1
Mathematics for Thermodynamics
Partial Differentiation
The first derivative of a function denotes the change of a function. The second derivative
denotes the change of the first derivative, etc… When the function depends only on one
variable, the change described by a derivative depends only on the change of that single
variable. What is the case when we have a function that depends on two or more variables?
How would be able to describe the change of the function? We would have to specify a
change of the function with respect to the some other change, namely a change of one of the
variables. Thus, the case presented is the basis for partial differentiation. A partial
derivative is the derivative of a multi-variable function.
When one does a partial derivative on a function, the other variables are considered as
constants. In the function, h  w 3 t 5 we find a derivative with respect to w or with respect to t.
If we want the change with respect to w, then we want to make sure that we are not changing t
at the same time. Therefore when we find the first derivative of h with respect to w, we will
hold t constant. The variables being held constant are indicated as subscripts. For example,
the partial derivative of h with respect to w is written as
 h 
2 5

  3w t

w

t
NOTATION: Note that the script “” is used rather than “d” for the differentials. “d” is used
for single variable ordinary differentials and “” multi-variable partial differentials.
Continuing, what is the partial derivative of h with respect to t?
 h 
3 4
   5w t

t
 w
 f 
Example: What is   for the function f  x 2 y  3y3z 2 ?
 x  y,z
   x 2 y  3y3z 2  
   x2y 
   3y3z 2  
 f 
 
 
  2xy  0  2xy
   

 x 


x
x
 x  y,z 
 y,z 
 y,z 
 y,z
 f 
Example What is   for the same function f  x 2 y  3y3z 2 ?
 y  x,z
   x 2 y  3y3z 2  
  x2y 
   3y3z 2  
 f 




  x 2  9y 2 z 2


  

 y 


y
y
 y  x,z 
 x,z 
 x,z 
 x,z
2
The chain rule for partial derivatives needs to be obeyed as well.
 G 
2
2
Example: Find 
 when G  ln  a  2ab  b 
 a b
  a 2  2ab  b 2 
 2a  2b   2  a  b   2
1
 G 




2
2
a
 a b  a  2ab  b 
 a 2  2ab  b2   a  b 2  a  b 
Implicit Differentiation
Sometimes we wish to take the derivative of a function with respect to a variable that is within
the function in a complicated way. For example, consider the following function.
ln x  3y2  5y  1
If we wanted to take the derivative of y with respect to x using the method with which we’re
accustom, we would need to algebraically rearrange the equation to solve for y.
y
5  252  12 1  ln x 
6
With enough patience we can certainly find dy/dx; however, it would be better to use the more
sophisticated technique of implicit differentiation. Rather than doing algebra and then
differentiation (which is messy for both on the above function) we can do the differentiation
first and then the algebra to find the derivative. We will take the derivative of the function as
we originally found it keeping in mind that y depends on x.
d
d
ln x 
3y 2  5y  1 

dx
dx

1
 dy   dy 
 6y    5    0
x
 dx   dx 
1  dy 
1
 dy 
    6y  5     
x  dx 
 dx  x  6y  5 
If we desired we could substitute what we know for y into the expression, but often this final
step is not necessary. The above derivative took three steps using implicit differentiation.
Using our traditional methods would have been much more time consuming.
3
Second Derivatives
A second derivative can be thought of as two consecutive first derivatives, i.e., the change of a
change. When we have a multi-variable function, we have a choice of partial first derivatives
as discussed above. For a partial second derivative, we have even more choices. For a
function that depends on three variables, we have nine possible second derivatives. Consider
the function Q(r, s, t). All of the following are possible second derivatives.
  2Q    2Q    2Q    2Q    2Q    2Q    2Q    2Q    2Q 
 2 ,
,
,
, 2 ,
,
,
, 2 
 r   rs   rt   sr   s   st   tr   ts   t 
  2Q       
  2Q       
where  2       Q , 
    Q,
 r   r s,t  r s,t
 rs   r s,t  s r,t
  2Q       

