Functions on Surfaces
Outline
1. Smooth Functions
Let S be a smooth surface, and let f : S → Rn be a real-valued function on S
Definition. We say that f is smooth if the composition f ◦ σ is smooth for every smooth
surface patch σ : U → S.
This definition involves every possible surface patch on a surface. However, it suffices to check
the surface patches in an atlas:
Theorem. Let {σ1 , . . . , σn } be an atlas of regular surface patches for S. If each of the
compositions f ◦ σi is smooth, then f is a smooth function.
2. Restrictions and Extensions
Let F : V → Rn be a function defined on an open subset V of R3 , and let S be a surface
contained in V . The restriction of F to S is the function f : S → Rn defined by
f (p) = F(p)
for all p ∈ S. If F is a smooth function, then so is the restriction f .
In the other direction, if f : S → Rn is any smooth function on a surface S and V is an
open subset of R3 containsing S, an extension of f to V is a smooth function F : V → Rn
whose restriction to S is f . Though it is far from obvious, any smooth function on a surface
can be extended:
Theorem. If S is a smooth surface and f : S → Rn is any smooth function, then f can be
extended to a smooth function F : R3 → Rn defined on all of R3 .
This is a basic theorem about smooth functions, but the book ignores it because the proof is
too difficult. However, extensions are very useful both theoretically and computationally, and
it makes sense to be aware of this theorem even if you don’t know how to prove it.
3. Derivatives
As we have seen, the derivative of a function F : Rm → Rn is an n × m matrix Dp F. Equivalently, we can think of Dp F as a linear transformation from the vector space Rm to the
vector space Rn . From this point of view, the “derivative matrix” is the matrix for this linear
transformation with respect to the standard bases.
If f : S → Rn is a smooth function defined on a surface, then the derivative of f at a
point p is again a linear transformation. However, the domain of this linear transformation
is the tangent space Tp S to the surface at the point p. Since the subspace Tp S does not have
a standard basis, we cannot think of this derivative as simply a matrix.
We will use the following definition, which is equivalent to the one in the book:
Definition. Let S be a smooth surface, let f : S → Rn be a smooth function, and let F be
an extension of f to an open subset of R3 . The derivative of f at a point p ∈ S is the
linear transformation Dp f : Tp S → Rn obtained by restricting the linear transformation
Dp F : R3 → Rn to the tangent space Tp S.
Though this definition of Dp f may seem to depend on the extension F of f , any smooth
extension of f gives the same result.
Given a basis {t1 , t2 } for the tangent space Tp S, the linear transformation Dp f can be
expressed as an n × 2 matrix with respect to this basis. Specifically, the columns of the matrix
for Dp f are the partial derivatives Dp F(t1 ) and Dp F(t2 ).
If σ : U → S is a surface patch for S, then σ gives us a basis {σu , σv } for the tangent
space at each point. In this case, the columns of the matrix for Df are the partial derivatives
(f ◦ σ)u and (f ◦ σ)v . That is, Df is the matrix


∂y1 ∂y1
 ∂u ∂v 


.. 
 ..
. 
 .


 ∂yn ∂yn 
∂u
∂v
where (y1 , . . . , yn ) = f σ(u, v) .
4. The Gradient
Given a surface S and a smooth function f : S → R, the gradient of f at a point p is the
tangent vector ∇f ∈ Tp S such that
(∇f ) · t = Dp f (t)
for all t ∈ Tp S. That is, the dot product of the gradient ∇f with a tangent vector t gives
the directional derivative of f in the direction of t. As with the gradient on the plane, ∇f is
perpendicular to the level curves for f on the surface S.
Given an extension F of f to an open set in R3 , the gradient of f is the projection of the
gradient of F onto the tangent space. That is,
∇f = ∇F − (n · ∇F ) n,
where n is a unit normal vector to the surface. In particular, the critical points for f are the
same as the points on the surface at which ∇F is parallel to the normal vector.
This leads to the method of Lagrange multipliers for finding critical points for a realvalued function f on a surface. Here is an outline of this method:
(i) Find an extension F of f to some open set in R3 , and compute the gradient ∇F .
(ii) Find a normal vector N at each point on the surface.
(iii) Find all points on the surface which satisfy the equation ∇F = λN, where λ is an
unknown scalar.
Step (iii) involves solving a system of four equations in four unknowns. Specifically, the
unknowns are x, y, z, and λ, and the system consists of three equations obtained from the
vector equation ∇F = λN, together with one equation that defines the surface.
5. Maps Between Surfaces
If S and Se are smooth surfaces, a function f : S → Se is called smooth if it is smooth when
viewed as a function from S to R3 . The derivative of such a function at a point p ∈ S is a
linear transformation between tangent spaces:
e
Dp f : Tp S → Tpe S.
e then the derivative
e = f (p). If we are given bases for the tangent spaces Tp S and Tpe S,
where p
Dp f can be expressed as a 2 × 2 matrix.
e → Se a surface patch for S,
e
Now let σ : U → S be a surface patch for S, and let σ̃ : U
e Then the columns of the
giving us a basis {σu , σv } for Tp S and a basis {σ̃ũ , σ̃ṽ } for Tpe S.
2 × 2 matrix for Dp f with respect to these bases are the partial derivatives (σ̃−1 ◦ f ◦ σ)u and
(σ̃−1 ◦ f ◦ σ)v of the composition σ̃−1 ◦ f ◦ σ. That is, the matrix for Df is
 ∂e
u
 ∂u


