Transcript Document

Chiang Ch. 10 Exponential and
Logarithmic Functions
10.1 The Nature of Exponential Functions
10.2 Natural Exponential Functions and
the Problem of Growth
10.3 Logarithms
10.4 Logarithmic Functions
10.5 Derivatives of Exponential and
Logarithmic Functions
10.6 Optimal Timing
10.7 Further Applications of Exponential and
Logarithmic Derivatives
1
10.1(a) Types of functions (Chiang, p.26)
functionname
specific form
Polynominals (algebraicfunctions)
economicuse
Constant
y  a0
fixed costs,interestrates
Linear
y  a0  a1 x
equilibria
Quadratic
y  a0  a1 x  a2 x 2
optimaand/or equilibria
Cubic
y  a0  a1 x  a2 x 2  a3 x 3
optima
Rational (algebraic) y  a / x
optima
T ranscendental (non- algebraic,independent var.associatedw/ exponent)
Exponentia
l y  bx
Logarithmic x  logb y (inverse)
compounding & discounting
growth,productionfunctions
2
Rules of exponents
  
1) x m  n  x m x n
m
x
2) x m  n  n
x
1
n
3) x  n
x
4) x 0  1
5) x
1
n
x0
n x
 
7)  xy   x  y 
6) x
mn
 x
m
m n
m
m
Chiang & Wainwright, pp.23 - 24
3
7.3.2 Inverse-function rule
• This property of one to one mapping is unique to the class
of functions known as monotonic functions:
• Definition of a function (p. 17)
– Function
– Monotonic function
– One x for each y, aka
one y for each x and
one x for each y
inverse function
A monotonicallyincreasingfunction,if
1)
x1  x 2 then f x1   f x2 
e.g., supplyfunction(whereb1  0)
2)
Q s  b 0  b1P
and an inversesupplyfunction
3)
P  - b 0 /b1  (1/b1 )Q s
4)
5)
y  ex
x  ln y
range0,  , domain - , 
range- , , domain 0,  
4
7.3 Rules of Differentiation
Involving Functions of Different Variables
Inverse- functionrule
Let y be a strictlymonotonicfunctionof x
1)
y  f ( x)
2)
dy dx  f x 
Then
3)
4)
x  f 1 ( y )
dx dy  f
1
 y   1 dy dx  1 f x 
5
10.1 The Nature of Exponential Functions
10.1(a) Simple exponential function
10.1(b) Graphical form
10.1(c) Generalized exponential function
10.1(d) A preferred base
7
10.1(a) Simple exponential function
y = f(x) = bx
where base b > 1, x is exponent, f(x) 
The term exponent (x) refers to the power to which a base
number (b) is raised.
Base exclusions:
• b  1 and b  0, because
f(x) = 1x = 1;
f(x) = 0x = 0, i.e., constants
• 0 < b < 1 excluded since they can be expressed as negative
exponents
• b<0 excluded because many values of f(x) from the domain
would be imaginary, e.g., (-b)½
• popular bases: e and 10
8
10.1(c) Generalized exponential function
• Where
y = dependent variable
b = base
t = independent variable
a = vertical scale factor
(directly related)
c = horizontal scale factor
(inversely related)
y  ab
ct
9
10.1(b) Graphical Exponential Functions y=bt where b=3,e,2
10
10.1(d) A preferred base (e)
Given
1)
y  bt
b t  t  b t
b t b t  b t
b t  1
t
2)
dy dt  f (t )  lim
 lim
 b lim
t 0
t 0
t 0 t
t
t
A logarithmis the power towhicha baseis raised to attaina particularnumber.Let
/


b t  1
ln b t  1
t ln b  ln1
3)
lim
 ln b  lim
 lim
t 0 t
t 0
t 0
t
t
Such that
4)
f / (t )  b t ln b
Is there a base e such that
5)
If
6)
ln e  1?
e  lim 1  1 m m  2.71828...
m 
Then
7)
8)
e t  1
ln e  lim
1
t 0 t
dy dt  f / (t )  e t ln e  e t  y
11
y1=e and y2=(1+1/m)m as m
12
10.5 Derivative of the inverse of y=et, i.e. y=ln t (p. 277)
Given theinverseof y  e t in which we reverse y and t
1)
2)
y  e t ln y  t alternatively
y  ln t
dy d
f (t )  f ( N )
ln t  ln N

