ENGI 3424
4.
4 – Partial Differentiation
Page 4-01
Partial Differentiation
For functions of one variable, y  f  x  , the rate of change of the dependent variable can
dy
 f   x  . In this chapter we explore
dx
rates of change for functions of more than one variable, such as z  f  x, y  .
be found unambiguously by differentiation:
Contents:
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Partial Derivatives - introduction, chain rule, practice
Higher Partial Derivatives, Clairaut’s theorem, Laplace’s PDE
Differentials; error estimation; chain rule [again]; implicit functions; partial
derivatives on curves of intersection
The Jacobian - implicit and explicit forms; plane polar; spherical polar
Gradient Vector, directional derivative, potential function, central force law
Extrema; Second Derivative Test for z  f  x, y 
Lagrange Multipliers; nearest point on curve of intersection to given point
Miscellaneous Additional Examples
ENGI 3424
4.1
4.1 – Partial Derivatives
Page 4-02
Partial Derivatives - introduction, chain rule, practice
Example 4.1.1
At a particular instant, a cone has a height of h = 2 m and a base radius of r = 1 m.
The base radius is increasing at a rate of 1 mm/s. The height is constant.
How fast is the volume V increasing at this time?
V 
1 2
r h
3
We need
dV
.
dt
h = const.  V is a function of r only.
dV
dV dr


dt
dr dt
 chain rule 

1 d 2
dr
2
dr
 r  h 
  rh 
3 dr
dt
3
dt

2
1 
3 1
 1 2  
 ms
3
1000


Therefore
dV


dt
750
m3s1  0.00419 m3s1
dr
dh
 1 mms1 and
  2 mms1,
dt
dt
dV
how do we find
?
dt
We shall return to this question later.
But if
4.1 – Partial Derivatives
ENGI 3424
Page 4-03
Graph of V against r and h :
Plotting z = V (where V = r 2h / 3) against both x = r and y = h yields
The cross-section of this surface in the
vertical plane h = 2 is
The cross-section of this surface in the
vertical plane r = 1 is
V
V
 lim
r  0 r
r
V
V
 lim
h  0 h
h
V  r  r , h   V  r , h 
r  0
r
 lim
1
3
r
2
r
r  0
 lim
r  0




 2r r   r  h  13  r 2 h
2
2  rh
3
4r
3

1  h r
3
at h  2 


V  r , h  h   V  r , h 
h  0
h
2
1

 3r
 lim
 lim
2  rh
3
4.1 – Partial Derivatives
ENGI 3424
Page 4-04
V
.
r
V
The tangent line to the surface in a cross-section (r = constant) has a slope of
.
h
At each point on the surface, these two tangent lines define a tangent plane.
The tangent line to the surface in a cross-section (h = constant) has a slope of
V is a function of r and h, each of which in turn is a function of t only.
In this case, the chain rule becomes
dV
V dr
V dh




dt
 r dt
 h dt
In example 4.1.1,
V
2
4
V
1
1
  rh   ,
  r2   ,
r
3
3
h
3
3
dV
4
1

2

dr
1
dh
2

,




 

m3 s1
dt
1000
dt
1000
dt
3 1000
3 1000
1500
r = 1, h = 2,
Alternative notations:
V

 Vr  Dr V 
V  r , h  
r
r
If w = f (x, y, z), then
 f  x, y  y, z   f  x, y, z  
w
 lim 
 , etc.

y

0
y
y


A Maple worksheet, used to generate the graph of V  r , h  
1  r 2h ,
3
is available at
"http://www.engr.mun.ca/~ggeorge/3424/demos/conevolume.mws".
Open this worksheet in Maple and click on the graph.
Then, by dragging the mouse (with left button down), one can change the direction of
view of the graph as one wishes. Other features of the graph may be changed upon
opening a menu with a right mouse click on the graph or by using the main menu at the
top of the Maple window.
4.1 – Partial Derivatives
ENGI 3424
Page 4-05
Example 4.1.2
f  x, y, z  
x2  y 2  z 2   x2  y 2  z 2 
1/2
Find f z  0,3, 4 
[the first partial derivative of f with respect to z, evaluated at the point (0, 3, 4)].
Apply the general differentiation rule
d
dx
 f  x    n  f  x 
n
1/2 
f
1
  x2  y 2  z 2 
  x2  y 2  z 2 
z
2
z


1
z
 0  0  2z  
2f
f
f z  0, 3, 4  
4
4

5
0  9  16
OR
f 2  x2  y 2  z 2
Using implicit differentiation,

 2
f 2 

 x  y2  z2 
z
z
f
 2f
 0  0  2z
z

fz 
z
4
 ,
f
5
 as before 
In this example,
f = (distance of the point (x, y, z) from the origin).
n1
f  x :
4.1 – Partial Derivatives
ENGI 3424
Page 4-06
Example 4.1.3
u  x y/ z
Find the three first partial derivatives, ux , u y , uz .
 
