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1. Problems
(1) Consider the surface S paramaterized by
φ : R2 −→ R3 , φ(x, y) = (x + y, 2x − y, ex
2 −y
+ x3 ).
Compute the unit normal vector to S at every point φ(x, y).
(2) (Exercises 2 of 2.2 of Do Carmo) Is the set S = {x2 + y 2 ≤ 1, z = 0}
a surface? Is the set S = {x2 + y 2 < 1, z = 0} a surface?
(3) (Exercises 9 of 2.2 of Do Carmo) Let V be an open set in the xyplane. Show that S = {(x, y, z) : z = 0, (x, y) ∈ V } is a surface.
(4) (Exercises 12 of 2.2 of Do Carmo) Show that
φ(u, v) = (a sin u, cos v, b sin u sin v, c cos u),
0 < u < π, 0 < v < 2π
is a chart for the ellipsoid S = {x2 /a2 + y 2 /b2 + z 2 /c1 = 1}.
Describe the curves u = const. on the ellipsoid.
(5) (Exercises 13 of 2.2 of Do Carmo) Find a parametrization of the
hyperboloid of two sheets S = {−x2 − y 2 + z 2 = 1}.
(6) (Exercises 1 of 2.4 of Do Carmo) Show that the equation of a tangent
plane at p = (x0 , y0 , z0 ) of S = f −1 (0), where ∇f 6= 0 on S is given
by
∂x f (p)(x − x0 ) + ∂y f (p)(y − y0 ) + ∂z f (p)(z − z0 ) = 0.
(7) (Exercises 2 of 2.4 of Do Carmo) Determine the tangent plane of
S = {x2 + y 2 − z 2 = 1} at the points (x, y, 0) and show that they
are parallel to the z-axis.
(8) (Exercises 10 of 2.4 of Do Carmo) Let α : I → R3 be a curve
parametrized by arclength and with nonzero ~k at every point. Consider the tube of radius r around α which has chart
φ(s, v) = α(s) + r(n(s) cos v + b(s) sin v),
1
r = const, s ∈ I, 0 < v < 2π,
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where n = ~k/|~k is the normal vector to α and b = α0 ×n the binormal
vector.
Show that φ is indeed a chart and the normal vector to the surface
is given by
N (s, v) = −(n(s) cos v + b(s) sin v).
(9) (Exercises 11 of 2.4 of Do Carmo) Show that the normal vectors to
the surface given by the chart
φ(u, v) = (f (u) cos v, f (u) sin v, g(u)),
f (u) 6= 0, g 0 6= 0
all pass through the z-axis.
(10) (Exercises 26 of 2.4 of Do Carmo) Show that if p is a point in a surface, by a convenient choice of coordinates it is possible to represent
a neighborhood of S as {z = f (x, y)} where
f (0, 0) = 0,
∇f (0, 0) = (0, 0).
(11) Let φ : U ⊂ R2 −→ R3 be a chart, h : W ⊂ R2 −→ U ⊂ R2 a
diffeomorphishm, and set ψ = φ ◦ h. Show that
∂ψ ∂ψ ∂φ
∂φ ∂u × ∂v (u, v) = ∂u0 × ∂v 0 (h(u, v))|det Dh|(u, v)
(12) Let S, M be two surfaces in R3 , F : S −→ M a smooth map, and
φ : U ⊂ R2 −→ R3 a chart with φ(0) = p. Recall that the definition
of
DFp : Tp S −→ TF (p) M
was
d(F ◦ α)
(0) where α : (−ε, ε) −→ S is such that α(0) = p.
DFp (α0 (0)) =
dt
a) Using this definition argue that
∂φ
∂(F ◦ φ)
DFp
(0) =
(0).
∂xi
∂xi
b) Use the previous exercise to compute DF0 when S = {(x, y, z) | z =
2x − y 3 } and F (x, y, z) = (z, y, 0).
2. Solutions
(1) We have
∂φ
∂φ
2
2
= (1, 2, 2xex −y + 3x2 ),
= (1, −1, −ex −y ).
∂x
∂y
Thus
∂φ ∂φ 2
2
×
= −2(x + 1)ex −y − 3x2 , (2x + 1)ex −y + 3x2 , −3 .
∂x ∂y
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Hence
+ 3x2 , −3
N (x, y) =
| −2(x + 1)ex2 −y − 3x2 , (2x + 1)ex2 −y + 3x2 , −3 |
−2(x + 1)ex
2 −y
− 3x2 , (2x + 1)ex
2 −y
(2) The first set is a NOT smooth surface because if we choose the point
v = (1, 0, 0) for instance, the surface is not the graph of function
defined on an open set. The second set is a surface because it is the
graph of a function defined over an open set of R2 (the function is
f (x, y) = 0 and the open set is the unit disc).
(3) The set in question is a surface because is the graph of a smooth
function defined over an open set of R2 .