      Q , etc …
 rt   r s,t  t r,s
Example: Calculate all of the partial second derivatives for the function
r 3s rv 2 s3 v
Q  r,s, v   2 
 4
v
s
r
  2Q         r 3s rv2 s3 v      3r 2s v2 4s3 v  6rs 20s3 v
 4    2   5  2  6
 2      2 
s
r   r s,v  v
s
r  v
r
 r   r s,v  r s,v  v
  2Q         r 3s rv2 s3 v      r 3 rv2 3s 2 v  3r 2 v 2 12s 2 v
 4     2  2  4  2  2  5

     2 
s
r   r s,v  v
s
r  v
s
r
 rs   r s,v  s r,v  v
  2Q         r 3s rv2 s3 v      2r 3s 2rv s3 
6r 2s 2v 4s3










 5

      2
 
s
r 4   r s,v  v3
s
r 4 
v3
s
r
 rv   r s,v  v r,s  v
  2Q         r 3s rv2 s3 v      r 3 rv2 3s 2 v  2v2 6sv
 4     2  2  4   3  4
 2      2 

s

s

s
v
s
r   s r,v  v
s
r  s
r






r,v
r,v 
  2Q         r 3s rv2 s3 v      3r 2s v2 4s3 v  3r 2 v 2 12s 2 v
 4     2   5   2  2  5

     2 
s
r   s r,v  v
s
r  v
s
r
 sr   s r,v  r s,v  v
  2Q         r 3s rv2 s3 v      2r 3s 2rv s3 
2r 3 2rv 3s 2









 2  4

      2
  
4 
3
4
3

s

v

s

v
v
s
r

s
v
s
r
v
s
r








r,v
r,s 

r,v 

  2Q         r 3s rv2 s3 v      2r 3s 2rv s3  6r 3s 2r
 4      3 
 4 4 
 2      2 
s
r   v r,s  v
s
r  v
s
 v   v r,s  v r,s  v
  2Q         r 3s rv2 s3 v      3r 2s v2 4s3 v 
6r 2s 2v 4s3
 4     2   5   3 
 5

     2 
s
r   v r,s  v
s
r 
v
s
r
 vr   v r,s  r s,v  v
  2Q         r 3s rv2 s3 v      r 3 rv2 3s 2 v 
2r 3 2rv 3s 2








 2  4

      2
 
s
r 4   v r,s  v 2 s 2
r 4 
v3
s
r
 vs   v r,s  s r,v  v
  2Q    2Q    2Q    2Q 
  2Q    2Q 
Note that 
;
and


 
 



 

 rs   sr   rv   vr 
 sv   vs 
This interchangeability of mixed partial derivatives is one of the properties of partial
derivatives.
4
Operations with Partial Derivatives
Interchangeability of Mixed Partial Derivatives
 2U   2U 



 TV   VT 
 2z   2z 



 xy   yx 
2
2
     
    f    f    f      f       
f



 
        f
   
    
 x  y  y x
 x  y  y x  xy   yx   y x  x  y  y x  x  y
Inversion of Partial Derivatives
 z 
1
   y
 y  x  
 
 z  x
1
 H 

  T

 T p 


 H p
Example: Let H  p 2T 4
 H     2 4
2 3

    p T  4p T

T

T

p  p
1
H  p 2T 4
 H 4
 T 2 
p 
H
 T4  2
p
1
1
 H 

  T

 T p 


 H p
1
3
 T      H  4 1  1  4  4


H
 

 
  2
4  p2 
 H p  H p  p 
1
3
1
1 3
3
3
1
2 3
2 4 4
4 4
2
4
2
2 2
 4p T 
 4p H  4p  p T   4p p  T   4p 2T 3
1
3
1  1 4 4
H
4  p 2 
Chain Rule
 z   z   w 
  

 
 y  x  w  x  y  x
 A 
 A   T 

 
  