 y/ z

x
x
d n
x   n x n1

dx

y  1

y 
  x z
z
 ux 

y
y
  x z  x 1
z
y  y z / z
yu
x

z
xz
y

u  x y / z  exp  ln x 
z

 uy 
1
y
u
 ln x  exp  ln x   ln x
z
z
z

uy 

1 y/ z
x ln x
z
OR
ln u 

y
ln x
z

1
1
 u y   ln x
u
z
ln u  yz 1 ln x


 y
 ln u    ln x 
y
y  z

 uy 
u
 ln x
z


1
 u z  y 1z 2 ln x
u
 uz  
uy
y
ln x   2 x y / z ln x
2
z
z
4.2 – Higher Partial Derivatives
ENGI 3424
4.2
Page 4-07
Higher Partial Derivatives, Clairaut’s theorem, Laplace’s PDE
ux 
u
x
  u 
 2u

 u xx
 
x  x 
 x2
  u 
 2u

 ux y
 
 y  x 
 y x
   2u 
 3u

 ux y z


z   y x 
z  y x
   2u 
 3u

 ux y y


 y   y x 
 y2  x
etc.
Example 4.2.1
u  x et sin y
Find the second partial derivatives uxy , u yx , uxx and the third partial derivative uttt .
ux  et sin y
ut   x et sin y   u
u y  x et cos y
 utt   u t   u
u xy 
 uttt   u t   u

 ux   et cos y
y
Therefore
u yx 
uxx  0

uy
x
   et cos y
 u xy
uttt   x et sin y
4.2 – Higher Partial Derivatives
ENGI 3424
Page 4-08
Clairaut’s Theorem
If, on a disk D containing the point (a, b), f is defined and both of the partial derivatives
f xy and f yx are continuous, (which is the case for most functions of interest), then
f xy  a, b   f yx  a, b 
that is, the order of differentiation doesn’t matter.
One of the most important partial differential equations involving second partial
derivatives is Laplace’s equation, which arises naturally in many applications, including
electrostatics, fluid flow and heat conduction:
 2u
 2u

 0
 x2
 y2
3
or its equivalent in
:
 2u
 2u
 2u


 0
 x2
 y2
z2
Example 4.2.2
Does u  ln x 2  y 2 satisfy Laplace’s equation?

u  ln  x 2  y 2 
 ux 
1/2
  12 ln  x  y 
2
2

 using

1 2x  0
x
 2
 2
2
2 x y
x  y2
 u xx 
1 x 2  y 2   x  2 x  0 
 x2  y 2 
2
 using ln  x n   n ln x 



y 2  x2
 x2  y 2 
2
f  x 
d
ln  f  x   

dx
f  x 
[using the quotient rule]
u is symmetric with respect to x, y.
Therefore, to find u yy , interchange x, y in u xx .
 uy y 
x2  y 2
y
2
x

2 2
  u xx
 uxx  u yy  0
Therefore u does satisfy Laplace’s equation (except at (0, 0)).
ENGI 3424
4.3
In
4.3 – Differentials
Page 4-09
Differentials; error estimation; chain rule [again]; implicit functions; partial
derivatives on curves of intersection
2
, let a curve have the Cartesian equation y = f (x).
The small change in y, ( y ), caused by travelling along
the curve for a small horizontal distance x , may be
approximated by the change dy that is caused by
travelling for the same horizontal distance x along the
tangent line instead.
The exact form is y  f  x  x   f  x  .
The approximation to y is
y  dy  f   x  dx
where the increment x has been replaced by the differential dx.
The approximation improves as x decreases towards zero.
Stepping up one dimension, let a surface have the Cartesian equation z = f (x, y).
The change in the dependent variable z caused by small changes in the independent
variables x and y has the exact value
z  f  x  x, y  y   f  x, y  .
The approximation to z is
 z  dz 
f
f
dx 
dy
x
y
Example 4.3.1
A rectangle has quoted dimensions of 30 cm for length and 24 cm for width.
However, there may be an error of up to 1 mm in the measurement of each dimension.
Estimate the maximum error in the calculated area of the rectangle.
Let A = area, L = length and W = width.
Length = (30 ± 0.1) cm  L = 30 and L = 0.1
Width = (24 ± 0.1) cm  W = 24 and W = 0.1
A = LW
dA 
A
A
dL 
dW  W dL  L dW
L
W
4.3 – Differentials
ENGI 3424
Example 4.3.1 (continued)
Let dL = dW = 0.1, then
max(error) = A ≈ dA
= 24  0.1 + 30  0.1
= 5.4 cm2 .
Compare this to a direct calculation:
max  error   max
 A
max
 A ,  A  Amin 

A  Amin  30  24   30  0.1 24  0.1
= 5.39 cm2 .
Amax  A   30  0.1 24  0.1  30  24
= 5.41 cm2 .
Therefore max(error) = 5.41 cm2 .
Relative error:
relative error in L =
L 0.1
1


 (one-third of one percent)
L
30 300
relative error in W =
W 0.1
1


W
24 240
A = LW

dA = W dL + L dW

dA
W dL
L dW


A
LW
LW

dA
dL
dW
1
1
45
3






A
L
W
300
240
1200
400
A
dA

 0.75%
A
A
and 0.75% of A = 720 cm2 is 5.4 cm2 .

Page 4-10
ENGI 3424
4.3 – Differentials
Page 4-11
Chain Rule
z  f  x, y  .
If x and y are both functions of t only, then, by the chain rule,
dz
 z dx
 z dy


dt
 x dt
 y dt
If z  f  x, y  and y in turn is a function of x only, then
replace t by x in the formula above.
dx
 1 and
dx
dz
z
 z dy


dx
x
 y dx
Note the distinction between the total derivative
z
dz
and the partial derivative
.
dx
x
4.3 – Differentials
ENGI 3424
Page 4-12
Example 4.3.2
In the study of fluid dynamics, one approach is to follow the motion of a point in the
fluid. In that approach, the velocity vector is a function of both time and position, while
position, in turn, is a function of time. v  v  r, t  and r  r  t  .
The acceleration vector is then obtained through differentiation following the motion of
the fluid:
dv
v
v d x
v d y
v d z
a t  




dt
t
x dt
 y dt
z dt
dv
v
or, equivalently,
a t  

  v  v
dt
t
[The gradient operator,  , will be introduced later, on page 4-21.]
Further analysis of an ideal fluid of density  at pressure p subjected to a force field F
leads to Euler’s equation of motion
v
p
  v  v  F 
t

