Download Partial Derivatives

Partial Derivatives
Limits
For a single variable function f (x), the limit lim f (x) exists only if the right-hand side limit
x→a
equals to the left-hand side limit, i.e.,
lim f (x) = lim+ f (x) .
x→a−
x→a
For a two variables function f (x, y), the limit
lim
(x,y)→(a,b)
f (x, y) exists only if the value of the
limit is unique for (x, y) approaches to (a, b) in any path.
If we can find any two different paths which give different values for
lim
(x,y)→(a,b)
f (x, y), then it
follows that the limit does not exist.
1. Find the limit, if exists.
(a)
x2 − y 2
(x,y)→(0,0) x2 + y 2
(b)
lim
x2 y 3
(x,y)→(0,0) 2x2 + y 2
lim
Continuity
A function f (x) is continuous at x = a if
lim f (x) = f (a) .
x→a
For the two variables function f (x, y), the function is continuous at (a, b) if
lim
(x,y)→(a,b)
f (x, y) exists and
lim
(x,y)→(a,b)
f (x, y) = f (a, b) .
1. Determine which of the following function is continuous at (0, 0).

 x2 y 3
if (x, y) 6= (0, 0)
(a) f (x, y) =
2x2 + y 2

if (x, y) = (0, 0)
1 2 3
 xy
if (x, y) 6= (0, 0)
(b) f (x, y) =
2x2 + y 2

0
if (x, y) = (0, 0)
First Principle
The derivative (rates of change) of the function f (x) at x = a, denoted by f ′ (a), is defined by
f (a + ∆x) − f (a)
,
∆x→0
∆x
f ′ (a) = lim
if the limit exists.
If f is a function of two variables, f (x, y), its partial derivatives at the point (a, b), denoted by
fx (a, b) and fy (a, b), are defined by
f (a + ∆x, b) − f (a, b)
f (a, b + ∆y) − f (a, b)
and fy (a, b) = lim
,
∆x→0
∆y→0
∆x
∆y
if the limits exist.
fx (a, b) = lim
1. Use the definition of partial derivatives to find fx (a, b) and fy (a, b) at the given point
(a, b), if exist.
1
(a) f (x, y) =
p
x2 + y 2 ; (0, 0)
(b) f (x, y) =
x
; (1, −1)
y
2. Use the First Principle to find fx (x, y) and fy (x, y).
(a) f (x, y) = 16 − 4x2 − y 2
(b) f (x, y) =
√
2x − 3y
Partial Derivatives
There are many alternative notations for partial derivatives. For instance, instead of fx we can
∂f
as the first partial derivative of f (x, y) with respect to x. If z = f (x, y), we write
write
∂x
fx (x, y) = fx =
∂
∂z
∂f
∂
∂z
∂f
=
f (x, y) =
= zx and fy (x, y) = fy =
=
f (x, y) =
= zy .
∂x
∂x
∂x
∂y
∂y
∂y
The rules for finding partial derivatives of z = f (x, y):
• To compute fx , regard y as a constant and differentiate f (x, y) with respect to x.
• To compute fy , regard x as a constant and differentiate f (x, y) with respect to y.
• Apply the Product Rule, Quotient Rule, or Chain Rule if necessary.
1. Find the first partial derivatives of the given functions.
4
(a) f (x, y) = x3 y 5
√
(b) f (x, y) = 2x − 3y
x3 + y 3
(c) f (x, y) = 2
x + y2
(d) f (x, y) = (3xy 2 − x4 + 1)
(e) f (x, y) = ln (x2 + y 2 )
(f) f (x, y) = ex tan (x − y)
Higher Derivatives
If f is a function of two variables, then its partial derivatives fx and fy are also functions of two
variables, so we can consider their partial derivatives (fx )x , (fx )y , (fy )x , and (fy )y , which are
called the second partial derivatives of f . If z = f (x, y), we use the following notation:
(fx )x = fxx
∂
=
∂x
∂f
∂x
∂ 2f
∂2z
=
=
,
∂x2
∂x2
(fy )x = fyx
∂
=
∂x
∂f
∂y
∂ 2f
∂2z
=
=
, and
∂x∂y
∂x∂y
(fx )y = fxy