ln t  lim
 lim
tN
tN
dt dt
tN
tN
lnt N 
1
 lim

lim lnt N 
tN
tN
t  N tN
Let
m  N t  N 
m N  1 t  N 
t  N  N  1 m
t N  N N 1 m
t N  11 m
dy m
1
1 1 1
1
m
8)

lim ln1  1 m  
lim ln1  1 m  
 or  t
dt N t  N
N tN
N t
y e
3)
4)
5)
6)
7)
13
10.5 Derivative of y=et using the Inverse function rule
1)
y  et
function
2)
ln y  t ln e
3)
t  ln y
inverse
derivativeof the inverse
d
4)
ln y  1 y
dy
derivativeof the function
d t
1
5)
e 
 y  et
dt
1y
14
10.5(d) The case of base b
derivativeof a transendentalfunctionin thebase b
exponentia
l function
logarithmic function
d logb t
dbt
1
t
 b ln b

dt
dt
t ln b
derivativeof a transendentalfunctionin thebase e
exponentia
l function
logarithmic function
de t
t
e
dt
d loge t d ln t 1


dt
dt
t
15
10.1(d) A preferred base (e)
(e) = 2.71828…, irrational number w/ beautifully simple characteristics
Derivativeof
1)
y  bx
2)
ln y  x ln b
3)
f ( x) 
/

x  x  ln b  x ln b
lim
x
x ln b
 lim
x 0 x
 ln b
x 0
What ln (baseb) willhavea slope f / ( 0 )  1 as x  0
4)
if
b  e,
5)
then
ln b  ln e  1
16
10.1(d) A preferred base (e)
(e) = 2.71828, irrational number w/ beautifully simple characteristics)
x
b
1
/
0
f (0)  b lim
 ln b
x 0
x
What base b will havea slope f / ( 0 )  1 as x  0
base b
x
2
2.718282
3
10
0.1
0.7177
1.0517
1.1612
2.5893
0.01
0.6956
1.0050
1.1047
2.3293
0.001
0.6934
1.0005
1.0992
2.3052
0.0001
0.6932
1.0001
1.0987
2.3029
0.00001
0.6931
1.0000
1.0986
2.3026
17
10.1(b) Graphic for f(x)=ex
f(x)  e x where b  e
domain of x : (-, )
range of y :
(0 ,  )
y - int ercept: 1
x - int ercept s: none
horizont alasympt ot e:
x - axis as x  -
At (0,1) t heslope of t he
t angent f / ( 0 )  1
18
10.1(d) A preferred base (e)
(e) = 2.71828, irrational number w/ beautifully simple characteristics)
At x  0, what base b will havea slope f / ( 0 )  1 as x  0
e Δx  1
lim
1
Δx0
Δx
Δx
e
1
f / (x)  e x lim
 e x( 1 )  e x
Δx0
Δx
f ( x)  e is theone whose tangentline at (0,1)has a slope f / ( 0 )
thatis exactly1 (45o ).
(Stewart,Calculus, 3rd , p. 355)
19
10.1(d) A preferred base (e)
(e) = 2.71828, an irrational number w/ beautifully simple characteristics)
Derivativeof y  et wrt t, thenaturalexponentia
l function
 
dy d t d et d t 
 e 
 et 1  et  y
dt dt
d t  dt
 
d2y d  d t  d t
t

e

e

e


dt 2 dt  dt  dt
Derivativeof V  Aert wrt t, thegeneralnaturalexponentia
l function


dV dAert d Aert d rt 


 r Aert  rV
dt
dt
d rt  dt


20
10.2 Natural Exponential Functions and the
Problem of Growth
• 10.2(a) The number e
• 10.2(b) An economic interpretation of e
• 10.2(c) Interest compounding and the
function Aert
• 10.2(d) Instantaneous rate of growth
• 10.2(e) Continuous vs. discrete growth
• 10.2(f) Discounting and negative growth
21
10.2(a) The number e
m
1