This application of partial differentiation will be explored in some majors in a later
As a simple example here, suppose that v   e x 1 10 1  t   , then
find the acceleration vector.
T
semester.
First note that the velocity vector is the derivative of the displacement vector, so that
 dx
dr
v 
 
dt
 dt
v

t
0

dz 
dt 
T
0
0 0
T
  e x 1 10 1  t  
v
  e x
x
0 10  ,
v
v


y
z
a t  
dy
dt
0 0 
T
T
T
dv
v
v d x
v d y
v d z




dt
t
 x dt
 y dt
 z dt
dv

dt
0
0 10 
T
  e x
0 0  e x  0  0   e2 x
T
[One can show that x  t   ln t  c , where c is a constant.]
0 10 
T
4.3 – Differentials
ENGI 3424
Page 4-13
Generalized Chain Rule
Let z be a function of n variables  x1, x2 , , xn  ,
each of which, in turn, is a function of m variables
 t1, t2 , , tm  , so that
z  f  x1  t1 , t2 ,
, tm  , x2  t1 , t2 ,
, tm   .
, xn  t1 , t2 ,
To find
, tm  ,
z
,
 ti
trace all paths that start at z and end at ti , via all of the
dz 
z
dx j

j 1  x j
n
and
z

ti
 x j  variables.
z xj
j 1
j  ti
n
 x
i  1, 2,
, m
Example 4.3.3:
u  x y  yz  zx , x  st , y  est and z  t 2 .
Find us in terms of s and t only. Find the value of us when s = 0 and t = 1.
u
u  x
u  y


s
 x s
 y s
[There is no z term because z is not a function of s.]
u
  y  0  z  t   x  z  0  t e st
s

 
 
 t est  t 2  st  t 2 est


 t 3  t 1  st  t 2 est
This derivative could also be found directly by replacing x, y and z by the respective
functions of s and t before differentiating u:
      
  st  t  est   t est  t   t  1  t
u   st  est  est t 2  t 2  st 

us
2
2
3
us  0,1  1  11  0  1 e0  1  2  3 .
3


 t 1  st  t 2 e st
4.3 – Differentials
ENGI 3424
Page 4-14
Implicit functions:
If z is defined implicitly as a function of x and y by F  x, y, z   c , then
dF  Fx dx  Fy dy  Fz dz  0
 dz  

1
Fx dx  Fy dy
Fz

provided Fz 
F
0
z
Example 4.3.4:
Find the change in z when x and y both increase by 0.2 from the point (1, 2, 2) on the
sphere x2  y 2  z 2  9 .
F  x 2  y 2  z 2  9  dz  
1
1
 2 x dx  2 y dy     x dx  y dy 
2z
z
Solution − Approximate motion on the sphere by motion on the tangent plane:
x = 1, y = z = 2,
 z  dz  
dx ≈ Δx = 0.2 , dy ≈ Δy = 0.2 .
1
1 0.2  2  0.2    0.3
2
Therefore z decreases by [approximately] 0.3
Exact:
zold 
9  12  22 
znew 
9  1.2    2.2  
2
4  2,
2
Therefore Δz = −0.3507...
2.72  1.649
4.3 – Differentials
ENGI 3424
Page 4-15
Curves of Intersection
Example 4.3.5:
Find both partial derivatives with respect to z on the curve of intersection of the sphere
centre the origin, radius 5, and the circular cylinder, central axis on the y-axis, radius 3.
Sphere:
f = x2 + y2 + z2 = 25
Cylinder:
g = x2

df = 2x dx + 2y dy + 2z dz = 0
and
dg = 2x dx
+ z2 = 9
+ 2z dz = 0
which leads to the linear system
 x y   dx 
1
 x 0    dy     1  z dz

  
 
1
 dx 
 x y  1
z  0 y  1
    
   z dz    

dz

x   1 
xy   x
 dy 
 x 0  1

z  y 
z  1

dz      dz

xy  0 
x 0
Therefore
z
dx   dz and dy  0 dz
x
x
z
y

 
and
 0
z
x
z
[The intersection is the pair of circles x2 + z2 = 9, y = 4 .
Because y never changes on each circle, x is actually a function of z only.]
4.3 – Differentials
ENGI 3424
Page 4-16
Example 4.3.6
A surface is defined by f  x, y, z   xz  y 2 z  z  1 .
Find
z
.
y
Implicit method:
df 
f
f
f
dx 
dy 
dz
x
y
z


0  z dx  2 yz dy  x  y 2  1 dz
But

f 1 
 x  y2  1 z
dz
  z dx  2 yz dy
z
1
 dz   z 2  dx  2 y dy 
In the slice in which the partial derivative

z
is evaluated, x is constant
y
 dx = 0.
z
  z 2  0  2 y   2 yz 2
y
Explicit method:
z   x  y 2  1
1

2
z
   x  y 2  1  0  2 y  0    2 yz 2
y
z2
In general, if a dx  b dy  c dz then
z
a
z
(because y is constant and dy = 0 in the slice in which
is evaluated) and

x
c
x
z
b
z
(because x is constant and dx = 0 in the slice in which
is evaluated).

