TOTAL DIFFERENTIALS
The differential was introduced in Section B-4. Recall that the differential of a function C œ 0 ÐBÑ is
.C œ 0 w ÐBÑ .BÞ Here .B is the differential with respect to the independent variable B and is equal to
?Bß the increment of BÞ
In Section B-4, we observed that .C is a pretty good approximation for ?C, which is the increment of
C, and defined as ?C œ 0 ÐB  ?BÑ  0 ÐBÑÞ
This differential .C approximates ?C with respect to one variable. Now suppose D œ 0 ÐBß CÑ is a
function of two variables. What is a good approximation for ?D if B changes from B to B  ?B and C
changes from C to C  ?C?
TOTAL DIFFERENTIAL
Definition: Let D œ 0 ÐBß CÑ be a function of B and C . Let .B and .C be real numbers. Then the total
differential of D is
.D œ 0B ÐBß CÑ † .B  0C ÐBß CÑ † .C
EXAMPLE 1 À Consider D œ 0 ÐBß CÑ œ "!B$  )B# C  %C$ Þ
(a) Find the total differential .DÞ
(b) Evaluate .D when B œ #ß C œ "ß .B œ Þ!"ß and .C œ  Þ!#Þ
(a) First we find 0B ÐBß CÑ and 0C ÐBß CÑÞ
0B ÐBß CÑ œ $!B#  "'BC
and
0C ÐBß CÑ œ  )B#  "#C#
So by definition,
.D œ a$!B#  "'BCb .B  a  )B#  "#C# b .C
(b) Putting in the values into the result of (a) we get
.D œ Ð))ÑÐÞ!"Ñ  Ð  #!ÑÐ  Þ!#Ñ
œ "Þ#)
The result here tells us that if you increase B by Þ!" and decrease C by .02, D œ 0 ÐBß CÑ will
increase
by approximately 1.28.
Just as in the one variable case, we can use .D as an approximation for ?DÞ
We define the increment of the dependent variable D as
?D œ 0 ÐB  ?Bß C  ?CÑ  0 ÐBß CÑ œ 0 ÐB  .Bß C  .CÑ  0 ÐBß CÑ
MHIE À For relatively small values of ?B and ?C, .D ¸ ?D
To see this, consider the following:
EXAMPLE 2: Let D œ 0 ÐBß CÑ œ #ÈB#  C#
(a) Use the total differential .D to approximate ?D when B changes from $ to #Þ*) and C changes
from % to %Þ!"Þ
(b) Calculate the actual change ?D .
.D œ 0B ÐBß CÑ † .B  0C ÐBß CÑ † .C œ Š ÈB#B
.B  Š ÈB#C
.C
# C # ‹
# C # ‹
(a) First we find .D À
Note that .B œ ?B œ #Þ*)  $ œ  Þ!# and .C œ ?C œ %Þ!"  % œ Þ!"Þ
So our approximation is
#Ð%Ñ
.D œ Š È#Ð$Ñ
‹a  Þ!#b  Š È$# %# ‹aÞ!"b œ  !Þ!!)
$# %#
(b) To calculate the actual change, we find ?D œ 0 ÐB  ?Bß C  ?CÑ  0 ÐBß CÑ:
?D œ 0 Ð#Þ*)ß %Þ!"Ñ  0 Ð$ß %Ñ œ #È#Þ*)#  %Þ!"#  #È$#  %#
œ  !Þ!!(*!$"#$
So our approximation for ?D is pretty good.
Now since 0 ÐB  .Bß C  .CÑ  0 ÐBß CÑ œ ?D ¸ .D œ 0B ÐBß CÑ † .B  0C ÐBß CÑ † .CÞ Then
0 ÐB  .Bß C  .CÑ ¸ 0 ÐBß CÑ  .D
EXAMPLE 3:
Approximate a"Þ*##  #Þ"# b
"Î$
Þ
$
First we note that D œ 0 ÐBß CÑ œ È
B#  C# œ aB#  C# b
.B œ  Þ!) and .C œ !Þ"Þ
Now we calculate .DÞ
.D œ "$ aB#  C# b
#Î$
"Î$
Þ We start with B œ #ß C œ #ß
† #B .B  "$ aB#  C# b
#Î$
† #C .C
Evaluating at .D at the above values yields
.D œ "$ a##  ## b
œ !Þ!!'(
#Î$
Ð#ÑÐ#ÑÐ  !Þ!)Ñ  "$ a##  ## b
#Î$
Ð#ÑÐ#ÑÐ!Þ"Ñ
So we approximate 0 Ð"Þ*#ß #Þ"Ñ ¸ 0 Ð#ß #Ñ  .D œ #  !Þ!!'( œ #Þ!!'(
Definition À
If A œ 0 ÐBß Cß DÑ is a function of three variables, the total differential .A is
.A œ 0B ÐBß Cß DÑ † .B  0C ÐBß Cß DÑ † .C  0D ÐBß Cß DÑ † .D
APPLICATION
EXAMPLE 4: Approximate the change in the volume of a beverage can in the shape of a right
circular cylinder as the radius changes from 3 to 2.5 and the height changes from 14 to 14.2.
