nus - ma1505 (2012) - chapter 7
TRANSCRIPT
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
1/42
Chapter 7. Functions of Several Variables
7.1 Introduction
In elementary calculus, we encountered scalar func-
tions of one variable, e.g. f = f(x). However,
many physical quantities in engineering and science
are described in terms of scalar functions of several
variables. For example,
(i) Mass density of a lamina can be described by
(x, y), where (x, y) are the coordinates of a point
on the lamina.
(ii) Pressure in the atmosphere can be described by
P(x,y,z) where (x,y,z) are the coordinates of a
point in the atmosphere.
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
2/42
2 MA1505 Chapter 7. Functions of Several Variables
(iii) Temperature distribution of a heated metal ball
can be described by T(x , y , z , t) where (x,y,z)
are the coordinates of a point in the ball and tis
the time.
7.1.1 Functions of Two Variables
A functionfoftwo variablesis a rule that assigns
to each ordered pair of real numbers (x, y) a real
number denoted by f(x, y).
We usually write z = f(x, y) to indicate that z is
a function ofx and y. Moreover, x, y are called the
independent variables and z is called the dependent
variable. The set of all ordered pairs (x, y) such that
f(x, y) can be defined is called the domainoff.
2
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
3/42
3 MA1505 Chapter 7. Functions of Several Variables
7.1.2 Example
(a) f(x, y) =x2y3.
This is a function of two variables which is defined
for any xandy. So the domain offis the set of
all (x, y) with x, y R.
(b) f(x, y) =
1 x2 y2.
This function is only defined when 1x2y2 0,
or equivalently x2 +y2 1.
So the domain off is the set
D= {(x, y) :x2 +y2 1}.
Note that D represents all the points in the xy
plane lying within (and on) the unit circle.
3
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
4/42
4 MA1505 Chapter 7. Functions of Several Variables
(c) We can also define fin pieces as a compound
function. For example
f(x, y) =
x y ifx > y,y x ifx < y,
1 ifx=y .
7.1.3 Functions of Three or More Variables
We can define functions ofthree variablesf(x,y,z),
fourvariables f(x , y , z , w), etc in a similar way.
7.2 Geometric Representation
7.2.1 Graphs of functions of two variables
The graph of a function f(x) of one variable is a curve
in the xy-plane, which can be regarded as the set of
all points (x, y) in the xy-plane such that y=f(x).
By analogy, we have the graphof a functionf(x, y)
4
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
5/42
5 MA1505 Chapter 7. Functions of Several Variables
of two variables is the set of all points (x,y,z) in the
three dimensional xyz-space such that z=f(x, y).
This set represents a surface in the xyz-space.
7.2.2 Example
The graph off(x, y) = 5 3x 2yis the plane with
equationz= 5 3x 2y (or 3x+ 2y+z= 5).
7.2.3 Example
The graph ofg(x, y) = 8x2 + 2y2 is the paraboloid
(see diagram below) with equation z= 8x2 + 2y2.
...............................................................................................................................................................................
................ y
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
..
.
.
..
.
.
.
.
.
.
.
.
.
.
..
.
.
..z
............
.............
..............
.............
..............
.............
..............
.............
..............
........................
................
x
..............
........
....
...
..
..
..
..
..
.
..
.
..
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
..
.
..
..
.
............................................................
...............................................................................................................................................................................................................
.........................
...............................................................................................................................................................
...........
................
.........................
...........
........................................................................................................................................................................
.......
.
.
.
.
.
.
..
.
.
..
..
....................................................................................
............
......................................................... .
.
.
.
.
.
..
..
..
......
......
......
......
....................................
............
......
..
....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
5
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
6/42
6 MA1505 Chapter 7. Functions of Several Variables
7.3 Partial Derivatives
Letf=f(x, y) be a function of two variables. Either
a change ofxor a change ofycan cause a change of
f. In order to measure the rate of change off with
respect to the variable x, we need to fix the variable
y, and vice versa. Note that when we fix one of the
variables off(x, y), then it becomes a function of one
variable.
7.3.1 Example
Let f(x, y) = x2 2xy+ 3y3. If we fix y = 2, say,
then
f(x, 2) =x2 4x+ 24
is a function in xalone.
