Ch5_calc

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Chapter 5: Basic Calculus

Derivatives

Derivatives can be evaluated in Mathematica using many different commands. The D

command requires three arguments inside the square brackets: the function, the variableand the number of derivatives. The last two arguments must be in a list.

Clear f

f =Sin x ;

firstDer =D f, x, 1 ;

Print "1st derivative of f = ", firstDer

1st derivative of f = Cos x

secDer =D f, x, 2 ;

Print "2nd derivative of f = ", secDer2nd derivative of f = - Sin x

Partial Derivatives

Another way of evaluating derivatives is to use the palettes. The symbol ¶ x f will

evaluate the partial derivative of the function f with respect to the variable x:

¶x f

Cos x

Try this command on a function containing two variables, x and y:

g =Sin x - Cos y ;

¶x g;

Print "The partial derivative of g with respect to x = ",

%DThe partial derivative of g with respect to x = Cos x

¶y g;

Print "The partial derivative of g with respect to y = ",

%

The partial derivative of g with respect to y = Sin y

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!ou can also evaluate the derivative of a function with respect to several variables using

the palettes. The e"pression ¶x ¶y h will allow you to evaluate the partial derivative of

h with respect to y and the partial derivative of the result with respect to x.

h =x2 y4;

¶x ¶y h

8 x y3

The e"pression above can also be written as ¶y,x h where the list of variables after the

symbol gives the order of differentiation:

¶y,x h

8 x y3

From Physical Chemistry A Molecular Approach, by Donald McQuarrie and John

D !imon:

"#ample $%&

"valuate the t'o (irst partial derivatives o( the molar pressure P (or the van der

)aals e*uation:

P + R T

V - b % a

V #

P depends on T and V so P $ P (T, V). Define an equation for P :

P = * T ! - - a ! 2

- a

V2 +

R T

- b +V

%sing D, take the partial derivative of P with respect to T at constant V :

der1 =D P, T, 1 ;

Print "¶P

¶T ! = ", der1

¶ P

¶ TV =

R

- b +V

%sing ¶V P , take the partial derivative of P with respect to V at constant T :

der2 =¶ ! P;

Print "¶P

¶ ! T = ", der2

¶ P

¶ VT =

2 a

V3 -

R T

- b +V 2

n(inite !eries

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The Taylor series of the function f at xo $ c can be e"pressed as:

f '"( $ f n c

n! x - c n $ f (c) )

c

+! x )c

#! x#)

f c $ -, we have the aclauren series:

f '"( $ f n 0

n! x n $ f (0) )

* 0

+! x )** 0

#! x )

The command /eries will e"ecute a Taylor series for any function. The command

requires four arguments: the function, the variable, the initial value of the variable and the

order of the e"pansion. The last three arguments must be in a list.

0or e"ample, e"ecute the Taylor series for c $ + to the th order for the function:

f (x) = ln x

Series #og x , x, 1, $

x - 1 - 1

2x - 1

2+

1

3x - 1

3-

1

x - 1

+

1

!x - 1

!+" x - 1

#

1ow e"ecute the aclauren series for c $ - to the 2th order for the function:

g(x) = +

++ x #

Series 1 1+x 2, x, , 4

1 - 2 x +3 x2 - x

3 +! x +" x

!

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From Physical Chemistry A Molecular Approach, by Donald McQuarrie and John

D !imon:

Problem $%-

"valuate¶U

¶V T (or an ideal .as and (or a van der )aals .as

¶ U

¶ V T + T

¶ P

¶T V % P

deal .as: P + R T

V

van der )aals .as: P + R T

V - b % a

V #

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Ans'ers: Problem $%-

0or an ideal gas:

Define an equation for the molar pressure P of an ideal gas:

Clear , T, !, P

Pideal = * T !

R T

V

%sing ¶T Pideal , take the partial derivative of Pideal with respect to T at constant V :

der&deal =¶T Pideal

RV

5valuate T ¶ P

¶T V 6 P to get

¶U

¶V T and print out the answer:

ans1 =T* der&deal - P;

Print "¶'

¶ ! T = ", ans1

¶ $

¶ V T = - P +

R T

V

7ecall that the molar pressure P was defined as Pideal . %se the substitution command 8.

and print out the actual answer:

ans2 =ans1 (P ®Pideal;

Print "¶'

¶ ! T = ", ans2, " for an ideal gas"

¶ $

¶ VT = % for an ideal gas

The internal energy U of an ideal gas is independent of V at constant T .

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0or a van der aals gas:

Define an equation for the molar pressure P of a van der aals gas:

Clear , T, !, P, a,

Pvdw = * T ! - - a ! 2

- a

V2 +

R T

- b +V

%sing ¶T Pvdw, take the partial derivative of Pvdw with respect to T at constant V :

dervdw =¶T Pvdw

R

- b +V

5valuate T ¶ P

¶T V 6 P to get

¶U

¶V T and print out the answer:

ans) =T* dervdw - P;

Print "¶'

¶ ! T = ", ans)

¶ $

¶ VT = - P +

R T

- b +V

7ecall that the molar pressure P was defined as Pvdw. %se the substitution command 8.and print out the actual answer:

ans4 =ans) (P ®Pvdw;

Print "¶'

¶ ! T = ", ans4, " for a van der *aals gas"

$

¶ VT =

a

V2 for a van der &aals gas

The internal energy U of a van der aals gas is dependent on V at constant T .

2-

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