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
(I.T. Leong)
=
=
∂w
∂x
∂w
∂x
∂x
∂w ∂y
∂w ∂z
+
⋅
+
⋅ ;
∂u
∂y ∂u
∂z ∂u
∂x ∂w ∂y ∂w ∂z
⋅
+
⋅
+
⋅ .
∂v
∂y ∂v
∂z ∂v
⋅
Math 200 in 2010
2010 c 9
22 F
1 / 28
Theorem. 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), (u, v), (u, v) ) is a
differentiable function in u, v and w, such that the partial functions are given by
∂S
∂u
∂S
∂v
∂S
∂w
(I.T. Leong)
=
=
=
∂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
⋅
Math 200 in 2010
2010 c 9
22 F
2 / 28
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 , one
can easily check that 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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
3 / 28
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 , one
can easily check that 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,
x +y +z
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
3 / 28
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 , one
can easily check that 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
= √
∂S
∂z
=
x +y +z
y
x2 +y2 +z2
√2 z2 2
x +y +z
(I.T. Leong)
= sin φ sin θ,
∂y
∂ρ
=
∂
∂ρ ( ρ sin φ sin θ )
= cos φ,
∂z
∂ρ
=
∂
∂ρ ( ρ cos φ )
Math 200 in 2010
= sin φ sin θ, and
= cos φ.
2010 c 9
22 F
3 / 28
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 , one
can easily check that 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
= sin φ sin θ,
x2 +y2 +z2
∂S
z
√
= cos φ,
∂z =
x2 +y2 +z2
∂S ∂x
∂S ∂y
And ∂S
∂ρ = ∂x ⋅ ∂ρ + ∂y ⋅ ∂ρ
(I.T. Leong)
+
∂S
∂z
⋅
∂y
∂ρ
=
∂
∂ρ ( ρ sin φ sin θ )
∂z
∂ρ
=
∂
∂ρ ( ρ cos φ )
= sin φ sin θ, and
= cos φ.
∂z
∂ρ
Math 200 in 2010
2010 c 9
22 F
3 / 28
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 , one
can easily check that 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,
= √
x +y +z
y
∂y
∂
= sin φ sin θ,
∂ρ = ∂ρ ( ρ sin φ sin θ ) = sin φ sin θ,
x2 +y2 +z2
∂S
∂z
∂
√ 2 z 2 2 = cos φ,
∂z =
∂ρ = ∂ρ ( ρ cos φ ) = cos φ.
x +y +z
∂S ∂x
∂S ∂y
∂S ∂z
And ∂S
∂ρ = ∂x ⋅ ∂ρ + ∂y ⋅ ∂ρ + ∂z ⋅ ∂ρ
= (sin φ cos θ )2 + (sin φ sin θ )2 + (cos φ)2
∂S
∂y
and
= (sin2 φ)(cos2 θ + sin2 θ ) + cos2 φ
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
3 / 28
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 , one
can easily check that 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,
= √
x +y +z
y
∂y
∂
= sin φ sin θ,
∂ρ = ∂ρ ( ρ sin φ sin θ ) = sin φ sin θ,
x2 +y2 +z2
∂S
∂z
∂
√ 2 z 2 2 = cos φ,
∂z =
∂ρ = ∂ρ ( ρ cos φ ) = cos φ.
x +y +z
∂S ∂x
∂S ∂y
∂S ∂z
And ∂S
∂ρ = ∂x ⋅ ∂ρ + ∂y ⋅ ∂ρ + ∂z ⋅ ∂ρ
= (sin φ cos θ )2 + (sin φ sin θ )2 + (cos φ)2
∂S
∂y
and
= (sin2 φ)(cos2 θ + sin2 θ ) + cos2 φ
= 1.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
3 / 28
Theorem. Chain Rule of 2-variables. Suppose that f (x, y)nd is real valued
function defined on the domain D which is part of R2 , and that x = x(t) and
y = y(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)) changing with respect to t, so
it traces out a path in domain D given by r(t) = x(t)i + y(t)j. Then we obtain a
real-valued function g(t) = f (x(t), y(t)). Then the chain rule tells us that
d
∂f
dx
∂f
dy
g′ (t) = ( f (x(t), y(t)) ) =
(r(t)) ⋅
+ (r(t)) ⋅
=
dt
∂x
dt
∂y
dt
fx (r(t))x′ (t) + fy (r(t))y′ (t).
