.
. Chain rules
2. Directional derivative
1
. Gradient Vector Field
4. Most Rapid Increase
3
. Implicit Function Theorem, Implicit Differentiation
6. Lagrange Multiplier
5
.
. Second Derivative Test
7
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Matb 210 in 2012
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Theorem. Suppose that w = f (x, y, z) is a differentiable function,
where x = x(u, v), y = y(u, v), z = z(u, v), where the coordinate
functions are parameterized by differentiable functions. Then the
composite function w(u, v) = f ( x(u, v), (u, v), (u, v) ) is a
differentiable function in u and v, such that the partial functions are
given by
.
∂w
∂u
∂w
∂v
=
=
∂w
∂x
∂w
∂x
∂x
∂w ∂y
∂w ∂z
·
· ;
+
+
∂u
∂y ∂u
∂z ∂u
∂x ∂w ∂y ∂w ∂z
·
+
·
+
· .
∂v
∂y ∂v
∂z ∂v
·
.
Remark. The formula stated above is very important in the theory of
.surface integral.
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Matb 210 in 2012
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Theorem (Chain Rule for Coordinate Changes). Suppose that
s = f (x, y, z) is a differentiable function, where
x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), where the coordinate
functions are parameterized by differentiable functions in variables
u, v and w. Then the composite function
S(u, v, w) = f ( x(u, v, w), (u, v, w), (u, v, w) ) is a differentiable
function in u, v and w, such that the partial functions are given by
.
∂S
∂u
∂S
∂v
∂S
∂w
=
=
=
∂S
∂x
∂S
∂x
∂S
∂x
∂x
∂S ∂y
∂S ∂z
+
·
+
· ;
∂u
∂y ∂u
∂z ∂u
∂x ∂S ∂y ∂S ∂z
·
+
·
+
· ;
∂v
∂y ∂v
∂z ∂v
∂x
∂S ∂y
∂S ∂z
·
+
·
+
·
.
∂w
∂y ∂w
∂z ∂w
·
.
Remark. The formula stated above is very important in the theory of
inverse
function theory and integration theory.
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Matb 210 in 2012
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Example. In spherical coordinates, we have the parameters (ρ, θ, ϕ)
to represent (x, y, z) as follows:
x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ,
with ρ ≥ 0, √
0 ≤ θ ≤ 2π, and 0 ≤ ϕ ≤ π. Define
S(x, y, z) = x2 + y2 + z2 . Evaluate the partial derivative ∂S
∂ρ in two
ways.
.
Solution. (i) Since S(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) = ρ, so ∂S
∂ρ = 1
for any choices of parameters involved.
(ii) The second method is to apply chain rule.
∂S
∂x
∂
√ 2 x 2 2 = sin ϕ cos θ,
∂x =
∂ρ = ∂ρ ( ρ sin ϕ cos θ ) = sin ϕ cos θ,
∂S
∂y
= √
x +y +z
y
x2 +y2 +z2
= sin ϕ sin θ,
∂y
∂ρ
=
∂
∂ρ ( ρ sin ϕ sin θ )
and
∂S
√
∂z =
And
z
∂z
∂
= cos ϕ,
∂ρ = ∂ρ ( ρ cos ϕ )
x2 +y2 +z2
∂S ∂x
∂S ∂y
∂S ∂z
∂S
∂ρ = ∂x · ∂ρ + ∂y · ∂ρ + ∂z · ∂ρ
= (sin ϕ cos θ )2 + (sin ϕ sin θ )2 + (cos ϕ)2
= sin ϕ sin θ,
= cos ϕ.
= (sin2 ϕ)(cos2 θ + sin2 θ ) + cos2 ϕ = 1.
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Matb 210 in 2012
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Theorem. (Chain Rule of 2-variables)
Suppose that f (x, y)nd is a real valued function defined on the planar
domain D, and that r(t) = x(t)i + y(t)j is a curve in the domain D.
Then we obtain a real-valued function g(t) = f (x(t), y(t)) which is a
function of t. Then the derivative of g is given by
d
∂f
dx
∂f
dy
g′ (t) = ( f (x(t), y(t)) ) =
(r(t)) ·
+ (r(t)) ·
=
dt
∂x
dt
∂y
dt
f.x (r(t))x′ (t) + fy (r(t))y′ (t).
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Matb 210 in 2012
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Theorem. Chain Rule of 3-variables Suppose that f (x, y, z) is real
valued function defined on the domain D which is part of R3 , and that
x = x(t), y = y(t) and z = z(t) is a curve in the domain D.
One can think of the a particle moving in domain D, and its position is
given by (x(t), y(t), z(t)) changing with respect to t, so it traces out a
path in domain D given by r(t) = x(t)i + y(t)j + z(t)k. Then we
obtain a real-valued function g(t) = f (x(t), y(t), z(t)). Then the chain
rule tells us that
d
g′ (t) = ( f (x(t), y(t), z(t)) )
dt
∂f
∂f
∂f
dy
dz
= ∂x
(chain rule)
(r(t)) · dx
dt + ∂y (r(t)) · dt + ∂z (r(t)) · dt
′
′
′
= fx (r(t))x (t) + fy (r(t))y (t) + fz (r(t))z (t).