(4) The map is clearly smooth. So we only need to see that it is one-toone and its differential is always injective. Let’s call the map φ to
make notation easier
If φ(u, v) = φ(s, t) then cos u = cos s and thus u = s. Using this
we then have sin v = sin t and cos v = cos t, which implies t = v.
Thus the map is one-to-one.
To compute the differential of the map we see that
∂φ
= (a cos u cos v, b cos u sin v, c sin u)
∂u
∂φ
= (−a sin u sin v, b sin u cos v, 0)
∂v
and these are the columns of the 3×2 matrix Dφ. Because 0 < u < π,
the third component of the first vector (c sin u) is always non-zero
which means that the two vectors above must always be linearly
independent. A moment of thought reveals that this is equivalent to
Dφ being injective because if this were not true then
Dφ(v1 , v2 ) = (0, 0, 0) =⇒ v1
∂φ
∂φ
+ v2
= 0 =⇒ v1 = 0, v2 = 0
∂u
∂v
(5) The parametrization consists of two charts φ1 : R2 −→ R3 , φ2 :
R2 −→ R3 given by
p
p
φ1 (x, y) = (x, y, 1 + x2 + y 2 ) and φ2 (x, y) = (x, y, − 1 + x2 + y 2 ).
It is simple to see that φ1 , φ2 are indeed two charts and that every
point in the hyperboloid lies in the image of one of these two charts.
(6) In class we saw that if we set S = f −1 (0), where ∇f 6= 0 on S,
then S is a surface and for every p ∈ S, ∇f (p) is orthogonal to the
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tangent plane Tp S. Hence if p = (x0 , y0 , z0 ) we have
Tp S = {~x ∈ R3 | ∇f (p).~x = 0}
= {(a, b, c) ∈ R3 | a∂x f (p) + b∂y f (p) + c∂z f (p) = 0}.
The tangent plane that do Carmo is considering is the set p + Tp S,
where the difference is that p = (x0 , y0 , z0 ) belongs to the second
set but not necessarily to the first. On the other hand, the set Tp S
contains always the origin, while that might not be true for the set
p + Tp S.
(7) The surface being considered is the level surface of the function
f (x, y, z) = x2 + y 2 − z 2 , i.e, S = f −1 (1). We have
∇f = (2x, 2y, −2z).
Thus, at the point p = (x, y, 0) we have
Tp S = {(a, b, c) ∈ R3 : ax + by = 0}.
Note that the vector (0, 0, 1) belongs to Tp S and so indeed Tp S is
parallel to the z-axis.
(8) Let’s call the map φ. We have
∂φ
= α0 (s) + r(n0 (s) cos v + b0 (s) sin v)
∂s
∂φ
= r(−n(s) sin v + b(s) cos v),
∂v
where n(s), b(s) are orthonormal vectors, both orthogonal to α0 (s),
and n(s) is parallel to α00 (s). A priori α00 (s) does not need to be
orthogonal to α0 (s) but, because we have parametrization by arc
length, we obtain by differentiating in t that
α0 (s).α0 (s) = 1 =⇒ α00 (s).α0 (s) = 0.
To check that N (s, v) is a unit normal vector we just need to check
that
• |N (s, v)| = 1. This is true because
N (s, v).N (s, v) = n(s).n(s) cos2 v + b(s).b(s) sin2 v = cos2 v + sin2 v = 1.
0
0
• N (s, v). ∂φ
∂s = 0. This is true because α (s).n(s) = α (s).b(s) = 0
and, by differentiating with respect to s, we have
n(s).n(s) = 1 and b(s).b(s) = 1 =⇒ n(s).n0 (s) = 0 and b0 (s).b(s) = 0
and
n(s).b(s) = 0 =⇒ b0 (s).n(s) + n0 (s).b(s) = 0
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Thus
N (s, v).
∂x
= N (s, v).α0 (s) + r cos vn0 (s).N (s, v) + r sin vb0 (s).N (s, v)
∂s
= r cos vn0 (s).N (s, v) + r sin vb0 (s).N (s, v)
= −r cos v sin v(b0 (s).n(s) + n0 (s).b(s)) = 0.
• N (s, v). ∂φ
∂v = 0. This follows from
N (s, v).
∂x
= r(sin v cos v − sin v cos v) = 0.
∂v
(9) There are many ways of doing this. Because g 0 6= 0 and z = g(u),
we can write u = g −1 (z) and so, setting h = f ◦ g −1 we get a new
parametrization
x(z, v) = (h(z) cos v, h(z) sin v, z).
Hence, the surface can be described as
S = {(x, y, z) ∈ R3 | x2 + y 2 = h2 (z)},
which is the level surface of f˜(x, y, z) = x2 + y 2 − h2 (z). Therefore
the normal line is spanned by ∇f˜ = (2x, 2y, −2h(z)h0 (z)) and so the
line do Carmo is looking for is
L = (x, y, z) + {t(2x, 2y, −2h(z)h0 (z)) | t ∈ R}.
