Chap 1 Advanced Differentiation

WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation

Mohd Nasir Mahmud@UniKL MICET Page 1 of 25 Wednesday, January 16, 2008

1.1 IMPLICIT DIFFERENTIATION

Procedure for Implicit Differentiation

Example 1

If )(xfy = is a differentiable function of x such that

yxyyx 232 32 +=+

find .dx

dy

Objectives: At the end of this lesson, the students should be able to:

i. differentiate the implicit function containing products and quotients. ii. find and evaluate a partial derivative (1st order) iii. determine the higher-order partial derivative (2nd order & mixed 2nd order) iv. verify that a function satisfies the given equation v. determine the total differential of the given function.

vi. use spital'oHl' rule to compute the indeterminate forms ∞∞

, 0

0 etc.

vii. apply the gradient function for optimization in the life sciences viii. find the rate of change of volume/area etc ix. determine the approximate error of a quantity caused by small changes in the

variables associated with the quantity.

CHAPTER 1 ADVANCED DIFFERENTIATION

Suppose an equation defines y implicitly as a differentiable function of x . To find

:dx

dy

Step 1. Differentiate both sides of the equation with respect to x . Remember that y is really a function of x for part of the curve and use the chain rule when differentiating terms containing y .

Step 2. Solve the differentiated equation algebraically for .dx

dy

WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation


Solution

Differentiate both sides of this equation term by term with respect to x :

( ) ( )yxdx

dyyx

dx

d232 32 +=+

( ) ( ) ( ) ( )ydx

dx

dx

dy

dx

dyx

dx

d232 32 +=+

( ) ( ) ( )dx

dyy

dx

dyx

dx

dyy

dx

dx 2)1(332

rulePower Extended

2

ruleProduct

22 +=

++4434421444 3444 21

dx

dy

dx

dyyxy

dx

dyx 2362 22 +=++

Finally, solve this equation for :dx

dy

( )

26

23

2326

22

22

−+−=∴

−=−+

yx

xy

dx

dy

xydx

dyyx

Example 2 Differentiate the following functions with respect to :x

a) 42y b) t3sin

Solution

a) Let 42yu = , then, by the function of a function rule:

dx

dyy

dx

dyy

dy

d

dx

dy

dy

du

dx

du

3

4

8

)2(

=

×=×=

b) Let tu 3sin= , then, by the function of a function rule:

dx

dtt

dx

dtt

dt

d

dx

dt

dt

du

dx

du

3cos3

)3(sin

=

×=×=

■

■

■

It is possible to differentiate an implicit function by using the function of a function rule, which may be

stated as dxdy

dydu

dxdu ×= .

i.e. f(y)u = .

WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation MTH1134/LN/01 Chap 1 Advanced Differentiation


Example 3 Differentiate the following functions with respect to :x

a) y5ln4 b) 2θ3

5

1 −e

Solution

a) Let yu 5ln4= , then, by the function of a function rule:

dx

dy

y

dx

dyy

dy

d

dx

dy

dy

du

dx

du

4

)5ln4(

=

×=×=

b) Let 2θ3

5

1 −= eu , then, by the function of a function rule:

dx

de

dx

de

d

d

dx

d

d

du

dx

du

θ

5

3

θ)

5

1(θ

θ

θ

2θ3

2θ3

−

−

=

×=×=

Differentiating implicit functions containing produ cts and quotients

Example 4 Determine ).2( 23 yxdx

d

Solution

In the product rule of differentiation let 32xu = and 2yv = . Thus

+=

+=

+

=

+=

ydx

dyxyx

yxdx

dyyx

xydx

dyyx

xdx

dyy

dx

dxyx

dx

d

322

64

)6)((2)2(

)2()()()2()2(

2

223

223

322323

■

■

■

WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation


Example 5 Find ).2

3(

x

y

dx

d

Solution

In the quotient rule of differentiation let yu 3= and xv 2= . Thus

−=

−=

−

=

−=

ydx

dyx

x

x

ydx

dyx

x

ydx

dyx

x

xdx

dyy

dx

dx

x

y

dx

d

2

2

2

2

2

34

66

4

)2)(3(3)2(

)2(

)2()3()3()2()