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
4 / 28
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
dy
∂f
df
= ∂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).
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
5 / 28
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, b + k) − f (a, b)
f (a + h, b) − f (a, b)
, and fy (a, b) = lim
.
fx (a, b) = lim
h
k
k →0
h→0
The limits are taken along the coordinate axes.
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
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
6 / 28
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, b + k) − f (a, b)
f (a + h, b) − f (a, b)
, and fy (a, b) = lim
.
fx (a, b) = lim
h
k
k →0
h→0
The limits are taken along the coordinate axes.
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
g′ (0) = lim
t→0
(I.T. Leong)
f ( r(t) )−f ( r(0) )
t
= lim
t→0
Math 200 in 2010
f (a+th,b+tk)−f (a,b)
.
t
2010 c 9
22 F
6 / 28
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, b + k) − f (a, b)
f (a + h, b) − f (a, b)
, and fy (a, b) = lim
.
fx (a, b) = lim
h
k
k →0
h→0
The limits are taken along the coordinate axes.
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) )
t
= lim f (a+th,b+ttk)−f (a,b) .
t→0
Suppose that f is continuously differentiable, then it follows from the
(multivariate) chain rule that
∂f dy
∂f
∂f
∂f dx
+
= (P)h + (P)k
g′ (0 ) =
∂x
dt
∂y
dt
∂x
∂y
(
)
= 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.
g′ (0) = lim
t→0
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
6 / 28
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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
7 / 28
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
(−1), if and only if ∇f (P) (
−∇f (P) ) is parallel to v. In this case, we have Dv (P) = ∥∇f ∥ ( −∥∇f ∥ ).
maximum (minimum) value if and only if cos θ = 1
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
8 / 28
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 given by
3j+6k
= ⟨12i + 14j − 12k, √2i−
⟩ = − 90
7 .
22 + 32 + 62
(b) Dv f (P) is maximum(minimum) ⇐⇒ v (−v) is parallel to
∇f (P) = 12i + 14j − 12k.
∇f (P)
(c) The maximum magnitude of Dv f (P) is given by ∇f ⋅ ∥∇f (P)∥ =
√
∥∇f (P)∥ = ∥12i + 14j − 12k∥ = 144 + 196 + 144 = 22.
Dv f = ∇ f ⋅
(I.T. Leong)
v
∥v∥
Math 200 in 2010
∥∇f (P)∥2
∥∇f (P)∥
2010 c 9
22 F
=
9 / 28
Example. 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.
Prove that gradient vector ∇f of f is always perpendicular to the tangent
vector r′ (t) at r(t) for all t i.e.
∇f (r(t)) ⊥ r′ (t)
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
0 = g′ (t ) =
∂f
∂x
at r(t) for all t.
⋅
dx
dt
+
∂f
∂y
⋅
dy
dt
+
∂f
∂z
⋅
dx
dt
= ⟨∇f (r(t)), r′ (t)⟩ for all t. So ∇f ⊥ r′ (t)
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).
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
10 / 28
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.
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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
11 / 28
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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
12 / 28
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).
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
13 / 28
The most familiar example of an implicitly defined function is provided by the
186
6 Implicit Functions
equation f (x, y) = x2 + y2 . The locus or level curve f (x, y) = k is a circle of
therefore be graphs of different implicit functions. Any point of the form (a, b) =
√ it as √
radius k if(a,k +>√0,
we can view
the
graph
of equally
two different
k−
k would
serve
well as a functions,
seed for ϕ+ ;
√a2), with − √k < a <
2
2
point
y = ϕ± (xlikewise,
) = ± anyk −
x (a,
. − k − a ) would serve for ϕ− .
y
y = ϕ +(x)
y
(0, √ k)
x2 + y2 = k
x
(√ k, 0)
x
y = ϕ−(x)
(0, −√ k)
√ (±√k, 0). As√the figure on the right shows, (√k, 0)
This leaves only the points
take
either
P
(
0,
+
k) or P 0, − k) as a fixed
point of the level
+
cannot
√ be a seed for a function of x,( because the circle
√has no y-value at all when
x
>
k
and
gives
two
y-values
near
y
=
0
when
x
<
k
and
x
is arbitrarily close
curve, so that
such√that
√ ϕ± defines a function respectively
√
to k. There is a similar problem for (− k, 0). (Of√
course, ( k, 0) serves perfectly
(i) the graph
passes
through
P±we
(0,concentrate
± k), and
well as
a seed for
a functionthe
x =point
ψ (y), but
on x as the independent
variable for the moment.) In a different way, there is no seed when k = 0. Certainly
(ii) the graph
of f lies completely on the level curve, i.e. all the points
there is a point on the locus—namely (a, b) = (0, 0)—but nothing can grow out of
it,
because
entire
locus
x2 + y2f=
0 isϕjust
point.
