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P1 Calculus II

Partial Differentiation & Multiple Integration

Prof David Murray

david.murray@eng.ox.ac.ukwww.robots.ox.ac.uk/∼dwm/Courses/1PD

4 lectures, MT 2017

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MotivationSo far in your explorations of the differential and integral calculus,you have considered only functions of one variableThis comes naturally to you ...

ddx

(ln(cosh

(sin−1 (1− 1

x))))

: x > 1 and∫

x(x2 + 1)n

dx .

But you do not have to look too hard to find quantities whichdepend on more than one independent variable

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News, good and ...

This lecture course explains how to extend the calculus to tomultiple independent variables.

There’s good newsMost changes kick in going from one to two independant variables ...

... so usually two will do f = f (x , y).

And even better newsWhereas calculus in 1 variable can feel somewhat mechanistic ...

... now you really have to think!

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Course ContentsThe material splits into four topics to be covered over the four lectureslots.

é Partial differentiation and differentialsé Relationships arising from the nature of the functioné Changing variables from one set to anotheré Multiple integration

The tutorial sheet directly associated with this course is

é 1P1H Calculus II: Partial Differentiation & Multiple Integrals

But, as ever, the sheet is not enough.• You must read, mark, learn and inwardly digest

— and not just these notes.• You must try worked examples from books.

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Reading

James, G. (2001) Modern Engineering Mathematics, Prentice-Hall,3rd Ed., ISBN: 0-13-018319-9 (paperback).

Stephenson, G. (1973) Mathematical Methods for Science Students,Longman Scientific & Technical, 2nd Ed., ISBN: 0-582-44416-0(paperback).

Kreyszig, E. (1999) Advanced Engineering Mathematics, John Wiley& Sons, 8th Ed., ISBN: 0-471-15496-2 (paperback).

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Course WWW Pages

Pdf copies of the notes, copies of the lecture slides, the tutorial sheets,corrections, answers to FAQs etc, will be accessible from

www.robots.ox.ac.uk/∼dwm/Courses

Only the notes and the tute sheets get put on weblearn.www.weblearn.ox.ac.uk

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1 Partial Derivatives & Differentials

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Lecture contents

In this lecture we are concerned with ...

1.1 Continuity/limits for functions of more than one variable1.2 Partial derivatives1.3 A cautionary tale: partial derivatives are not fractions1.4 The total or perfect differential

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1.1 Continuity and limits

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Revision of single variable continuity

Function y = f (x) is continuous atx = a if, for every positive number ε(however small), one can find aneighbourhood

|x − a| < η

in which

|f (x) − f (a)| < ε .

f(x)

x

f(a)

a

ηη

<ε

<ε

Two useful facts about continuous functions are:The sum, difference and product of two continuous functions iscontinuous.The quotient of two continuous functions is continous at every pointwhere the denominator is not zero.

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Revision of single variable derivative

The total derivative isdefined as

dfdx

= limδx→0

[f (x + δx) − f (x)

δx

].

f (x) is differentiable if thislimit exists, and existsindependent of how δx → 0.

x+ δx

f(x+ δx)

f(x+ δx)

δx

δxas tends to zeroTends to tangent

f(x)

x

f=δ − f(x)

Note thatA differentiable function is continuous.A continuous function is not necessarily differentiable. f (x) = |x | isan example of a function which is continous but not differentiable(at x = 0).

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Moving to more than one variable

The big changes take place in going from n = 1 variables to n > 1variables. So for much of the time we can deal with functions of only twovariables — eg, f (x , y).

f (x , y) is conveniently represented as a surface, as opposed to a planecurve for a one variable function.

−4−2

02

4

−5

0

5−20

−10

0

10

20

xy

f(x,y

)

−4−2

02

4

−5

0

5−1

−0.5

0

0.5

1

xy

f(x,y

)

(x + y)(x − y) exp[−(x2 + y2)/10] sin(2x) cos(4y)

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Continuity for functions of several variablesFunction f (x , y) iscontinuous at a point(a, b) in R if, for everypositive number ε(however small), it ispossible to find a positiveη such that

project onto

surface

circle, radius

at (x,y)=(a,b)x

y

ε

ε

f(x,y)

f(a,b)

η

|f (x , y) − f (a, b)| < ε for all points (x − a)2 + (y − b)2 < η2

Thenlim

(x ,y)→(a,b)f (x , y) = f (a, b).

