288
GUIDED PRACTICE
9.
for Examples 2, 3 and 4
WHAT IF? Use the function in Example 4 to
estimate the wind speed near a tornado’s center if
its path is 150 miles long.
ANSWER
The wind speed near the tornado’s center is about 267
miles per hour.
Standard 11
Laws of Logarithms
Definition of b
b
p
q
p
q
q
 ( b)  b p
q
p
Examples :
16
25
3
4
3
 (4 16 ) 3  2 3  8
2

25
3
1
1
 3 

3
125
5
( 25 )
1
1
2
Standard 11
Laws of Logarithms
LAWS OF LOGARITHMS
1. log b MN  log b M  log b N
M
2. log b
 log b M  log b N
N
3. log b M k  k log b M
LOGARITHMI C THEOREM
a loga x  x , for any positive number x and
log a a x  x , for any number x.
Standard 11
Examples using Laws of Logarithms
Simplify
Answers
log 6 6 2
2
log x x y
y
log 2 16
4
log 10 100
2
1
9
1
9
3
log3
Standard 11
Examples using Laws of Logarithms
Express each log in terms
of log M and log N
Answers
log 6 M 2 N 3
2 log 6 M  3 log 6 N
M2
log 3 ( 3 )
N
2 log 3 M  3 log 3 N
log 4 (
1
N
3
M
)
1
 3 log 4 N  log 4 M
2
Standard 11
Examples using Laws of Logarithms
Express as a single logarithm
Answers
log a 3  log a 4
log a 12
4 log a 2
log a 16
1
log a 36
2
log a 6
log b 3  log b 5  log b 2
log b 30
1
1
log a r  log a s
2
2
log a (rs )
1
2
or log a rs
Standard 11
Examples using Laws of Logarithms
Let c = log3 10 and d = log3 5
Answers
log 3 50
cd
log 3 500
d  2c  2c  d
log 3 250
log 3 2
2d  c  c  2d
cd
Chapter 2
Preliminary calculus
2.1.1 Differentiation from first principles
f ' ( x) 
df ( x )
f ( x  x )  f ( x )
 lim
x 0
dx
x
the limit does exist at a
point x=a
f (a  x )  f (a )
f (a  x )  f (a )
lim

lim
ax function
at x
0
x  0
xis continuous
differentiable
at x=a
x=a
P:
differentiable
A:
undifferentia
ble
Chapter 2
Preliminary calculus
high order derivative:
( n 1 )
( n 1 )
'
'
f
(
x


x
)

f
( x)
n
f
(
x


x
)

f
(
x
)
''
f
(
x
)

lim
f ( x )  lim
x  0
x
x  0
x
useful formulas:
d
x
(sin 1 ) 
dx
a
1
a2  x x
d
x
(cos 1 ) 
dx
a
1
a2  x x
d
x
a
(tan 1 )  2
dx
a
a  x2
Hin
t:   sin 1
x
x
dx
 sin    cos d 
a
a
a
d
1
1
1





2
2
dx a cos  a 1  sin 2 
a 1 x a
1
a2  x2
Chapter 2
Preliminary
calculus
2.1.2 Differentiation of product
products
f ( x )  u( x )v( x ) 
df
d
dv( x ) du( x )

[u( x )v( x )]  u( x )

v( x )
dx dx
dx
dx
2.1.3 The chain rule
f ( x )  g( u( x )) 
df
df du

dx du dx
2.1.4 Differentiation of quotients
u( x )
1 '
u'v  uv '
'
' 1
f ( x) 
 f  u( )  u ( ) 
v( x )
v
v
v2
2.1.5 Implicit differentiation
Ex : x 3  3 xy  y 3  2
dx 3
d
dy 3
d


( 3 xy ) 

2
dx dx
dx
dx
dy
dy
y  x2
2
2 dy
 3 x  (3 x
 3 y)  3 y
 0

dx
dx
dx y 2  x
Chapter 2
Preliminary calculus
2.1.6 Logarithmic differentiation
Ex : y  a x  ln y  ln( a x )  x ln a

