Chapter 8: Functions of Several Variables

Section 8.4Differentials

Written by Richard Gill

Associate Professor of Mathematics

Tidewater Community College, Norfolk Campus, Norfolk, VA

With Assistance from a VCCS LearningWare Grant

To begin this section on differentials in three variables, we will begin with a review of differentials in two variables. Consider the function y = f(x).Now consider a generic value of x with a tangent to the curve at (x, f(x)).

Compare the initial value of x to a value of x that is slightly larger.

))(,( xxfxx

xx x

))(,( xfx

).(' xfdx

dy

The slope of the tangent line is:

It was at this point that we first saw dx defined as .x

xdx

Since and since dx is

being defined as “run” then dy becomes “rise” by definition.

dx

dy

run

rise

y

dy

For small values of dx, the rise of the tangent line, was used as an approximation for , the change in the function.y

dxxfdy )('

z

x y),( yx

)),(,,( yxfyx

Now consider the extension of the differential concept to functions of several variables. Consider an input (x,y) for this function. Its outputs z = f(x,y) create a set of points (x,y,z) that form the surface you see below.

z = f(x,y)

The differential dz will have two parts: one part generated by a change in x and the other part generated by a change in y.

Consider a new point in the domain generated by a small change in x.

x

),( yxx

Now consider the functional image of the new point.

)),(,,( yxxfyxx

We can use the function to calculate , the difference between the two z-coordinates.

z

By subtracting the z-coordinates:

zyxfyxxf ),(),(

z

x y),( yx

)),(,,( yxfyx

Now consider the extension of the differential concept to functions of several variables. Consider an input (x,y) for this function and its output (x,y,z).

z = f(x,y)

The differential dz will have two parts: one part generated by a change in x and the other part generated by a change in y.

x

),( yxx

)),(,,( yxxfyxx

By subtracting the z-coordinates:

zyxfyxxf ),(),(

Since the y-coordinate is constant, we can use that cross section to draw the tangent to )),(,,( yxfyx

Remember that, dz is the change in the height of the tangent line, and can be used to estimate the change in z.

dz

z

x y

)),(,,( yxfyx

z = f(x,y)

x

),( yxx

)),(,,( yxxfyxx

Since there has been no change in y we can express the differential so far in terms of the change in x:

dxx

zdz

),( yx

Now track the influence on z when a new point is generated by a change in the y direction.

y

),( yyxx

)),(,,( yyxxfyyxx

Now that the change in z is generated by changes in x and y, we can define the total differential:

dyyxfdxyxfdz

ordyy

zdxx

zdz

yx ),(),(

z

x y

)),(,,( yxfyx

z = f(x,y)

x

),( yxx

)),(,,( yxxfyxx

),( yx

y

),( yyxx

)),(,,( yyxxfyyxx dyydxxdz

dyy

zdxx

zdz

46

z.for aldifferenti total theFind .23 :equation by the

generated is with ngbeen worki have graph we theSuppose 1. Example22 yxz

Try this on your own first.

Example 2.

hundredth.nearest the toRound

z. it to compare and dz aldifferenti total thecalculate

),08.1,1.2( and )1,2( evaluate ,cos),( If

ffxyyxf

Hint: first calculate dx and dy.

Solution: dx = 2.1 – 2 = 0.1 and dy = 1.08 – 1 = 0.08

129.0

))1.2(cos(08.1))1.2(cos(08.1

)1,2()08.1,1.2(

416.0))1.2(cos(08.1)1,2(

545.0))1.2(cos(08.1)08.1,1.2(

124.0.....0332.0....0909.0

)08.0(2cos)1.0(2sin1

8.0,1.0,)1,2(),(

cossin

cos

ffz

f

f

dz

dydxyx

dyxdxxy

dyy

zdxx

zdz

xyz

zfor ion approximat good a is dz

Definition of Differentiability

).0,0(),( as 0 and where

),(),( form in the expressed

becan if b)(a,at abledifferenti isfunction the,),(For

21

21

yx

yxybafxbafz

zyxfz

yx

z. eapproximat todz use weerror when the

be to and consider can you ,),(),( Since 21

yxdyyxfdxyxfdz yx

The following theorem is presented without proof though you can usually find the proof in the appendix of a standard Calculus textbook.