∂e
v
∂u
where (e
u, ve) = (σ̃−1 ◦ f ◦ σ)(u, v).
∂e
u
∂v 
,

∂e
v
∂v
Practice Problems
1. Let S be the cylinder x2 + y 2 = 1 in R3 , and let f : S → R2 be the function
f (x, y, z) = (xz, yz).
For each point (x, y, z) ∈ S, let t1 , t2 be the tangent vectors t1 = (−y, x, 0) and t2 = (0, 0, 1).
(a) What does the image of S look like under the function f ?
(b) Find the 2 × 2 matrix for Df with respect to the basis {t1 , t2 }.
(c) Which points on the surface S are critical points for f ?
2. Let S 2 be the unit sphere in R3 , and let σ : (0, 2π) × (0, π) → S 2 be the surface patch
σ(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ).
Define a function f : S 2 → R by the formula f (x, y, z) = y.
(a) Find the formula for the composition f ◦ σ : (0, 2π) × (0, π) → R.
(b) Compute the 1 × 2 matrix for Df with respect to the basis {σθ , σφ }.
(c) Find all critical points for f that lie in the image of σ.
3. Let S be the paraboloid z = x2 + y 2 , and let p be the point (1, 0, 1).
(a) Find a unit normal vector to S at the point p.
(b) Let F : R3 → R be the function F (x, y, z) = x2 + y 2 . Compute the gradient vector for F
at the point p.
(c) Now let f be the restriction of F to the surface S. Compute the gradient vector ∇f ∈ Tp S
for f at the point p.
4. Let S be the plane x+y +z = 6, and let f : S → R be the function f (x, y, z) = 2x2 +3y 2 +6z 2 .
Use the method of Lagrange multipliers to find the critical point for f on S.
5. Let S be the unit sphere in R3 with the points (0, 0, 1) and (0, 0, −1) removed, and let C be
the cylinder x2 + y 2 = 1 in R3 . Define a map f : S → C by
y
x
f (x, y, z) = √
,√
,z .
1 − z2
1 − z2
Let σ : (0, 2π) × (0, π) → S be the surface patch σ(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ), and
let σ̃ : (0, 2π) × R → C be the surface patch σ̃(u, v) = (cos u, sin u, v)
(a) Find a simple formula for the composition σ̃−1 ◦ f ◦ σ.
(b) Compute the 2 × 2 matrix for Df with respect the bases {σθ , σφ } and {σ̃u , σ̃v }.
Solutions
1. (a) The image is the double cone x2 + y 2 = z 2 .
(b) Let F : R3 → R2 be the extension of f defined by F(x, y, z) = (xz, yz). Then
z 0 x
DF =
.
0 z y
In particular,
 
−y
−yz
z 0 x  
x =
DF(t1 ) =
xz
0 z y
0
 
0
x
z 0 x  
0 =
and DF(t1 ) =
y
0 z y
1
so
−yz x
Df =
.
xz y
(c) Using the basis t1 , t2 , we have
det(Df ) = (−yz)(y) − (xz)(x) = −y 2 z − x2 z = −z.
Thus, the critical points occur when z = 0 , which corresponds to a circle of critical
points on the cylinder S.
2. (a) We have (f ◦ σ)(θ, φ) = sin θ sin φ .
(b) Using the basis σθ , σφ , the matrix for Df is the same as the derivative matrix for f ◦ σ.
Thus Df = cos θ sin φ sin θ cos φ .
(c) Since sin φ 6= 0, the critical points will occur when cos θ = 0 and cos φ = 0, which gives
(θ, φ) = (π/2, π/2) and (θ, φ) = (3π/2, π/2). This corresponds to the points (0, 1, 0)
and (0, −1, 0) on the unit sphere.
3. (a) The paraboloid is a level surface for the function g(x, y, z) = x2 + y 2 − z. Since the
gradient of g at the point (1, 0, 1) is ∇g = (2x, 2y, −1) = (2, 0, −1), the unit normal is
1
n = √ (2, 0, −1) (or its negative).
5
(b) ∇F = (2x, 2y, 0) = (2, 0, 0) .
4
(c) ∇f = ∇F − (n · ∇F )n = (2, 0, 0) − √ n =
5
2
4
, 0,
5
5
.
4. Let F : R3 → R be the function F (x, y, z) = 2x2 + 3y 2 + 6z 2 . Then
∇F (x, y, z) = (4x, 6y, 12z).
Let N = (1, 1, 1). Then N is normal to each point on the surface S. We wish to find solutions
to the equation ∇F = λN that lie on the plane. This gives us the equations:
(4x, 6y, 12z) = λ(1, 1, 1)
and
x + y + z = 6,
which can be written
4x = λ,
6y = λ,
12z = λ,
x + y + z = 6.
Solving for x, y, and z in terms of λ and plugging this into the fourth equation gives λ = 12,
and therefore (x, y, z) = (3, 2, 1) is the only critical point.
Note: See Stewart’s Calculus, Concepts & Contexts, section 11.8 for further discussion and
examples of Lagrange multipliers.
5. (a) (f ◦ σ)(θ, φ) =
!
cos θ sin φ
sin θ sin φ
p
,p
, cos φ = (cos θ, sin θ, cos φ), and therefore
1 − cos2 φ
1 − cos2 φ
σ̃−1 ◦ f ◦ σ)(θ, φ) = (θ, cos φ)
(b) Df =
1
0
0 −sin φ