Note that for functions of more variables f (x1, x2, x3, ...) theneighbourhood would be defined by

(x1 − a)2 + (x2 − b)2 + (x3 − c)2 + . . . < η2 .

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1.2 The partial derivative

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The partial derivative

Problem!

The slope of the f (x , y) surface at(x , y) depends on which directionyou move off in!

We have to think about slope in aparticular direction.

The obvious directions are thosealong the x− and y−axes.

Solution! To move off from (x , y) in the x direction, keep y fixed.

Define the partial derivative wrt x :

fx =

(∂f∂x

)y= limδx→0

[f (x + δx , y) − f (x , y)

δx

]

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More than two independent variables

If we are dealing with a function of more variables ...

... simply keep all but the one variable constant.

Eg for f (x1, x2, x3, ...) we have

The partial derivative again ...(∂f∂x3

)x1x2x4...

= limδx3→0

[f (x1, x2, x3 + δx3, x4, ...) − f (x1, x2, x3, x4, ...)

δx3

]

Given that you know the list of the variables, and know the one beingvaried, the “held constant” subscripts are superfluous and are oftenomitted.

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Geometrical interpretation of the partial derivative

plane

= constantyf(x,y)

yx

x

fSlope is

fx y

plane

= constantxf(x,y)

yx

fSlope is

y

fy x

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The mechanics of evaluating partial derivatives

Operationally, partial differentiation is exactly the same as normaldifferentiation with respect to one variable, with all the others treatedas constants.

♣ ExampleSuppose

f (x , y) = x2y3 − 2y2

First assume y is a constant:

∂f∂x

= 2xy3

Then x is a constant:∂f∂y

= 3x2y2 − 4y

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Examples♣ ExampleFind the 1st partial derivatives of f (x , y) = e−(x2+y2) sin(xy2)

∂f∂x

= e−(x2+y2)[−2x sin(xy2) + y2 cos(xy2)]

∂f∂y

= e−(x2+y2)[−2y sin(xy2) + 2xy cos(xy2)]

♣ Example

If f (x , y) = ln(xy), find the product(∂f∂x

)(∂f∂y

)in terms of f .

f (x , y) = ln(x) + ln(y)

⇒∂f∂x

= 1/x and∂f∂y

= 1/y

⇒(∂f∂x

)(∂f∂y

)= 1/xy = e−f (x ,y) .

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Higher partial derivatives∂f /∂x and ∂f /∂y are probably perfectly good functions of (x , y), so wecan differentiate again.

∂2f∂x2 =

∂

∂xfx = fxx

♣ Example.

f (x , y) = x2y3 − 2y2

∂f /∂x = 2xy3

∂f /∂y = 3x2y2 − 4y

Hence

∂2f /∂x2 = 2y3

∂2f /∂y2 = 6x2y − 4 .

But we should also consider∂

∂y

(∂f∂x

)=

∂2f∂y∂x

= 6xy2 and∂

∂x

(∂f∂y

)=

∂2f∂x∂y

= 6xy2.

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Higher partial derivatives∂f /∂x and ∂f /∂y are probably perfectly good functions of (x , y), so wecan differentiate again.

∂2f∂x2 =

∂

∂xfx = fxx

♣ Example.

f (x , y) = x2y3 − 2y2

∂f /∂x = 2xy3

∂f /∂y = 3x2y2 − 4y

Hence

∂2f /∂x2 = 2y3

∂2f /∂y2 = 6x2y − 4 .

But we should also consider∂

∂y

(∂f∂x

)=

∂2f∂y∂x

= 6xy2 and∂

∂x

(∂f∂y

)=

∂2f∂x∂y

= 6xy2.

Oooo!∂2f∂x∂y

=∂2f∂y∂x

— is that always true?