Why does L intersect the z-axis? Well, just take t = −1/2 in the
above expression and we get that the point (0, 0, z + h(z)h0 (z)) lies
in L.
(10) We can always choose coordinates so that p corresponds to the origin
and the tangent plane corresponds to the xy-plane. From Proposition 3, Chapter 2.2, we see that S is given by the graph of a function
z = f (x, y), x = f (z, y), or y = f (z, x) near the origin. Note that
in any of these cases we must have f (0, 0) = 0 (otherwise the origin
would not belong to the surface). Because we are assuming the tangent plane is actually the xy-plane then only the first case is possible
and so we have that, locally, the surface is given by {z = f (x, y)}.
∂f
We need to argue that ∂f
that the vec∂x (0) = ∂y (0) = 0. I will show
∂f
tor perpendicular to the tangent plane at the origin is − ∂f
(0),
−
(0),
1
.
∂x
∂y
But we are assuming that the tangent plane is the xy-plane, in which
case the normal vector must be parallel to (0, 0, 1). Hence the only
∂f
way this can happen is if ∂f
∂x (0) = ∂y (0) = 0.
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∂f
Why is − ∂f
(0),
−
(0),
1
a normal vector? If we set
∂x
∂y
h(x, y, z) = z − f (x, y)
then, locally, we have {h(x, y, z) = 0} = {z = f (x, y)}, i.e, the
surface is the zero level set of h. From a Lemma we saw in class we
know that ∇h(0) is perpendicular to the tangent plane at the origin.
The claim follows from the fact that
∂f
∂f
∇h(0) = − (0), − (0), 1 .
∂x
∂y
(11) Let φ : U ⊂ R2 −→ R3 be a chart, h : W ⊂ R2 −→ U ⊂ R2 a
diffeomorphishm, and set ψ = φ ◦ h. Show that
∂φ
∂ψ ∂ψ ∂φ
∂u × ∂v (u, v) = ∂u0 × ∂v 0 (h(u, v))|det Dh|(u, v)
From the cain rule, and using the notation (u0 , v 0 ) = h(u, v) we
have
∂ψ
∂φ ∂u0
∂φ ∂v 0
∂ψ
∂φ ∂u0
∂φ ∂v 0
=
+
and
=
+
.
∂u
∂u0 ∂u
∂v 0 ∂u
∂v
∂u0 ∂v
∂v 0 ∂v
∂φ
I am keeping the notation simple but where is ∂u
0 you should really
0
0
∂φ
∂u
see ∂u0 (h(u, v)), where is ∂u you should really see ∂u
∂u (u, v), and so
on.
Hence
0 0
∂u ∂v
∂v 0 ∂u0 ∂φ
∂φ
∂φ
∂φ
∂ψ ∂ψ
×
=
−
× 0 = (detDh) 0 × 0 .
0
∂u
∂v
∂u ∂v
∂u ∂v ∂u
∂v
∂u
∂v
Therefore, taking the norm on both sides, we obtain
∂ψ ∂ψ ∂φ
∂φ ∂u × ∂v (u, v) = ∂u0 × ∂v 0 (h(u, v))|det Dh|(u, v).
(12) a) Set e1 = (1, 0), e2 = (0, 1). Consider the path αi (t) = tei in U
and the path γi (t) = φ ◦ αi (t) in S. From the chain rule we have
∂φ
dγi
(0) =
(0).
dt
∂xi
Thus
∂φ
dγi
d(F ◦ γi )
DFp
(0) = DFp
(0) =
(0)
∂xi
dt
dt
d((F ◦ φ) ◦ αi )
∂(F ◦ φ)
=
(0) =
(0),
dt
∂xi
where the last identity follows from the chain rule.
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b) Use the previous exercise to compute DF0 when S = {(x, y, z) | z =
2x − y 3 } and F (x, y, z) = (z, y, 0).
In this case M = {z = 0} and F : S −→ M ⊂ R3 . Because
F (0) = 0 we need to compute T0 S and T0 M . The latter one is easy
because M is a vector space and so T0 M = {z = 0}. The first one
is not hard as well because we have the chart
φ : R2 −→ S,
φ(x, y) = (x, y, 2x − y 3 )
and so
T0 S = span
∂φ
∂φ
(0),
(0)
∂x
∂y
= span{(1, 0, 2), (0, 1, 0)} = {z = 2x}.
To determine DF0 it suffices to determine DF0 (1, 0, 2) and DF0 (0, 1, 0).
Note that F ◦ φ = (2x − y 3 , y, 0) and so, using the previous exercise,
we have
∂(F ◦ φ)
∂φ
(0) =
(0) = (2, 0, 0)
DF0 (1, 0, 2) = DF0
∂x
∂x
and
∂φ
∂(F ◦ φ)
DF0 (0, 1, 0) = DF0
(0) =
(0) = (0, 1, 0).
∂y
∂y