.
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Matb 210 in 2012
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Partial Derivatives
.
Suppose that z = f (x, y) is a function defined in a domain D, and
P(a, b) is a point in D. Recall that the partial derivatives
f (a + h, b) − f (a, b)
fx (a, b) = lim
, and
h
h→0
f (a, b + k) − f (a, b)
fy (a, b) = lim
.
k
k→0
The
limits are taken along the coordinate axes.
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Matb 210 in 2012
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Directional Derivative
.
Through the point P(a, b) we choose any direction
u = (h, k) = hi + kj, then we consider the straight line through the
point P along the direction u given by r(t) = (a + ht, b + kt), and the
rate of change of g(t) = f (r(t)) at t = 0 is
f ( r(t) )−f ( r(0) )
g′ (0) = lim
= lim f (a+th,b+ttk)−f (a,b) .
t
t
→
0
t→0
.
z
T
P(x¸, y¸, z¸)
y
3
x
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Matb 210 in 2012
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Directional Derivative
.
Through the point P(a, b) we choose any direction
u = (h, k) = hi + kj, then we consider the straight line through the
point P along the direction u given by r(t) = (a + ht, b + kt), and the
rate of change of g(t) = f (r(t)) at t = 0 is
f ( r(t) )−f ( r(0) )
g′ (0) = lim
= lim f (a+th,b+ttk)−f (a,b) .
t
t
→
0
t→0
.
.
Suppose that f is differentiable, then it follows from the (multivariate)
chain rule that
∂f dx
∂f dy
∂f
∂f
g′ (0) =
+
= (P)h + (P)k
∂y dt
∂x
∂y
(∂x dt
)
= fx (a, b)i + fy (a, b)j · (hi + kj) = ∇f (a, b) · u,
where ∇f is the vector-valued function fx i + fy j, called the gradient of
f at the point (x, y). Note g′ (0) only depends of the choice of the
′
.curve through P(a, b) with tangent direction r (0) only.
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Matb 210 in 2012
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In order to simplify the notation more, one requires the directional
vector u to be an unit vector.
.
Definition (Directional derivative) The resulting derivative g′ (0) is
called the directional derivative Du f of f along the direction u, and
hence we write Du f = ∇f · u to represent the rate of the change of f
in
. the unit direction u.
Remark. In general, if f = f (x1 , · · · , xn ) is a function of n variables,
one define
∂f
∂f
(i) the gradient of f to be ∇f = ( ∂x , · · · , ∂xn ), and
1
(ii) the directional derivative Du f by Du f = ∇f · u.
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Matb 210 in 2012
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Example. Evaluate the directional derivative
f (x, y) = xey + cos(xy) at the point P0 (2, 0) in
the
. direction 3i − 4j.
Remark. The blue curve is the level curve of f at different values.
Solution. Let u = √3i2−4i 2 = 35 i − 45 j. And ∇f (2, 0) =
3 +4
(fx (2, 0), fy (2, 0)) = (ey − y sin(xy), xey − x sin(xy))|(x,y)=(2,0) = (1, 2).
It follows that Du f (2, 0) = ∇f (2, 0) · u = 35 − 85 = −1.
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Matb 210 in 2012
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Proposition. The greatest rate of change of a scalar function f , i.e.,
the maximum directional derivative, takes place in the direction of,
and
has the magnitude of, the vector ∇f .
.
Proof. For any direction v, the directional derivative of f along the
direction v at a point P in the domain of f , is given by
Dv (P) := ⟨∇f (P), ∥vv∥ ⟩ = ∥∇f ∥ cos θ, where θ is the angle between
the vectors ∇f (P) and v. Hence Dv (P) attains maximum (minimum)
value if and only if cos θ = 1 (−1), if and only if ∇f (P) ( −∇f (P) ) is
parallel to v. In this case, we have Dv (P) = ∥∇f ∥ ( −∥∇f ∥ ).
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Matb 210 in 2012
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Example. (a) Find the directional derivative of f (x, y, z) = 2x3 y − 3y2 z
at P(1, 2, −1) in a direction v = 2i − 3j + 6k.
(b) In what direction from P is the directional derivative a maximum?
(c)
. What is the magnitude of the maximum directional derivative?
Solution.
(a) ∇f (P) = 6x2 yi + (2x3 − 6yz)j − 3y2 k|(1,2,−1) = 12i + 14j − 12k at
P. Then the directional derivative of f along the direction v is
2i−3j+6k
given by Dv f = ∇f · ∥vv∥ = ⟨12i + 14j − 12k, √ 2 2 2 ⟩ = − 90
7 .