 S V  T V  S V
Note that we’ve already had an example above. Here’s another example.
5
 x 
 f 
Example: Find   for the function f (x, y)  ln sin  2
2 
 x  y
x y 
x
This function has the form f  g  h  x, y   . h(x, y)  2
, g(x, y)  sin h , f (x, y)  ln g
x  y2






1
2x 2 
 f   f   g   h  
 cos  x    1 




        
 x     x 2  y 2    x 2  y 2  x 2  y 2 2 
 x  y  g  y  h  y  x  y


 sin  2
2 
  x  y 
  x    x 2  y 2  2x 2    x    y 2  x 2 

  cot 


 cot  2
2 
2
2 
2
2 2 
2
2 2


x

y
x

y
  x  y 

  x  y 
 

 


Euler’s Rule
- For functions that depend only on three variables.
- common situation in thermodynamics.
 x   z   y 
       1
 y  z  x  y  z  x
 H   T   P 

 
    1
 P T  H P  T H
- Note the cyclic relationship between the variables.
Example: Let H  p 2T 4
 H     2 4
4

    p T  2pT
 p T  p T
1
H  p 2T 4
H
 T4  2
p
 H 4
 T 2 
p 
H  p 2T 4
H
 p2  4
T
 H 2
 p 4 
T 
1
1
3
 T      H  4 1  1  4  4


H
 

 
  2
4  p2 
 H p  H p  p 
1
1
1
1



1
 p 
    H  2 1 2  1  2  4   2H 2 
      4   H  4   5 
3
2
 T H  T H  T 
 T   T   T 


1
1


3
1
1
1
1
4


 H   T   p 
2H 2 
4 1  1 
2 4 4
4
2
4
2



 
     2pT    2  H   3   p TH  p T  p T   1
4p 
T
 p T  H p  T H


 
1
 H 2
p 4 
T 
6
Total Differentials
An aside: There is a difference between a derivative and a differential.
Differential – an infinitesimal change in a quantity.
Derivative – a ratio of infinitesimals
– the change of a quantity with respect to another quantity.
Total differential – an infinitesimal change of a multi-variable function that accounts for
changes in all the variables.
The total differential of the function R  , , ,   is
 R 
 R 
 R 
 R 
dR  
d  
d  
d
 d  



  , ,
   ,,
  , ,
  ,,
The total differential describes how the quantities change when all of the variables are able to
change.
nRT n 2a
Example: What is the total differential for the following function p  n, V,T  : p 
?

V  nb V 2
 p 
 p 
 p 
dp    dn  
 dV    dT
 n V,T
 V n,T
 T n,V
2
RT
nRTb
2na
 p 
    nRT n a 
 2 

 2
    
2
V
 n V,T  n V,T  V  nb V  V  nb  V  nb 
2
nRT
2n 2a
 p 
    nRT n a 
 2 


 
 
2
 V n,T  V  n,T  V  nb V 
 V  nb  V3
2
nR
 p 
    nRT n a 



 
  
2 
 T n,V  T n,V  V  nb V   V  nb 
 p 
 p 
 p 
dp    dn  
 dV    dT
 n V,T
 V n,T
 T n,V
 RT

 nR 
nRTb
2na 
RT
2n 2a 



dn





 dV  
 dT
2
2
2
3
 V  nb  V  nb 

  V  nb 

V
V
V

nb








Note how the total change, dp; can be written as the sum of three changes:
 RT

nRTb
2na 
RT
2n 2a 
 p 
 p 

 2  dn , 
 3  dV ,
  dn  
 dV   
2
2
V 
V 
 n V,T
 V n,T
 V  nb  V  nb 
  V  nb 
 nR 
 p 
 dT

 dT  
 T n,V
  V  nb  
7
Exact Differentials
Consider a general expression for the differential of a three-variable function, f (x, y)
 f 
 f 
df    dx    dy  M  x, y  dx  N  x, y  dy
 x  y
 y  x
df is an exact differential, if and only if, the following condition is met.
 M  x, y    N  x, y  