6
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
7/42
7 MA1505 Chapter 7. Functions of Several Variables
Similarly, if we fix x=
1, then
f(1, y) = 1 + 2y+ 3y3
is a function in y alone.
7.3.2 First order partial derivatives
Let f(x, y) be a function of two variables. Then the
(first order) partial derivative of f with re-
spect to xat the point (a, b) is
d
dxf(x, b)
x=a
= limh0
f(a+h, b) f(a, b)h
We say that the partial derivative does not exist at
(a, b) if the limit on the RHS above does not exist.
When the above partial derivative exists, we denote
it by f
x(a,b)
or fx(a, b).
7
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
8/42
8 MA1505 Chapter 7. Functions of Several Variables
Similarly, the (first order) partial derivative of
f with respect to y (instead of x) at the point
(a, b) is:
d
dyf(a, y)
y=b = limh0f(a, b+h)
f(a, b)
h
and is denoted by
f
y (a,b)
or fy(a, b).
If we let z=f(x, y), we also write
fx=z
x, and fy=
z
y.
In practice, when we compute fx(a, b) (resp. fy(a, b)),
we simply treat they(resp. x) variable off(x, y) as
constant and differentiate f with respect to x (re-
spectivelyy) before substituting x=aand y=b.
8
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
9/42
9 MA1505 Chapter 7. Functions of Several Variables
7.3.3 Example
Let f(x, y) = (x3 +y)cos(y2).
Find fx(2, 0), and fy(2, 0).
Solution: Treat y as a constant and compute
fx = d
dxf(x, y) =
d
dx(x3 +y)cos(y2)
= 3x2 cos(y2)
Then fx(2, 0) = 3(2)2 cos(02) = 12.
Treatxas a constant and compute
fy = d
dyf(x, y) =
d
dy(x3 +y)cos(y2)
= cos(y2) (x3 +y)sin(y2) 2y.
Thenfy(2, 0) = cos (02) (23 +0)sin(02) 2(0) = 1.
9
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
10/42
10 MA1505 Chapter 7. Functions of Several Variables
7.3.4 Geometric interpretation
Geometrically, fx(a, b) measures the rate of change
offin the direction of vector iat the point (a, b). If
we consider the liney =bon thexy-plane parallel to
the x-axis and passing through the point (a, b), the
image of this line underfis a curveC1on the surface
z=f(x, y). Thenfx(a, b) is just the gradient of the
tangent line to C1at (a, b).
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
................ y
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
..
.
..
.
.
.
.
.
.
.
.
.
.
.
..
.
..z
...........
..............
..............
..............
..............
..............
..............
..............
..............
..............
.
..................................
..............
......................
................
x
..............................................................................................
......................
.....................................
.........................................................................................................................................................................................................................................................................................
....................
..................
.................
.................
.................
.................
....................
.........................
...........................................................................................................................................................
...............
........................................................................................................
...................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
f(a, b)C1
(a, b)
.
.
.
.
..
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
..
..
.
.
.
.
.
.
.
.
.
.
.
..
..
.
..
.
.
..
..
..
.
..
..
the tangent line to thecurve C1 has gradient
fx(a, b)
.
..
.
.
.
..
.
.
.
..
.
.
.
..
.
.
.
..
.
.
.
..
.
.
..
.
..
.
..
..
.
..
..
..
.
..
.
..
..
.
.................................
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
...
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
.
..
..
.....................................
...............................................................
the line y= b
z= f(x, y)
10
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
11/42
11 MA1505 Chapter 7. Functions of Several Variables
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
................ y
..............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
..
.
.
..
.
.
.
.
.
.
.
.
.
.
..
.
.
..
z
...........
..............
..............
..............
..............
..............
..............
..............
..............
..............
..............
..............
..............
.............................
................
x
..............................................................................................
......................
......................................
........................................................................................................................................................................................................................................................................................
....................
..................
.................
.................
.................
.................
....................
.........................
...................................................................................................................................................................
...............................................................................................................
...................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..........
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
f(a, b)C2
(a, b)
..
.............
.............
.............
.............
.............
.............
.............
.............