b).(a,at abledifferenti is then b)(a,at

continuous are and b)(a, containingregion a

inexist and sderivative partial theIf

f

ff yx

),(),(limb)(a,y)(x,

bafyxf

Theorem: If a function of x and y is differentiable at (a,b) then it is continuous at (a,b).

Solution: the objective is to show that

).,(),( as 0 and 0 and,, where

)),(()),(( definitionby Then,

b).(a,at abledifferenti be ),(Let

21

21

bayxybyxax

ybafxbafz

yxfz

yx

proof. thecompleteswhich

),(),(),(),(0),(),(

0. togoes expression above the,),(),( aslimit theTaking

))(),(())(),((

)),(()),((),(),(

above, From

.z that know also We

221

221

bayxasyxfbafyxfbaf

bayx

ybbafxabaf

ybafxbafyxfbafz

,y)f(a,b)-f(x

x

x

Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2 ft. These measurements have possible errors in accuracy as laid out in the table below. Complete the table below. Comment on the relationship between dV and for the indicated errors.V

0.1 ft. 0.1 ft.

0.01 ft. 0.01 ft.

0.001 ft. 0.001 ft.

r h dV VSolution:

dhrdrrhdV

dhh

Vdr

r

VdV

hrhrfV

2

2

2

),(

3

2

3

2

3

2

0.0750.024

)001.0)(2()001.0)(5)(2(2001.0

754.24.004.2.0

)01.0)(2()01.0)(5)(2(201.0

540.74.24.02

)1.0)(2()1.0)(5)(2(21.0

ft

dVhr

ft

dVhr

ft

dVhr

7.54 cu ft

.754 cu ft

.075 cu ft

Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2 ft. These measurements have possible errors in accuracy as laid out in the table below. Complete the table below. Comment on the relationship between dV and for the indicated errors.V

0.1 ft. 0.1 ft.

0.01 ft. 0.01 ft.

0.001 ft. 0.001 ft.

r h dV VSolution:

dhrdrrhdV

dhh

Vdr

r

VdV

hrhrfV

2

2

2

),(

7.54 cu ft

.754 cu ft

.075 cu ft

075.0)5()2()001.5()001.2()5,2()001.5,001.2(

757.0)5()2()01.5()01.2()5,2()01.5,01.2(

826.7)5()2()1.5()1.2()5,2()1.5,1.2(

),(),(

22

22

22

ffV

ffV

ff

hrfhhrrfV

7.826 cu ft

0.757 cu ft

0.075 cu ft

smaller.get and asbetter gets Vfor ofion approximat The hrdV

Example 4. A right circular cylinder is constructed with a height of 40 cm. and a radius of 25 cm. What is the relative error and the percent error in the surface area if the possible error in the measurement of each dimension is ½ cm.

Solution: If the measurements are correct the surface area will be

20508001250

)40)(25(2)25(2

222

2

rhrA The total differential will generate an estimate for the possible error.

115)5(.50)5)(.80100(

)5)(.25(2)5))(.40(2)25(4(

2)24(

22 2

dhrdrhrdhh

Adrr

AdA

rhrA

5.6% iserror percent The

056.02050

115

A

dA iserror relative The

This concludes the material for Lesson 8.4. This also concludes the lessons on Blackboard and the material for the semester.

As was the case in previous Bb lessons, we have posted three sets of exercises on Bb. Any exercises worked correctly will add to your thinkwell exercise totals.

The practice exam should be up and running on Bb. Good luck preparing for the exam and congratulations on getting to the end of one of the toughest courses in the curriculum.