2 +3 +6
(b) Dv f (P) is maximum(minimum) ⇐⇒ v (−v) is parallel to
∇f (P) = 12i + 14j − 12k.
(c) The maximum magnitude of Dv f (P) is given by
∇f (P)
f (P)∥
∇f · ∥∇
= ∥∇
= ∥∇f (P)∥ = ∥12i + 14j − 12k∥ =
f (P)∥
∥∇f (P)∥
√
144 + 196 + 144 = 22.
2
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Matb 210 in 2012
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Proposition. Let C : r(t) = x(t)i + y(t)j + z(t)k be a curve lying on the
level surface S : f (x, y, z) = c for some c, i.e.
c = f (r(t)) = f ( x(t), y(t), z(t) ) for all t. Then the gradient vector ∇f
of f is always perpendicular to the tangent vector r′ (t) at r(t) for all t
∇f (r(t)) ⊥ r′ (t)
.i.e.
Proof. Define the composite function g(t) = f ( x(t), y(t), z(t) ), it
follows from the given condition that g(t) = f ( x(t), y(t), z(t) ) = c is a
constant function, so one can differentiate the identity
c = g(t) = f ( x(t), y(t), z(t) ), so
∂f
∂f dy
∂f dx
′
0 = g′ (t) = ∂x · dx
dt + ∂y · dt + ∂z · dt = ⟨∇f (r(t)), r (t)⟩ for all t. So
∇f ⊥ r′ (t) at r(t) for all t.
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Matb 210 in 2012
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Proposition. Let f (x, y, z) be a differentiable function defined in R3 ,
and S : f (x, y, z) = c be a level surface for some constant c, i.e.
S = { (x, y, z) | f (x, y, z) = c }. Suppose that P(a, b, c) be a point on S
such that the gradient vector ∇f (a, b, c) of f at point P(a, b, c) is not
zero, then the equation of the tangent plane of S at P is given by
< ∇f (a, b, c), (x − a, y − b, z − c) >= 0, i.e.
fx (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − c) = 0.
.
.
Remark. For any given level surface S defined by a scalar function f ,
the tangent plane of S at any P of S is spanned by the tangent vector
of the curve contained in S. The result above tells us that the normal
direction to the tangent plane of S at any point P of S is parallel to
.∇f (P).
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Matb 210 in 2012
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Let f (x, y) be a differentiable function defined on xy-plane. For any
real number k, recall the level level of f for k is given by the set
{ (x, y) | f (x, y) = k }. When the value k changes, the level curve
changes
gradually.
.
.
Let f (x, y) = x2 − 7xy + 2y2 defined on
xy-plane. The blue curves represent the
level curves Ck : f (x, y) = k of various values
c. And the red arrows represent the gradient
vector field ∇f (a, b) = ( fx (a, b), fy (a, b) )
which is normal to the tangent vector to level
curve
Ca at P(a, b) of various values k.
.
.
Proposition. Let Ck : f (x, y) = k be a fixed level curve with a point
P(a, b) in C − k. If ∇f (a, b) ̸= (0, 0), then the equation of the tangent
line of Ck at P is given by ∇f (a, b) · (x − a, y − b) = 0, i.e.
.fx (a, b)(x − a) + fy (a, b)(y − b) = 0.
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Matb 210 in 2012
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Example. Let F(x, y, z) = xα + yα + zα , where α is non-zero number.
Determine the equation of the tangent plane of the level surface
S : F(x, y, z) = k of some point P(a, b, c) in S, where k is a positive
constant.
.
Solution. The normal direction N of the tangent plane of S at P is
parallel to ∇F(x, y, z) = α(xα−1 , yα−1 , zα−1 ) evaluated at P(a, b, c). So
N = (aα−1 , bα−1 , cα−1 ), it follows that the equation of the tangent
plane of S at P(a, b, c) is given by
0 = ⟨N, (x − a, y − b, z − c)⟩
= aα −1 ( x − a ) + bα −1 ( y − b ) + cα −1 ( z − c ) ,
So the equation of the tangent plane of S at P is given by
aα−1 x + bα−1 y + cα−1 z = aα + bα + cα = F(a, b, c) = k.
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Matb 210 in 2012
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Example. Show that the surface S : x2 − 2yz + y3 = 4 is perpendicular
to any member of the family of surfaces Sa : x2 + 1 = (2 − 4a)y2 + az2
at
. the point of intersection P(1, −1, 2).
Solution. Let the defining equations of level surfaces S, Sa be
F(x, y, z) = x2 − 2yz + y3 − 4 = 0 and
G(x, y, z) = x2 + 1 − (2 − 4a)y2 − az2 = 0. Then
∇F(x, y, z) = 2xi + (3y2 − 2z)j − 2yk, and
∇G(x, y, z) = 2xi − 2(2 − 4a)yj − 2azk.
