Bai giang ham so kha vi va vi phan cua ham nhieu bien

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Chapter 8: Functions of Several Variables Section 8.4 Differentials Written by Richard Gill Associate Professor of Mathematics Tidewater Community College, Norfolk Campus, Norfolk, VA With Assistance from a VCCS LearningWare Grant

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Transcript of Bai giang ham so kha vi va vi phan cua ham nhieu bien

Page 1: Bai giang ham so kha vi va vi phan cua ham nhieu bien

Chapter 8: Functions of Several Variables

Section 8.4Differentials

Written by Richard GillAssociate Professor of Mathematics

Tidewater Community College, Norfolk Campus, Norfolk, VA

With Assistance from a VCCS LearningWare Grant

Page 2: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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

).(' xfdxdy

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

dxdy

runrise

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 )('

Page 3: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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 ),(),(

Page 4: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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

Page 5: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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:

dxxzdz

),( 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

ordyyzdx

xzdz

yx ),(),(

Page 6: Bai giang ham so kha vi va vi phan cua ham nhieu bien

z

x y

)),(,,( yxfyx

z = f(x,y)

x

),( yxx

)),(,,( yxxfyxx

),( yx

y

),( yyxx

)),(,,( yyxxfyyxx dyydxxdz

dyyzdx

xzdz

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.

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Example 2.

hundredth.nearest the toRoundz. 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(2sin18.0,1.0,)1,2(),(

cossin

cos

ffzff

dzdydxyxdyxdxxy

dyyzdx

xzdz

xyz

zfor ion approximat good a is dz

Page 8: Bai giang ham so kha vi va vi phan cua ham nhieu bien

Definition of Differentiability

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

),(),( form in the expressedbecan if b)(a,at abledifferenti isfunction the,),(For

21

21

yx

yxybafxbafzzyxfz

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

Page 9: Bai giang ham so kha vi va vi phan cua ham nhieu bien

),(),(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

ybafxbafzyxfz

yx

proof. thecompleteswhich ),(),(),(),(0),(),(

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

)),(()),((),(),( above, From

.z that know also We

221

221

bayxasyxfbafyxfbafbayx

ybbafxabafybafxbafyxfbafz

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

x

x

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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

dhhVdr

rVdV

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

Page 11: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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

dhhVdr

rVdV

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

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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

dhrdrhrdhhAdr

rAdA

rhrA

5.6% iserror percent The

056.02050115

AdA iserror relative The

Page 13: Bai giang ham so kha vi va vi phan cua ham nhieu bien

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.