.........................
..
..............
the tangent line to thecurve C2 has gradientfy(a, b)
...................................................................................................
...........................................................................................
..............................................................................................
Similarly,fy(a, b) is just the gradient of the tangent
line at (a, b) of the curveC2traced out as the image
of the line x=aunder f.
7.3.5 Higher order partial derivatives
We have seen that the partial derivativesfxandfyof
a function of two variablesfare also functions of two
variables. Hence, we can study the partial derivatives
offxand fy.
11
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
12/42
12 MA1505 Chapter 7. Functions of Several Variables
Thesecond order partial derivativesoffare:
fxx= (fx)x=2f
x2 fxy= (fx)y=
2f
yx
fyx = (fy)x= 2f
xy fyy = (fy)y =
2f
y2.
Ifz=f(x, y), we also have the following notation:
fxx=2z
x2 fxy=
2z
yx fyx =
2z
xy fyy =
2z
y2.
7.3.6 Example
Find the second partial derivatives of
f(x, y) = 4x3 +x2y3 6y2.
Solution: We have fx= 12x2
+ 2xy3
, so
fxx= 24x+ 2y3, fxy = 6xy
2.
We have fy = 3x2y2 12y, so
fyx = 6xy2, fyy = 6x
2y 12.12
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
13/42
13 MA1505 Chapter 7. Functions of Several Variables
7.3.7 Mixed Derivatives
For most functions in practice, we have
fxy(a, b) =fyx(a, b). (1)
7.3.8 Example
Let f(x, y) =xy + ey
y2 + 1. Find fyx.
Solution: The notationfyx means differentiating
first with respect to y and then with respect to x.
This is the same as fxy, i.e. we can postpone the
differentiation with respect toyand differentiate first
with respect to x:
fx=y = fxy= (fx)y = 1.
This is much easier than to first differentiate with
respect to y!
13
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
14/42
14 MA1505 Chapter 7. Functions of Several Variables
7.3.9 Functions of Three or More Variables
For functions of three or more variables, we have sim-
ilar definitions and notations for partial derivatives.
For example, for functions of three variables f(x,y,z),
we fix two of the variables and differentiate with re-
spect to the third one.
fx, fy, fz, or fx
, fy
, fz
.
7.4 Chain Rule
Suppose the length, widthwand heighthof a box
change with time. At time t0, the dimensions of the
box are = 2 m,w= 3 m,h = 4 m, andand ware
increasing at a rate of 5 ms1 while h is decreasing
14
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
15/42
15 MA1505 Chapter 7. Functions of Several Variables
at a rate of 6 ms1. What is the rate of change of
the volume of the box at time t0?
In the above problem, the volume Vof the box is a
function of three variables in , w, h:
V =V(,w,h)
while these three variables are in turn functions of
time t: = (t), w = w(t), h= h(t).Clearly, a
change oftwill cause a change ofV.
We say thatV is acompositefunction oft and write
V(t) =V((t), w(t), h(t)).
The rate of change ofV is given bydV
dt.
Can we expressdV
dt in terms of
d
dt,
dw
dt, and
dh
dt?
The answer is given by Chain rule.
15
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
16/42
16 MA1505 Chapter 7. Functions of Several Variables
7.4.1 Chain rule for one independent vari-
able on f(x, y)
Supposez=f(x, y) is a function of two variables x
and y, and x= x(t), y = y(t) are both functions of
t. Then z is a function of t: z(t) = f(x(t), y(t))
and
dz
dt =
f
x
dx
dt+
f
y
dy
dt.
7.4.2 Example
Let z = 3xy2 +x4y, where x = sin2t, y = cos t.
Finddz
dt .