y
c
y
4.4 – The Jacobian
ENGI 3424
4.4
Page 4-17
The Jacobian - implicit and explicit forms; plane polar; spherical polar
The Jacobian is a conversion factor for a differential of area or volume between one
orthogonal coordinate system and another.
Let
 x, y  ,  u, v 
be related by the pair of simultaneous equations
f  x, y, u, v   c1
g  x, y, u, v   c2

df  f x dx  f y dy  fu du  fv dv  0
and
dg  g x dx  g y dy  gu du  gv dv  0

 


fx
f y   dx 
 f
  u


g y   dy 
  gu
gx
A
 fv   du 

 gv   dv 
B
 dx 
1  du 
 
  A B

 dy 
 dv 
which leads to
dA  dx dy 
  x, y 
du dv
 u, v 
where the Jacobian is
  x, y 
det B

 u, v 
det A
The Jacobian for the transformation from  x, y  to  u, v  is also the magnitude of the
cross product of the tangent vectors that define the boundaries of the element of area, so
that
  x, y 
r r


 u, v 
u  v
4.4 – The Jacobian
ENGI 3424
Page 4-18
Example 4.4.1
Transform the element of area dA  dx dy to plane polar coordinates.
x  r cos , y  r sin  .
f  x  r cos  0
g  y  r sin   0

df  1 dx  0 dy  cos dr  r sin  d  0
and
dg  0 dx  1 dy  sin  dr  r cos d  0
 1 0  dx 
  cos  r sin   dr 
 

  


 0 1  dy 
  sin   r cos   d 
B
A

 x, y 

 r ,  
B
A

r cos 2   r sin 2 
1
 r
 r
if r  0 
Therefore
dA  r dr d
Geometrical derivation of dA :
element of arc length  r d
If x, y can be written as explicit functions of (u, v), then an explicit form of the Jacobian
is available:
  x, y 

 u, v 
x
u
y
u
x
v
y
v
The Jacobian can also be used to express an element of volume in terms of another
orthogonal coordinate system:
 x, y, z 
dV  dx dy dz 
du dv dw
 u, v, w
4.4 – The Jacobian
ENGI 3424
Page 4-19
Spherical Polar Coordinates
The “declination” angle  is the angle
between the positive z axis and the
radius vector r . 0     .
The “azimuth” angle  is the angle on
the x-y plane, measured anticlockwise
from the positive x axis, of the shadow
of the radius vector. 0    2 .
z  r cos 
The shadow of the radius vector on the
x-y plane has length r sin  .
It then follows that
x  r sin  cos  and y  r sin  sin  .
Example 4.4.2
Express the element of volume dV
x
 r sin  cos 


r   y    r sin  sin 
 z 
 r cos 
Using the explicit form,
  x, y , z 

 r ,  ,  
xr
x
x
yr
y
y
zr
z
z

in spherical polar coordinates,




sin  cos 
sin  sin 
cos 
r cos  cos 
r cos  sin 
r sin 
r sin  sin 
r sin  cos 
0
Expanding along the last row of the determinant,
  x, y , z 
 cos  r 2 sin  cos  cos2   sin 2    r sin   r sin 2   cos2   sin 2    0
  r,  ,  
 r 2 sin   cos2   sin 2    r 2 sin 
Note that sin   0 because 0     .
Therefore
dV  r 2 sin  dr d d
ENGI 3424
4.4 – The Jacobian
For a transformation from Cartesian to plane polar coordinates in
 x, y 
 r so that dA  dx dy  r dr d
 r ,  
Page 4-20
2
,
For cylindrical polar coordinates in 3 ,
  x, y , z 
 r  dV  r dr d dz
 r ,  , z 
Explicit method for plane polar coordinates:
x  r cos , y  r sin 

 x

 r
  x, y 


 ABS det 

 r ,  
 y

 r


x 
 

y
 


 



cos 
sin 
r sin 
r cos 
 r cos2   r sin 2   r .
Implicit method for plane polar coordinates [Example 4.4.1]:
f  x  r cos  0
 df  dx  cos dr  r sin  d  0
g  y  r sin   0
 dg  dy  sin  dr  r cos d  0
 1 0   dx 
 cos  r sin    dr 
 
   



 0 1   dy 
 sin  r cos    d 
A
B
B
 x, y 
r cos 2   r sin 2 



 r  r
 r ,  
A
1
Note that for spherical polar coordinates  r , ,   there is no agreement among textbooks
as to which angle is  and which angle is  . In this course we adopt  as the angle
between the positive z axis and the radius vector r .
You are likely to encounter
textbooks and software in which that same angle is labelled as  .
4.5 – The Gradient Vector
ENGI 3424
4.5
Page 4-21
Gradient Vector, directional derivative, potential function, central force law
Let F  F  r  and r  r  t  .
dF
 F dx
 F dy
 F dz



dt
 x dt
 y dt
 z dt
 
 
  x


y
T
 dr
 
F

 dt
z 

dF
dr
 F
dt
dt
 = “del” or “nabla” = gradient operator
F = gradient vector
Directional Derivative
The rate of change of the function F at the point Po in the direction of the vector
a  aaˆ
 a  0 is
Daˆ F P  F P aˆ
o
o
The maximum value of the directional derivative of F at any point Po occurs when
the vector a is parallel to the gradient vector F .
Also, the vector Ν  F is normal
to the surface defined by F  r   constant.
[Interpretation of the gradient vector:
The gradient vector is in the plane of
the dependent variables.
Its direction at any point is the direction in which one must travel in order to experience
the greatest possible rate of increase of the dependent variable at that point.
Its magnitude is that greatest possible rate of increase.]
4.5 – The Gradient Vector
ENGI 3424
Page 4-22
Example 4.5.1
The temperature in a region within 10 units of the origin follows the form
T  r er , where r  x 2  y 2  z 2 .
Find the rate of temperature change at the point (−1, −1, +1) in the direction of the vector
1
0 0 .
T
T and r are symmetric with respect to
 x, y, z  .
T
dT  r