 


y

 x  x
y
This condition is the same as requiring that
  2f    2f 




x

y

  yx 
Example: Is the following differential exact:
df  M  x, y  dx  N  x, y  dy   y 2  2x  4  dx   2xy  4y  5  dy ?
 M  x, y     
2

     y  2x  4   2y
y

 x  y  x
 N  x, y     

     2xy  4y  5   2y
 x
 y  x  y
Therefore df is exact.
Aside: Since df is exact, when can find the function that produces both differentials:
f  xy2  x 2  4x  2y2  5y .
Example: Is the following differential exact?
dR  M  p, q  dp  N  p, q  dq   p3  3p 2q 2  7q  dp   2pq 2  4p 2q  5p  dq ?
 M  p, q      3
2 2
2

     p  3p q  7q   6p q  7

q

q

p  p
 N  p, q     
2
2
2

     2pq  4p q  5p   2q  8pq  5

p

p

q  q
 M  p, q    N  p, q  

 
 , therefore dR is not exact.

q

p

p 
q
Advantages of a differential being exact.
1. Line integral of differential does not depend on the path of integration.
2. If df is exact, f is known as a state function.
3. Changes in state functions do not need to be found using integration.
8
Line Integrals
Integrals are often pictured as the area under a curve. However, not all integrals are areas.
Sometimes integrals are used to calculate the length of a path. If the path is not flat, then an
integral can be used to find the “effort” needed to traverse the path. Such an integral is called a
line integral.
The simplest conception of a line integral is work. Mechanical work is defined as the energy
needed to move a force along a specified path. As an equation, work is defined as
w   F  dr
where dr is the differential path element. The path of integration is very important. In
general, different paths will yield different results.
Illustration: Consider the graphic at the right. Note that the
work against gravity going up and down the hill (red path)
will be different from the work against gravity going
around the hill (blue path).
- Note the change of energy is the same, but the amount
of work is different. We will be considering such
details later in the course.
A
Example: Calculate the work done against the force
F  xy ˆi  y 2 ˆj going from point (0,0) to the point (2,1) for the following paths:
a) straight line: y = ½ x
b) y = ¼ x2
c) taxicab path: (0,0) to (2,0) to (2,1)
d) x = 2t3, y = t2 t: 0  1
X( t )  2 t
3
Y( t )  t
2
y ( x) 
1
4
2
x
1
1
2
t
y( x)
0.5
Y( t)
0
0
1
t  x  X ( t)
2
B
9
Surface plot of F  xy ˆi  y 2 ˆj
Contour plot of F  xy ˆi  y 2 ˆj
M
M
For all paths, the solution of the problem begins as follows:
w   F  dr where dr  dx ˆi  dy ˆj



w   F  dr   xy ˆi  y 2 ˆj  dx ˆi  dy ˆj    xy dx  y 2 dy 
a) straight line: y = ½ x
y = ½ x  dy = ½ dx
w    xy dx  y dy  
2
(2,1)
  xy dx  y
2
dy 
(0,0)
Using the relationship between x and y given by the path, we can substitute for one or the
other. Let us first substitute ½ x for y and ½ dx for dy
w    xy dx  y 2 dy  
2
2
 1 
 1   1 
2
x
y
dx

y
d
y

x
x
dx

x

    2   2   2 dx 


( 0,0 )
0 

( 2 ,1)
1 
3
3 1
1
1
1
   x 2  x 2  dx   x 2 dx   x 3 |02  23  03  1
2
8 
8
8 3
8
8
0
0
2
2
As mentioned above, we could have just as easily substituted 2y for x and 2 dy for dx
w    xy dx  y 2 dy  
1
( 2 ,1)
  xy dx  y
( 0,0 )
1
1
2
dy     2y  y  2 dy   y 2 dy 
0
1
   4y 2 dy  y 2 dy    3y 2 dy  3  y3 |10  13  03  1
3
0
0
10
b) y = ¼ x2
Let’s substitute as to do the line integral as an integration of the x variable. Thus, substitute ¼
x2 for y and ½ xdx for dy
w    xy dx  y 2 dy  
2
2
 1 2