Solution:
dz
dt =
z
x
dx
dt+
z
y
dy
dt
= (3y2 + 4x3y)(2cos2t) + (6xy+x4)( sin t).16
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
17/42
17 MA1505 Chapter 7. Functions of Several Variables
7.4.3 Chain rule for two independent vari-
ables on f(x, y)
Supposez=f(x, y) is a function of two variables x
and y, and x = x(s, t), y = y(s, t) are both func-
tions of two variables s and t. Then z is a function
ofsand t: z(s, t) =f(x(s, t), y(s, t)) and
z
s
=f
x
x
s
+f
y
y
s
and z
t
=f
x
x
t
+f
y
y
t
7.4.4 Example
Let z=e2x cos3y, wherex=st2, y=s2t.
zs = zx xs +zy ys
= (2e2x cos3y)t2 + (3e2x sin3y)(2st).z
t =
z
x
x
t+
z
y
y
t
= (2e2x cos3y)(2st) + (3e2x sin3y)(s2).17
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
18/42
18 MA1505 Chapter 7. Functions of Several Variables
7.4.5 Chain rule for one independent vari-
able on f(x,y,z)
Chain rules can be extended in a similar way for func-
tions of three or more variables.
For example, supposew=f(x,y,z) is a function of
three variables x, y and z, and x = x(t), y = y(t),
z= z(t) are functions oft. Then w is a function of
t: w(t) =f(x(t), y(t), z(t)) and we have
dw
dt =
f
x
dx
dt+
f
y
dy
dt+
f
z
dz
dt.
7.4.6 Example
Back to the problem at the beginning of this section.
The volume
V(,w,h) = w h.18
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
19/42
19 MA1505 Chapter 7. Functions of Several Variables
Hence
V
=wh,
V
w=h,
V
h =w.
By chain rule,
dVdt
= V
ddt
+Vw
dwdt
+Vh
dhdt
= whd
dt+h
dw
dt +w
dh
dt
We are given, at time t0,
= 2 m, w= 3 m, h= 4 m,
whiled
dt= 5 ms1,
dw
dt= 5 ms1 and
dh
dt = 6 ms1.
Hence
dV
dt = (3)(4)(5)+(2)(4)(5)+(2)(3)(6) = 64 m3s1.
19
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
20/42
20 MA1505 Chapter 7. Functions of Several Variables
7.4.7 Chain rule for two independent vari-
ables on f(x,y,z)
Now suppose w = f(x,y,z) is a function of three
variables x, y and z, and x = x(s, t), y = y(s, t),
z = z(s, t) are functions of two variables s and t.
Then wis a function ofsand t:
w(s, t) =f(x(s, t), y(s, t), z(s, t)) and we have
w
s =
f
x
x
s+
f
y
y
s+
f
z
z
s
w
t =
f
x
x
t+
f
y
y
t+
f
z
z
t
7.5 Directional Derivatives
7.5.1 Extension of Partial Derivatives
We have seen earlier that, given a function f(x, y),
the partial derivatives give the rates of change off
20
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
21/42
21 MA1505 Chapter 7. Functions of Several Variables
with respect to x and y, i.e. along the directions of
x- and y-axes.
A natural question to ask is, what is the rate of
change offalong an arbitrary direction? This gives
rise to the notion of directional derivatives.
Let fbe a function ofxand y.
The directional derivative off at (a, b) in the
direction of a unit vector u=u1 i +u2 jis
Duf(a, b) = limh0
f(a+hu1, b+hu2) f(a, b)h
if this limit exists.
Note that Dif(a, b) =fx(a, b) and Djf(a, b) =fy(a, b),
whereiandjare the standard unit vectors of thexy-
plane.
21
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
22/42
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
23/42
23 MA1505 Chapter 7. Functions of Several Variables
7.5.3 A formula
Since Duf(a, b) is also the rate of change off(x, y)
at (a, b) in the direction ofu, and the coordinates x
and y refer to points on the line L:
x=a+u1t, y=b+u2t, z= 0,
it follows that the chain rule can be used, as follows:
df
dt =
f
x dx
dt +
f
y dy
dt .
Thus,
Duf(a, b) = fx
(a, b)
u1
+ fy(a, b)
u2.
7.5.4 Example
Letf(x, y) =x23xy2+2y3. FindDuf(2, 1), where
u= 32
i +12j.
23
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
24/42
24 MA1505 Chapter 7. Functions of Several Variables
Solution: First,fx= 2x
3y2,fy=
6xy + 6y2.
Thusfx(2, 1) = 1 and fy(2, 1) = 6.
Therefore,
Duf(2, 1) = (1)(
3
2 ) + (6)(1
2) =
3
6
2 .