∂
=
∂y
∂f
∂x
=
∂2f
∂2z
=
,
∂y∂x
∂y∂x
(fy )y = fyy
∂
=
∂y
∂f
∂y
=
∂2f
∂2z
=
.
∂y 2
∂y 2
A function f is said to be of class C k if all the k-th order partial derivatives of f exist and
continuous in the domain of the function.
If z = f (x, y) is a C 2 -function, then
fxy = fyx or
∂2f
∂2f
=
.
∂y∂x
∂x∂y
1. Find all the second partial derivatives for the following functions and identify which one
is a C 2 -function.
2
(a) f (x, y) = x5 y 4 − 3x2 y 3 + 2x2
√
(b) f (x, y) = x2 y + x y
(c) f (x, y) = sin (x + y) + cos (x − y)
(d) f (x, y) = sin−1 (xy 2 )
2. Find the indicated partial derivative.
2
(c) f (x, y, z) = x5 + x4 y 4 z 3 + yz 2 ; fxyz
(a) f (x, y) = exy ; fxxy
(b) z = x sin y;
∂ 3z
∂y 2 ∂x
(d) z = ln sin (x − y);
3. Verify that the function u = e−α
ut = α2 uxx .
2 k2 t
∂3z
∂y∂x2
sin kx is a solution of the heat conduction equation
4. Show that the following functions are solutions of the wave equation utt = a2 uxx .
(b) u = (x − at)6 + (x + at)6
(a) u = sin (kx) sin (akt)
Chain Rule
Recall the Chain Rule for functions of single variable gives the rule for differentiating a composite
function: If y = f (x) and x = g (t), where f and g are differentiable functions, then y is indirectly
a differentiable function of t , i.e., y = f (g (t)) and
dy dx
dy
=
.
dt
dx dt
Extending this idea, there are several versions of the Chain Rule for functions of more than one
variable:
Case I: Suppose that z = f (x, y) is a differentiable function of x and y, where x = g (t) and
y = h (t) are both differentiable functions of t. Then
∂z dx ∂z dy
dz
=
+
.
dt
∂x dt
∂y dt
Case II: Suppose that z = f (x, y) is a differentiable function of x and y, where x = g (s, t) and
y = h (s, t), and the partial derivatives gs , gt , hs , and ht exist. Then
∂z ∂x ∂z ∂y
∂z
∂z ∂x ∂z ∂y
∂z
=
+
and
=
+
.
∂s
∂x ∂s ∂y ∂s
∂t
∂x ∂t
∂y ∂t
Case III: Suppose that z = f (x) is a differentiable function of x, where x = g (s, t), and the
partial derivatives gs and gt exist. Then
∂z
dz ∂x
∂z
dz ∂x
=
and
=
.
∂s
dx ∂s
∂t
dx ∂t
dz
.
dt
(a) z = x2 + y 2 , x = t3 , y = 1 + t2
1. Use the chain rule to find
2. Use the chain rule to find
(b) z = 6x3 − 3xy + 2y 2 , x = et , y = cos t
∂z
∂z
and
.
∂s
∂t
3
(b) z = sin x cos y, x = (s − t)2 , y = s2 − t2
(a) z = x2 sin y, x = s2 + t2 , y = 2st
3. If z = f (x − y), show that
∂z ∂z
+
= 0. [Hint: let u = x − y]
∂x ∂y
4. If z = f (x, y), where x = r cos θ, y = r sin θ and f is a C 2 -function,
∂z
∂z
and
and
∂r
∂θ
(b) show that
(a) find
∂z
∂x
2
+
∂z
∂y
2
=
∂z
∂r
2
1
+ 2
r
∂z
∂θ
2
and
∂ 2z ∂ 2z
∂2z
1 ∂ 2 z 1 ∂z
+
=
+
+
.
∂x2 ∂y 2
∂r2 r2 ∂θ2 r ∂r
5. Use the substitution u = x + ct and v = x − ct to reduce the wave equation c2
to the form
∂2w
= 0.
∂u∂v
6. Suppose g (u) is a C 1 -function of u and let w =
differentiation to evaluate the expression x
√
∂ 2w
∂2w
=
∂x2
∂t2
x
xy · g
. Use the chain rule of
y
∂w
∂w
+y
− w.
∂x
∂y
(2007 Quiz)
7. Let
u = f (x, y) = x2 + xy − y 2 ,
v = g (x, y) = x3 y 2 + 1
(∗)
so that (u, v) = (1, 2) when (x, y) = (1, 1).
(a) Find the partial derivatives fx , fy , gx , gy at (x, y) = (1, 1).
(b) Let
x = F (u, v) ,
y = G (u, v)
be the inverse function of (∗).