2

3(

Do it yourself

1. In Problems (a) – (e) differentiate the given functions with respect to x .

a) 53y

b) θ4cos2

c) k

d) t3ln2

5

e) 12

4

3 +ye

f) y3tan2

2. Differentiate the following with respect to y .

a) θ2sin3

b) 34 x c)

te

2

3. Differentiate the following with respect to u .

a) )13(

2

+x

b) θ2sec3

c) 3

2

y

4. Determine

a) )3( 32 yxdx

d

b) )4

3(

v

u

du

d

c) dy

dz given yxz ln2 3=

5. Find )5

2(

x

y

dx

d

■



1.2 PARTIAL DIFFERENTIATION

1.2.1 FIRST ORDER PARTIAL DERIVATIVES Example 6 If ,),( 223 yxyxyxf += find

a) xf b) yf

Solution a) For xf , hold y constant and find the derivative with respect to :x

22 23),( xyyxyxf x += ■

b) For yf , hold x constant and find the derivative with respect to :y

yxxyxf y23 2),( += ■

Example 7 Finding and evaluating a partial derivative Let ).3sin( 32 yxxz += Evaluate:

a) )0,

3

π(x

z

∂∂

b) yz at (1, 1).

Solution

a) )3)(3cos()3sin(2 323 yxxyxxx

z +++=∂∂

)3cos(3)3sin(2 323 yxxyxx +++= Thus,

=∂∂

)0,3

π(x

z

3

π)1(

3

π)0(

3

π2πcos

3

π3πsin

3

π2

222

−=−+=

+

■

For ),,( yxfz = the partial derivatives and yx ff are denoted by

)(),(),( fDzyxfxx

z

x

fyxf xxx ==

∂∂=

∂∂=

∂∂=

and

)(),(),( fDzyxfyy

z

y

fyxf yyy ==

∂∂=

∂∂=

∂∂=

The values of the partial derivatives of ),( yxf at the point ),( ba are denoted by

),(),(

bafx

fx

ba

=∂∂

and ),(),(

bafy

fy

ba

=∂∂



b) )3cos(3)3)(3cos( 322232 yxyxyyxxzy +=+= so that

4cos3)13cos()1()1(3)1,1( 22 =+=yz ■

Example 8 Partial derivative of a function of three variables Let ;2),,( 322 yzxyxzyxf ++= determine:

a) xf b) yf c) zf

Solution a) For xf , think of f as a function of x alone with y and z treated as constants:

222),,( yxzyxf x += ■

b) 34),,( zxyzyxf x += ■

c) 23),,( yzzyxf z = ■ Example 9 Partial derivative of an implicitly defined function Let z be defined implicitly as a function of x and y by the equation

xyzzx =+ 32

Determine x

z

∂∂

and y

z

∂∂

.

Solution Differentiate implicitly with respect to x , treating y as a constant:

132 22 =∂∂+

∂∂+

x

zyz

x

zxxz

Then solve the equation for x

z

∂∂

:

22 3

21

yzx

xz

x

z

+−=

∂∂

Similarly, holding x constant and differentiating implicitly with respect to y , we find

03 232 =∂∂++

∂∂

y

zyzz

y

zx

so that

22

3

3yzx

z

y

z

+−=

∂∂

■



Do it yourself

1. If yyxxz 325 234 −+= find

a) x

z

∂∂

and b) .y

z

∂∂

2. Given ,2cos3sin4 txy = find

a) x

y

∂∂

and b) .t

y

∂∂

3. If xyz sin= show that =∂∂x

z

y

1.

1

y

z

x ∂∂

4. In Problems (a) – (f), find x

z

∂∂

and .y

z

∂∂

a) xyz 2=

b) 23 2 yxyxz +−=

c) y

xz =

d) )34sin( yxz +=

e) yx

yyxz

12

23 +−=

f) yxz 4sin3cos=

1.2.2 HIGHER-ORDER PARTIAL DERIVATIVES (2ND ORDER & MIXED 2 ND ORDER)

Given ).,( yxfz = 2nd order partial derivatives

xxxx ffx

f

xx

f ==

∂∂

∂∂=

∂∂

)(2

2

yyyy ffy

f

yy

f ==

∂∂

∂∂=

∂∂

)(2

2

Mixed 2nd order partial derivatives

yxxy ffy

f

xyx

f ==

∂∂

∂∂=

∂∂∂

)(2

xyyx ffx

f

yxy

f ==

∂∂

∂∂=

∂∂∂

)(2



Example 10 Higher-order partial derivatives of a function of two variables For ,325),( 32 yxyxyxfz +−== determine these higher-order partial derivatives.