Thus, the normals to the two surfaces at P(1, −1, 2) are given by
N1 = 2i − j + 2k, N2 = 2i + 2(2 − 4a)j − 4ak. Since
N1 · N2 = (2)(2) − 2(2 − 4a) − (2)(4a) = 0, it follows that N1 and N2
are perpendicular for all a, and so the required result follows.
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Matb 210 in 2012
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Implicit Functions
.
Given a relation between two variables expressed by an equation of
the form f (x, y) = k, we often want to “ solve for y.” That is, for each
given x in some interval, we expect to find one and only one value
y = φ(x) that satisfies the relation. The function φ is thus implicit in
the relation; geometrically, the locus of the equation f (x, y) = k is a
level curve in the (x, y)-plane that serves as the graph of the function
.y = φ(x).
.
Example. Let C : x2 + y2 = 1 be a level curve defined by a function
f (x, y) = x2 + y2 . Is it possible that this level curve C in xy-plane is
.given by the graph of some "nice" function?
2
2
2
2
If y = g(x), then
√ we have 1 = x + (g(x)) , and hence g(x) = 1 − x ,
so g(x) = ± 1 − x2 . Though we find out a possible representation of
y = g(x), which is usually called "explicit function," in fact g not
differentiable at x = ±1. On the contrary, we call y is defined implicitly
by f (x, y) = c.
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Matb 210 in 2012
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The most familiar example of an implicitly defined function is provided
by the equation f (x, y) = x2 + y2 . The locus or level curve f (x, y) = k
is a circle of radius k if k > 0, we can
√ view it as the graph of two
different functions, y = φ± (x) = ± k − x2 .
√
√
we can take either P+ (0, + k) or P( 0, − k) as a fixed point of the
level curve, so that φ± defines a function respectively
such that
√
(i) the graph passes through the point P± (0, ± k), and
(ii) the graph of f lies completely on the level curve, i.e. all the points
(x, φ± (x)) lies on the level curve, f (x, φ± (x)) = k for all x ∈dom(f ).
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Matb 210 in 2012
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The explicit functions φ± defined by means of implicit function
f (x, y) = k, satisfy
√
(i) the graph passes through the point P± (0, ± k), and
(ii) the graph of f lies completely on the level curve, i.e. all the points
.(x, φ± (x)) lies on the level curve, f (x, φ± (x)) = k for all x ∈dom(f ).
.
√
Thing completely fails if we chose the point P( k, 0), the reason is
that a function
√ can takes√on only one
√ value, though we can write
down y = + k − x2 for k ≤ x ≤ k, but the graph can not be
extended to any bigger domain
√ to meet the second condition (ii).
k − x2 does not have any derivative at
Moreover,
the
function
y
=
+
√
.x = k, which checked directly.
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Matb 210 in 2012
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Implicit Function Theorem I. Let C : f (x, y) = k be a level curve
defined by a differentiable scalar function f of 2 variables. Suppose
∂f
P(a, b) is a point in the domain of f such that ∂y (P) ̸= 0, then there
exists δ > 0 and a differentiable function g defined in an interval
I = (a − δ, a + δ) such that
(i) f (x, g(x)) = c for all x ∈ I with g(a) = b; i.e.
y = g(x) is an explicit function defined by the level curve C; and
fx (x, g(x))
(ii) g′ (x) = −
for all x ∈ I.
f
y (x, g(x))
.
Remark. (i) In general, we can’t write down the explicit function g.
∂f
(ii) one can interchange the role of x and y, if ∂x (P) ̸= 0.
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Matb 210 in 2012
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Remark. Recall that the level surface associated to a scalar function f
and a fixed number k, is the set { (x, y, z) | f (x, y, z) = k }. In general,
this set is not expected to have any nice condition. However, we have
the following important
.
Implicit Function Theorem II. Let S : F(x, y, z) = k be a level surface
defined by a differentiable scalar function F, and suppose that
P(a, b, c) is a point on the level surface, i.e. F(a, b, c) = k. Suppose that
∂F
∂z (P) ̸ = 0, then there exists δ > 0 and a differentiable function
z = g(x, y) defined on the open disc B( (a, b), δ) such that
(i) F(x, y, g(x, y)) = k for all (x, y) ∈ B( (a, b), δ), with g(a, b) = c; and
Fy (x, y, g(x, y) )
∂g
Fx (x, y, g(x, y) ) ∂g
(ii)
(x, y) = −
(x, y) = −
for all
∂x
Fz (x, y, g(x, y) ) ∂y
Fz (x, y, g(x, y) )
.(x, y) ∈ B( (a, b), δ).