1 dy
dy
 ln a 
 y ln a  a x ln a
y dx
dx
2.1.7 Leibnitz theorem
f  uv
 f '  uv '  u'v
 f ''  uv ''  2u'v '  u''v
 f '''  uv '''  3u'v ''  3u''v '  u'''v
 f ( 4 )  uv ( 4 )  4u'v '''  6u''v ''  4u'''v '  u( 4 ) v
n
f
(n)
n
n!
( r ) ( n r )

u v
  C rn u( r ) v ( n r )
r  0 r ! ( n  r )!
r 0
Chapter 2
Preliminary calculus
3
f
(
x
)

x
sin x
the
Ex: Find the third derivative of
3
function
( 3)
u( x )  x 3 , v( x )  sin x
f
  C 3 u ( r ) v ( 3 r ) ,and
r 0
r
set
f '''  uv '''  3u'v ''  3u''v '  u'''v
  x 3 cos x  3( 3 x 2 )(  sin x )  3(6 x ) cos x  6 sin x
 3( 2  3 x 2 ) sin x  x (18  x 2 ) cos x
2.1.8 Special points of a function
Q
stationary points:
df
0
dx
(1) local
maximum: Q
(2) local
minimum: B
(3) stationary
S
B
Chapter 2
Preliminary calculus
d2 f
0
2
dx
d2 f
0
2
dx
(1) for a local
minimum
(2) for a local
maximum
d2 f
(3) for a stationary point of dx 2  0
d2 f
inflection
an
changes sign through the
dx 2
d3 f
0
dx 3
d
point, so
3
2
f
(
x
)

2
x

3
x
 36 x  2
Ex:
df
 6 x 2  6 x  36  0  x 2  x  6  0
dx
d2 f
 12 x  6
 ( x  3)( x  2)  0  x  3, x  2
dx 2
(1)
(2)
2
d
f
forx  3, 2  30  0  x  3 is a
dx
d2 f
minimum


30

0

x


2
x


2
,
for
is a maximum
dx 2
Chapter 2
Preliminary calculus
general points of
inflection:
df
d2 f
(1) dx  ( )0, dx 2  0
f ( x)
G
at G
d2 f
dx 2
changes sign from
the left (concave
upwards) to the
right (concave
note: a stationary
point of
downwards).
inflection with
is a special case
(
2)
x
df
0
dx
Chapter 2
Preliminary calculus
2.1.9 Curvature of a function
(1) at
point P
(2)  0
 tan  
df
dx
CP  CQ  
s
ds

  0  
d
  lim
:
the radius of
curvature f ( x )
1
: the curvature
at P

of
  0 : the curve is locally concave
  0 : upwards
the curve is locally concave
downwards
Chapter 2
Preliminary calculus
the radius of curvature in terms
of x and f(x):
ds
1
2
1/ 2

d
s
d
x
θ
 sec  (1  tan  )
dx cos 
df
 tan   sec2 d  f '' ( x )dx
dx
dx sec2  1  [ f ' ( x )]2
 ''

d
f ( x)
f '' ( x )
 [1  (
df 2 1 / 2
) ]
dx
' 2
ds
ds dx
[1  ( f ' )2 ]3 / 2
' 2 1/ 2 1  ( f )


 (1  ( f ) )

d dx d
f ''
f ''
 for a stationary point of
inflection
and the curvature is zero
df 2
0   
2
dx
Chapter 2
Preliminary calculus
Ex: Show that the radius of curvature at the point (x,y)
x 2 ellipse
y2
(a 4 y 2  b 4 x 2 ) 3 / 2
on the
 2  1 has
and the
2
a 4b4
a
b
opposite
magnitude
sign to y. Check the special case b=a, for which the ellipse
becomes a circle.
So differentiating the
2 y dy
dy  b 2 x
l: 2 x equation
0 

a2
b 2 dx
dx
a2 y
d 2 y  b2 d x
 b 2 y  xy '
 b4
 2
( ) 2 (
) 2 3
2
2
dx
a dx y
a
y
a y

y 3 ( or y ) determines the sign
[1  b 4 x 2of
/( a 4 y 2 )]3 / 2 [a 4 y 2  b 4 x 2 ]3 / 2
 

4
2 3
 b /( a y )
a 4b4
for
b=a
  a 2 ( x 2  y 2 ) 3 / 2  a 2 a 3  a
the function is a
circle.
Chapter 2
Preliminary calculus
2.1.9 Theorem of differentation
Rolle’s
a xc
theorem:
(1) f(x) is
a  x  c , and f (a )  f (c )
continuous
(2) f(x) is for
'
x

b
,
where
a

b

c
,
f
(b)  0
at least one for
differentiable
Proopoint
f(x)
f
(
x
)

f
(
a
),
x

[
a
,
c
],
f
(
x
)
f:(1)
 f ' ( x)  0
if is a
x  (a , c )
( constant

2)if
f ( x )  f (a )  f ( x )maximum
f ( x )  f (a )  f ( x )minimum
f ' ( x)  0
a
c
b
f ' (b)  0
x
Chapter 2