1)
base e  lim 1  
m 
 m
A Maclaurinseries approx.of e x when x0  0
2)
f(x)  f(x0 )  f '(x0 )(x  x0 )  f "(x0 )/ 2! (x  x0 )2 
  f (n)(x0 )/n! (x x0 )n  R
Given e x  f(x)  f /(x)  f //(x) , then
e x0 2
e x0 n
3)
f(x)  e  e (x) 
(x)   
(x)  R
2!
n!
Because x 0  0  e x0  e 0  1
x0
x0
1
1
e x  1  x  (x)2  (x)3    R
2
6
At x  1 theMaclaurinapprox.convergeson e
1 1 1
5)
e  11  
   2.71828 
2 6 24
4)
22
10.2(b&c) Interest compounding and the function Aert
• y = Aert is the value of an $A investment at
nominal interest rate r / compounding period
compounded t times over the investment period (#
days, months, or years) (growth in an investment)
• As m  , e  $2.71828 …
• $1 = initial investment
• $1 = 100% return
(if m = 1 at the end of the year, then e = 2
if m > 1, then r = 100% is a nominal rate and e > 2)
• $.71828 = interest on the interest received during
the year as m  , r = 171.8% (effective rate)
23
10.2(e) Continuous vs. discrete growth


Find the value for continuouscompounding lim V m 
m 
given initialinvestmentA overt years at r interestpaid m periods / year
t
m


r
r
r  mr 



 
V m   A1    A1     A1   
 m
 m  
 m  
mt
(1)
rt
w
m
 1
Let w 
and given e  lim 1  
w 
r
 w
T hecontinuouscompounding equation
(2)
 1 
lim V m   A lim 1  
m 
w
 w 
w rt

rt
  Ae

24
10.2(e) Continuous vs. discrete growth
Given initialinvestmentA overt years at i interestpaid m periods / year
i 

T hediscretecompounding equationin base b  1   and base e where b m  e r
m

mt
mt
m
m
i 
i 


rt
(1)
V m   A1    Aert ;
1    e ;
 m
 m
Discretecompounding equation over one year (t  1)
i 

m t ln1    rt
 m
i 
i 
i 



r
(2)
V m   A1    Aer ;
m ln1    r
1    e ;
 m
 m
 m
Discreteannualcompounding equation over one year (t  1, m  1)
(3)
V m   A1  i   Aer ;
1  i  er ;
ln1  i   r
Solving for i and r for effective(i ) and nominal(r ) rateequivalence
(4)
i  e r  1;
r  ln(1  i )
25
Relation between effective (i) and nominal (r) rates of interest
i : effectiverateof interestper year compounded discretely
r : nominalrateof interestper year compounded continuously
m : number of compounding periodsper year
m
i 

(1)
e r  1  
 m
let m  1, compoundedyearly
(2)
er  1  i
Given i, solve(2) for r
(3)
r  ln ( 1  i)
Given r, solve(2) for i
(4)
i  e r -1
let m  , compounding continuously
m
(5)
i 

lim 1    e r
m 
 m
ir
26
10.2(d) Instantaneous rate of growth of a stock
at two points in time (Chiang, pp. 278-289)
V is thefuture value of an initialinvestment(A)
over time(t) at interestrate(r) w/ compounding
V  Ae
rt
dV
 rAert  rV
dt
dV / dt
r
V
dV / dt
V
r
Future value (V)
Flow of returnsdV dt 
T herateof growth (r)
Relationof stock toflow
27
10.2(d) Instantaneous rate of growth of a stock
at two points in time (Chiang, pp. 278-289)
V is t hefut ure value of an init ialinvest ment(A)
over t ime(t ) at int erestrat e(r) w/ compounding
V  Ae
Fut ure value (V)
A  Ve -rt
P resent value (A)
rt
dA dt   rVe  rt   rA
dA / dt
-r 
A
dA / dt
A
r
Flow of ret urnsdA dt 
T herat eof growt h (r)
Relat ionof st ock t oflow
28
T hestock (V) - flow (v) problem:
Given interestrate(r) provethat
v
v
v
v


...


(1  r ) (1  r ) 2
(1  r ) n r
v
v
v
V


...

 ...
(1  r ) (1  r ) 2
(1  r ) n
V
 1

1
1

 v


...


...
2
n
(
1

r
)
(
1

r
)
(
1

r
)


1
Let x 
and applyingthegeometricseries
1 r
 1

V  v x  x2    xn
 v
 1
1 x 



 1

 v
 1
1
1



 1 r

1 r 
 v
 1
 r




 1

 v
 1
1

r

1




 1 r

1 r  r 
1
 v
  v 
 r 
r
 1
 v
1
1
 
V  v


...