a) yx

z

∂∂∂ 2

b) xy

f

∂∂∂ 2

c) 2

2

x

z

∂∂

d) )2 ,3(xyf

Solution a) First differentiate with respect to y ; then differentiate with respect to .x

292 yxy

z +−=∂∂

( ) 292 22

−=+−∂∂=

∂∂

∂∂=

∂∂∂

yxxy

z

xyx

z ■

b) Differentiate first with respect to xand then differentiate with respect to :y

yxx

z210 −=

∂∂

( ) 22102

−=−∂∂=

∂∂

∂∂=

∂∂∂

yxyx

z

yxy

z ■

c) Differentiate first with respect to x twice:

( ) 102102

2

=−∂∂=

∂∂

∂∂=

∂∂

yxxx

z

xx

z ■

d) Evaluate the mixed partial found in part (b) at the point (3, 2):

2)2 ,3( −=xyf ■

Example 11 Partial derivatives of functions of two variables Determine ,,, xxyxxy fff and ,xxyf where yyexxf 2)y ,( = .

Solution We have the partial derivatives

yx xyef 2= yy

y yexexf 22 +=

The mixed partial derivatives (which must be the same by the previous theorem) are

yyyxxy xyexeff 22)( +== ■ yy

xyyx xyexeff 22)( +== ■

Finally, we compute the second- and higher-order partial derivatives:

yxxxx yeff 2)( == ■ and yy

yxxxxy yeeff 22)( +== ■



Example 12 Verifying that a function satisfies the heat equation

Verify that c

xetxT t cos) ,( −= satisfies the heat equation, .

2

22

x

Tc

t

T

∂∂=

∂∂

Solution

c

xe

t

T t cos−−=∂∂

and

c

xe

c

c

xe

cxx

T

t

t

cos1

sin1

2

2

2

−

−

−=

−∂∂=

∂∂

Thus, T satisfies the heat equation 2

22

x

Tc

t

T

∂∂=

∂∂

■

1.3 TOTAL DIFFERENTIAL

Example 13 Determine the total differential of the given functions:

a) zyxzyxf 652),,( 43 −+= b) )23ln(),( 22 xyxyxf −= Solution

a) dzdyydxxdzz

fdy

y

fdx

x

fdf 6206 32 −+=

∂∂+

∂∂+

∂∂= ■

b) dyy

fdx

x

fdf

∂∂+

∂∂=

dyxy

yxdx

xy

xxyx

dyxy

yxdx

xyxxyx

23

6

23

2)23ln(2

23

6

23

2)23ln(2

2

2

2

22

22

222

−+

−−−=

−+

−−+−=

■

The total differential of the function ),( yxf is

dyyxfdxyxfdyy

fdx

x

fdf yx ),(),( +=

∂∂+

∂∂=

where dx and dyare independent variables. Similarly, for a function of three variables

),,( zyxfw = the total differential is

dzz

fdy

y

fdx

x

fdf

∂∂+

∂∂+

∂∂= ( أ ).……………



Example 14 If ),( yxfz = and ,1232 ++=y

xyxz determine the total differential, .dz

Solution The total differential is the sum of the partial differentials, i.e.

222

3

23

22

y

xyx

y

z

yxy

x

z

dyy

zdx

x

zdz

−=∂∂

+=∂∂

∂∂+

∂∂=

Hence dyy

xyxdx

yxydz

−+

+=

2223 2

32

2 ■

Example 15 If ),,( wvufz = and 232 423 vwvuz +−= find the total differential, .dz Solution The total differential

22

3

12

82

6

vww

z

vwv

z

uu

z

dww

zdv

v

zdu

u

zdz

=∂∂

+−=∂∂

=∂∂

∂∂+

∂∂+

∂∂=

Hence dwwvdvvwududz )12()28(6 223 +−+= ■

Do it yourself

1. In Problems (a) – (f), find the total differential .dz

a) 23 yxz +=

b) xxyz cos2 −=

c) yx

yxz

+−=

d) yxz ln=

e) 4−+=y

xxyz

2. If ),,( cbafz = and ,32 2 abccbabz +−= find the total differential .dz

(i.e. y is kept constant)

(i.e. x is kept constant)

(i.e. wv and are kept constant)

(i.e. wu and are kept constant)

(i.e. vand u are kept constant)



1.4 L’HOPITAL’S RULE

Example 17 Using l’Hopital’s rule to compute a familiar trigonometric limit

Evaluate .sin

lim0 x

xx→

Solution Note that this is an indeterminate form because sin x and x both approach 0 as .0→x This means that l’Hopital’s rule applies:

11

1

1

coslim

)(

)(sinlim

sinlim

000====

→→→

x

xdx

d

xdx

d

x

xxxx

■

Example 18 l’Hopital’s rule with a 0/0 form

Evaluate .8

128lim

3

7

2 −−

→ x

xx

Solution

For this example, 128)( 7 −= xxf and 8)( 3 −= xxg , and the form is 0/0.