x
dr  x
dT
d

r e r   1  r  e r

dr
dr
1/ 2
r
1
x
  x2  y 2  z 2   2x  0  0 
x
2
r

T
x
 1  r  e r
x
r
By symmetry, it then follows that
T
y
T
z
 1  r  e r
and
 1  r  e r
y
r
z
r
1  r r
T
 T 
e x y z
r
At  1, 1, 1 , r 
111 
3  T 
1 3  3
T
e  1 1 1 
3
aˆ  ˆi
 DaT
P
 1 
1

3
 1 

3
 T ˆi 
e


3
 1 
1
0

 0

 


3 1  3
e
 0.0748 Km 1
3
[Note that, in this example, the temperature distribution is spherically symmetric about
the origin, with a minimum of zero at the origin, rising to a maximum of 1/e = 0.368...
one metre away from the origin and falling asymptotically back to zero as r  ∞.
Outside the r = 1 sphere, the gradient vector points radially inwards. At (−1, −1, 1) the
gradient vector has a magnitude of approximately 0.104 K m−1.]
4.5 – The Gradient Vector
ENGI 3424
Page 4-23
Central Force Law
If   r  is the potential function for some force per unit mass or force per unit charge
F  r  , then F    (or, in some cases, F   ). For a central force law, the potential
function is spherically symmetric and is dependent only on the distance r from the
origin. When the central force law is a simple power law,   k r n .
r 

x2  y 2  z 2

d  r

x
dr  x
 k n r n1 
 kn x 
1/2
1 2
x  y2  z 2   2x  0  0

2
r n1
 kn x r n2
r
By symmetry,

 kn y r n2
y
   kn r
and
x
 y
 z
n 2 

 kn z r n2
z

  kn r n2 r


But r  r rˆ
Therefore
   F   kn r n1 rˆ
Examples include the inverse square laws, for which n = −1:
Electromagnetism:
Q
Q
Q
k 
,  
, F    
rˆ
4 
4  r
4  r 2
Gravity:
k   GM ,  
GM
GM
, F    
rˆ
r
r2
4.5 – The Gradient Vector
ENGI 3424
Page 4-24
Some Applications of the Gradient Vector:
(1)
The directional derivative D at the point Po  xo , yo , zo  is maximized by
 F  r  P .
choosing a to be parallel to F at Po , so that Daˆ F
o
Po
(2)
A normal vector to the surface F  x, y, z   c at the point Po  xo , yo , zo  is
N  F  r  P .
o
(3)
The equation of the line normal to the surface F  x, y, z   c at the point

Po  xo , yo , zo  is r  OPo  t F
Po
(4)
o
Po
The equation of the tangent plane to the surface F  x, y, z   c at the point

Po  xo , yo , zo  is r  OPo
(5)
  r  t F 
 F P   r  r  F P   0
o
o
o
If the point Po  xo , yo , zo  lies on both of the surfaces F  x, y, z   c and
G  x, y, z   k , then the acute angle of intersection  of the surfaces at the
point is given by
cos  
 F   G 
Po
F
Po
Po
 G
Some of these features are applied in the problem sets.
Po
ENGI 3424
4.6
4.6 – Extrema; Second Derivative Test
Page 4-25
Extrema; Second Derivative Test for z  f  x, y 
Much of differential calculus in the study of maximum and minimum values of a function
of one variable carries over to the case of a function of two (or more) variables. In order
to visualize what is happening, we shall restrict our attention to the case of functions of
two variables, z  f  x, y  .
For a function f  x, y  defined on some domain D in
2
, the point P  xo , yo  is a
critical point [and the value f  xo , yo  is a critical value] of f if
1)
P is on any boundary of D; or
f  xo , yo  is undefined; or
2)
3)
f x and/or f y is undefined at P; or
4)
f x and f y are both zero at P (  f  0 at P).
A local maximum continues to be
equivalent to a “hilltop”,
while a local minimum continues to be
equivalent to a “valley bottom”.
“Local extremum” is a collective term for local maximum
or local minimum.
Instead of tangent lines being horizontal at critical points
of type (4), we now have tangent planes being horizontal
at critical points of type (4).
At any local extremum at which f  x, y  is differentiable,
the tangent plane is horizontal, f x  f y  0 and f  0 .
The converse is false.
f  0 does not guarantee a local extremum. There could be a saddle point (the higher
dimensional equivalent of a point of inflection) instead.
4.6 – Extrema; Second Derivative Test
ENGI 3424
Page 4-26
Example 4.6.1
Find and identify the nature of the extrema of f  x, y   x 2  y 2  4 x  6 y .
Polynomial functions of x and y are defined and are infinitely differentiable on all of
Therefore the only critical points are of type (4).
 2x  4 
f  