1 2 1
2
xy
dx

y
dy

x
x
dx

x

    4   4   2 x dx 


(0,0)
0 

(2,1)
1
1
1
1 1
1 1
1

   x 3  x 5  dx   x 3 dx   x 5 dx   x 4 |02   x 6 |02
4
32 
40
32 0
4 4
32 6
0
2

2
2
1 4 4  1 1 6 6  16 1 64
1 2
2  0      2  0      1  

16
3 3
 32 6
 16 32 6
Now let’s substitute as to do the line integral as an integration of the y variable. Thus,

1
substitute x  2 y and dx  y 2 dy
w    xy dx  y 2 dy  
1 
 dy  2 
2
xy
dx

y
dy

2
y
y


  y dy 
 

y

(0,0)
0 




(2,1)

1
1 3
 2
   2y dy  y 2 dy    2y dy   y 2 dy  y 2 |10  y3 |10  13  03   1  03  
3
3
 3
0
0
0
1
1
1
c) taxicab path: (0,0) to (2,0) to (2,1)
We’ll breakup the integration path into two pieces: (0,0) to (2,0), that is y = 0 and
(2,0) to (2,1), that is, x = 2.
Path 1: (0,0) to (2,0)  y = 0
w    xy dx  y dy  
2
(2,0)
  xy dx  y
(0,0)
2
2
2
dy     x  0  dx   0   0    0


0
Path 2: (2,0) to (2,1)  x = 2
1
y3
1
w    xy dx  y dy     xy dx  y dy    2y  0   y dy   


3 0
3
3
(2,0)
(2,0)
(2,1)
1
(2,1)
2
2
3
2
d) x = 2t3, y = t2 t: 0  1
Do line integral using t as the integration variable. Thus, dx = 6t2 dt and dy = 2t dt
w    xy dx  y 2 dy  
1
( 2 ,1)
1
2
 2t 3  t 2  6t 2dt    t 2 2  2t dt  
x
y
dx

y
dy




0 

( 0,0 )
1
1
12 2 3 1 9 2 7
  12t 7 dt  2t 5dt   12  t 8 |10 2  t 6 |10       
8
6
8 6 2 3 6 6 6
0
11
Example: Evaluate the following line integral
x
2
 y 2 dx  2xy dy using each of the
following paths from (0,0) to (1,2):
a) y = 2x2
x
2
 y  dx  2xy dy 
2
1,2
 x
 
0,0
1
2


 y 2  dx  2xy dy   x 2   2x 2  dx  2x  2x 2  4x dx
0
1
1
1
0
0
0
   x 2  4x 4  dx  16x 4 dx   x 2dx   20x 4dx 
2
x3 1
x5
1
11
|0 20  |10   4  
3
5
3
3
b) x = t2, y = 2t
2
2
  x  y  dx  2xy dy 
1
1,2

 0,0
1


 x 2  y2  dx  2xy dy    t 2    2t  2t dt  2  t 2   2t  2dt
0
1
2
2
1
t6 1
x4 1 1
11
   2t  8t  dt  8t dt   2t dt   16t dt  2 |0 16  |0   4  
6
4
3
3
0
0
0
5
3
3
5
3
c) straight line from (0,0) to (2,0), then straight line from (2,0) to (1,2)
Path 1: (0,0) to (2,0) i.e., y = 0
x
2
 y  dx  2xy dy 
2
 2,0
 x
 
2
2
0,0
2
  x 2 dx 
0


 y 2  dx  2xy dy   x 2   0  dx  2xy  0 
2
0
x 3 2 23 8
|0  
3
3 3
Path 2: (2,0) to (1,2) i.e., y = -2x + 4
2
2
  x  y  dx  2xy dy 
1