)8(

)128(lim

8

128lim

3

7

23

7

2−

−=

−−

→→x

dx

d

xdx

d

x

xxx

Let f and g be differentiable functions with 0)( ≠′ xg on an open interval containing

c (except) possibly at c itself). Suppose )(

)(lim

xg

xfcx→

produces an indeterminate form 0

0 or

∞∞

and that

Lxg

xfcx

=′′

→ )(

)(lim

where L is either a finite number, ,∞+ or .∞− Then

Lxg

xfcx

=→ )(

)(lim

The theorem also applies to one-sided limits and to limits at infinity (where +∞→x and −∞→x ).

l’Hopital’s rule



3

112

3

)2(7

3

7lim

3

7lim

4

4

2

2

6

2

==

=

=

→

→

x

x

x

x

x

Example 19 Limit is not an indeterminate form

Evaluate .sec

cos1lim

0 x

xx

−→

Solution You must always remember to check that you have an indeterminate form before applying l’Hopital’s rule. The limit is

01

0

seclim

)cos1(lim

sec

cos1lim

0

0

0==

−=−

→

→

→ x

x

x

x

x

x

x ■

ATTENTION: If you blindly apply l’Hopital’s rule in Example 19, you obtain the WRONG answer:

xx

x

x

xxx tansec

sinlim

sec

cos1lim

00 →→=−

11

1

sec

coslim

0===

→ x

xx

Example 20 l’Hopital’s rule applied more than once

Evaluate .sin

lim30 x

xxx

−→

Solution This is a 0/0 indeterminate form, and we find that

2030 3

cos1lim

sinlim

x

x

x

xxxx

−=−→→

This is still the indeterminate form 0/0, so l’Hopital’s rule can be applied once again:

6

1)1(

6

1sinlim

6

1

6

)sin(lim

3

cos1lim

0020===−−=−

→→→ x

x

x

x

x

xxxx

■

Example 21 l’Hopital’s rule with an ∞∞ form

Evaluate .253

132lim

2

2

−++−

+∞→ xx

xxx

Solution

We could compute this limit by multiplying by ( ) ( )22 11 xx . Instead, we note that this is of

the form ∞∞ and apply l’Hopital’s rule:

56

34lim

253

132lim

2

2

+−=

−++−

+∞→+∞→ x

x

xx

xxxx

■

Simplify

Limit of a quotient

Apply l’Hopital’s rule again



3

2

6

4lim ==

+∞→x ■

Example 22 Using l’Hopital’s rule with other limit properties

Evaluate .cos

4sin)cos1(lim

30 xx

xxx

−→

Solution The limit has the form 0/0, but direct application of l’Hopital’s rule leads to a real mess. Instead, we compute the given limit by using the product rule for limits first, followed by two simple applications of l’Hopital’s rule. Specifically, using the product rule for limits, we have

−=−→→→→ xx

x

x

x

xx

xxxxxx cos

1lim

4sinlim

)cos1(lim

cos

4sin)cos1(lim

002030

( )( ) 2142

1

1

1

1

)1(4

2

coslim

cos

1lim

1

4cos4lim

2

sinlim

0

000

=

=

=

=

→

→→→

x

x

x

x

x

x

xxx

Example 23 Hypothesis of l’Hopital’s rule are not satisfied

Evaluate .cos

sinlim

xx

xxx −

++∞→

Solution This limit has the indeterminate form ∞∞ . If you try to apply l’Hopital’s rule, you find

x

x

xx

xxxx sin1

cos1lim

cos

sinlim

++=

−+

+∞→+∞→.