Mean value
(1) f(x) is
theorem
continuous
(2) f(x) is for
Preliminary calculus
an f (a ) 
a xd
c , f ' (b)  0
a  x  c,
differentiable
at least onefor
value b ( a < b < c)
such that
Pro the equation of the line
g( x )  f (a )  ( x  a )[ f (c )  f (a )] /( c  a )
of:
AC is
h( x )  f ( x )  g ( x )
 f ( x )  ( x  a )[ f (c )  f (a )] /( c  a )
f ' (b) 
f (c )  f (a )
ca
f(x)
C
f(c)
f(
h(a )  h(c )  0 by Rolle’s theorem, at a)
'
x

b

(
a
,
c
)

h
(b)  0
poi
least one
nt h' ( x )  f ' ( x )  f (c )  f (a )
for
ca
f (c )  f (a )
h' (b )  0  f ' (b ) 
ca
f (c )
A
g(x)
x
a
b
c
Chapter 2
Preliminary calculus
Ex: What semi-quantitative results can be deduced by
Rolle’s theorem to 2the following
function,3 with2 a and
2
( i ) sin x ( ii ) cos x ( iii) x  3 x  2 ( iv) x  7 x  3 (v )2 x  9 x  24 x  k
c are chosen so that f(a)=f(c)=0?
So (i)sin x  0  x  n , n  0,1,2,3........
d sin x
l:
 cos x  n  x  ( n  1 / 2)  ( n  1)  cos x  0
dx
(i cos x  0  x  (n  1 / 2) ,n  0,1,2,3...
d cos x
i) dx   sin x  (n  1 / 2)  x  (n  1)  (n  3 / 2)  sin x  0
2
(ii f ( x )  x  3 x  2  0  ( x  2)( x  1)  0
i) df / dx  23x  3 2 0  x  3 / 2 ' 1  x  23 / 2  2
(v)f ( x )  2 x  9 x  24 x  k  f ( x )  6 x  18 x  24  0
 x 2  3 x  4  0  x  1, x  4
f ( x)  0  1 , 2 are two different 1   2
 1  1   2 o 1  4   2
roots if
f ( x )  0, 1 ,  2 ,  3 are three different 1   2   3
 1  1   2 r4   3
roots if
Chapter 2
Preliminary calculus
2.2 Integration
I   f ( x )dx
b
f(x)
a
integration
from
a  x  b  a   0   1   2  .....   n  b
principles:
n
 S   f ( xi )( i   i 1 )  i 1  xi   i
i 1
integration
x
as the inverse of
F ( x )   f ( u)du a is
differentiation:
a
x  x
x
x  x
F ( x  x )  
f ( u)duarbitrary
  f ( u)du  
a
a
 F ( x)  
x  x
x
x
a
b
f ( u)du
f ( u)du
F ( x  x )  f ( x )
1 x  x

f ( u)du
x
x x
dF
1
let x  0, LHS 
, RHS 
f ( x )x  f ( x )
dx
x
x
Chapter 2
Preliminary calculus

definite
b
b
f
(
x
)
dx

integral:
a
x
0

b
a
f ( x )dx  
a
x0
f ( x )dx  F (b)  F (a )
dF ( x )
dx  F (b)  F (a )
dx

integration by
 a ln cos bx
I

a
tan
bxdx

c
(1) 
inspection:
b
sin bx
a
1
dx  
[
d (cos bx )]
cos bx
cos bx b
 a ln cos bx

c
b
a sinn1 bx
n
(2)I   a cos bx sin bxdx  b(n  1)  c
Sol:I   a
(
3)
I
Hi
a
dx  tan 1 ( x / a )  c
2
2
a x
set
x / a  tan 
1  tan 2   sec2 
Chapter 2
Preliminary calculus
 a cos n1 bx
I   a sin bx cos bxdx 
c
b( n  1)
(
4)
(5) I  
(6) I  
n
1
x
dx  cos 1 ( )  c
a
a2  x2
1
x
dx  sin 1 ( )  c
a
a2  x2
Hin x / a  sin  dx  a cosd
t:
 integration of sinusoidal
2
5
4
2
I

sin
xdx

sin
x
sin
xdx


(
1

cos
x
)
d (cos x )
(1) 
function:


   (1  2 cos 2 x  cos 4 x )d (cos x )
  cos x  2 cos 3 x / 3  cos 5 x / 5  c
4
2
2
2
I

cos
xdx

(cos
x
)
dx

[(
1

cos
2
x
)
/
2
]
dx
(2) 