Find the partial derivatives Fu , Fv , Gu , Gv at (u, v) = (1, 2).
(2007 Test)
p
1
8. Let f be a C 2 function and w (x, y, z, t) = f (r − t) where r = x2 + y 2 + z 2 .
r
(a) Find all the first order partial derivatives.
(b) Show that
∂2w
∂2w ∂2w ∂2w
+
+
=
.
∂x2
∂y 2
∂z 2
∂t2
(2008 Exam)
4
Total Differential
Recall that for a function of one variable, y = f (x), we defined the increment of y as
∆y = f (x + ∆x) − f (x)
and the differential of y as
dy = f ′ (x) dx and ∆y ≈ dy.
If z = f (x, y) is a function of two variables, then the increment of z is
∆z = f (x + ∆x, y + ∆y) − f (x, y) ,
where ∆x and ∆y are the increments of x and y.
The differential dz, also called the total differential, is defined by
dz = fx (x, y) dx + fy (x, y) dy =
∂z
∂z
dx +
dy and ∆z ≈ dz.
∂x
∂y
Similarly, the relative error and percentage error are defined as
Relative Error =
dz
dz
and Percentage Error =
× 100%.
z
z
If we take
dx = ∆x = x1 − x0 and dy = ∆y = y1 − y0
for computing the increment of z for (x, y) changes from (x0 , y0 ) to (x1 , y1 ), then
∂z ∂z ∆x +
∆y.
dz = fx (x0 , y0 ) ∆x + fy (x0 , y0 ) ∆y =
∂x (x0 ,y0 )
∂y (x0 ,y0 )
For maximum increment,
∂z ∂z |dz| = |fx (x, y) dx + fy (x, y) dy| ≤ |fx (x, y)| |∆x| + |fy (x, y)| |∆y| = |∆x| + |∆y| .
∂x
∂y
1. Find the total differential of the given functions.
(a) z = x2 y 3
(c) w = x sin (yz)
p
(d) w = ln x2 + y 2 + z 2
(b) z = x4 − 5x2 y + 6xy 3 + 10
2. If z = 5x2 + y 2 and (x, y) changes from (1, 2) to (1.05, 2.1), compute the value of ∆z and
dz.
3. If z = x2 − xy + 3y 2 and (x, y) changes from (3, −1) to (2.96, −0.95), compute the value
of ∆z and dz.
4. The dimensions of a closed rectangular box are measured as 80cm, 60cm, and 50cm,
respectively, with a possible error of 0.2cm in each dimension. Use differentials to estimate
the maximum error in calculating the surface area of the box.
5
5. If R is the total resistance of three resistors, connected in parallel, with resistances R1 ,
R2 , R3 , then
1
1
1
1
=
+
+
.
R
R1 R2 R3
If the resistances are measured in ohms as
R1 = 25Ω, R2 = 40Ω, and R3 = 50Ω, with a
dRi = 0.5%, where i = 1, 2, 3, estimate the
possible error of 0.5% in each case, i.e., Ri maximum error in the calculated value of R.
Linear Approximation
By using the total differential, we can estimate the value of f (x1 , y1 ) when f (x0 , y0 ) is known
and (x1 , y1 ) is close to (x0 , y0 ). This is called the linear approximation.
Since ∆z ≈ dz, then
f (x1 , y1 ) = f (x0 + ∆x, y0 + ∆y) = f (x0 , y0 ) + ∆z ≈ f (x0 , y0 ) + dz.
1. Use differentials to approximate the value of f at the given point.
p
(b) f (x, y, z) = xy 2 sin (πz), (3.99, 4.98, 4.03)
(a) f (x, y) = x2 − y 2 , (5.01, 4.02)
2. Use linear approximation to evaluate the following numbers.
q
√ 4
√
(a) 8.94 9.99 − (1.01)3
(b)
99 + 3 124
q
(c) (3.02)2 + (1.97)2 + (5.99)2
Relative Maxima and Minima
A function of two variables has a relative maximum at (a, b) if f (x, y) ≤ f (a, b) for all points
(x, y). The value of f (a, b) is called a relative maximum value. If f (x, y) ≥ f (a, b) for all (x, y),
then (a, b) is a relative minimum point and f (a, b) is a relative minimum value.
Procedure for finding relative maxima and minima:
I. Find all critical point(s) by solving the system of equations
fx (x, y) = 0
.
fy (x, y) = 0
f (x , y ) fxy (x0 , y0 )
II. For each critical point (x0 , y0 ), compute the value of H = xx 0 0
fyx (x0 , y0 ) fyy (x0 , y0 )
III. (a) If H > 0 and fxx (x0 , y0 ) > 0, then f (x0 , y0 ) is a relative minimum.
.
(b) If H > 0 and fxx (x0 , y0 ) < 0, then f (x0 , y0 ) is a relative maximum.
(c) If H < 0, then f (x0 , y0 ) is not a relative extremum and (x0 , y0 ) is a saddle point.
(d) If H = 0, then we do not have enough information about the critical point.
1. Find all the relative maximum and minimum values and saddle points of the given function.
(a) f (x, y) = x2 + y 2 + 4x − 6y
(c) f (x, y) = 3x2 y + y 3 − 3x2 − 3y 2 + 2
(b) f (x, y) = x3 − 3xy + y 3
(d) f (x, y) = x sin y
6
Taylor’s Formula
Recall the Taylor’s Formula for a one variable function f (x), where f is a C n+1 -function defined
on an open interval I containing x0 ,
n
n
X
X
f (i) (x0 )
f (n+1) (ζ)
f (i) (x0 )
i
(n+1)
≈
(∆x) +
(∆x)
(∆x)i ,
f (x) =
i!
(n + 1)!
i!
{z
} i=0
{z
} |
|i=0
Error Term
Taylor Series
where ∆x = x − x0 and ζ is a point between x and x0 .
Suppose now z = f (x, y) is a C 3 -function defined in a domain D of the xy-plane, and let (x0 , y0 )
be a point in D. For any ∆x = x−x0 and ∆y = y −y0 , the Taylor’s Formula for f (x, y) becomes
f (x, y) = f (x0 , y0 ) + fx (x0 , y0 ) ∆x + fy (x0 , y0 ) ∆y +R2 ,
|
{z
}
Three Terms Formula
f (x, y) =f (x0 , y0 ) + fx (x0 , y0 ) ∆x + fy (x0 , y0 ) ∆y
1 +
fxx (x0 , y0 ) (∆x)2 + 2fxy (x0 , y0 ) ∆x∆y + fyy (x0 , y0 ) (∆y)2 +R3 , etc . . . ,
{z
}
| 2!
Six Terms Formula
where R2 and R3 are the error terms.
1. Find the first six terms of the Taylor’s Formula at the given point for the following functions:
√
1
(b) 2x + y; (5, −1)
; (2, 1)
(a)
1+x−y
Vectors
Given the points A (x1 , y1 , z1 ) and B (x2 , y2 , z2 ) in the three-dimensional space, the vector r (or
−→
−
→
r ) represents a directed line segment AB from point A to point B and
r = [x2 − x1 , y2 − y1 , z2 − z1 ] = (x2 − x1 ) i + (y2 − y1 ) j + (z2 − z1 ) k,
where i, j and k are the unit vectors in the x, y and z direction, respectively.
Given a = [a1 , a2 , a3 ] and b = [b1 , b2 , b3 ] are two non-zero vectors in the three-dimensional space,
and c is a constant, the length of vector a, (or the norm), is defined by
q
kak = a21 + a22 + a23 ,
the vector a + b is defined by
a + b = [a1 + b1 , a2 + b2 , a3 + b3 ] = (a1 + b1 ) i + (a2 + b2 ) j + (a3 + b3 ) k, and
the vector ca is defined by
ca = [ca1 , ca2 , ca3 ] = ca1 i + ca2 j + ca3 k.
If u is a unit vector of a, then u is a vector which has the same direction as a with length equals
to 1, i.e.,
[a1 , a2 , a3 ]
a
=p 2
.
u=
kak
a1 + a22 + a23
If b is a vector parallel to a, then b = ca, where c 6= 0 and
b1 = ca1 , b2 = ca2 , b3 = ca3 .
−→
1. Find a vector r with representation given by the directed line segment AB.
7
(a) A (1, 3), B (4, 4)
(b) A (0, 3, 1), B (2, 3, −1)
2. Find kak, a + b, a − b, 2a, and 3a + 4b
(a) a = [5, −12], b = [−2, 8]
(b) a = i + j + k, b = 2i − j + 3k
3. Find a unit vector for the given vector.
(a) [3, −5]
(b) 2i − 4j + 7k
The Dot Product