(x, ϕ± (x)) lies onthe
the
level
curve,
(x,
)) single
= k for
all x ∈dom(f ).
± (xthis
Although there is nothing wrong with having two different parts of the locus be
the graph of two different implicit functions, we do require that only one implicit
function ϕ should be able to grow out of a given seed on that locus. This is a sigx
nificant restriction, and places yet another impediment in the way of obtaining ϕ .
(I.T.
Mathquadratic
200 in 2010 equation f (x, y) = y2 − x2 =
2010
c9
22 F
WeLeong)
can illustrate the problem with the
0. The
the locus
we can
ot seeds
14 / 28
The explicit functions ϕ± defined by means of implicit function f (x, y) = k,
satisfy
√
186 passes through the point P± (0, ± k), and
(i) the graph
6 Implicit Functions
(ii) the graph
of fbelies
completely
on
level curve,
i.e.ofallthethe
therefore
graphs
of √
different implicit
Any point
formpoints
(a, b) =
√
√ thefunctions.
+ k − a ), with − √
k < a < k would serve equally well as a seed for ϕ+ ;
(x, ϕ± (x))(a,
lies on the level curve,
f (x, ϕ± (x))
= k for all x ∈dom(f ).
likewise, any point (a, − k − a2) would serve
for ϕ .
2
−
y
y = ϕ +(x)
(0, √ k)
y
x2 + y2 = k
(√ k, 0)
x
x
y = ϕ−(x)
the locus
Thing
ot seeds
(0, −√ k)
√
√ on the right shows, (√k, 0)
This leaves only the points (± k, 0). As the figure
completely
fails
if
we
chose
the
point
P
(
k, 0has
), no
they-value
reason
that a
cannot be a seed for a function of x, because the circle
at alliswhen
√
√
x >√ takes
k and gives
two y-values
near y =
x < can
k and
x is arbitrarily
function can
on only
one value,
though
write
down close
√0 when we
√
√ to k. There√is a similar problem
√ for (− k, 0). (Of course, ( k, 0) serves perfectly
2
− xas for
k ≤a function
x ≤ xk,=but
the
can not
to any
y = + kwell
a seed for
ψ (y),
butgraph
we concentrate
on xbe
as extended
the independent
variable for the moment.) In a different way, there is no seed when k = 0. Certainly
bigger domain
to meet the second condition
(ii). Moreover,
the function
(a, b) = (0, 0)—but√
nothing can grow out of
√ there is a point on the locus—namely
2
2
2
the entire
locus xany
+ y derivative
= 0 is just thisatsingle
y = + kit,−because
x does
not have
x =point.
k, which checked directly.
Although there is nothing wrong with having two different parts of the locus be
of two different implicit Math
functions,
we do require that only one implicit
200 in 2010
2010 c 9
theLeong)
graph
(I.T.
22 F
15 / 28
Implicit Function Theorem I. Let C : f (x, y) = k be a level curve defined by a
differentiable scalar function f of 2 variables.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
16 / 28
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 P(a, b) is a point in the
domain of f such that
(I.T. Leong)
∂f
∂y (P)
∕= 0, then
Math 200 in 2010
2010 c 9
22 F
16 / 28
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 P(a, b) is a point in the
∂f
∂y (P)
∕= 0, then there exists δ > 0 and a differentiable
function g defined in an interval I = (a − δ, a + δ) such that
domain of f such that
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
16 / 28
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 P(a, b) is a point in the
∂f
∂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
domain of f such that
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
16 / 28
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 P(a, b) is a point in the
∂f
∂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.
fy (x, g(x))
domain of f such that
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
16 / 28
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 P(a, b) is a point in the
∂f
∂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.
fy (x, g(x))
domain of f such that
Remark. (i) In general, we can’t write down the explicit function g.
(ii) one can interchange the role of x and y, if
(I.T. Leong)
Math 200 in 2010
∂f
∂x (P)
∕= 0.