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♣ A more complicated random example

f (x , y) = e−(x2+y2) sin(xy2)

∂f∂x

= e−(x2+y2)[−2x sin(xy2) + y2 cos(xy2)]

∂f∂y

= e−(x2+y2)[−2y sin(xy2) + 2xy cos(xy2)]

⇒ fyx = e−(x2+y2)[−4x2y cos() + 2y cos() − 2y3x sin()− 2y [−2x sin() + y2 cos()]]

= e−(x2+y2)[sin()[−2y3x + 4xy ] + cos()[−4x2y + 2y − 2y3]]

and fxy = e−(x2+y2)[−2y3 cos() + 2y cos() − 2xy3 sin()− 2x [−2y sin() + 2xy cos()]]

= e−(x2+y2)[sin()[−2xy3 + 4xy ] + cos()[−2y3 + 2y − 4x2y ]]

So they are equal in this case too.

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Yes, they are equal

When both sides exist, and are continuous at the point of interest,

∂2

∂x∂y≡ ∂2

∂y∂x

Although the order is unimportant — a little thought can save time!♣ Example.Q: Find

∂3

∂t∂y∂x[(y5 + xy) cosh(cosh(x2 + 1/x)) + y2tx

].

A: You have three seconds ...

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1.3 Partial derivatives are not fraction-like

A severe warning

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Partial derivatives are not fractions

You know from the chain rule that total derivatives have fraction-likequalities ...

♣ Example.Given y = u1/3, u = v3 and v = x2, find dy/dx .

dydx

=dydu· dudv· dvdx

=13u−2/3 3v2 2x = 2x =

131v2 3v2 2x = 2x

which you can check by finding y = x2 explicitly.

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This is NOT the case for partials

♣ Example.Given the perfect gas law pV = RT , determine the product(

∂p∂V

)T

(∂V∂T

)p

(∂T∂p

)V

Of course those tempted to divide would guess the answer = 1

But you’d be DEAD WRONG!

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If pV = RT then p =RTV

, V =RTp

, and T =VpR

.

Thus(∂p∂V

)(∂V∂T

)(∂T∂p

)=

(−RTV 2

)(Rp

)(VR

)=

−RTpV

= −1.

In fact if we have any function f (x , y , z) = 0, then(∂x∂y

)(∂y∂z

)(∂z∂x

)= −1

Partial derivatives do not behave as fractions.Although we can write down expressions like df = . . .

We can NEVER write down ∂f = . . ..

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1.4 Total and partial differentials

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The total differential

A differential is a different from a derivative.

Suppose that we have a continuous function f (x , y) in some region, andboth (∂f /∂x) and (∂f /∂y) are continuous in that region.

The differential tells one by how much the value of the function changesas one moves infinitesimal amounts dx and dy in the x- and y -directions.

The total or perfect differential of f (x , y)

df =∂f∂x

dx +∂f∂y

dy

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Explanation (Proof in the notes)

For changes δx and δy the smallchange in f is

δf = f (x +δx , y +δy)− f (x , y) .

Make the move in two steps —one keeping y fixed, the otherkeeping x fixed.

Function surface

dxfx

dyfy

x x+dx

y

y

x

f

f

f+df

y+dy

The small change in f along x is (∂f /∂x)δx .But! The step over δy is made at x + δx not x .In the limit the correction becomes negligible.

The total or perfect differential sums the partial differentials

df =∂f∂x

dx +∂f∂y

dy

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[**] Taylor’s expansion in 2 variables to 1st order

Note that our expression for df is EXACT in the limit as dx and dytend to zero.

If δx and δy are just small rather than infinitesimally small then

δf ≈ ∂f∂xδx +

∂f∂yδy .

Recalling the expression for Taylor’s series in one variable, you may spotthat this is a Taylor’s expansion to 1st order in two variables.In other words

f (x + δx , y + δy) ≈ f (x , y) +∂f∂xδx +

∂f∂yδy .

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[**] Taylor expansion in 2 variable to higher order

We might as well see how it continues!f (x + δx , y + δy)

≈ f (x , y)

1st order: +∂f∂xδx +

∂f∂yδy

2nd order: +12!