 (1 / 4) (1  2 cos 2 x  cos 2 2 x )dx
 3 x / 8  sin 2 x / 4  sin 4 x / 32  c
Chapter 2
Preliminary calculus

logarithmic
f ' ( x)
integration:
dx  ln f ( x )  c

f ( x)
6 x 2  2 cos x
3 x 2  cos x
dx  2 3
dx  2 ln( x 3  sin x )  c
Ex:I   3
x  sin x
x  sin x
integration using partial
1
1
1
1
fractions:
E I   x 2  x dx   x( x  1)dx   ( x  x  1 )dx
x:  ln x  ln( x  1)  c  ln( x /( x  1))  c

integration by
1
substitution:
Ex:I   a  b cos x dx



Hint:
set
2
t  tan( x / 2)
a2  b
x
tan [
tan ]  c
ab
2
a 2  b2
2
1
1
b a
2
2
ln[
for a  b
a  b  b 2  a 2 tan( x / 2)
a  b  b  a tan( x / 2)
2
2
] for a  b
Chapter 2
Preliminary calculus
2
dx
1  3 cos x
2
2
t

tan(
x
/
2
)

dt

(
1
/
2
)
sec
(
x
/
2
)
dx

[(
1

t
) / 2]dx
set
Ex: I  
2
2
(
 1  3[(1  t 2 )(1  t 2 )1 ] 1  t 2 )dt
2
2
1
1

dt

dt

(

 ( 2  t )( 2  t )

2 t2
2
2t
I

1
2t
1
2  tan( x / 2)
ln(
)
ln[
]
2
2t
2
2  tan( x / 2)
Ex:I 
set
I
1
)dt
2t


1
dx 
x2  4x  7
1
 ( x  2)2  3dx
y  x  2  dx  dy
1
1
y
1
1
1 x  2
dy

tan
(
)

c

tan
(
) c
2
y 3
3
3
3
3
Chapter 2
Preliminary calculus

integration by
d
dv du
parts:
( uv )  u

v
dx
dx
dx
d
dv
du
(
uv
)
dx

u
dx

 dx
 dx  dx vdx
dv
du
dv
du
 uv   u dx   vdx   u dx  uv   vdx
dx
dx
dx
dx

Ex:I   ln xdx
dv
du
u

ln
x

1


set
dx
dx
I  x ln x  
1
xdx  x ln x  x  c
x
Ex: I   x sin xdx
dv
set u  x 2 dx  xe  x
I
1
vx
x
 1 2  x2
x e
2
du
 1  x2
 2x v 
e
dx
2
2
1
 1 2  x2 1  x2
  2 xe  x dx 
x e
 e
c
2
2
2
2

Chapter 2
Ex: I   e
set u  e ax
ax
Preliminary calculus
cos bxdx
dv
du
1
 cos bx 
 ae ax v  sin bx
dx
dx
b
1
a
I  e ax sin bx   e ax sin bxdx
b
b
dv
du
1
I 1   e ax sin bxdx set u  e ax
 sin bx 
 ae ax v 
cos bx
dx
dx
b
 1 ax
a
 1 ax
a

e cos bx   e ax cos bxdx 
e cos bx  I
b
b
b
b
1
a  1 ax
a
I  e ax sin bx  (
e cos bx  I )
b
b b
b
a
a2
ax 1
 e ( sin bx  2 cos bx )  2 I  c
b
b
b
e ax
I 2
(b sin bx  a cos bx )  c
a  b2
Chapter 2
Preliminary calculus

reduction
1
1
3 n
I 2   (1  x 3 )2 dx  ?
formula
Ex: I n  0 (1  x ) dx to
0
1
I n   (1  x 3 )(1  x 3 ) n1dx evaluate
0
  [(1  x )
1
0
3 n 1
3 n 1
 x (1  x )
3
]dx  I n1   x 3 (1  x 3 )n1 dx
1
0
dv
du
1
 x 2 (1  x 3 ) n 1 
1 v 
(1  x 3 ) n
dx
dx
3n
x
1 1
1
3 n 1
3 n
I n  I n 1 
(1  x ) |0 
(
1

x
)

I

In
n 1
3n
3n 0
3n
3n
 In 
I n 1
3n  1
1
3
3
 I 0   dx  1  I 1  I 0 
0
4
4
3 2
6 3
9
 I2 
I1   
3 2  1
7 4 14
set u  x
Chapter 2
Preliminary calculus

infinite

b
I   f ( x )dx  lim  f ( x )dx  limF (b)  F (a )
integrals:
a
b a
b

Ex:I  0
b
x
x
dx

lim
dx
b   0 ( x 2  a 2 ) 2
( x 2  a 2 )2
 lim[
b
1 2
1
1
1
1
( x  a 2 ) 1  c ]b0  lim
[ 2