 2y  6 
f  0
But

 x, y 
  2, 3 only.
f  x, y    x  2   4   y  3  9
2
2
 0  4  0  9   13  f  2, 3
Therefore the only local extremum is a minimum value of −13 at (−2, 3).
It is also an absolute minimum.
Second Derivative Test
To determine the nature of a critical point:
1)
Examine the values of f in the neighbourhood of P; or
2)
[First derivative test:] Examine the changes in f x and f y at P; or
3)
Use the second derivative test:
At all points (a, b) where f  0 , find all second partial derivatives, then find
f xx f xy
D 
(the “Hessian” determinant)
f yx f yy
and evaluate D at (x, y) = (a, b).
D  a, b   0 and f xx  a, b   0  a relative minimum of f is at (a, b)
D  a, b   0 and f xx  a, b   0  a relative maximum of f is at (a, b)
D  a, b   0  a saddle point of f is at (a, b)
D  a, b   0  test fails (no information).
2
.
4.6 – Extrema; Second Derivative Test
ENGI 3424
 
Note that D  a, b   0 
Therefore D  0 
f xx f yy  f xy f yx and f xy f yx  f xy
Page 4-27
2
0
f xx f yy  0 
f xx , f yy must have the same sign at an extremum.
One can therefore check the sign of whichever of f xx or f yy is easier to evaluate in the
second derivative test, when determining whether an extremum is a maximum or a
minimum.
Example 4.6.1 (again):
Find all extrema of
f  x, y   x 2  y 2  4 x  6 y .
Any critical points will be of type (4) only.
 2x  4 
f  
  0
 2y  6 

D 
f xx
f xy
f yx
f yy


 x, y 
2 0
0 2
  2, 3 only.
 4  0   x, y 
D  0 and f xx  0 at (2, 3)  there is a relative minimum at (2, 3) and
the minimum value is f (2, 3) = 13.
As there are no other critical points, f  x, y  has an absolute minimum value of 13 at
(2, 3) and has no maxima.
[ z  f  x, y  is a circular paraboloid, vertex at (2, 3, 13) and axis of symmetry parallel
to the z-axis.]
4.6 – Extrema; Second Derivative Test
ENGI 3424
Page 4-28
Example 4.6.2
Find all extrema of f  x, y   2 x3  x y 2  5x 2  y 2 .
f  x, y  is a polynomial function of x and y
 any critical points will be of type (4) only.
 6 x 2  y 2  10 x 
f  

 2 xy  2 y 
f  0 
 2
6 x 2  10 x  y 2  0
and
1
2 y  x  1  0
 2
 y  0 or x   1
y  0 in 1  2 x  3x  5  0
 x  0 or x  
5
3
x   1 in 1  6  10  y 2  0
 y2  4  y   2
The only critical points are
 x, y 
 5 
  0, 0  ,   , 0  ,  1,  2  and  1,  2  .
 3 
f x  6 x 2  y 2  10 x and f y  2 y  x  1
D 
f xx
f xy
f yx
f yy

12 x  10
2y
2y
2  x  1
4.6 – Extrema; Second Derivative Test
ENGI 3424
Page 4-29
Example 4.6.2 (continued)
z  f  0,0   0
At (0, 0):
D 
10 0
0
 20  0
2
D  0 and f xx  10  0 


 
At  53 ,0 :
z  2  53
3
minimum
 
 0  5  53
   2   53   5 
  53
D 
12   53   10
2
0
2   53  1
0

2
 0
25  10  15 
125

 
9 
3
27

 10
0
0
 43
 
40
 0
3
D  0 and f xx  0  maximum
At  1,  2  :
z  f  1,  2   2  1   1 2   5  1   2 
2
 2  4 5 4  3
D 
2 4
4
0
 0   4    16  0
2
D  0  saddle point
Therefore the critical points are
Minimum at  0, 0, 0 


Maximum at  53 , 0, 125
27
Saddle points at  1, 2, 3 and  1, 2, 3
Slices
y=0
and
x=0
2
2
4.6 – Extrema; Second Derivative Test
ENGI 3424
Example 4.6.3
Find all extrema of z  f  x, y   x 2  y 2
Again this function is a polynomial function of x and y
so that the only critical points occur where f  0 .
 2x 
f  

 2 y 
f  0 
 x, y    0, 0
only.
f  0, 0   0
D 
f xx
f xy
f yx
f yy

2
0
0 2
 0
Therefore the only critical point is a saddle point at (0, 0, 0).
The surface is an hyperbolic paraboloid.
Page 4-30
ENGI 3424
4.6 – Extrema; Second Derivative Test
Page 4-31
Example 4.6.4
Find and identify the nature of the extrema of
f  x, y  
1  x2  y 2 .
The domain of f  x, y  is the circle x 2  y 2  1 and all points inside it.
f  x, y   1  x 2  y 2 
1/ 2
1
2 x
 f  x, y   
2  1  x2  y 2



1  x 2  y 2 
2 y
T
 
x
 
1  x2  y 2  y 
1
Inside the domain, f and its partial derivatives are well defined everywhere,
except that the gradient diverges everywhere on the boundary.
Therefore the only critical points to consider are types (1), (3) and (4).
Type (4) [zero gradient]:
Inside the domain, f  0 
 x, y    0, 0 only.
f  0, 0   1
However, 0  1  x2  y 2  1 everywhere else in the domain


f  x, y   f  0, 0  everywhere else in the domain.
there is an absolute maximum at (0, 0, 1).
Types (3) [undefined gradient] and (1) [the boundary]:
Everywhere on the boundary, 1  x2  y 2  1  1  0
and 0  f  x, y  everywhere else in the domain.