[(∂2f∂x2

)(δx)2 + 2

(∂2f∂x∂y

)(δx)(δy) +

(∂2f∂y2

)(δy)2

]higher orders: + . . . think binomial . . .

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♣ Example

Q: A material with a temperature coefficient of α is made into a block ofsides x , y , z measured at some temperature T . The temperature israised by a finite δT . (a) Derive the new volume of the block.A:

V + δV = x(1+ αδT )y(1+ αδT )z(1+ αδT ) = V (1+ αδT )3 .

Q: (b) Now let δT → dT and find the change using the total differential.A: The volume of the block is V = xyz .

dV = yzdx + xzdy + xydz= yzx(αdT ) + xzy(αdT ) + xyz(αdT )

= 3VαdT⇒V + dV = V (1+ 3αdT ) .

This is exact in the limit as dT tends to zero.

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When is an expression a total differential?Suppose we are given some expression p(x , y)dx + q(x , y)dy .Is it total differential of some function f (x , y)?Now, if it is,

df = p(x , y)dx + q(x , y)dy .

But then we must have that

p(x , y) =∂f∂x

and q(x , y) =∂f∂y

and using fxy = fyx

∂

∂yp(x , y) =

∂2f∂y∂x

=∂

∂xq(x , y) =

∂2f∂x∂y

The t.d. test: p(x , y)dx + q(x , y)dy is a total differential when

∂p∂y

=∂q∂x

.

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Example

♣ Example.Q: Show that there is NO function having continous second partialderivatives whose total differential is xydx + 2x2dy .

A: We know that p(x , y)dx + q(x , y)dy is a total differential iff

∂p∂y

=∂q∂x

.

Set p = xy and q = 2x2. Then

∂p∂y

= x 6= ∂q∂x

= 4x .

Clang.

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Recovering the function from its total differential

Suppose we found p(x , y)dx + q(x , y)dy to be total differential using theabove test. Could we recover the function f ?

To recover f we must perform the reverse of partial differentiation.As ∂f /∂x = p(x , y):

f =

∫p(x , y)dx + g(y) + K1

where g is a function of y alone and K1 is a constant.

Similarly,

f =

∫q(x , y)dy + h(x) + K2

We now need to resolve the two expressions for f , and this is possible, upto a constant K , as the following example shows.

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♣ Eg: f is xy3 + sin x sin y + 6y + 10 — but pretend we don’t know it.

Q: Is this a perfect differential, and if so of what function f (x , y)?

(y3 + cos x sin y)dx + (3xy2 + sin x cos y + 6)dy

A: Using the ∂p/∂y = ∂q/∂x test ...∂

∂y(y3 + cos x sin y) = 3y2 + cos x cos y

∂

∂x(3xy2 + sin x cos y + 6) = 3y2 + cos x cos y

SAME! so it is a perfect differential. Integrating and resolving:

From∂f /∂x : f = y3x + sin x sin y + + g(y) + K1l l l l

From∂f /∂y : f = xy3 + sin x sin y + h(x) + 6y + K2

Sof = xy3 + sin x sin y + 6y + K1 .

We’d need an extra piece of info to recover K1.

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Round-up

The differentialdf =

∂f∂x

dx +∂f∂y

dy

is the value through which a function changes as one moves infinitesimalamounts dx and dy in the x− and y−directions.

Given p(x , y)dx + q(x , y)dyTest whether: ∂p/∂y = ∂q/∂x . If good:Integrate p(x , y) wrt x , remembering g(y) is a const ofintegrationIntegrate q(x , y) wrt y , remembering h(x) is a const ofintegrationResolve the two expressions.

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Summary

In this lecture we have1.1 Extended notions of Continuity and limits to functions of more than

one variable1.2 Defined Partial derivatives and discussed higher partial derivatives1.3 Been told in no uncertain terms that partial derivatives are not

fractions1.4 Introduced the total or perfect differential as the sum of partial

differentials

Please noteDerivatives are different from differentials.That δ is different from d is different from ∂. Use them properly!