24
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
25/42
25 MA1505 Chapter 7. Functions of Several Variables
Gradient Vector
In view of the expression of the directional derivative
in terms of partial derivatives, it is convenient and
useful to introduce the notion of a gradient vector.
The gradientoff(x, y) is the vector (function)
f = fx i + fyj.
For a given unit vector u = u1i +u2j, we obtain
f(a, b) u
= (fx(a, b) i + fy(a, b)j) (u1i +u2j)
= fx(a, b) u1 + fy(a, b) u2
= Duf(a, b).
Thus,
Duf(a, b) = f(a, b) u.25
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
26/42
26 MA1505 Chapter 7. Functions of Several Variables
Noting that
f(a, b) and uare vectors, let be the
angle (0 ) between them. Then
Duf(a, b) = f(a, b) u = ||f(a, b)|| cos .
Since 1 cos 1, we obtain some useful prop-
erties of the above formula when f(a, b) =0:
Facts.
(1) The function f increases most rapidly in the di-
rection f(a, b).
(2) The function fdecreases most rapidly in the di-
rection f(a, b).
26
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
27/42
27 MA1505 Chapter 7. Functions of Several Variables
Example. Let f(x, y) = 9 x2 y2. Find thelargest possible value ofDuf(2, 1).
Solution. The surface z = f(x, y) is the upper
hemisphere of the sphere of radius 3 and centred at
(0, 0, 0). First compute
fx = x
9 x2
y2
, fy = y
9 x2
y2
.
The largest possible value of Duf(2, 1) is obtained
when uis in the direction of
f(2, 1) = fx(2, 1)i + fy(2, 1)j
= i 12j.
Now,
||f(2, 1)|| = (1)2 + 122
= 52 .27
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
28/42
28 MA1505 Chapter 7. Functions of Several Variables
Let
u = f(2, 1)||f(2, 1)|| =
25
i 15j.
Thus, the largest possible value ofDuf(2, 1) is
f(2, 1) u
= (1) 2
5
+
1
2
1
5
=
5
2
.
7.5.5 Physical meaning
As we mentioned at the beginning of this section, the
directional derivativeDuf(a, b) measures the change
in the valuedfof a functionfwhen we move a small
distance dt from the point (a, b) in the direction of
28
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
29/42
29 MA1505 Chapter 7. Functions of Several Variables
the vector u:
df=Duf(a, b) dt.
7.5.6 Example
Let f(x, y) =x2y3 + 1.
Estimate how much the value of f will change if a
point Qmoves 0.1 unit from (2, 1) towards (3, 0).
Solution: Q moves in the direction (3 i+ 0 j)
(2 i + j) =i j.
The unit vectorualong this direction is 1
2i 1
2j.
Nowfx= 2xy3, fy = 3x
2y2.
Thusfx(2, 1) = 4 and fy(2, 1) = 12.
29
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
30/42
30 MA1505 Chapter 7. Functions of Several Variables
Therefore,
Duf(2, 1) = (4)( 1
2) + (12)( 1
2) = 8
2.
So
df=Duf(2, 1) dt= ( 82
)(0.1) 0.57.
So the value of f decreases by approximately 0.57
unit.
7.5.7 Functions of Three Variables
We can also define directional derivatives for func-
tions of three variables.
Let fbe a function ofx, y and z. The directional
derivative off at (a,b,c) in the direction of a unit
30
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
31/42
31 MA1505 Chapter 7. Functions of Several Variables
vector u=u1 i +u2 j +u3 kin the xyzspace is
Duf(a,b,c)
= limh0
f(a+hu1, b+hu2, c+hu3) f(a,b,c)h
if this limit exists.
Similarly, we have the formula
Duf(a,b,c) =fx(a,b,c)u1+fy(a,b,c)u2+fz(a,b,c)u3
7.6 Maximum and Minimum Values
7.6.1 Local maximum and minimum
(1)f(x, y) has a local maximumat (a, b) if
f(x, y) f(a, b) for all points (x, y) near (a, b).
The numberf(a, b) is called alocal maximum
value.
31
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
32/42
32 MA1505 Chapter 7. Functions of Several Variables
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...................