the absolute minima are all points on the circle x2 + y2 = 1, z = 0.
[The surface is the upper half of the sphere x2 + y2 + z2 = 1 .]
ENGI 3424
4.7
4.7 – Lagrange Multipliers
Page 4-32
Lagrange Multipliers; nearest point on curve of intersection to given point
In order to obtain an appreciation for the geometric foundation for the method of
Lagrange multipliers, we shall begin with an example that could be solved in another
way.
Example 4.7.1
A farmer wishes to enclose a rectangle of
land. One side is a straight hedge, more than
30 m long. The farmer has a total length of
12 m of fencing available for the other three
sides. What is the greatest area that can be
enclosed by the available fencing?
The function to be maximized is the area enclosed by the fencing and the hedge:
A  x, y   x y
The constraint is the length of fencing available:
L  x, y   x  2 y  12
There are additional constraints. Neither length may be a negative number.
Therefore any solution is confined to the first quadrant of the  x, y  graph.
Superimpose the graph of the constraint function L(x, y) = 12 on the contour graph of the
function z = A(x, y) :
The least value of A(x, y) is zero, on the coordinate axes.
At (0, 6), two 6-metre segments of fence are touching, enclosing zero area.
At (12, 0), all of the fence is up against the hedge, enclosing zero area.
4.7 – Lagrange Multipliers
ENGI 3424
Page 4-33
Example 4.7.1 (continued)
As one travels along the constraint line L = 12 from one absolute minimum at (0, 6) to
the other absolute minimum at (12, 0), the line first passes through increasing contours of
A(x, y),


A  6  2 7, 3  7   4
A  6  18, 3  4.5   9
A 6  34, 3  8.5  1
A(4, 4) = 16
until, somewhere in the vicinity of (6, 3), the area reaches a maximum value, then
declines, (for example, A(8, 2) = 16), back to the other absolute minimum at (12, 0).
At the maximum, the graph of L(x, y) = 12 just touches a contour of A(x, y).
The two graphs share a common tangent there.

gradients are parallel
 A   L
 y
1
     
x
2
Therefore solve the system of simultaneous equations
y = 
x = 2
x + 2y = 12
Solution:
(the constraint)
(x, y) = (6, 3).
From the graph, A is a maximum at (6, 3).
Amax = A(6, 3) = 63 = 18 m2 .
 is the Lagrange multiplier.
ENGI 3424
4.7 – Lagrange Multipliers
Page 4-34
To maximize f  x, y  subject to the constraint g  x, y   k :
As one travels along the constraint curve g  x, y   k , the maximum value of f  x, y 
occurs only where the two gradient vectors are parallel. This principle can be extended
to the case of functions of more than two variables.
General Method of Lagrange Multipliers
To find the maximum or minimum value(s) of a function f  x1 , x2 ,
, xn  subject to a
constraint g  x1 , x2 , , xn   k , solve the system of simultaneous (usually non-linear)
equations in (n + 1) unknowns:
f   g
g = k
where  is the Lagrange multiplier.
Then identify which solution(s) gives a maximum or minimum value for f.
This can also be extended to the case of more than one constraint:
In the presence of two constraints g  x1 , x2 , , xn   k and h  x1 , x2 ,
the system in (n + 2) unknowns:
f   g   h
g = k
h = c
, xn   c , solve
4.7 – Lagrange Multipliers
ENGI 3424
Page 4-35
Example 4.7.2
Find the points, on the curve of intersection of the surfaces x  y  2 z  6 and
2x2  y 2  z 2 , that are nearest / farthest from the origin.
The maximum and minimum values of distance d occur at the same place as the
maximum and minimum values of d 2 . [Differentiation of d 2 will be much easier than
differentiation of d.]
The function to be maximized/minimized is
f  x, y, z   d 2  x 2  y 2  z 2
The constraints are
g  x, y, z   x  y  2 z  6
and
h  x, y, z   2 x 2  2 y 2  z 2  0
f   g   h


and
 2x
2 y 2z

T
   1 1 2 
T
2 x 1  2   
(1)
2 y 1  2   
(2)
2 z 1      2
(3)
x  y  2z  6
(4)
2 x2  2 y 2  z 2  0
(5)
   4 x 4 y 2 z
(1)  (2)  2  x  y 1  2   0

y  x or 2  1 .
Case 2 = 1:
(1) or (2)    0
(3)  z  0 [or    1 , which contradicts 2  1 .]

T
Note that f, g and h are all
symmetric with respect to
(x, y). It then follows
that, at any critical point,
y = x.
Equation (2) is redundant,
which allows for a faster
solution than is shown
here.
4.7 – Lagrange Multipliers
ENGI 3424
Example 4.7.2 (continued)
(5)  2  x2  y 2   0 
But
 x, y    0, 0
(4)  0 + 0  0 = 6
only.
- impossible!
Therefore 2  1 .
Case y = x:
(4)  2 x  2 z  6  z  x  3
(6)
(5)  4 x 2   x  3  0
2
 4 x2  x2  6 x  9  0
 3  x 2  2 x  3  0
 3  x  3 x  1  0
 x   3 or
x  1.
x  3  y  3
(6)  z   6
f  3, 3, 6    3   3   6 
2
2
2
 d 2  9 1  1  4   54 .
x 1  y 1
(6)  z   2
f 1, 1, 2   12  12   2 
2
 d 2 11 4  6 .
These are the only critical points. Therefore
The nearest point is (1, 1, 2) and the farthest point is (3, 3, 6).
[The surfaces are a plane and a right circular cone.
The curve of intersection is an ellipse.]
Page 4-36
4.8 – Additional Examples
ENGI 3424
4.8
Page 4-37
Miscellaneous Additional Examples
Example 4.8.1
A window has the shape of a rectangle surmounted by a semicircle. Find the maximum
area of the window, when the perimeter is constrained to be 8 m.
The function to be maximized is
A  r, h   2rh  12  r 2
The constraint is the perimeter function
P  r, h   h  2r  h   r  2h    2  r  8
Single variable method:
P 8
 h 
8    2  r
2