1,2 

 2,0 
1


 x 2  y2  dx  2xy dy   x 2   2x  4  dx  2x  2x  4  2 dx 

2
2
1
  x   4x  16x  16  dx   8x  16x  dx    11x 2  32x  16  dx
2
2
2
2
2
3
2
x 1
x
11
32
|2 32 |12 16x |12   13  23   12  22   16 1  2 
3
2
3
2
11
32
77
77 96
19
    7     3  16  1 
 48  16 


3
2
3
3
3
3
 11
Path 1 + Path 2 =
8 19
11
 
3 3
3
12
The line integrals for all of the paths are identical. This happy coincidence is because
df   x 2  y 2  dx  2xy dy is an exact differential.
To confirm this fact, let us assume that M  x, y    x 2  y 2  and N  x, y    2xy  and then
 M  x, y    N  x, y  
 M  x, y    N  x, y  
calculate 
 . If 
 
 , then df is exact and
 ,
y
y
y

 x  x
y

 x  x
the path of the line integral will not change its value depending on the path. Then also, the
function f is a state function.
 M  x, y  
  
2
2

     x  y   2y

y

y
 x

x
 N  x, y  
  

     2xy   2y
 x  y
 x
y
 M  x, y    N  x, y  

 
  df is exact. Though we will not need to do the following
y

 x  x
y
x3
operation is our studies, the state function f can be found to be f   xy 2 .
3
Since f is a state function, all of the detailed line integrations done above are unnecessary to
find the change in f. We only need to find f at the end point and subtract f at the beginning
point, i.e., f = f(1,2) – f(0,0).
f  f 1, 2   f  0, 0 
1

  0 3
 1
11
2
 1 2   
  0  0     4  
3
3
 3
 3
3
2
Series Expansions
In the pursuit to construct a relationship between physical quantities, we often find that using
simple functions is inadequate. To express the relationship as a mathematical equation, we
must resort to using approximations. One of the most common ways to approximate a function
is to use a Taylor series.
One-dimensional Taylor series
2
3
f  x 0 
1  f  x0 
1  f x0 
2
3
x

x

x

x



x  x0  
0
0
2
3
x
2! x
3! x
n

1  f  x0 
n
 x  x0 

n
x
n  0 n!
f  x   f  x0  
Using a Taylor series, we can fit k data points to a kth order polynomial.
Note: If x0 = 0, then series is called a Maclaurin series.
13
Uses of Taylor series
1) We can approximate a function as a polynomial series.
Example 1: Expand the function cos t about the point t  0 .
f  0  cos 0  1
 d 2f  t 0  
2
2
2

   cos t |t 0   cos 0  
2
 dt

4
 d f  t0  
4
4
4

   cos t |t 0   cos 0  
4
 dt

 df  t 0  

   sin t |t 0   sin 0  0
 dt 
 d 3f  t 0  
3
3

   sin t |t 0   sin 0  0
3
 dt

2
3
4
f  t 0 
1  f  t0 
1  f  t0 
1  f  t0 
2
3
4
t

t

t

t

t

t

 0
 0
 0
t  t0  
2
3
4
t
2! t
3! t
4! t
1
1
1
2
3
4
cos t  1   0  t  0    2   t  0    0  t  0    4   t  0  
2!
3!
4!
2
4
 2  4
 1
t 
t 
2
24
f  t   f  t0  
Example 2: Expand the function ln p about the point p  1 .
f 1  ln1  0
 d 3f  p 0  
1