We note the volume of the can is Z œ 1<# 2Þ So our function is in two variables, i.e.
Z Ð<ß 2Ñ œ 1<# 2Þ
We now calculate .Z À
.Z œ Z< Ð<ß 2Ñ † .<  Z2 Ð<ß 2Ñ † .2
œ #1<2 † .<  1<# † .2
Given that < œ $ß 2 œ "%ß .< œ  !Þ&ß and .2 œ !Þ#ß we get
.Z œ #1Ð$ÑÐ"%ÑÐ  !Þ&Ñ  1Ð$# ÑÐ!Þ#Ñ
¸  "#'Þ#*#!
Thus by decreasing the radius by 0.5 and increasing the height by 0.2 , the volume of the can
decreases by approximately "#'Þ#*#! cubic units.
PRACTICE
If D œ 0 ÐBß CÑ or A œ 0 ÐBß Cß DÑ, 0 ind the total differential .D or .A of the following functions:
1.
#Þ
0 ÐBß CÑ œ $B#  &BC  'B
0 ÐBß CÑ œ ÈB$  C#
3.
0 ÐBß CÑ œ lnaB#  C# b
4.
0 ÐBß CÑ œ B/$C  $B(
5.
0 ÐBß CÑ œ C# /B  B ln C
6.
0 ÐBß Cß DÑ œ B# C$ D
7.
0 ÐBß Cß DÑ œ
8.
0 ÐBß Cß DÑ œ B/CD  BCD
D
BC
Find the approximate change in D when the point ÐBß CÑ changes from ÐB! ß C! Ñ to ÐBß CÑÞ
9.
0 ÐBß CÑ œ BC  &B# à from Ð#ß $Ñ to Ð#Þ!"ß $Þ!#Ñ
10.
0 ÐBß CÑ œ 2B"Î# C#Î$ à from Ð%ß )Ñ to Ð$Þ*)ß )Þ!$Ñ
11.
0 ÐBß CÑ œ ÈB#  C$ à from Ð$ß $Ñ to Ð$Þ!"ß #Þ)*Ñ
C
BC à
"#Þ
0 ÐBß CÑ œ
from Ð  #ß "Ñ to Ð  #Þ!&ß "Þ!&Ñ
13.
0 ÐBß CÑ œ $C/B à from Ð!ß %Ñ to Ð!Þ!$ß %Þ!"Ñ
14.
0 ÐBß CÑ œ C#  B/BC à from Ð"ß  "Ñ to Ð!Þ*)ß  !Þ!)Ñ
15.
0 ÐBß CÑ œ B ln C  C ln Bà from Ð$ß &Ñ to Ð#Þ*&ß &Þ!%Ñ
16.
Use the total differential to approximate È'Þ!(#  (Þ*&# Þ
17.
The monthly profit (in dollars) at a fancy department store depends on the level of inventory B
(in thousands of dollars) and the floor space C (in thousands of square feet) available for display of the
merchandise, as given by the equation
T ÐBß CÑ œ  !Þ!#B#  "&C#  BC  $*B  #&C  #!ß !!!
Currently, the level of inventory is $4,000,000 (B œ %!!!Ñ, and the floor space is 150,000 square feet
(C œ "&!Ñ. Find the expected change in monthly profit if management increases the level of inventory
by $500,000 and decreases floor space for display of merchandise by 10,000 square feet.
18.
The :<9.?->398 function for one country is D œ B!Þ'& C!Þ$& ß where B stands for units of labor
and C for units of capital. At present, 50 units of labor and 29 units of capital are available. Use
differentials to estimate the change in production as the number of units is increased to 52 and capital
is decreased to 27 units.
Answers
1.
.D œ Ð'B  &C  'Ñ .B  Ð  &BÑ .C
2.
$B#
C
.D œ Œ
.B  Œ
.C

ÈB$  C# 
#ÈB$  C#
3.
.D œ Œ
4.
.D œ Š/$C  #"B' ‹ .B  $B/$C .C
5.
.D œ ŠC# /B  68 C‹ .B  Š#C/B  BC ‹ .C
6.
.A œ #BC$ D .B  $B# C# D .C  B# C$ .D
7.
.A œ
8.
.A œ Š/CD  CD ‹ .B  ŠBD/CD  BD ‹ .C  ŠBC/CD  BC‹ .D
9.
 !Þ"$
10.
!
11.
 !Þ#%#&
12.
 !Þ!!&'
13.
 !Þ$$
14.
 #Þ"()%&
15.
 !Þ!"(!)
16.
"!Þ!!#
17.
an increase of $19,250 per month
18.
0.0769 units
#B
#C
.B


Œ
 .C
B#  C#
B#  C#
D
D
"
.B 
.C 
.D
#
#
ÐB  CÑ
ÐB  CÑ
ÐB  CÑ