]

b 2
2
b  a2 a2
2a 2

improper
x  c  [a , b ] 
integrals:

b
a
f ( x)  
f ( x )dx  lim 
c 
 0 a
f ( x )dx  lim 
b
 0 c 
f ( x )dx
I   ( 2  x )1 / 4dx
2
Ex:
0
f ( x )  ( 2  x ) 1 / 4  f ( 2 )  
2 
4
I  lim  ( 2  x ) 1 / 4 dx  lim
( 2  x ) 3 / 4 |02
 0 0
 0 3
 4 3/ 4
4
 lim
[
 23 / 4 ]  23 / 4
 0 3
3
Chapter 2
Preliminary calculus

integration in plane
polar

1 2
1 2
dA


d


A

coordinates:
 2 d
2
2
1
A 
2
y
1 2
a d  a 2
2
the area of a
0
circle is:
Ex:Find the
area2 of an ellipse with
2
1 cos  sin 

anequation
2
2

a
2
2
t

tan


dt

sec

d


d


cos
dt
se


dt
t  2a 2b 2 0 2 dt 2 2  2b 2 0
A
b a t
(b / a ) 2  t 2
 2b 2 [1 /( b / a )  tan 1 ( t /( b / a ))]0  ab
d 
 (  d )
dA
 ( )
b
1 2 2
1 2
a 2b 2
A    d  
d
2 0
2 0 b 2 cos 2   a 2 sin 2 
2
1
d
2 2
 2a b 
(
)
0 b 2  a 2 tan 2  cos 2 
C
0
B
x
1
1
1 x
dx

tan
( ) c
 a2  x2
a
a
Chapter 2
Preliminary calculus

finding the length of a
curve:
s  x 2  y 2  x  0  ds 

ds
dy
 1  ( )2  S 
dx
dx

b
a
1 (
dx 2  dy 2
dy 2
) dx
dx
in plane polar
coordinate:
s
x
y=f(x)
y
x
x  r cos   dx  dr cos   r sin d
y  r sin   dy  dr sin   r cos d
 dx 2  dy 2  (dr ) 2  ( rd ) 2
 ds  (dr )  ( rd )  S  
2
f(
x)
2
r2
r1
1  r2(
d 2
) dr
dr
Chapter 2
Preliminary calculus
surface
b
area:
S
  2yds  (ds )2  (dx )2  (dy )2
a
 S   2y 1  (
b
a
dy 2
) dx
dx
E Find the surface area of a
x: cone formed by rotating
abouth the x-axis the line y=2x
h
2
S   2 
2 x and
1  2 x=h.
dx  4 5  xdx
S between
x=0
0
0
ol:  2 5x 2 |0h  2 5h2