  2 3 |p 1  2
3
p
 dp 
 df  p0   1

  |p1  1
 dp  p
 d 4f  p0  
1

  6 4 |p 1  6
4
p
 dp 
 d 2f  p0  
1

   2 |p 1  1
2
p
 dp

2
3
4
f  p 0 
1  f  p0 
1  f  p0 
1  f  p0 
2
3
4
p

p

p

p

 p  p0  




 p  p0  
0
0
2
3
4
p
2! p
3! p
4! p
1
1
1
2
3
4
ln p  0  1 p  1   1 p  1   2  p  1   6  p  1 
2!
3!
4!
1
1
1
2
3
4
  p  1   p  1   p  1   p  1 
2
3
4
f  p   f  p0  
14
2) Create a polynomial function that fits a set of data points.
Example: Find an expression that relates the Darth Vader temperature scale to the Kelvin
scale given the data below.
Vader
1V
45 V
1291 V
Kelvin
0K
310 K
10000 K
We need to find V = f(K).
First question: What point do we choose to expand about, i.e., what is x0?
We can choose any point, so let’s choose the most convenient. How about x0 = 0?
For the first data point
2
f  K 0 
1  f  K0 
2
 K  K0  
K  K0 
2
K
2! K
2
f  0 
1  f 0
2
 f  0 
K  0
 K  0 
2 
K
2! K
2
f  0 
1  f 0
2
1V  f  0  
0

0

0  0  f 0


2 
K
2! K
f  0   1V
V  f  K   f  K0  
For the second data point
2
f  0 
1  f 0
2
V  f  0 
310  0 
 310  0  
2 
K
2! K
2
f  0 
1  f  0
45 V  1V 
310 
3102
2
K
2! K
2
f  0 
 f  0
44 V 
310 
48050
K
K 2
For the third data point
2
f  0 
1  f  0
2
V  f 0 
10000  0 
10000  0  
2 
K
2! K
2
f  0 
1  f  0
1291V  1V 
10000 
100002
2
K
2! K
2
f  0 
 f 0
1290 V 
10000 
50, 000, 000
K
K 2
15
We now have two equations and two unknowns. The two unknowns are the coefficients of the
f  0   2 f  0 
,
Taylor series,
K
K 2
 2f  0 
f  0 
 2.670  106
 0.1423 and
2
K
K
Thus our function for temperature conversion between the Vader and Kelvin scale is
V  1V  0.1423V  K 1  T   2.670 106 V  K 2  T 2 
What is the temperature in Vader for the melting point of steel, 3695 K? (Using a light saber,
of course!)
V  1V  0.1423 V  K 1  3695 K   2.670 106 V  K 2  3695 K 
2
 1V  525.8 V  36.5 V  490.3V
What is the temperature in Vader for the coldest temperature ever on the ice planet Hoth 90.7 K?
V  1V  0.1423 V  K 1  90.7 K   2.670 10 6 V  K 2  90.7 K 
2
 1V  12.9 V  0.02 V  13.9 V
Two-dimensional Taylor series
 f  x 0 , y 0  
 f  x 0 , y 0  
f  x, y   f  x 0 , y 0   
  x  x0   
  y  y0 
x
y