...
2
n
(
1

r
)
(
1

r
)
(
1

r
)

 r
WalterNicholson,Microeconomic T heory,1972,p. 381,
29
10.2(f) Discounting as negative growth or rate
of decay
future V=
present A=
f(compounding the present A) f(discounting the future V)
V  Ae
rt
A  Ve
 rt
30
10.3 Logarithms
•
•
•
•
10.3(a) The meaning of logarithm
10.3(b) Common log and natural log
10.3(c) Rules of logarithms
10.3(d) An application
31
10.3(a) The meaning of logarithm
• Exponents
• (solve for y given t)
t
3
2
1
0
-1
-2
-3
y=10t
1000
100
10
1
0.1
0.01
0.001
Common logs
(solve for t given y)
Log10 1000  3
Log10 10  1
Log10 1  0
Log10 0.1  1
Log10 0.01  2
Log10 0.001  3
32
10.3(b) Common log and natural log
• Exponent
Y e
Natural log
t
• Exponent
Y b
t  loge Y  ln Y
Common log
t
t  logb Y
33
10.3(c) Rules of logarithms in the land
of exponents
•
•
•
•
Product
Quotient
Power
Base
inversion
• Base
conversion
lnuv  ln u  ln v
lnu / v   lnu  lnv
lnu  alnu
a
1
1
logb e 

loge b ln b
ln(u )
logb u  logb e loge u  
ln(b)
34
10.4 Logarithmic Functions
• 10.4(a) Log functions and exponential
functions
• 10.4(b) The graphical form
• 10.4(c) Base conversion
35
10.4(a) Log functions and exponential functions
• Exponential function
Log function
– Dependent variable on left
– Monotonically increasing functions
• If ln y1 = ln y2, then y1 = y2
Y e
t
t  loge Y  ln Y
36
10.4(b) The graphical form
y=et (blue), y=2t (red-top),
y=ln(t) (red), 45o (green)
37
10.4(c) Base conversion
• Let er = bc
• Then ln er = ln bc
r = ln bc = c ln b
• Therefore
er = e c ln b
• And
y = Abct = Ae(c ln b)t =Aert
38
10.4- 6 (b)
Find the cont inuous- compounding nominalinterest
rateper annum (r) thatis equivalent to a
discret e - compounding rate(i) of 5% per annum,
compounded semiannually.
y  abct  aert
i
where a  1, i  .05, c  2, t  1, b  1   1.025
c
let e r  b c
r ln e  c ln b
r  c ln b  2 ln 1.025  4.94%
y  e c lnb t   e 2 ln1.025 1  1.050625
39
Derivationof continuousfromdiscrete,
(Chiang& Wainwright, pp.263& 265)
compounding equation: V  A(1 i) t  V  Aert
mt
r

V (m)  A1  
 m
Let i  r m , and b  1  i
V (m)  A1  i 
mt
discretecompounding equation
V (m)  Abmt
Let b m  e r , such that
 
 
ln b m  ln e r
m lnb   r
V m   Aert
continuouscompounding equation
40
10.5 Derivatives of Exponential
and Logarithmic Functions
•
•
•
•
•
10.5(a) Log-function rule
10.5(b) Exponential-function rule
10.5(c) The rules generalized
10.5(d) The case of base b
10.5(e) Higher derivatives
41
Given
1)
2)
Let
3)
4)
5)
6)
7)
8)
10.5 Derivative of y=ln t
y  ln t
d
f (t )  f ( N )
ln t  ln N
ln t  lim
 lim
tN
tN
dt
tN
tN
lnt N 
1
 lim

lim lnt N 
tN
tN
t  N tN
m  N t  N 
m N  1 t  N 
t  N  N  1 m
t N  N N 1 m
t N  11 m
d
m
1
1 1
m
ln t 
lim ln1  1 m  
lim ln1  1 m  

dt
N tN
N tN
N t
42
10.5(a) Log-function rule
• Derivative of a log function with base e
Simple
d ln t 1

dt
t
General
d ln f t  f t 

dt
f t 
43
10.5(b) Exponential-function rule
• Derivative of an exponential function with base e
Simple
General
t
dy de
t