if the surface is formed by rotating a line
about the
y-axisdx 2
b
S

a
2x 1  (
dy
) dy
Chapter 2
Preliminary calculus

volumes of
b
dV  y 2dx  V   y 2dx
revolution:
a
Ex: Find the volume of a cone enclosed by
the surface formed by rotating about
the x-axis the line y=2x between x=0
So and x=h.
l:
V    ( 2 x ) dx   4x 2dx 
h
0
2
h
0
4 3
h
3
The General Analytical Problem
Select sample
Extract analyte(s) from matrix
Separate analytes
Detect, identify and
quantify analytes
Determine reliability and
significance of results
Errors in Chemical Analysis
Impossible to eliminate errors.
How reliable are our data?
Data of unknown quality are useless!
•Carry out replicate measurements
•Analyse accurately known standards
•Perform statistical tests on data
Mean
N
 xi
Defined as follows:
x =
i=1
N
Where xi = individual values of x and N = number of replicate
measurements
Median
The middle result when data are arranged in order of size (for even
numbers the mean of middle two). Median can be preferred when
there is an “outlier” - one reading very different from rest. Median
less affected by outlier than is mean.
Illustration of “Mean” and “Median”
Results of 6 determinations of the Fe(III) content of a solution, kn
contain 20 ppm:
Note: The mean value is 19.78 ppm (i.e. 19.8ppm) - the median value
Precision
Relates to reproducibility of results..
How similar are values obtained in exactly the same way?
Useful for measuring this:
Deviation from the mean:
di  xi  x
Accuracy
Measurement of agreement between experimental mean and
true value (which may not be known!).
Measures of accuracy:
Absolute error: E = xi - xt (where xt = true or accepted value)
Relative error:
x x
t  100%
E  i
r
x
t
(latter is more useful in practice)
Illustrating the difference between “accuracy” and “precision”
Low accuracy, low precision
Low accuracy, high precision
High accuracy, low precision
High accuracy, high precision
Some analytical data illustrating “accuracy” and “precision”
HN
H
S
NH3+ClH
Benzyl
isothiourea
hydrochloride
O
OH
N
Analyst 4: imprecise, inaccurate
Analyst 3: precise, inaccurate
Analyst 2: imprecise, accurate
Analyst 1: precise, accurate
Nicotinic acid
Types of Error in Experimental
Data
Three types:
(1) Random (indeterminate) Error
Data scattered approx. symmetrically about a mean value.
Affects precision - dealt with statistically (see later).
(2) Systematic (determinate) Error
Several possible sources - later. Readings all too high
or too low. Affects accuracy.
(3) Gross Errors
Usually obvious - give “outlier” readings.
Detectable by carrying out sufficient replicate
measurements.
Sources of Systematic Error
1. Instrument Error
Need frequent calibration - both for apparatus such as
volumetric flasks, burettes etc., but also for electronic
devices such as spectrometers.
2. Method Error
Due to inadequacies in physical or chemical behaviour
of reagents or reactions (e.g. slow or incomplete reactions)
Example from earlier overhead - nicotinic acid does not
react completely under normal Kjeldahl conditions for
nitrogen determination.
3. Personal Error
e.g. insensitivity to colour changes; tendency to estimate
scale readings to improve precision; preconceived idea of
“true” value.
Systematic errors can be
constant (e.g. error in burette reading less important for larger values of reading) or
proportional (e.g. presence of given proportion of
interfering impurity in sample; equally significant
for all values of measurement)
Minimise instrument errors by careful recalibration and good
maintenance of equipment.
Minimise personal errors by care and self-discipline
Method errors - most difficult. “True” value may not be known.
Three approaches to minimise:
•analysis of certified standards
•use 2 or more independent methods
•analysis of blanks
Statistical Treatment of
Random Errors
There are always a large number of small, random errors
in making any measurement.
These can be small changes in temperature or pressure;
random responses of electronic detectors (“noise”) etc.
Suppose there are 4 small random errors possible.
Assume all are equally likely, and that each causes an error
of U in the reading.
Possible combinations of errors are shown on the next slide:
Combination of Random Errors
Total Error
No.
Relative Frequency
+U+U+U+U
+4U
1
1/16 = 0.0625
-U+U+U+U
+U-U+U+U
+U+U-U+U
+U+U+U-U
+2U
4
4/16 = 0.250
-U-U+U+U
-U+U-U+U
-U+U+U-U
+U-U-U+U
+U-U+U-U
+U+U-U-U
0
6
6/16 = 0.375
+U-U-U-U
-U+U-U-U
-U-U+U-U
-U-U-U+U
-2U
4
4/16 = 0.250
-U-U-U-U
-4U
1
1/16 = 0.01625
The next overhead shows this in graphical form
Frequency Distribution for
Measurements Containing Random Errors
4 random uncertainties
A very large number of
random uncertainties
10 random uncertainties
This is a
Gaussian or
normal error
curve.
Symmetrical about
the mean.
Replicate Data on the Calibration of a 10ml Pipette
No.
Vol, ml.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
9.988
9.973
9.986
9.980
9.975
9.982
9.986
9.982
9.981
9.990
9.980
9.989
9.978
9.971
9.982
9.983
9.988
Mean volume 9.982 ml
Spread
0.025 ml
No.
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
9.975
9.980
9.994
9.992
9.984
9.981
9.987
9.978
9.983
9.982
9.991
9.981
9.969
9.985
9.977
9.976
9.983
Vol, ml.
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
No.
9.976
9.990
9.988
9.971
9.986
9.978
9.986
9.982
9.977
9.977
9.986
9.978
9.983
9.980
9.983
9.979
Median volume
Standard deviation
9.982 m
0.0056
Calibration data in graphical form
A = histogram of experimental results
B = Gaussian curve with the same mean value, the same precision
and the same area under the curve as for the histogram.
SAMPLE = finite number of observations
POPULATION = total (infinite) number of observations
Properties of Gaussian curve defined in terms of population.
Then see where modifications needed for small samples of data
Main properties of Gaussian curve:
Population mean (m) : defined as earlier (N  ). In absence of syst
m is the true value (maximum on Gaussian curv
x ) defined
Remember, sample mean
(
for small values of N.
(Sample mean  population mean when N  20)
Population Standard Deviation (s) - defined on next overhead
s : measure of precision of a population of data,
given by:
N
s
2
(
x