If a = [a1 , a2 , a3 ] and b = [b1 , b2 , b3 ], then the dot product (or the inner product) of a and b is
the scalar a • b given by
b
a • b = a1 b 1 + a2 b 2 + a3 b 3 .
If θ, 0 ≤ θ < π, is the angle between the vectors a and b, then
a • b = kak kbk cos θ.
a and b are orthogonal, θ =
π
,
2
a
if and only if a • b = 0.
θ
proja b
The vector projection of b onto a is defined by
proja b = kproja bk ·
| {z }
scalar
a
kak
|{z}
= kbk cos θ ·
direction
(unit vector)
a•b
a
a•b
a
= kbk ·
·
=
a.
kak
kak kbk kak
kak2
1. Find a • b.
(a) a = [2, 5], b = [−3, 1]
(b) a = 2i + 3j − 4k, b = i − 3j + k
2. Find the angle between the given vectors.
(a) a = [1, 2, 2], b = [3, 4, 0]
(b) a = i + j + 2k, b = 2j − 3k
3. Determine whether the given vectors are orthogonal, parallel, or neither orthogonal nor
parallel.
(a) a = [2, −4], b = [−1, 2]
(b) a = [2, −4], b = [4, 2]
(c) a = [2, 8, −3], b = [−1, 2, 5]
4. Find the projection vector of b onto a.
(a) a = [2, 3], b = [4, 1]
(b) a = 2i − 3j + k, b = i + 6j − 2k
The Cross Product
If a = [a1 , a2 , a3 ] and b = [b1 , b2 , b3 ], then the cross product of a and b is the vector a × b given
by
i j k a × b = a1 a2 a3 = [a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ] .
b1 b2 b3 8
The vector a × b is orthogonal to both a and b.
If θ, 0 ≤ θ < π, is the angle between the vectors a and b, then
ka × bk = kak kbk sin θ.
Two non-zero vectors a and b are parallel, θ = 0, if and only if a × b = 0.
The length of the cross product a × b is equal to the area of the
parallelogram determined by a and b:
b
θ
Area = A = ka × bk .
The volume of the parallelepiped determined by the vectors a, b, and
c is the magnitude of their scalar triple product:
Volume = V = |a • (b × c)| .
a
c
b
a
1. Find the cross product a × b.
(a) a = [1, 0, 1], b = [0, 1, 0]
(b) a = j + 4k, b = 6i − 5k
2. Find two unit vectors orthogonal to both a = [1, −1, 1] and b = [0, 4, 4].
3. Find the area of the parallelogram which bounded by the following vertices.
(a) A (0, 1), B (3, 0), C (5, −2), and D (2, −1)
(b) P (0, 0, 0), Q (5, 0, 0), R (2, 6, 6), and S (7, 6, 6)
4. Find the volume of the parallelepiped determined by the vectors a, b, and c.
(a) a = [1, 0, 6], b = [2, 3, −8], and c = [8, −5, 6]
(b) a = 2i + 3j − 2k, b = i − j, and c = 2i + 3k
Gradient Vector
If f is a function of two variables x and y, then the gradient of f is the vector function ∇f
defined by
∂f
∂f
i+
j.
∇f (x, y) = [fx (x, y) , fy (x, y)] =
∂x
∂y
Similarly, if g is a function of three variables x, y, and z, then the gradient of g, ∇g is defined
by
∂g
∂g
∂g
i+
j + k.
∇g (x, y, z) = [gx (x, y, z) , gy (x, y, z) , gz (x, y, z)] =
∂x
∂y
∂z
1. Find the gradient vector for the following multiple variables functions.
(a) f (x, y) =
√
(b) f (x, y, z) = xeyz + xyez
x−y
2. Find the gradient vector for the following functions at the given point.
9
(a) f (x, y) = xe−y + 3y; (1, 0)
(b) f (x, y, z) =
x y
+ ; (4, 2, 1)
y z
Tangent Plane and Normal Line
Let S: f (x, y, z) = C be a level surface for a function f (x, y, z).
If point P (x0 , y0 , z0 ) is on S, then ∇f (x0 , y0 , z0 ) is a vector normal to S at P , i.e., ∇f (x0 , y0 , z0 ) is the normal vector. Since