e
dt
dt
f t 
dy d e

dt
dt
 f t e f t 
44
10.3(d) An application
The common use of the logarithmic transformation in
economic research is when estimating production
functions and other multiplication nonlinear
theoretical specifications
Transforming a multiplicative production function into a
logarithmic one suitable for estimation by linear
regression. Let Q = output and L & K are labor and
capital inputs respectively
Q  AL K 
lnQ  ln A   ln L   ln K
45
Total differential of Q using logs
Q  AK  L
lnQ  ln A   lnK   lnL
d ln Q   d ln A  d ln K    d ln L 
dQ 1


 dA  dK  dL
Q
A
K
L

 
1
dQ  Q dA  dK  dL 
K
L 
A
46
Exponent example
11.6(c)Input decisionsof a firm
(Example5, pp.337- 38)
1) π  R-C  PQ-wL-rK
2) Q  Lα K α


α  .5
3) π  P L K -wL-rK
α
α


 α L K P-r  0
α 1
4) π L  α L K P-w  0
5) π K
α
*
α
α 1
Solve for L and K
*
47
11.6(c)Input decisionsof a firm (Example5, p. 337)
Solving (4) for K and substituting theresultsinto(7)
1α
6)
7)
 w 1α 
K 
L 
 Pα

α
α 1
Pαα K  r  0
 w 1α 
8)
Pαα 
L 

 Pα
Solve for L*
1α
α
9)

r 0

 1  1 1 2 
L  Pw
*



α 1
r
48
10.5(d) The case of base b
derivativeof a transendentalfunctionin thebase b
exponentia
l function
logarithmic function
d logb t
dbt
1
t
 b ln b

dt
dt
t ln b
derivativeof a transendentalfunctionin thebase e
exponentia
l function
logarithmic function
de t
t
e
dt
d loge t d ln t 1


dt
dt
t
49
10.5(e) Higher derivatives
derivatives of transendentalfunctionsin thebase e
exponentia
l
logarithmic
de t
 et
dt
d 2e t
t

e
dt 2
d 3e t
t

e
dt 3
d 4e t
t

e
dt 4
d loge t d ln t  1


dt
dt
t
d 2 loge t d t 1  1

 2
2
dt
dt
t
d 3 loge t d  t  2
2

 3
3
dt
dt
t
d 4 loge t d 2t 3
6

 4
4
dt
dt
t
 


 
50
10.6 Optimal Timing
• 10.6(a) A problem of wine storage
• 10.6(b) Maximization conditions
• 10.6(c) A problem of timber cutting
51
10.6(a) A problem of wine storage
A(t )  Ve  rt
P resent value
V  ke
Growth in value
t
A(t)  ke e
t1 2
 rt
t 1 2  rt
 ke
ln At   lnk  lne
 lnk  t
 ln k  t
½
½
t ½  rt

 rt 
 rt lne
10.6(a) A problem of wine storage
ln At 
dA 1
dt A
dA
dt
dA
dt


 ln k  t ½  rt
1 1 2
 t r
2
 1 1 2

 A t  r   0
2

1 1 2
 t r 0
2
53
10.6(b) Maximization conditions
1
2 t
r
1
 2r
t
1
t 2
4r
1
t
2r
let (r )  10%
1
t
 25 years
2
4( 0.10 )
t ½  rt
A(t)  ke
let ( k )  $1 / bot tle
A(t)  e 5(.1 )(( 25 )
A(t)  e 2 .5  $12.18 / bot tle
V  Aert
V  $12.18e(.1 )( 25 )
V  $148.38 / bot tle
54
10.6(a) A problem of wine storage
plot of .5(t)-.5=r, r=.10, t=25
55
10.6(c)T he timbercut tingproblem(p.285)
At   Ve  rt
V 2
t
present value
2
t1 2
fut ure value
At   2 e  rt
t1 2
ln At   ln(2)
t1 2
 ln(e)  rt  t 1 2 ln(2)  rt
1 dA 1 1 2
 t ln 2  r
A dt
2
dA
 1 1 2