The limit on the right does not exist, because both sin x and cos x oscillate between -1 and 1

as .+∞→x Recall that l’Hopital’s rule applies only if Lxg

xfcx

=′′

→ )(

)(lim or is .∞± This does

not mean that the limit of the original expression does not exist or that we cannot find it; it simply means that we cannot apply l’Hopital’s rule. To find this limit, factor out an x from the numerator and denominator and proceed as follows:

101

01cos

1

sin1

limcos

1

sin1

limcos

sinlim =

−+=

−

+=

−

+=

−+

+∞→+∞→+∞→

x

xx

x

x

xx

x

xx

xx

xxxxx

■

OTHER INDETERMINATE FORMS

Example 24 Limit of the form ∞1

■



Show that .1

1lim ex

x

x=

++∞→

Solution

Note that this limit is indeed of the indeterminate form ∞1 . Let

x

x xL

+=+∞→

11lim

Take the logarithm of both sides:

x

x

xx

x

xL

x

x

x

x

x

x

1

11ln

lim

11lnlim

11lnlim

11limlnln

+=

+=

+=

+=

+∞→

+∞→

+∞→

+∞→

101

1

11

1lim

1

1

11

1

lim

2

2

=+

=

+=

−

−+=

+∞→

+∞→

x

x

xx

x

x

Thus, 1ln =L and eeL == 1 ■ Example 25 l’Hopital’s rule with the form ∞⋅0

Evaluate .tan2

πlim

2πxx

x

−−→

Solution The limit has the form ∞⋅0 , because

+∞==

−−− →→

xxxx

tanlim and 02

πlim

2π2π

ln x is continuous

Property of logarithms

Form 0/0


Simplify



Write x

xcot

1tan = to obtain

x

xxx

xx cot2

π

limtan2

πlim

2π2π

−=

−−− →→

1)sin(lim

csc

1lim

2

2π

22π

−=−=−

=

−

−

→

→

x

x

x

x

Example 26 Limit of the form 00

Find x

xxsin

0lim

+→.

Solution

This is a 00 indeterminate form. We begin by using properties of logarithms.

x

x

xx

x

xL

xL

x

x

x

x

x

x

x

x

cscln

lim

ln)(sinlim

lnlim

limlnln

lim

0

0

sin

0

sin

0

sin

0

+

+

+

+

+

→

→

→

→

→

=

=

=

=

=

0)0)(1(

cossinsin

lim

cossin

lim

cotcsc1

lim

0

2

0

0

==

−

=

−=

−=

+

+

+

→

→

→

x

x

x

xxx

x

xx

x

x

x

x

Thus, 10 == eL ■

Example 27 Limit of the form 0∞ Find x

xx1lim

+∞→.

Solution

This is a limit of the indeterminate form 0∞ .

If ,lim 1 x

xxL

+∞→= then

Form 0/0


■

ln x is continuous This is ∞.0 form

This is ∞∞ form




01

1

lim

lnlim

ln1

lim

lnlim

limlnln

1

1

=

=

=

=

=

=

+∞→

+∞→

+∞→

+∞→

+∞→

x

x

x

xx

x

xL

x

x

x

x

x

x

x

Thus, we have ;0ln =L therefore, 10 == eL ■ Example 28 l’Hopital’s rule with the form ∞−∞

Evaluate

−+→ xxx sin

11lim

0.

Solution

As it stands, this has the form ∞−∞ , because +∞→x

1 and +∞→

xsin

1 as 0→x from

the right. However, using a little algebra, we find

xx

xx

xx xx sin

sinlim

sin

11lim

00

−=

−++ →→

This limit is now of the form 0/0, so the hypotheses of l’Hopital’s rule are satisfied. Thus,

xxx

x

xx

xxxx cossin

1coslim

sin

sinlim

00 +−=−

++ →→

02

0

cos)sin(cos

sinlim

0==

+−+−=

+→ xxxx

xx

■

ATTENTION: Not all limits that appear indeterminate actually are indeterminate. For example,

form 0

0ln

tanlim

form )(-- )ln(csclim

form 0 0)(sinlim

0

0

1

0

∞=

∞∞++∞=−

=

+

+

+

→

→

∞

→

x

x

xx

x

x

x

x

x

Other such “false indeterminate forms” include ,0 ),( ∞+∞+∞+ and ,.∞∞ which are all actually infinite.

Do it yourself

1. An incorrect use of l’Hopital’s rule is illustrated in the following limit computations. In each case, explain what is wrong and find the correct value of the limit.

