Chapter 3: Surfaces in R^3 Flashcards
DEF 3.1 open subset
A subset U of R^n is open if, whenever P is a point in U,
there is a positive number ε such that every point Q ∈ R^n within a distance ε of P is also in U
- Q∈ R^n and ||PQ|| < ε implies Q∈U ( for p∈U)
- R^n and ∅ are trivial e.g any ε , e.g. no points in the set so no points to worry about
open subset examples
- R^n and ∅ are trivial e.g any ε , e.g. no points in the set so no points to worry about
- Products of open intervals : ]α_1,β_1 [ x…x]α_n, β_n[ for α_i less than β_i for i=1,…,n
eg n=1 [α, β] on real line
eg n=2 rectangle not including boundary ( can always find epsilon for any point)
eg n=3 cuboid without boundary
- The open ball eg B_R(P) = {Q∈R^n : ||PQ|| less than R} for P∈ R^n and R bigger than 0
disc without boundary
EXAMPLE: [α, β] is it open
Take [α, β] ⊂ R.
{x∈R: α ≤x≤ β}. If we take p=β then for any ε>0 we will have points not within the set.
e.g. take Q= p + 0.5ε. Then |QP| = 0.5ε less than ε. But Q= β. So Q not in [α, β]. This means [α, β] isn’t open
DEF: continuous at P in X
F is continuous at P ∈ X if, given any number ε bigger than 0, there is a number δ bigger than 0 such that
Q ∈ X and ||PQ|| < δ ⇒
||F(P)F(Q)|| < ε.
DEF continuous map F
The map F is continuous if it is continuous at every point in X
For subsets X ⊂R^m , Y⊂ R^n map F:X → Y is continuous at p ∈X if points near p are mapped by F to points in Y near F(P)
differentiability and continuity
all differentiable functions defined on an open set are continuous
DEF: 3.2.1
The partial derivative of a vector-valued function
σ : R^2 → R^3
σ(u, v) = (x(u, v), y(u, v), z(u, v))
Consider a vector-valued function σ : R^2 → R^3
σ(u, v) = (x(u, v), y(u, v), z(u, v))
The partial derivative of σ(u, v) with respect to u/ u-derivative,
is
∂σ/∂u= (∂x/∂u, ∂y/∂u, ∂z/∂u)
or σ_u(u, v) = (x_u(u, v), y_u(u, v), z_u(u, v))
which is another vector-valued function of the same form.
similarly wrt v
component by component
2nd order partial derivatives of a vector-valued function
σ : R^2 → R^3
σ(u, v) = (x(u, v), y(u, v), z(u, v))
∂^2σ/∂u^2,
∂^2σ/∂u∂v,
∂^2σ/∂v∂u,
∂^2σ/∂v^2,
or σuu, σuv, σvu, σvv.
usually functs..
In practice, we will always deal with functions σ : U → R^3
(where U ∈ R^2
is open) that are differentiable in both variables infinitely many times. In this case, σuv = σvu.
DEF: 3.2.3 dot product
Given two vector-valued functions σ_1, σ_2 : R^2 → R^3
σ_1(u, v) = (x_1(u, v), y_1(u, v), z_1(u, v))
σ_2(u, v) = (x_2(u, v), y_2(u, v), z_2(u, v))
their dot product σ_1 · σ_2 is the scalar-valued function (σ_1 · σ_2)(u, v) = x_1(u, v)x_2(u, v) + y_1(u, v)y_2(u, v) + z_1(u, v)z_2(u, v)
in R^3 dot product encodes lengths of vectors and angles for tangent vectors of surfaces.
PROP 3.2.4
a product rule
The partial derivatives of the dot product satisfy a product
rule:
(σ1 · σ2)_u = (σ1)_u · σ2 + σ1 · (σ2)_u;
(σ1 · σ2)_v = (σ1)v · σ2 + σ1 · (σ2)_v.
DEF 3.2.5: cross product
the cross product of σ_1(u, v) and σ_2(u, v) is the vector-valued function σ_1 × σ_2 given by (σ1 × σ2)(u, v) = (y1(u, v)z2(u, v) − z1(u, v)y2(u, v), z1(u, v)x2(u, v) − x1(u, v)z2(u, v), x1(u, v)y2(u, v) − y1(u, v)x2(u, v))
= det {
[ i j k ]
[x_1 y_1 z_1]
[x_2 y_2 z_2]
}
Example:
u derivative of dot product of σ(u, v) · σ(u, v)
Given a vector-valued function σ(u, v), the u-derivative of
σ(u, v) · σ(u, v) is
(σ · σ)u = σu · σ + σ · σu
= 2 σ_u · σ
and similarly its v-derivative is 2 σv · σ.
||σ|| constant
||σ|| constant⇒
σ_u(u, v) ⊥ σ(u, v),
σ_v(u, v) ⊥ σ(u, v)
for all (u, v) ∈ R^2.