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Matb 210 in 2012
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Implicit Function Theorem II. Let S : F(x, y, z) = k be a level surface
defined by a differentiable scalar function F, and suppose that
P(a, b, c) is a point on the level surface, i.e. F(a, b, c) = k. Suppose that
∂F
∂z (P) ̸ = 0, then there exists δ > 0 and a differentiable function
z = g(x, y) defined on the open disc B( (a, b), δ) such that
(i) F(x, y, g(x, y)) = k for all (x, y) ∈ B( (a, b), δ), with g(a, b) = c; and
Fy (x, y, g(x, y) )
∂g
Fx (x, y, g(x, y) ) ∂g
(ii)
(x, y) = −
(x, y) = −
for all
∂x
Fz (x, y, g(x, y) ) ∂y
Fz (x, y, g(x, y) )
.(x, y) ∈ B( (a, b), δ).
Remark. Differentiate F(x, y, g(x, y)) = k with respect to x and y
respectively by means of chain rule, we have
∂F
∂F
∂g
∂
(x, y, g(x, y)) + (x, y, g(x, y)) · (x, y) = (k) = 0, and the
∂x
∂z
∂x
∂x
result follows.
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Matb 210 in 2012
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.
Let z = z(x, y) be implicitly defined by zexz = 2z + y + 1. Find zx at the
.point (x, y, z) = (0, 0, −1).
Solution. Write z(x, y) instead of z, and then differentiate the identity
z(x, y)exz(x,y) = 2z(x, y) + y + 1
with respect to x, we have
zx exz + zexz (xzx + z) = (zexz )x = (2z + y + 1)x = 2zx ,
hence
zx · (exz + xzexz − 2) = −z2 exz .
At (x, y, z) = (0, 0, −1), we have
zx · (1 + 0 − 2) = −(−1)2 ,
i.e. zx (0, 0) = 1.
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Matb 210 in 2012
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Example. Suppose that the implicit function given by the level surface
S : F(x, y, z) = 0 defines the following explicit functions:
x = x(y, z), y = y(x, z) and z = x(x, y), where F is a differentiable
∂x ∂y ∂z
function. Then
·
·
=
.
∂y ∂z ∂x
.
Fy
∂x
=− ,
∂y
Fx
Fz
Fx
∂y
∂z
for all (x, y, z) in S. Similarly, we have
= − , and
= − , for
∂z
Fy
∂x
Fz
all (x, y, z) in S.(It follows
that
) (
) (
)
Fy
∂x ∂y ∂z
Fz
Fx
·
·
= −
· −
· −
= −1, for all (x, y, z) in S.
∂y ∂z ∂x
Fx
Fy
Fz
Solution. It follows from the implicit function theorem that
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Matb 210 in 2012
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Theorem. Let r(t) = (x(t), y(t), z(t)) be a curve on the level surface
S : f (x, y, z) = c, prove that the tangent vector r′ (t) of the curve r(t) is
normal to the gradient ∇f at the point of S. Consequently, ∇f is the
normal vector of the tangent plane of level surface S at any point
P
. (x, y, z) of S.
Proof. The result follows easily from differentiate the identity
c = f ( x(t), y(t), z(t) ) for all the t in the domain of r with the help of
d
d
chain rule, so 0 = (c) = ( f (x(t), y(t), z(t)) ) =
dt
dt
∂f dy ∂f dz
dx dy dz
∂f dx
+
+
= ∇f · ( , , ) = ∇f · r′ (t), so ∇f is
∂x dt
∂y dt
∂z dt
dt dt dt
normal to the tangent vector r′ (t) of the curve.
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Example. Determine the extremum of the function z = z(x, y) defined
implicitly
by the equation 3x2 + 2y2 + z2 + 8yz − z + 8 = 0.
.
Solution. Define F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8, so the
function z = z(x, y) is in fact the graph of the level surface S
associated to the equation F(x, y, z) = 0, or sometimes we just denote
it by S : F(x, y, z) = 0. It follows that F(x, y, z(x, y)) = 0, for all (x, y) in
the (unspecified) domain of z(x, y), in fact we just think of the equality
as an identity in x and y. So differentiate with respect to x and y
respectively by means of chain rule, we have
∂
∂
∂F ∂x ∂F ∂z
∂z
0=
(0) = ( F(x, y, z(x, y)) ) =
+
= Fx + Fz , so
∂x
∂x
∂x ∂x
∂z ∂x
∂x
∂z
Fx (x, y, z(x, y))
Fx
one has
(x, y) = −
= − and
∂x
Fz (x, y, z(x, y))
Fz
Fy (x, y, z(x, y))
Fy
∂z
(x, y) = −
= − . One should notice that the
∂y
Fz (x, y, z(x, y))
Fz
assumption Fz ̸= 0 for all (x, y) in the domain of z = z(x, y) is
necessary, which one can obtain explicitly if Fz is known.
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Example. Determine the extremum of the function z = z(x, y) defined
implicitly
by the equation 3x2 + 2y2 + z2 + 8yz − z + 8 = 0.
.