2010 c 9
22 F
16 / 28
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 has 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
∕= 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) = −
and
(x, y) = −
for all
∂x
Fz (x, y, g(x, y) )
∂y
Fz (x, y, g(x, y) )
(x, y) ∈ B( (a, b), δ).
level surface, i.e. F(a, b, c) = k. Suppose that
∂F
∂z (P)
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 result
∂x
∂z
∂x
∂x
follows.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
17 / 28
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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
18 / 28
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 function. Then
∂x ∂y ∂z
⋅
⋅
=
.
∂y ∂z ∂x
Solution. It follows from the implicit function theorem that
Fy
∂x
= − , for all
∂y
Fx
Fz
∂z
Fx
∂y
= − , and
= − , for all (x, y, z) in S.
(x, y, z) in S. Similarly, we have
∂z
F
∂x
(
) (y ) (
)Fz
Fy
∂x ∂y ∂z
Fz
Fx
It follows that
⋅
⋅
= −
⋅ −
⋅ −
= −1, for all (x, y, z)
∂y ∂z ∂x
Fx
Fy
Fz
in S.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
19 / 28
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 chain rule,
d
d
∂f dx
∂f dy ∂f dz
so 0 = (c) = ( f (x(t), y(t), z(t)) ) =
+
+
=
dt
dt
∂x dt
∂y dt
∂z dt
dx dy dz
∇f ⋅ ( , , ) = ∇f ⋅ r′ (t), so ∇f is normal to the tangent vector r′ (t) of the
dt dt dt
curve.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
20 / 28
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 one has
∂x
∂x
∂x ∂x
∂z ∂x
∂x
Fy (x, y, z(x, y))
Fy
∂z
Fx (x, y, z(x, y))
Fx
∂z
(x, y) = −
= − and (x, y) = −
=− .
∂x
Fz (x, y, z(x, y))
Fz
∂y
Fz (x, y, z(x, y))
Fz
One should notice that the 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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
21 / 28
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
∂z
6x
=−
,
∂x
2z + 8y − 1
4y + 8z
∂z
=−
. To locate the extremum value of z = z(x, y), one
∂y
2z + 8y − 1
need its two partial derivatives zx and zy vanish, i.e. (6x, 4y + 8z) = (0, 0)
and
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 ). So far, we only locate the 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 = − 78 .
Remark. In the last part, we skip some details, but the gap can be filled in
after we learn the second derivative test.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
22 / 28
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 lambda.
Remarks. (i) The last condition just means that these two vectors ∇f (P) and
∇g(P) are parallel.
(ii) The last equation ∇f (P) = λ∇g(P) gives a necessary condition for finding
the point P, though λ is also an unknown.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
23 / 28
Steps of implementing methods of 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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
24 / 28
Steps of implementing methods of 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.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
24 / 28
Steps of implementing methods of 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.
(I.T. Leong)
(4)
Math 200 in 2010
2010 c 9
22 F
24 / 28
Steps of implementing methods of 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.
2
(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(I.T.of
f.
Leong)
Math 200 in 2010
2010 c 9
22 F
24 / 28
Example. Use Lagrange multipliers to find the point (x, y, z) at which
x2 + y2 + z2 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
(2x, 2y, 2z) = λ(1, 2, 3), and so
λ, i.e.
1 = x + 2y + 3z =
λ
3λ
+ 2λ + 3 ×
= 7λ,
2
2
i.e. λ = 17 . 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
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
25 / 28
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
(fxy (a, b))2 − fxx (a, b)fyy (a, b) > 0 for (h, k) ∕= (0, 0).
Solution. (i) 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 .
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 B2 − AC > 0, then for s ∈ R, the quadratic function
ℓ(s) = As2 + 2Bs + C =
B2A C
A
1
A
⋅ (A2 s2 + 2AB + B2 ) + C − B2 /A =
B2 −AC
1
2
≥ A > 0. It
A (As + B) +
12
h
2
0 < ℓ( k ) = k (Ah + 2Bhk + Ck2 ),
for all (h, k) ∈ R2 with k ∕= 0. If k =
(I.T. Leong)
follows that
and hence g′′ (0) = Ah2 + 2Bhk + Ck2 > 0
0, then g′′ (0) = Ah2 > 0 for all h ∕= 0.
Math 200 in 2010
2010 c 9
22 F
26 / 28
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 fxx (a, b) > 0, then f (a, b) is a local minimum;
2
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 minimum.
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
27 / 28
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).
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 .
(I.T. Leong)
Math 200 in 2010
2010 c 9
22 F
28 / 28