In words, if a vector-valued function has constant magnitude, it is always orthogonal to its partial derivatives.
Definition 3.3.1. A parametrized surface in R^3
A parametrized surface in R^3
consists of an open set U ⊆
R^2 and a continuous map
σ : U → R^3
U is some open set on R^2 such as a product of open intervals
DEF:
surface
parametrization of S
The image S = σ(U) is called a surface and σ is called a parametrization
of S.
applying σ to open set gives some surface, image in R^3
EXAMPLE: For the plane P in R^3 that contains the point (1, 0, 0) and is parallel to the vectors (−1, 1, 0) and (−1, 0, 1) a parametrization is
a parametrization is
σ_1 : R^2 → R^3
(u, v) → (1 − u − v, u, v).
image is {(x,y,z) : x+y+z = 1}
EXAMPLE: For the cylinder C := {(x, y, z) | x^2 + y^2 = 1}
a parametrization is
a parametrization is
σ_2 : ]−10, 10[ × R → R^3
(u, v) → (cos(u), sin(u), v).
centred at origin, surface of revolution of a curve parametrized by γ: R → R^3, γ(t)= (1,0,t)
or by (θ,t) → (cos(θ), sin(θ), t)
EXAMPLE: For the paraboloid {(x, y, z) | z = x^2 + y^2} a parametrization is
a parametrization is
σ_3 : R^2 → R^3
(u, v) → (u, v, u^2 + v^2)
2d to 3d cup
EXAMPLE: For a function f : R^2 → R, the graph of f is the set Γ_f := {(x, y, z) | z − f(x, y) = 0}; a parametrization is
a parametrization is
σ_4 : R^2 → R^3
(u, v) → (u, v, f(u, v)).
EXAMPLE:
Find a parametrization of the plane P through the three
points (0, 0, 1), (0, 2, 3) and (1, 0, 0).
Any point on plane reached to from c by linear combo of vectors CA and CB
ie if x ∈ P then x=(1,0,0) + u(-1,0,1) +v(-1,2,3) for u,v ∈ R. So we can parametrize p by σ:R^2 → R^3 σ(u,v) =(1-u-v, 2v, u+3v)
parametrized surface of revolution
Let γ: ]α, β[ → R^3, t → (x(t), 0, z(t)) be a regular parametrization of a curve C in the xz-plane. The parametrized surface of revolution of C is
σ : R^2 → R^3, (θ, t) →(x(t) cos θ, x(t) sin θ, z(t))
curve is in plane x-z and revolves about z axis
LATITUDES
The circles of constant t are called latitudes
LONGITUDES
the lines of constant θ are
called longitude
EXAMPLE: SPHERICAL COORDS
The sphere with the north and south poles removed
parametrized surface of revolution
The sphere with the north and south poles removed is the surface of revolution of the (open) semicircle
parametrize semicircle by
γ: ]−π/2, π/2[→ R^3
φ → (cos φ, 0, sin φ).
The resulting parametrized surface of revolution is
σ : ]−π/2, π/2[ × R → R^3
σ(φ, θ) = (cos φ cos θ, cos φ sin θ, sin φ)
*open so not including end points to stop being irregular
parametrized curve regular
parametrized curve regular if tangent vector is non zero everywhere
DEF 3.4.1:
A parametrized surface σ : U → R^3 is regular if…
A parametrized surface σ : U → R^3 is regular if for each
(u, v) ∈ U, the vectors σu(u, v) and σv(u, v) are linearly independent,
in
other words, if σ_u(u, v) and σ_v(u, v) span a plane in R^3
EXAMPLE:
Consider the parametrized surface σ : R^2 → R^3,
σ(u, v) = (u + v,(u − v)^2 , (u − v)^3) for u, v ∈ R.
Is this surface smooth? Is it regular? Identify the parameters (u, v) ∈ R2
such that σ(u, v) is on the “sharp edge
diagram duck mouth
- σ is smooth because each of the coordinate functions are polynomials in u and v so are infinitely differentiable in u and v
- σ_u(u,v) = (1, 2(u-v), 3(u-v)^2) ≠0 for all u, v∈R
- σ_v(u,v) = (1, -2(u-v), -3(u-v)^2) ≠0 for all u, v∈R
- These are linearly dependent if σ_u= λ*σ_u for some λ∈R
first components implies λ=1 implies 2(u-v) = -2(u-v) ie u-v =0 so u=v.
indeed σ_u = σ_v when u=v and this is the only place where linearly dependent. This is where the surface σ will not be regular. σ(u,w) = (2u,0,0) for u∈R ie along x axis
DEF 3.4.3: tangent plane
Let σ : U → R^3 be a regular parametrized surface and
suppose (u_0, v_0) ∈ U.