The limit of a log is the log of the limit

This is ∞∞


This is 0/0 form



a) 01

sinlim

cos1lim

ππ==−

→→

x

x

xxx

b) 01

coslim

sinlim

2π2π==

→→

x

x

xxx

2. Find each of the limits in Problems (a) – (g).

a) 1

1lim

2

3

1 −−

→ x

xx

b) 1

1lim

10

1 −−

→ x

xx

c) 67

43lim

2

2

1 +−−+

→ xx

xxx

d) sin

cos1lim

3

2

0 x

xx

−→

e) xx

xxx −

−→ tan

sinlim

0

f) 20

sinlim

x

xxx

−→

g) x

xx tan2

sec3lim

2π +→

h)

+−−→

xxx

tan2π

1lim

)2π(

i) 2

11lim

3x

++∞→ xx

j) )ln(ln

limx

xx +∞→

k) ( ) limx1

0xex

x+

+→

l) ln1

lim20

−+→

xxx

1.5 APPLICATION OF DIFFERENTIATION

1.5.1 GRADIENT FUNCTIONS (OPTIMIZATION IN THE LIFE SCIENCES)

Example 29 Maximization Applied to Enzymes An enzyme is a protein that acts as a catalyst for increasing the rate of a chemical reaction that occurs in cells. In a certain reaction, an enzyme is converted to another enzyme called the product. The product acts as a catalyst for its own formation. The rate R at which the product is formed (with respect to time) is given by

),( plkpR −= where l is the total initial amount of both enzymes, p is the amount of the product enzyme, and k is a positive constant. For what value of p will Rbe a maximum? Solution We can write ).( 2pplkR −= Setting 0=dpdR and solving for p gives

.2

,0)2(l

pplkdp

dR =∴=−=

Now, .222 kdpRd −= Since ,0>k the second derivative is always negative. Hence,

2lp = gives a relative maximum. Moreover, since R is a continuous function of ,p

we conclude that we indeed have an absolute maximum when 2lp = ■ Example 30 Modeling Problem: Maximum concentration of a drug Let )(tC denote the concentration in the blood at time t of a drug injected into the body intramuscularly. In a classic paper by E. Heinz (“Probleme bei der Diffusion



kleiner Substanzmengen innerhalb des menschlichenrpersoK && ”, Biochem., Vol. 319 (1949), pp. 482-492), it was observed that the concentration may be modeled by

0 t)()( ≥−−

= −− btat eeab

ktC

where ba , (with ),ab > and k are positive constants that depend on the drug. At what time does the largest concentration occur? What happens to the concentration as

?+∞→t Solution To locate the extrema, we solve .0)( =′ tC

[ ] )()()(

)()(

atbtbtat

btat

aebeab

kebea

ab

k

eeab

k

dt

dtC

−−−−

−−

−−

=−−−−

=

−−

=′

We see that 0)( =′ tC when

atbtatbt

atbt

eeea

b

aebe

−−

−−

==

=

a

batbt ln=−

a

b

abt ln

1

−= ■

( )0

00

1lim

1lim

)(lim)(lim

=

−−

=

−−

=

−−

=

+∞→+∞→

−−

+∞→+∞→

ab

k

eeab

k

eeab

ktC

bttatt

btat

tt

Example 31 The production of blood cells plays an important role in medical research involving leukemia and other so-called dynamical diseases. In 1977, a mathematical model was developed by A. Lasota that involved the cell production function

rsxseAxxP −=)( where rsA and ,, are positive constants and x is the number of granulocytes (a type of white blood cell) present. Find the granulocytes level x that maximizes the production function P .

■



Solution

rx

rx

rx

x

xr

x

esAxexr

sA

sAxeer

sAxxP

eAxxP

ss

s

s

ss

rsxsrsxs

srsxrsxs

rsxs

=∴=

=

=

=

=+

−=′

=

+−

−

−

−−−

−−−

−

1

1

1

1

1 0.)(

)(

Example 32 Beehives are formed by packing together cells that may be modeled as regular hexagonal prisms open at one end. It can be shown that a cell with hexagonal side of length s and prism height h has surface area