Solution. Let F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8. so
Fy
∂z
Fx
6x
∂z
4y + 8z
=− =−
, and
=− =−
. To
∂x
Fz
2z + 8y − 1
∂y
Fz
2z + 8y − 1
locate the extremum value of z = z(x, y), one need its two partial
derivatives zx and zy vanish, i.e. (6x, 4y + 8z) = (0, 0) where (x, y, z)
is a point of the level surface S : F(x, y, z) = 0. Hence, x = 0, and
y = −2z. Then 0 = F(0, −2z, z) = 2(−2z)2 + z2 + 8(−2z)z − z + 8 =
−7z2 − z + 8 = −(7z + 8)(z − 1) so z = 1 or − 78 . Hence P(0, −2, 1) or
8
Q(0, 16
7 , − 7 ) are the only critical point of the function z = z(x, y),
however, z = z(x, y) is not explicitly determined yet. One can
determine use the quadratic formula to express z in terms of x and y,
and then one can see that zmax = 1 and zmin = − 87 .
Remark. In the last part, we skip some details, but the gap can be
filled in after we learn the second derivative test.
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Theorem. (Lagrange multiplier). Let f (x, y) and
g(x, y) be functions with continuous first-order
partial derivatives. If the maximum (minimum)
value of f subject to the condition (constraint)
given by a level curve C : g(x, y) = 0 occurs at
a point P where ∇f (P) ̸= 0, then
.∇f (P) = λ∇g(P) for some real number λ.
Remarks.
1. The last condition just means that these two vectors ∇f (P) and
∇g(P) are parallel, in other words, at the point where f attains
maximum, the level curve of f will tangent to the constraint curve
2. The last equation ∇f (P) = λ ∇g(P) gives a necessary condition
for finding the point P, though λ is also an unknown:
fx (x, y) = λgx (x, y),
fy (x, y) = λgy (x, y),
g(x, y) = 0.
. The similar condition ∇f (P) = λ∇g(P) works for functions of any
variables, and the constant λ is called a multiplier.
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Example. Determine min. value of x2 + y2 subject to the constraint
xy
. = 1.
Solution. Let f (x, y) = x2 + y2 be the objective function, and
g(x, y) = xy be the constraint with the level curve given
C : g(x, y) = 1. Though C is not a bounded set, one can put more
restriction x2 + y2 ≤ M with the result curve CM which is closed and
bounded. As CM is closed and bounded in R2 , then the continuous
function f attains its minimum on CM at some point in CM . In fact, the
minimum value always occurs exactly at the same two points.
One apply the Lagrange multiplier to locate
the minimum that ∇f (x, y) = λ∇g(x, y) at
those extremum points.
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Example. Determine min. value of x2 + y2 subject to the constraint
xy
. = 1.
Solution. (One should know that it is only a necessary condition, but
not sufficient one.) Hence we have: (2x, 2y) = (λy, λx), and xy = 1.
From the last equation, one knows that x ̸= 0 and y ̸= 0, so 2x = λy,
and then λ = 2x/y. Substituting, we have 2y = (2x/y)x and hence
y2 = x2 , i.e. y = ±x. But xy = 1, so x = y = ±1 and the possible
points for the extreme values of f are (1, 1) and (−1, −1). The
minimum value is f (1, 1) = f (−1, −1) = 2.
Remark. Here there is no maximum value for f , since the constraint
xy = 1 allows x or y to become arbitrarily large, and hence
f (x, y) = x2 + y2 can be made arbitrarily large.
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Steps of implementing Lagrange multipliers
.
To find the maximum and minimum values
of f (x, y, z) subject to the constraint defined
by
. the level surface S : g(x, y, z) = k.
Suppose that these extreme values exist
and on the surface S, which is related to the
condition of S.
1. Find all values of x, y, z and λ such that



fx (x, y, z)
∇f = λ ∇g
fy (x, y, z)


f (x, y, z)
z
= λgx (x, y, z) (1)
= λgy (x, y, z) (2)
= λgz (x, y, z) (3)
and
g(x, y, z) = k.
(4)
. Evaluate f at all the points (x, y, z) that result from step (a). The
largest of these values is the maximum value of f ; the smallest is
the minimum value of f .
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Example. Use Lagrange multipliers to find the point (x, y, z) at which
2
2
2
.x + y + z is minimal subject to x + 2y + 3z = 1.
Solution. Let f (x, y, z) = x2 + y2 + z2 , and g(x, y, z) = x + 2y + 3z be
the objective function and the constraint function respectively. We
want to locate the point P(x, y) on the plane x + 2y + 3z = 1, such that
∇f = λ∇g for some λ,
i.e.
(2x, 2y, 2z) = λ(1, 2, 3), and so
1 = x + 2y + 3z =
3λ
λ
+ 2λ + 3 ×
= 7λ,
2
2
i.e. λ = 71 . And hence
(x, y, z) = ( λ2 , λ, 3λ
2 ) = (1/14, 1/7, 3/14).