(a) The tangent plane to σ at (u_0, v_0) is the plane through the point P = σ(u_0, v_0) and with directions the vectors σ_u(u_0, v_0) and σ_v(u_0, v_0).
* tangent space is like tangent plane but translated to origin
DEF 3.4.3: tangent vectors
tangent space
2D vector subspace of R^3 , plane through o
(b) We say that σ_u(u_0, v_0) is a tangent vector in the u-direction. Similarly,
σ_v(u_0, v_0) is a tangent vector in the v-direction. More generally,vectors of the form
aσ_u(u_0, v_0) + bσ_v(u_0, v_0) for a, b ∈ R
are called the tangent vectors of σ at the point P = σ(u_0, v_0).
We call the vector space spanned by σ_u(u0, v0) and σ_v(u0, v0) the tangent
space of σ at (u0, v0), and denote it
T_(u0,v0)σ (or TPS if that’s unambiguous).
*the tangent plane to σ at (u0, v0) is a 2-dimensional plane through P = σ(u0, v0), and that the tangent space T(u0,v0)σ is a 2-dimensional vector
subspace of R^3 (or a plane through the origin).
*σ_u and σ_v linearly indep implies regular
EXAMPLE:
Show that for the parametrization σ of a part of S^2 in Example 3.3.6 (w\o poles)
T(u,v)σ = the plane orthogonal to the vector σ(u, v)
σ(u, v) = (cos u cos v, cos u sin v, sin u)
TS spanned by σ_u and σ_v. To show its orthogonal to σ show its orthogonal to σ_u and to σ_v.
σ_u(u,v) = (-sinucosv, -sinu sinv, cosu)
showing orthogonal by σ_u dot σ = 0
σ_v(u,v) = (-sinvcosu, cosvcosu,0)
showing orthogonal by σ_vdot σ = 0
a parametrized curve in σ
If σ : U → R^3
is a parametrized surface for U ⊂ R^2 an open set, then a parametrized curve in σ is a parametrized curve σ ◦γ: ]α, β[ → R^3, for some α, β ∈ R, where γ: ]α, β[ → U is a parametrized
curve in R^2
LEMMA: tangent vector of parameterizations
A tangent vector of a parametrized surface σ : U → R^3 at a point (u, v) ∈ U is the same as a tangent vector of some parametrized curve
σ ◦ γ: ]α, β[ → R^3 on σ at t where γ(t) = (u, v).
LEMMA: tangent vector of parameterizations proof
Suppose we have a parametrized curve γ~ = σ ◦ γ we want to show that (γ~^dot) is in the tangent space T_(u(t),v(t)) σ ie (γ~^dot)(t) = aσ_u + bσ_v for some a,b in R.
(γ~^dot)(t) = (σ ◦ γ)’ (t)
= (dσ/du) (du/dt) + (dσ/dv)(dv/dt) where γ(t) = (u(t), v(t))
So (γ~^dot)(t) = u’(t)σ_u + v’(t)σ_v which is of the form we wanted: linear combination of σ_u and σ_v
CONVERSELY:
suppose that we have
X = aσ_u(u_0,v_0)+bσ_v(u_0,v_0)
∈T_(u_0,v_0) σ.
we need to find some parametrized curve in σ that this is the tangent vector of
interval on real line [-ε,ε] mapped to straight line through (u_0,v_0) with slope a/b inside open rectangle in R^2
because U is open and (u_0,v_0) in U then there exists ε bigger than 0 such that
γ:[-ε,ε]→ U, t→ (u_0,v_0) + t(a,b). we have γ(0)= (u_0,v_0)
γ’(0)= (a,b )=(u’(0),v’(0)).
Take γ~ = σ ◦ γ then (γ~^dot)(0)= aσ_u(u_0,v_0)+bσ_v(u_0,v_0)=X
Tangent plane construct
Collect all curves on σ passing through (u,v) and take their tangent vectors at (u,v)
EXAMPLE: Let γ: ]α, β[ → R^3,
t→(x(t), 0, z(t)) be a smooth and regular parametrized curve and let σ : ]α, β[ × R → R^3, σ(t, θ) = (x(t) cos θ, x(t) sin θ, z(t)) be the surface of revolution obtained by rotating the image of γ around the z-axis (see Example 3.3.4).
Show that σ is smooth. Show that σ is regular if and only if x(t) ≠ 0 for all t.
- σ is smooth as components x(t)cosϴ, x(t)sint, z(t) are infinitely differentiable
- σ is regular if and only if σ(t,θ)≠0 for all (t,θ) in ]α,β[ x R
x(t)cosϴ ≠ 0 , x(t) is regular hence cosϴ ≠0 implies ϴ≠ (pi/2) + 2npi and sinϴ≠0
never both 0