the normal vector is perpendicular to any vector on the tangent
plane, then the equation of the tangent plane for f (x, y, z) = C
at P (x0 , y0 , z0 ) is
∇f (x0 , y0 , z0 )
(x, y, z)
r
b
b
P (x0 , y0 , z0 )
∇f (x0 , y0 , z0 ) • r = fx (x0 , y0 , z0 ) (x − x0 ) + fy (x0 , y0 , z0 ) (y − y0 ) + fz (x0 , y0 , z0 ) (z − z0 ) = 0,
where r is a vector lies on the tangent plane (or a line segment from P to any point (x, y, z) on
the tangent plane.)
The normal line to S at P is the line passing through P and perpendicular to the tangent plane
and S or parallel to the normal vector. The equation for the normal line is
r = k∇f (x0 , y0 , z0 ) , where k 6= 0 =⇒
x − x0
y − y0
z − z0
=
=
.
fx (x0 , y0 , z0 )
fy (x0 , y0 , z0 )
fz (x0 , y0 , z0 )
1. Find equations of (i) the tangent plane and (ii) the normal line to the given surface at the
specified point.
(a) 4x2 + y 2 + z 2 = 24, (2, 2, 2)
(b) xyz = 6, (1, 2, 3)
2. Find the point on the ellipsoid x2 + 2y 2 + 3z 2 = 1 where the tangent plane is parallel to
the plane 3x − y + 3z = 1.
3. Find the point on the hyperboloid x2 − y 2 + 2z 2 = 1 where the normal line is parallel to
the line that joins the points (3, −1, 0) and (5, 3, 6).
Directional Derivatives
For a differentiable function f (x, y), the directional derivative of f at P (x0 , y0 ) in the direction
of any unit vector u = [a, b], denoted by Du f (x0 , y0 ), is
Du f (x0 , y0 ) = ∇f (x0 , y0 ) • u = fx (x0 , y0 ) a + fy (x0 , y0 ) b.
The directional derivative is the rate of change of f at P in the direction of u.
Similarly, for g (x, y, z) at Q (x0 , y0 , z0 ) in the direction of any unit vector v = [a, b, c], the
directional derivative is
Dv g (x0 , y0 , z0 ) = gx (x0 , y0 , z0 ) a + gy (x0 , y0 , z0 ) b + gz (x0 , y0 , z0 ) c.
Since
Du f (x, y) = ∇f (x, y) • u = k∇f (x, y)k kuk cos θ = k∇f (x, y)k cos θ,
then the maximum value of Du f (x, y) is k∇f (x, y)k when θ = 0, i.e., u has the same direction
as ∇f (x, y).
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1. Find the directional derivative of the function at the given point in the direction of the
vector w.
x
(a) f (x, y) = , (6, −2), w = [−1, 3]
y
−1 y
(b) g (x, y, z) = x tan
, (1, 2, −2), w = [1, 1, −1]
zπ x
(c) f (x, y) = e cos y, 1, 6 , w = i − j
(d) g (x, y, z) = z 3 − x2 y, (1, 6, 2), w = 3i + 4j + 12k
2. Find the direction that the following functions increase most rapidly at the given point
and find the maximum rate of increase.
p
(a) f (x, y) = x2 + 2y, (4, 10)
(b) f (x, y) = cos (3x + 2y), π6 , − π8
Lagrange Multipliers
If we want to find the maximum and minimum value of function f under condition(s) or constraint(s), then we can apply the method of Lagrange multipliers.
Suppose we want to minimize the function f (x, y) under the constraint, function g (x, y) = k,
then we define the Lagrange function as the following:
L (x, y, λ) = f (x, y) − λ [g (x, y) − k] .
If (x0 , y0 ) is a minimum for the original constrained problem, then there exists a λ such that
(x0 , y0 , λ) is a stationary point or critical point for the Lagrange function. Stationary points are
those points where the partial derivative of L are zero, i.e.,
Lx = 0, Ly = 0, and Lλ = 0.
Since
and