...................................................................................................................................................................................
............
.................................................................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
................
.
.
..
.
.
.
..
.
.
..
.
.
..
.
..
.
..
..
..
..
..
.
..
..
..
................
.
.
.
.
..
..
..
.........................................
.............
..................
.................................................................................................................................................
..
..
.
.
.
..
.
..
....................................................................................................................................................
..............................................................................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
......................................................................................
...........
...............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
............
.
.
.
.
..
..
.
..
.
.
..
..
.
..
....
............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
............
.
.
.
..
.
.
..
.
..
.
..
..
.
..
..
..
......
.
..
...
......
............
..........................................
....
(2)f(x, y) has a local minimumat (a, b) if
f(x, y) f(a, b) for all points (x, y) near (a, b).
The numberf(a, b) is called alocal minimum
value.
.
..............................................................................................................................................................................
..................................................................................................................................................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
..
.
.
.
..
.
.
..
.
.
..
..
.
..
..
..
..
..................................
.
.
..
.
..
..
......................................
...........
...............
............................
.....................................................................................................................................
...........................................................................................................................................................
...........................................................................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
................
.
..
....................................................................................
............
...............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
........
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
..
.
.
..
..
..
......
......
.
..
...
......
............
..........................................
....
7.6.2 Critical Points
A functionfmay have a local maximum or minimum
at (a, b) if
(i)fx(a, b) = 0 and fy(a, b) = 0; or
(ii) fx(a, b) or fy(a, b) does not exist.
32
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
33/42
33 MA1505 Chapter 7. Functions of Several Variables
A point offthat satisfies (i) or (ii) above is called a
critical point.
7.6.3 Example
Letf(x, y) =x2 + y2 + 4x 8y+ 24. Find the localmaxima and local minima off, if any.
Solution: The partial derivatives exist for any
point.
So we solve fx(x, y) = 0 and fy(x, y) = 0.
fx= 2x+ 4 = 0 x= 2
fy= 2y 8 = 0 y= 4.
This gives a solution (x, y) = (2, 4).
Is this point a local maximum, a local minimum, or
none?
33
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
34/42
34 MA1505 Chapter 7. Functions of Several Variables
By completing squares, we have
f(x, y) = 4 + (x+ 2)2 + (y 4)2 4.
Therefore, we conclude that (2, 4) is a local mini-
mum offwith minimum value 4.
7.6.4 Example
The following diagram shows the graph of a function
f(x, y) which has a local maximum at a point (a, b)
but fx(a, b) and fy(a, b) does not exist.
.
.
.
.
..
.
..
..
............................................................
....................................................................................................
...........
.....................................................................................................................................
.
.
.
.
.
.
..
.
..
.
..
..
...............................................
.........
............................................................................................................................................................................................................................................................................................
....................................................................................................................................................
.
.
.
.
.
.
..
..
..
......
......
......
......
......
............
......................................................
............
............
......
......
.
.
7.6.5 Saddle points
Let (a, b) be a point of f with fx(a, b) = 0 and
fy(a, b) = 0. We say (a, b) is a saddle point of
34
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
35/42
35 MA1505 Chapter 7. Functions of Several Variables
f if there are some directions along which f has a
local maximum at (a, b) and some directions along
which fhas a local minimum at (a, b).
7.6.6 Example
Find the local maximum or local minimum off(x, y) =
2y2 3x2, if any.
Solution: As before, we solve
fx= 6x= 0 and fy = 4y= 0.
The only solution is (x, y) = (0, 0). So this is the
only critical point off.
However, this point (0, 0) is neither a local maximum
nor a local minimum off.
To see this, consider the function falong the x-axis
35
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
36/42
36 MA1505 Chapter 7. Functions of Several Variables
which has equationy = 0. Substituting this equation
intof(x, y), we have
f(x, 0) = 3x2 0.
So along x-axis, fhas a local maximum at (0, 0).
On the other hand, if we consider falong they-axis
which has equation x= 0, we have
f(0, y) = 2y2 >0.
So along y-axis, fhas a local minimum at (0, 0).
Thereforefhas a saddle point at (0, 0).
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
..