y

x

2
 2f  x 0 , y0 
 2f  x 0 , y0 
1   f  x 0 , y0 
2
2
x

x

2
x

x
y

y






 y  y0  

0
0
0
2
2
2! 
x
xy
y

  3f  x 0 , y 0 

 3f  x 0 , y 0 
3
2
x

x

3


 x  x 0   y  y0  

0
3
2
x
x y
1

 
3

3!   3f  x 0 , y 0 

f
x
,
y


2
3
0
0
 3
x

x
y

y

y

y


0 
0
0 
xy 2
y3


Note that the coefficients in front of the partial derivatives are the binomial coefficients.
16
Numerical Integration
Many times in science, an analysis leads to the need to do an integration that does not have a
closed form. That is, we can’t find a solution to the integral by remembering our calculus
techniques or by finding the solution in a table of integrals. Nonetheless, we need the value of
the integral to continue our analysis. The integration could be of data that we have collected
that does not conform to a simple function, or it could be a function that does not have a known
solution. In both cases, to find the value of the function, we’ll need to perform a numerical
integration (as opposed to an analytical integration).
Trapezoidal Rule
There are many different methods for performing numerical integrations. The trapezoidal rule
is among the simplest. When used with a sufficient number of data points, the results are
reasonably accurate.
The geometric interpretation is an area under of a curve. The basic idea behind numerical
integration is break up the area under of a curve into shapes whose areas are easily calculated.
A common shape used is a trapezoid, whose area is ½(top length + bottom length) times the
height, i.e., A = ½(a + b) h
a
h
b
To find the area under a curve drawn by discrete points, the area is divided into separate
trapezoids.
Let us use an example finding the enthalpy of a material by measuring the heat capacity at
various temperatures. More specifically we’ll want to do the following integral,
Tf
H   Cp dT
Ti
Below is a plot that shows the how the heat capacity of a particular substance changes with
temperature.
Cp
T
17
This particular data set has 10 points. Thus, we want to do an integration from T1 to T10. The
integration is the area under the curve between T1 to T10. However, we only have points; thus,
we don’t even have a curve to integrate. Nonetheless, we can still find an area if we create
trapezoids with the points given. (Specifically, with 10 points, we’ll make 9 trapezoids.) In the
graphic below, we can see how to create the first trapezoid with the first two data points.
Cp
Cp(2)
Cp(1)
T1
T2
T3 T4
T 5 T6 T7
The area of this trapezoid is calculated as A1 
T8 T9 T10
T
1
 Cp  2   Cp 1   T2  T1  .
2
Cp
Cp(3)
Cp(2)
T1
T2
T3 T4
T 5 T6 T7
The area of this trapezoid is calculated as A 2 
T8 T9 T10
T
1
 Cp  3  Cp  2    T3  T2 
2
18
Thus we can find the area “under the curve” by adding together the area of all nine trapezoids.
Tf
H   Cp dT  A1  A 2  A3  A 4  A5  A 6  A 7  A8  A9
Ti
Cp
Cp(4)
Cp(3)
Cp(2)
Cp(1)
T1
T2
T3 T4
T 5 T6 T7
T8 T9 T10
T
The last comment on numerical analysis is that a numerical integration is an approximation,
albeit, a reasonably accurate one. The accuracy of a numerical integration increases as more
points are used in the integration.
Dimensional Analysis
Dimension is not the same as unit, though it is similar
- Dimension is broader object than unit
We classify the dependence of physical quantities upon a few, fundamental “dimensions” or
variables.
Dimension
mass – M
length – L
time – T
temperature – 
charge – Q
Examples of associated units
kilograms, amu, slugs
meter, inches, Angstroms
seconds, years
Kelvin, degrees Celsius, degrees Fahrenheit
coulomb, abcoulomb, statcoulomb
An arbitrary physical quantity is proportional to M  LT  Q
Examples:
Volume – L3
Momentum – M L/T
Energy – M L2/T2
Dipole moment – Q L
Entropy – M L2/ T2
Dimensional analysis is very helpful to check equations after a complicated derivation or
ensuring that a coefficient has the correct units.
19
Rules when dealing with quantities with dimensions
1.) Only quantities with the same dimension can be added or subtracted.
2.) Arguments of transcendental functions must be dimensionless.
Examples of transcendental functions
sin x, cos x, ex, ln x
Why is this true?
Transcendental functions can be written as series expansions.
x 2 x3 x 4
e x  1  x     (A Maclaurin expansion)
2 6 24
- x must be dimensionless to add members of series together.
3.) If x = y, then x and y must have the same dimension.
4.)
dx
x
has the same dimension as .
dy
y
5.)
 x dy has the same dimension as xy.
Example: What is the dimension of the gas constant, R?
F
ML 1 3
V
 2 L
2
PV A
ML2
PV  nRT  R 

 R T L
 2
nT
nT

T
note: amount, i.e., number of moles, is dimensionless.
- Consider why this is true?
V
, dimensionally consistent?
x
ML2
2
V
ML
ML
 T  2
F 2
x
L
T
T
Example: Is the equation, F  
Yes, the equation is dimensionally consistent.