)θcsc3θcot(5.16)θ( 2 +−+= sshS

for .2

πθ0 << What is the angle θ (to the nearest degree) that minimizes the surface

area of the cell (assuming that s and h are fixed)? Solution

oo

sS

sshS

557.54θ

3

1θcos

θsin

θcos3

θsin

1

θcot3θcsc

θcotθcsc3θcsc

0)θcotθcsc3θcsc(5.1)θ(

)θcsc3θcot(5.16)θ(

2

22

2

≈=∴

=

=

=

=

=−−=′

+−+=

1.5.2 RATES OF CHANGE Sometimes it is necessary to solve problems in which different quantities have different rates of change. From equation (the total differential)

dzz

fdy

y

fdx

x

fdf

∂∂+

∂∂+

∂∂= ,

the rate of change of f , dt

df is given by:

■

■



Example 33 The height of a right circular cone is increasing at 3 mm/s and its radius is decreasing at 2 mm/s. Determine, correct to 3 significant figures, the rate at which the volume is changing (in cm3/s) when the height is 3.2 cm and the radius is 1.5 cm. Solution

Volume of a right circular cone, hrV 2π

3

1=

Using equation (ب), the rate of change of volume,

2π

3

1 and π

3

2r

h

Vrh

r

Vdt

dh

h

V

dt

dr

r

V

dt

dV

=∂∂=

∂∂

∂∂+

∂∂=

Since the height is increasing at 3 mm/s, i.e. 0.3 cm/s, then 3.0+=dt

dh

and since the radius is decreasing at 2 mm/s, i.e 0.2 cm/s, then 2.0−=dt

dr.

Hence

( ) ( )

2

2

π1.0π3

4.0

3.0π3

12.0π

3

2

rrh

rrhdt

dV

+−=

+

+−

=

However, h = 3.2 cm and r = 1.5 cm. Hence

scm

dt

dV

/ 304.1

)5.1(π)1.0()2.3)(5.1(π3

4.0

3

2

−=

+−=

Thus the rate of change of volume is 1.304 cm3/s decreasing.

Example 34 The area A of a triangle is given by ,sin2

1BacA = where B is the

angle between sides a and c. If a is increasing at 0.4 units/s, c is decreasing at 0.8 units/s and B is increasing at 0.2 units/s, find the rate of change of the area of the triangle, correct to 3 significant figures, when a is 3 units, c is 4 units and B is 6π radians. Solution Using equation(ب), the rate of change of area,

dt

dz

z

f

dt

dy

y

f

dt

dx

x

f

dt

df

∂∂+

∂∂+

∂∂= (ب)……

■



dt

dB

B

A

dt

dc

c

A

dt

da

a

A

dt

dA

∂∂+

∂∂+

∂∂=

Since Bca

ABacA sin

2

1 ,sin

2

1 =∂∂= ,

units/s 2.0 and units/s 8.0 units/s, 4.0

cos2

1 and sin

2

1

=−==

=∂∂=

∂∂

dt

dB

dt

dc

dt

da

BacB

ABa

c

A

Hence

)2.0(cos2

1)8.0(sin

2

1)4.0(sin

2

1

+−

+

= BacBaBcdt

dA

When a = 3, c = 4 and B = 6

π then:

s

dt

dA

/units 839.0

)2.0(6

πcos)4)(3(

2

1)8.0(

6

πsin)3(

2

1)4.0(

6

πsin)4(

2

1

2=

+−

+

=

1.5.3 SMALL INCREMENT AND APPROXIMATION It is often useful to find an approximate value for the change (or error) of a quantity caused by small changes (or errors) in the variables associated with the quantity. If ),,( zyxfw = and

zyx δ ,δ ,δ denote small changes in zyx ,, respectively, then the corresponding approximate change fδ in f is obtained from equation ( أ ) by replacing the differentials by the small changes. Thus Example 35 Pressure p and volume V of a gas are connected by the equation

k pV =4.1 . Determine the approximate percentage error in k when the pressure is increased by 4% and the volume is decreased by 1.5%. Solution Using equation ( � ), the approximate error in k,

VV

kp

p

kk δδδ

∂∂+

∂∂≈

Let p, V and k refer to the initial values.

Since 4.04.14.1 4.1 and then VV

kV

p

k pVk =

∂∂=

∂∂= .

Since the pressure is increased by 4%, the change in pressure .04.0100

4δ ppp =×=

■ (correct to 3 significant figures)

zz

fy

y

fx

x

ff δδδδ

∂∂+

∂∂+

∂∂≈ …………….( � )



Since the volume is decreased by 1.5%, the change in volume

.015.0100

5.1δ VVV −=×−=

Hence the approximate error in k,

[ ].