.
.
.
..
.
.
.
.
.
.
.
.
.
..
.
.
.
..
.
z
.........................................................................................................................................................................................................................................
................ y................................
..............
...............
..............
..............
..............
...............
..............
..............
........................
.
...............
x
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
..
.
..
.
..
..
..
................................................................................................
.......................................................................................................................................
............................................................................................................................................... ................................................................................................................................................................
....................................................................................................................
....................
.....................................................................................................................................................................................................
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
..
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
..
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
..
.
..
..
..
..
.
.
.
.
.
.
............
............
........................................................................
.................
.............
............
............
............
...............
..
..
.
.
..
.
..
.
..
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
..
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
..
.
..
..
.
..
.
..
.
..............................
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
z= 2y2 3x2
36
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
37/42
37 MA1505 Chapter 7. Functions of Several Variables
7.6.7 Second Derivative Test
When the partial derivatives exist, we can determine
the type of critical point using the following system-
atic approach:
Let fx(a, b) = 0 and fy(a, b) = 0.
D=fxx(a, b)fyy(a, b) fxy(a, b)2.
(a) IfD > 0 and fxx(a, b) > 0, then fhas a local
minimum at (a, b).
(b) IfD > 0 and fxx(a, b) < 0, then fhas a local
maximum at (a, b).
(c) IfD
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
38/42
38 MA1505 Chapter 7. Functions of Several Variables
7.6.8 Example
Find the local maximum, local minimum and saddle
points (if any) off(x, y) =x3 + y3 + 3x2 3y2 8.
Solution: First, we solve fx= 3x2 + 6x= 0 and
fy = 3y2 6y= 0.
On solving, we obtain x= 0 or 2 and y= 0 or 2.
Therefore, the critical points are (0, 0), (0, 2), (2, 0)and (2, 2).
To apply the second derivative test, we compute the
second order partial derivatives.
fxx= 6x+ 6, fyy = 6y 6, fxy = 0.
ThusD(x, y) =fxxfyyf2xy = 36xy36x+36y36.
At (0, 0), D(0, 0) = 36
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
39/42
39 MA1505 Chapter 7. Functions of Several Variables
Hence,fhas a saddle point at (0, 0).
At (0, 2), D(0, 2) = 36 > 0 and fxx(0, 2) = 6 > 0.
Hencefhas a local minimum at (0, 2).
At (
2, 0), D(
2, 0) = 36 > 0 and fxx(
2, 0) =
6
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
40/42
40 MA1505 Chapter 7. Functions of Several Variables
7.6.9 Lagrange Multipliers
Many optimization models are subject to certain con-
straints. For example, production levels depend on
labour input and capital expenditure. With a given
budget (constraint), a manufacturer aims to maxi-
mize production. The method of Lagrange mul-
tipliersis illustrated below.
7.6.10 Example
Find relative extrema of
z = f(x, y) = 1 2x 16y+ 50
subject to the constraint x2 +y2 = 25.
40
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
41/42
41 MA1505 Chapter 7. Functions of Several Variables
Solution. The constraint is written as
g(x, y) = x2 +y2 25.
The following function is constructed:
F(x,y,) = f(x, y) g(x, y)
= 12x 16y+ 50 (x2 +y2 25).
Then set Fx= 0, Fy = 0, F= 0 to get respectively
12 2x = 0
16 2y = 0
x2 y2 + 25 = 0
Writing the first two equations as x = 6
, y =
8
41
-
8/10/2019 NUS - MA1505 (2012) - Chapter 7
42/42
42 MA1505 Chapter 7. Functions of Several Variables
and substituting into the third equation, we obtain
362 64
2 + 2 5 = 0
2 = 100
25 = 4
= 2
If= 2, then x=6
2= 3, y=
82
= 4 and
z = f(3,4) = 12(3) 16(4) + 50 = 150.
If= 2, then x= 62= 3, y=82= 4 and
z = f(3, 4) = 12(3) 16(4) + 50 = 50.
Thus, subject to the constraint x2 +y2 = 25, the
function z = f(x, y) = 1 2x 16y+ 50 attains:
(1) a local maximum ofz=f(3,4) = 150; and