Remark. Why does the point (x, y, z) = (1/14, 1/7, 3/14) give the
minimum of f ? One can consider the moving point
(x, y, z) = (3t + 1, 0, −t) lying on the plane x + 2y + 3z = 1, then
f (2t + 1, 0, −t) = (2t + 1)2 + 02 + (−t)2 = (2t + 1)2 + t2 ≥ t2 which
does not have any maximum value. However, one can prove that the
absolute minimum value of f does exist by means of Cauchy’s
inequality, and we skip the proof of this fact.
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Example. A rectangular box is placed on the xy-plane so that one
vertex at the origin, and the opposite vertex lies in the plane
Ax + By + Cz = 1, where A, B and C are positive. Find the maximum
volume
of such a box.
.
Solution. It follows from the given condition that the box has
dimension x × y × z, with x, y, z > 0 and satisfy Ax + By + Cz = 1.
Then the volume V (x, y, z) = xyz, subject to the constraint
D = { (x, y, z) | Ax + By + Cz = 1, and x, y, z ≥ 0 }, which is a closed
and bounded subset of R3 , hence the volume function V attains both
maximum and minimum. The minimum volume is obviously 0; and we
use Lagrange multiplier to find the maximum volume as
follows.(yz, xz, xy) = ∇V (x, y, z) = λ∇(Ax + By + Cz − 1) =
(λA, λB, λC). If λ = 0, then one of x, y, and z will be zero, in this case,
V (x, y, z) = 0, which is not maximum.
λ ̸= 0 so
√ Assume
√
xy·xz
λC·λB
BC
BC
2
x = yz = λA = λ A , i.e. x =
λ. Similarly, we have
A
√ √
√ √
λ, and z = AB
λ. At last, we have 1 = Ax + By + Cz =
y = AC
C
( √ B
√
√ )√
√
√
AC
AB
1
λ = 3 ABC λ, so λ = 9ABC
. Then
A BC
A +B
B +C
C
.
1
1
1
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Example. Let r(t) = (a + ht, b + kt) be the line in xy-plane passing
through the point P(a, b). Let f be a function defined in a domain D
containing P with continuous second order partial derivatives, and
that P is a critical point of f i.e. ∇f (P) = 0. Let g(t) = f (r(t)),
(i) evaluate the second derivatives of g at t = 0; and (ii) the sign of
g′′ (0) provided that fxx (a, b) > 0 and fxx (a, b)fyy (a, b) − (fxy (a, b))2 > 0
.for (h, k) ̸= (0, 0).
Solution. (i) Let A = fxx (a, b), C = fyy (a, b), B = (fxy (a, b). It follows
from chain rule that g′ (t) = fx (a + ht, b + kt)h + fy (a + ht, b + kt)k, and
hence g′′ (t) = fxx (a + ht, b + kt)h2 + fxy (a + ht, b + kt)hk + fyx (a +
ht, b + kt)kh + fyy (a + ht, b + kt)k2 . In particular, at t = 0,
g′′ (0) = fxx (a, b)h2 + 2fxy (a, b)hk + fyy (a, b)k2 = Ah2 + 2Bhk + Ck2 .
(ii) As A = fxx (a, b) > 0, and AC − B2 > 0, then for s ∈ R, then
ℓ(s) = As2 + 2Bs + C = A1 · (A2 s2 + 2ABs + B2 ) + C − B2 /A =
2
1
AC−B2
2
≥ ACA−B > 0. So , and hence
A (As + B) +
A
g′′ (0) = Ah2 + 2Bhk + Ck2 = k2 · (A(h/k)2 + 2Bh/k + C)
for all (h, k) ∈
h ̸= 0.
R2
with k ̸= 0. If k = 0, then
g′′ (0)
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Matb 210 in 2012
=
Ah2
.
= k2 ℓ( hk ) > 0
> 0 for all
.
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Proposition. (Maximum-Minimum Test for Quadratic Functions) Let
g(x, y) = Ax2 + 2Bxy + Cy2 , where A, B, C are constants.
1. If AC − B2 > 0, and A > 0, [respectively A < 0], then g(x, y) has
.
a minimum [respectively maximum] at (0, 0).
.2 If AC − B2 < 0, then g(x, y) takes both positive and negative
values near (0, 0), so (0, 0) is not a local extremum for g.
Proof. To prove these assertions, we consider the two cases
separately.
(1)(If AC − B2 > 0, then
cannot be zero (why?), so g(x,)y) =
) A(
2
2
C 2
B 2C 2
B 2
A x2 + 2B
= A x2 + 2B
A xy + A y
A xy + A2 y A y − A2 y
(
)2
= A x + AB y + A1 (AC − B2 )y2 . Both terms above have the same
B
sign as A, and they are both zero only when x + A
y = 0 and y = 0,
i.e. (x, y) = (0, 0). Thus (0, 0) is a minimum point for g if A > 0 (since
g(x, y) > 0 if (x, y) ̸= (0, 0)) and a maximum point if A < 0 (since
g(x, y) < 0 if (x, y) ̸= (0, 0) ). This completes the proof of (1)
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Proposition. (Maximum-Minimum Test for Quadratic Functions) Let
g(x, y) = Ax2 + 2Bxy + Cy2 , where A, B, C are constants.