  4  2
A  r   r 8    2  r   12  r 2  8r  
r
 2 

dA
 8    4  r
dr
dA
8
 0  r 
only.
dr
 4
The only other critical points are on the domain boundaries, at
r = 0 (when A = 0, clearly an absolute minimum) and at h = 0 (when A  12  r 2 ).
Inside the domain, r > 0 and h > 0  A > 0 (and A is differentiable throughout).
Therefore A must be at either a maximum or a point of inflexion at r 
8
.
 4
4.8 – Additional Examples
ENGI 3424
Example 4.8.1 (continued)
d2A
8
 0    4   0  maximum at r 
.
2
dr
 4
At the maximum, h 
8    2  r
8    4     2  
8
 
 
2
2
 4
 4

88
32
   4  8 

 
 4.48 m2

 
 4
 4
 2    4 
2

Amax
The maximum area occurs when
r  h 
8
 1.12 m .
 4
Lagrange Multiplier Method:
A  r, h   2rh  12  r 2
P  r, h   h  2r  h   r  2h    2  r  8
A   P
 2h   r 
  2 
 
 


2r


 2 

2h   r     2 
(1)
2r  2
(2)
and 2h    2  r  8
(3)
Page 4-38
4.8 – Additional Examples
ENGI 3424
Page 4-39
Example 4.8.1 (continued)
(2)
 r 
(1) becomes
2h   r   r  2r  2h  2r
 hr.
(3)

Therefore
  4  r  8
r  h 
8
 1.12 m .
 4
88
32
   4  8 
 
 4.48 m2

 
 4
 4
 2    4 
2
At this critical point, A 
The ends of the constraint line are at
At B,
 r, h 
 8

 B  0, 4  , C 
, 0 .
  2 
A  Amin  0
1  8 
32
At C ,
A  
 3.80 m2
 
2
2   2 
  2 
A > 0 and differentiable at all points between B and C.
2
Therefore the maximum area of
8
32
 1.12 m .
 4.48 m2 occurs at r  h 
 4
 4
4.8 – Additional Examples
ENGI 3424
Page 4-40
Example 4.8.2
Find the local and absolute extrema of the function
f  x, y  
3
x2  y2 .
f   x2  y 2 
1/3

2/3
f
1
2x
  x2  y 2   2 x  0 
2
x
3
3 3  x2  y 2 
By symmetry,
f
2y

2
y
3 3  x2  y2 
f is continuous on all of 2 ,
(therefore there is no boundary to check and no critical points of types (1) or (2)), but
f x   and f y   as  x, y    0,0  (critical point of type (3)).
f x and f y are not both zero everywhere else (no critical points of type (4)).
Therefore (0, 0) is the only critical point.
But f  0,0   0 and f  x, y   0 for all  x, y  except (0, 0).
Therefore the only critical point is an absolute minimum value of 0 at (0, 0).
[There is a cusp at the origin.]
[Any vertical cross-section containing the z axis.]
ENGI 3424
4.8 – Additional Examples
Example 4.8.3
Find the maximum and minimum values of the function
V  x, y   48x y  32 x3  24 y 2
in the unit square 0  x  1, 0  y  1.
[The full solution to this question is available on the course web site, at
"www.engr.mun.ca/~ggeorge/3424/demos/ex483.html".]
 32  V  x, y    2 ,
with the absolute maximum of V at
 x, y    12 , 12 
and the absolute minimum of V at
 x, y   1, 0 .
Page 4-41
4.8 – Additional Examples
ENGI 3424
Page 4-42
Example 4.8.4
A hilltop is modelled by the part of the elliptic paraboloid
x2
y2
h  x, y   4000 

1000
250
that is above the x-y plane. At the point P(500, 300, 3390), in which direction is the
steepest ascent?
 x  y 
h  
 500 125 
T
12 

 h P   1 
5 

T
Therefore the steepest ascent is in the direction of the vector
 5 
 12 


Note that this vector does not point directly at the summit.
The summit is on the z axis.
From the point P(500, 300, 3390), the summit is in the direction [5, 3] T.
4.8 – Additional Examples
ENGI 3424
Page 4-43
Example 4.8.5
2
 2u
1
2  u
Show that u  x, t    f  x  ct   f  x  ct   satisfies
.
 c
t 2
 x2
2
Let r = x  ct
and
s = x + ct
Then
u
u  r
u  s




t
 r t
 s t
1  df
1
df 
c  df df 

 0    c    0 


   c    

2  dr
2
ds 
2  dr ds 

ut  r
ut  s
 utt 



 r t
 s t




c  d2 f
d2 f 
c2  d 2 f
d2 f 
    2  0    c    0  2     c   



2   dr
ds 
2  dr 2
ds 2 



u
u  r
u  s
1 df
1 df





1 
1
x
r  x
s  x
2 dr
2 ds
 u xx 
Therefore
1 d2 f
1 d2 f
1
2
2
1

1  2 utt


2
2  
2 dr
2 ds
c
utt  c 2uxx .
1
 f  x  ct   f  x  ct   satisfies the wave equation
2
- and f  r  can be any twice differentiable function of r !
[Thus u  x, t  
This function models a pair of identical wave forms, moving at speed c in opposite
directions without any attenuation.]
[END OF CHAPTER 4]