 Lx = fx (x, y) − λgx (x, y)
Ly = fy (x, y) − λgy (x, y)

Lλ = − [g (x, y) − k]
∇L = [Lx , Ly , Lλ ] = [fx (x, y) − λgx (x, y) , fy (x, y) − λgy (x, y) , g (x, y) − k] = 0,
then


 fx (x, y) − λgx (x, y) = 0
 fx (x, y) = λgx (x, y)
fy (x, y) − λgy (x, y) = 0 =⇒ fy (x, y) = λgy (x, y) =⇒ ∇f = λ∇g with g (x, y) = k.


g (x, y) − k
=0
g (x, y) = k
The above system of equations can be extended to higher dimensions and more constraints.
Procedure for the method of Lagrange multipliers:
Suppose we want to find the maximum and minimum value of f (x, y, z) subject to the constraint
g (x, y, z) = k, then

fx (x, y, z)



fy (x, y, z)
I. Solve the system of equations
fz (x, y, z)



g (x, y, z)
=
=
=
=
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λgx (x, y, z)
λgy (x, y, z)
and find all sets of (x0 , y0 , z0 , λ).
λgz (x, y, z)
k
II. Evaluate f at all the points (x0 , y0 , z0 ) from Step I. The largest value is the maximum
value of f ; the smallest is the minimum value of f .
If there are two constraints, g and h, then we have to solve the following system of equations:

fx (x, y, z) = λgx (x, y, z) + µhx (x, y, z)




 fy (x, y, z) = λgy (x, y, z) + µhy (x, y, z)
fz (x, y, z) = λgz (x, y, z) + µhz (x, y, z) .


g (x, y, z) = k1



h (x, y, z) = k2
1. Use Lagrange multipliers to find the maximum and minimum values of the function subject
to the given constraint(s).
(a) f (x, y) = x2 − y 2 ;
(b) f (x, y) = 2x + y;
(c) f (x, y, z) = x + 2y;
(d) f (x, y, z) = yz + xy;
x2 + y 2 = 1
x2 + 4y 2 = 1
x + y + z = 1,
y 2 + z 2 = 1,
y2 + z2 = 4
xy = 1
2. Find the point on the cycle x2 + y 2 = 4 that has the shortest distance to the point (3, 1).
3. Find the shortest distance from the point (2, −2, 3) to the plane 6x + 4y − 3z = 2.
4. Find three positive numbers whose sum is 100 and whose product is a maximum.
5. A flat circular plate has the shape of a circular disc x2 + y 2 ≤ 1. The plate, including
the boundary where x2 + y 2 = 1, is heated so that the temperature at any point (x, y), is
T = x2 − x + 2y 2 . Find the hottest and coldest points on the plate and the temperature
at each of these points.
6. Let f (x, y) = 6x2 y − 3x2 − 2y 2 , find the maximum and minimum values of f on the region
D = {(x, y) : 3x2 + 2y 2 ≤ 1}.
(2008 Exam)
Implicit Differentiation
Suppose that z is given implicitly as a function z = f (x, y) by an equation of the form
F (x, y, z) = 0. This means that F (x, y, f (x, y)) = 0 for all (x, y) in the domain of f . By
the Implicit Function Theorem,
∂F
∂F
∂z
Fx
Fy
∂z
∂y
∂x
=−
= − ∂F
= − ∂F = − .
and
∂x
Fz
∂y
Fz
∂z
∂z
1. Find
∂z
∂z
and
for xyz = cos(x + y + z), where z = f (x, y).
∂x
∂y
2. Find
∂2z
∂ 2z
and
for x2 + y 2 − z 2 = 2x (y + z), where z = f (x, y).
∂x2
∂y∂x
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