 A t ln 2  r   0
dt
2

1 1 2
r  t ln 2
solvefor t*
2
56
10.6(c)T he timbercuttingproblem(p.285)
At   Ve
 rt
V 2
1 1 2
r  t ln 2,
2
t
1 2
t
2
2r

ln 2
t1 2
 ln 2 
t 

 2r 
2
2
 ln 2 
  48 years
t  
 2.05 
Let r  5%
V
2
48
 $121.92 thou. (future value)
At   121.92e .05 48   $11.06 thou. (present value)
57
10.6(c) Timber cutting problem
plot of .5(t)-.5ln(2)=r, r=.05, t=48
58
10.7 Further Applications of Exponential and
Logarithmic Derivatives
• 10.7(a) Finding the rate of growth
• 10.7(b) Rate of growth of a combination of
functions
• 10.7(c) Finding the point elasticity
59
10.7(a) Finding the rate of growth
dy
ry 
dt  f ' (t )  marginalfunction
y
f (t )
totalfunction
d ln f t  f t 

dt
f t 
V  Aert
ln(V )  ln(A)  rt ln(e)
d
d
d
ln(V )  A  rt
dt
dt
dt
1 dV
dV dt
r
V dt
V
60
10.7(b) Rate of growth of a combination of
functions; Example 3 consumption & pop.
C  A1e t
ln(C )  ln(A1 )  t ln(e)
d
d
d
ln(C )  ln(A1 )  t  
dt
dt
dt
H  A2 e t
rC 
ln(H )  ln(A2 )   t ln(e)
d
d
d
rH  ln(H )  ln(A2 )  t  
dt
dt
dt
d C d
d
rC H  ln   ln(C )  ln(H )    
dt  H  dt
dt
61
10.7(b) Rate of growth of a combination of
functions; Example 3 consumption & pop.
z uv
u  f (t )
ln(z )  ln(u  v)
v  g (t )
d
d
ln(z )  (u  v)
dt
dt
1 d
rz 
(u  v)
u  v dt
1 d
rz 
( f (t )  g (t ))
u  v dt
1
rz 
( f ' (t )  g ' (t ))
uv
f ' (t )
if (ru ) 
, then f ' (t )  uru , and g ' (t )  vrv
f (t )
u
v
ru  v 
ru 
rv  su ru  sv rv
uv
uv
rz 
d ln f t  f t 

dt
f t 
62
10.7(c) Finding the point elasticity
M d  f  y (t ), i (t )
dM d
dy
di
 f y '  fi '
dt
dt
dt
dM d dt f y ' dy f i ' di
rM d 


Md
f dt
f dt
rM d
rM d
rM d
f y ' y dy f i ' i di


f y dt
f i dt
f y ' y dy f i ' i di


f ydt
f idt
  M d y ry   md i ri
63
10.7- 11Given theproductionfunctionbelow, find a generalexpressionfor therate
of growth of Q in termsof thegrowth in K and L.
(T herateof growth is definedin eq 10.23,p. 302)
Given Q  Q( K , L), where
K  K (t )
L  L(t )
show that
rQ   QK rK   QL rL
Method1 takedifferentials, thensubstitute, divide, and substitute
Q  QK , L ,
where K  K (t )
Q
Q
dK
dQ 
dK 
dL
dK 
dt
K
L
dt
Q dK
Q dL
dQ 
dt 
dt
K dt
L dt
dQ Q dK Q dL


dt K dt L dt
1 dQ 1 Q K dK 1 Q L dL


Q dt Q K K dt Q L L dt
L  L(t )
dL
dL 
dt
dt
1 dQ  K Q  1 dK   L Q  1 dL 





Q dt  Q K  K dt   Q L  L dt 
rQ   QK rK   QL rL
64
10.7- 11Given theproductionfunctionbelow, find a generalexpressionfor therate
of growth of Q in termsof thegrowth in K and L.
(T herateof growth is defined in eq 10.23,p. 302)
Given Q  Q( K , L),
K  K (t )
where
L  L(t )
show that
rQ   QK rK   QL rL
Method2 makelogarithmic transformation,then takedifferentials and substitute
ln Q  ln Qln K , ln L ,
where
  ln Q 
  ln Q 
d ln Q  
d ln K  
d ln L

ln
K

ln
L




ln K  ln K (t )
d ln K 
1 dK
dt
K dt
ln L  ln L(t )
d ln L 
1 dL
dt
L dt
 K Q  1 dK 
 L Q  1 dL 
1


dQ  
dt  
dt
Q
Q

K
K
dt
Q

L
L
dt






1 dQ  K Q  1 dK   L Q  1 dL 





Q dt  Q K  K dt   Q L  L dt 
rQ   QK rK   QL rL
65