m
)
 i
i 1
N
Where m = population mean; N is very large.
The equation for a Gaussian curve is defined in terms of m and
y
e
 ( x  m ) 2 / 2s 2
s 2
Two Gaussian curves with two different
standard deviations, sA and sB (=2sA)
General Gaussian curve plotted
units of z, where
z = (x - m)/s
i.e. deviation from the mean of
datum in units of standard
deviation. Plot can be used for
data with given value of mean,
Area under a Gaussian Curve
From equation above, and illustrated by the previous curves,
68.3% of the data lie within s of the mean (m), i.e. 68.3% of
the area under the curve lies between s of m.
Similarly, 95.5% of the area lies between s, and 99.7%
between s.
There are 68.3 chances in 100 that for a single datum the
random error in the measurement will not exceed s.
The chances are 95.5 in 100 that the error will not exceed s.
Sample Standard Deviation, s
The equation for s must be modified for small samples of data, i.
N
s
2
(
x

x
)
 i
i 1
N 1
Two differences cf. to equation for s:
1.
Use sample mean instead of population mean.
2.
Use degrees of freedom, N - 1, instead of N.
Reason is that in working out the mean, the sum of the
differences from the mean must be zero. If N - 1 valu
known, the last value is defined. Thus only N - 1 degr
of freedom. For large values of N, used in calculating
Alternative Expression for s
(suitable for calculators)
N
s
N
(  xi ) 2
i 1
N
(  xi 2 ) 
i 1
N 1
Note: NEVER round off figures before the end of the calcul
Reproducibility of a method for determ
the % of selenium in foods. 9 measurem
were made on a single batch of brown
Sample
Selenium content (mg/g) (xI)
xi2
1
0.07
0.0049
2
0.07
0.0049
3
0.08
0.0064
4
0.07
0.0049
5
0.07
0.0049
6
0.08
0.0064
7
0.08
0.0064
8
0.09
0.0081
9
0.08
0.0064
2
Mean = Sxi/N= 0.077mg/g
(Sxi) /N = 0.4761/9 = 0.0529
Standard Deviation of a Sample
0.0533  0.0529
Sx
=
0.69
s

i
Standard deviation:
9 1
Sxi = 0.0533
 0.00707106
 0.007
2
Coefficient of variance = 9.2% Concentration = 0.077 ± 0.007 mg/
Standard Error of a Mean
The standard deviation relates to the probable error in a single measu
If we take a series of N measurements, the probable error of the mean
the probable error of any one measurement.
The standard error of the mean, is defined as follows:
sm  s
N
Pooled Data
To achieve a value of s which is a good approximation to s, i.e. N 
it is sometimes necessary to pool data from a number of sets of mea
(all taken in the same way).
Suppose that there are t small sets of data, comprising N1, N2,….Nt
The equation for the resultant sample standard deviation is:
s pooled 
N1
N2
N3
i 1
i 1
i 1
2
2
2
(
x

x
)

(
x

x
)

(
x

x
)
 i 1  i 2  i 3 ....
N 1  N 2  N 3 ......t
(Note: one degree of freedom is lost for each set of data)
Pooled Standard Deviation
Analysis of 6 bottles of wine
for residual sugar.
Bottle Sugar % (w/v) No. of obs.
Deviations from mean
1
0.94
3
0.05, 0.10, 0.08
2
1.08
4
0.06, 0.05, 0.09, 0.06
3
1.20
5
0.05, 0.12, 0.07, 0.00, 0.08
4
0.67
4
0.05, 0.10, 0.06, 0.09
5
0.83
3
0.07, 0.09, 0.10
6
0.76
4
0.06, 0.12, 0.04, 0.03
(0.05) 2  (010
. ) 2  (0.08) 2
0.0189
s1 

 0.0972  0.097
2
2
and similarly for all sn .
Set n  ( x  x )
1
0.0189
2
0.0178
3
0.0282
4
0.0242
5
0.0230
6
0.0205
Total
0.1326
2
i
sn
0.097
0.077
0.084
0.090
0.107
0.083
spooled
01326
.