1. If AC − B2 > 0, and A > 0, [respectively A < 0], then g(x, y) has
.
a minimum [respectively maximum] at (0, 0).
.2 If AC − B2 < 0, then g(x, y) takes both positive and negative
values near (0, 0), so (0, 0) is not a local extremum for g.
Proof. (2). If AC − B2 < 0 and A ̸= 0, then formula (1) still applies, but
now the terms on the right-hand side have opposite signs. By suitable
choices of x and y (try it!), we can make either term zero and the
other nonzero. If A = 0, then g(x, y) = y(2Bx + Cy), so we can again
achieve both signs.
Remark. In case (2), (0, 0) is called a saddle point for g(x, y).
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Theorem. (Second Derivatives Test) Suppose the second partial
derivatives of f (x, y) are continuous on a disk with center (a, b), and
suppose that ∇f (a, b) = (0, 0) i.e. (a, b) is a critical point of f . Let
D = D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]2
1. If D > 0 and f
xx (a, b) > 0, then f (a, b) is a local minimum;
. If D > 0 and fxx (a, b) < 0, then f (a, b) is a local maximum;
3. If D < 0, then f (a, b) is neither a local maximum nor a local
2
.
minimum.
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Example. Determine the nature of the critical points of
f. (x, y) = x3 + y3 − 6xy.
Solution. As ∇f (x, y) = (3x2 − 6y, 3y2 − 6x), so x2 = 2y and y2 = 2x.
It follows that x4 = 4y2 = 4 × 2x = 8x, i.e.
0 = x4 − 8x = x(x3 − 23 ) = x(x − 2)(x2 + 2x + 4).
As (x2 + 2x + 4) = (x + 1)2 + 3 > 0, we have x = 0 or x = 2.
So 2y = 02 or 22 , so the critical points of f are (0, 0) and (2, 2).
Next we need to apply the 2nd derivative test.And
fxx = 6x, fyy = 6y, fxy = −6, and the discriminant
2 = (6x)(6y) − (−6)2 = 36(xy − 1).
∆(x, y) = fxx fyy − fxy
And ∆(0, 0) = −36 < 0, ∆(2, 2) = 36(4 − 1) = 108.
Hence (0, 0) is a saddle point of f , where (2, 2) is a local minimum
point of f .
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Taylor’s Formula for f (x, y) at the Point (a, b)
.
Theorem. Suppose f (x, y) and its partial derivatives through order
n + 1 are continuous throughout an open rectangular region R
centered at a point (a, b). Then, throughout R,
f (a + h, b + k) = f (a, b) + (hfx + kfy )|(a,b) + · · ·
|
{z
}
Linear or 1st order approximation
f (a + h, b + k)
1
= f (a, b) + (hfx + kfy )|(a,b) + (h2 fxx + 2hkfxy + k2 fyy )|(a,b) + · · ·
2!
|
{z
}
2nd order approximation
f (a + h, b + k)
= f (a, b) + (hfx + kfy )|(a,b) + 2!1 (h2 fxx + 2hkfxy + k2 fyy )|(a,b)
+ 3!1 (h3 fxxx + 3h2 kfxxy + 2hk2 fxyy + k3 fyyy )|(a,b) + · · ·
)n
)n+1
(
(
∂
∂
∂
∂
+ k ∂y
f |(a,b) + (n+1 1)! h ∂x
+ k ∂y
f |(a+ch,b+ck)
+ n!1 h ∂x
for
. some c ∈ (0, 1).
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Taylor’s Theorem. Suppose f (x, y) and its partial derivatives through
order n + 1 are continuous throughout an open rectangular region R
centered at a point (a, b). Then, throughout R,
= f (a, b) + (hfx + kfy )|(a,b) + 2!1 (h2 fxx + 2hkfxy + k2 fyy )|(a,b)
+ 3!1 (h3 fxxx + 3h2 kfxxy + 2hk2 fxyy + k3 fyyy )|(a,b) + · · ·
(
)n
(
)n+1
∂
∂
∂
∂
+ n!1 h ∂x
+ k ∂y
f |(a,b) + (n+1 1)! h ∂x
+ k ∂y
f |(a+ch,b+ck)
for
. some c ∈ (0, 1).
.
Remarks.
1. The proof just applies the chain rule and the trick of n-th Taylor
polynomial to the function g(t) = f (a + ht, b + kt) one variable.
.2 If one have an estimate the last term (in blue), for example an
upper bound, then we can estimate the given function by means
of polynomials in 2 variables.
3. The theorem can be easily generalized to function of n variables
for n ≥ 1. Though this topics is not treated in this book, but its
application is important in other courses, so we put the result in
this notes for the sake of students.
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