100

9.1

100

9.1019.0

)015.0(4.104.0

)015.0)(4.1()04.0()(δ

4.14.1

4.1

4.04.1

kpVpV

pV

VpVpVk

≈≈≈

−≈

−+≈

i.e. the approximate error in k is a 1.9% increase. Example 36 Modulus of rigidity ,)θ( 4 LRG = where R is the radius, θ the angle of twist and L the length. Determine the approximate percentage error in G when R is increased by 2%, θ is reduced by 5% and L is increased by 4%. Solution

Using LL

GGR

R

GG δδθ

θδδ

∂∂+

∂∂+

∂∂≈ .

Since .θ

and θ

,θ4

,)θ(

2

4434

L

R

L

G

L

RG

L

R

R

G

L

RG

−=∂∂=

∂∂=

∂∂=

Since R is increased by 2%, RRR 02.0100

2δ == . Similarly,

.04.0Lδ and θ05.0δθ L=−= Hence

( ) ( )

GGL

R

LL

R

L

RR

L

RG

100

1δ i.e. ,

θ01.0

04.0θ

θ05.0)02.0(θ4

δ

4

2

443

−≈−≈

−+−

+

≈

Hence the approximate percentage error in G a 1% decrease.

1.5.3 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Definition

Let f be a function of two variables, and let juiuu 21 += be a unit vector. The

directional derivative of f at ) ,( ooo yxP in the direction of u is given by

h

yxfhuyhuxfyxfD oooo

hoou

) ,() ,(lim) ,( 21

0

−++=→

provided the limit exists.

■

■



Example 37 Find the directional derivative of 3223),( yxyxf +−= at the point

)2 ,1(P in the direction of the unit vector jiu23

21 −= .

Solution

First, find the partial derivatives xyxf x 4) ,( −= and .3) ,( 2yyxf y = Then since

21

1 =u and 23

2 −=u , we have

4.123622

3)2(3

2

1)1(4

23

)2 ,1(21

)2 ,1()2 ,1(

2 −≈−−=

−+

−=

−+

= yxu fffD

The Gradient

Let f be a differentiable function at ),( yx have partial derivatives ) ,( yxfx and

). ,( yxf y Then the gradient of f , denoted by f∇ (pronounced “del eff”), is a vector

given by

jyxfiyxfyxf yx ) ,() ,() ,( +=∇

The value of the gradient at the point ) ,( ooo yxP is denoted by

jyxfiyxff ooyooxo ) ,() ,( +=∇

Example 38 Find ) ,( yxf∇ for 32),( yyxyxf +=

Solution

Begin with the partial derivatives:

xyyyxx

yxfx 2)() ,( 32 =+∂∂= and 2232 3)() ,( yxyyx

yyxf y +=+

∂∂=

Then jyxixyyxf )3(2) ,( 22 ++=∇ ■

■



Example 39 Find the directional derivative of )ln(),( 32 yxyxf += at )3- ,1(oP in the

direction of .32 jiv −=

Solution

32

2) ,(

yx

xyxfx +

= , so 262

)3- ,1( −=xf

32

23) ,(

yx

yyxf y +

= , so 2627

)3- ,1( −=yf

jiffo 2627

262

)3,1( −−=−∇=∇

A unit vector in the direction of v is

)32(13

1

)3(2

3222

jiji

v

vu −=

−+

−==

Thus,

3381377

13

3

26

27

13

2

26

2.) ,(

=

−

−+

−=∇= ufyxDu

Example 40 Heat flow application

The set of points ),( yx with 50 ≤≤ x and 50 ≤≤ y is a square in the first quadrant

of the xy-plane. Suppose this square is heated in such a way that 22),( yxyxT += is the temperature at the point ).,( yxP In what direction will heat flow from the point

?)4 ,3(oP

Solution

The heat flow TkH ∇−=ˆ where =k the thermal conductivity (a positive constant)

From 22),( yxyxT += , we have jyixT 22 +=∇ .

oT∇ is the gradient at oP , therefore .86 jiTo +=∇

Thus, the heat flow at oP satisfies )86(ˆ jikTkH oo +−=∇−= .

■



Because the thermal conductivity k is positive, we can say that heat flows from oP in

the direction of the unit vector u given by

jiji

u5

4

5

3

)8()6(

)86(22

−−=−+−

+−= ■

Documents

Chap 1 Advanced Differentiation