 0.088%
23  6
Two alternative methods for measuring the precision of a set of results
VARIANCE: This is the square of the standard deviation:
N
s2 
2
2
(
x

x
)
 i
i 1
N 1
COEFFICIENT OF VARIANCE (CV)
(or RELATIVE STANDARD DEVIATION):
Divide the standard deviation by the mean value and express as a per
s
CV  ( )  100%
x
x ) tovalue
the true
How can we relate the observed mean
( mean (m)?
The latter can never be known exactly.
The range of uncertainty depends how closely s corresponds to s
x thataround
m must lie,
We can calculate the limits (above and below)
with a given degree of probability.
Define some terms:
CONFIDENCE LIMITS
interval around the mean that probably contains m.
CONFIDENCE INTERVAL
the magnitude of the confidence limits
CONFIDENCE LEVEL
fixes the level of probability that the mean is within the confid
Examples later.
First assume that the known s is a good
approximation to s.
Percentages of area under Gaussian curves between certain limits of z
50% of area lies between
0.67s
80%
“
1.29s
90%
“
1.64s
95%
“
1.96s
99%
2.58s
What this means,
for example,“is that 80 times
out of 100 the true me
between 1.29s of any measurement we make.
Thus, at a confidence level of 80%, the confidence limits are 1.29s
For a single measurement: CL for m = x  zs (values of z on next o
x ), the equivalent
For the sample mean of N measurements
(
expression is:
CL for m  x  zs
N
Values of z for determining
Confidence Limits
Confidence level, %
z
50
0.67
68
1.0
80
1.29
90
1.64
95
1.96
96
2.00
99
2.58
3.00
Note: 99.7
these figures assume that an excellent
approximation
99.9
to the real standard deviation is3.29
known.
Confidence Limits when s is known
Atomic absorption analysis for copper concentration in aircraft
engine oil gave a value of 8.53 mg Cu/ml. Pooled results of many
analyses showed s  s = 0.32 mg Cu/ml.
Calculate 90% and 99% confidence limits if the above result
(a)(a) 1, (b) 4, (c) 16 measurements. (b)
were based on
(164
. )(0.32)
 8.53  0.52 mg / ml
1
i.e. 8.5  0.5mg / ml
(164
. )(0.32)
 8.53  0.26mg / ml
4
i.e. 8.5  0.3mg / ml
90% CL  8.53 
(2.58)(0.32)
99% CL  8.53 
 8.53  0.83mg / ml
1
i.e. 8.5  0.8mg / ml
90% CL  8.53 
(2.58)(0.32)
 8.53  0.41mg / ml
4
i.e. 8.5  0.4 mg / ml
99% CL  8.53 
90% CL  8.53 
(c)
(164
. )(0.32)
16
 8.53  013
. mg / ml
i.e. 8.5  01
. mg / ml
(2.58)(0.32)
 8.53  0.21mg / ml
16
i.e. 8.5  0.2 mg / ml
99% CL  8.53 
If we have no information on s, and only have a value for s the confidence interval is larger,
i.e. there is a greater uncertainty.
Instead of z, it is necessary to use the parameter t, defined as follo
t = (x - m)/s
i.e. just like z, but using s instead of s.
By analogy we have:
CL for m  x  ts
N
(where x = sample mean for N measurements)
The calculated values of t are given on the next overhead
Values of t for various levels of probability
Degrees of freedom 80% 90% 95% 99%
(N-1)
1
3.08 6.31 12.7 63.7
2
1.89 2.92 4.30 9.92
3
1.64 2.35 3.18 5.84
4
1.53 2.13 2.78 4.60
5
1.48 2.02 2.57 4.03
6
1.44 1.94 2.45 3.71
7
1.42 1.90 2.36 3.50
8
1.40 1.86 2.31 3.36
9
1.38 1.83 2.26 3.25
Note: (1) 19 As (N-1) 1.33
, so t1.73
 z 2.10 2.88
1.30
1.67< ,2.00
2.66greater uncerta
(2) 59 For all values
of (N-1)
t > z, I.e.

1.29 1.64 1.96 2.58
Confidence Limits where s is not known
Analysis of an insecticide gave the following values for % of the chem
7.47, 6.98, 7.27. Calculate the CL for the mean value at the 90% confi
xi%
7.47
6.98
7.27
Sxi = 21.72
xi2
55.8009
48.7204
52.8529
Sxi2 =
157.3742
(  xi ) 2
(2172
. )2
x  N
157.3742 
3
s

N 1
2
 0.246  0.25%
2
i
x

x
2172
.

 7.24
N
3
i
(2.92)(0.25)
 7.24 
N
3
 7.24  0.42%
90% CL  x  ts
90% CL  x  zs
If repeated analyses showed that s s = 0.28%:
 7.24 
N
 7.24  0.27%
(164
. )(0.28)
3