Download MATH 31BH HOMEWORK 8 SOLUTIONS Problem 3.1.2 The set M

MATH 31BH HOMEWORK 8 SOLUTIONS
Problem 3.1.2
x
2
2
2
The set M =
∈ R : x + x + y = 2 is a smooth curve.
y
Proof. Consider the function F : R2 → R defined by
x
F
= x + x2 + y 2 − 2.
y
x
Now M is exactly the set of points for which F
= 0.
y
Notice that
[DF ] = [1 + 2x, 2y].
This matrix is onto so long as not both entries are equal to 0. This occurs if and only if
x = − 12 and y = 0.
But if x = − 12 and y = 0, then
2
1
1
1
2
2
+ (0)2 = − 6= 2,
x+x +y =− + −
2
2
4
so this point does not lie on M .
It follows that [DF ] is onto for every point in M , and therefore by Theorem 3.1.10 M
is a 1-dimensional smooth manifold.
Problem 3.1.8 (a) Let
Xa = {(x, y, z) ∈ R3 : x − y 2 = a}
and
Yb = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = b}.
Let F (x, y, z) = x − y 2 − a. Then
[DF ] = [1, −2y, 0],
Notice that this matrix is onto regardless of the values of x, y, z, or a. Therefore, Xa is
a smooth surface.
Next, let G(x, y, z) = x2 + y 2 + z 2 − b. Then
[DG] = [2x.2y, 2z],
which is onto for every point in R3 except for (0, 0, 0). But if b 6= 0, this point does not
belong to Yb . By Theorem 3.1.10 we conclude that Yb is a smooth surface for b 6= 0.
When b = 0, notice that Yb = {(0, 0, 0)}. So Y0 is also a smooth manifold.
1
2
MATH 31BH HOMEWORK 8 SOLUTIONS
(b) We now ask when Xa ∩ Yb is a smooth manifold. Define
x − y2 − a
F(x, y, z) =
.
x2 + y 2 + z 2 − b
Then
1 −2y 0
DF] =
.
2x 2y 2z
We may reduce this matrix to row echelon form as
1
−2y
0
.
0 2y − 4xy 2z
This matrix is surjective except for when
2y − 4xy = 0 and 2z = 0.
That is, the matrix is surjective for all points except those in
1
{(x, 0, 0) : x ∈ R} ∪
, y, 0 : y ∈ R .
2
Now notice that if a point in Xa ∩ Yb may have the form (x, 0, 0) if and only if a2 = b.
Similarly, one finds that a point in Xa ∩ Yb will have the form 12 , y, 0 if and only if
a + b = 43 .
Therefore Xa ∩ Yb is a smooth curve if a2 6= b and a + b 6= 34 .
Problem 3.1.12
Let X ⊂ R3 be the set of midpoints of segments joining a point of the curve C1 of
equation y = x2 , z = 0 to a point of the curve C2 of equation z = y 2 , x = 0.
(a)
For all real t, C1 is parameterized by the function γ1 defined by
 
t
γ1 (t) = t2  .
0
Similarly, C2 is parameterized by the function γ2 defined by
 
0
γ2 (s) =  s 
s2
for all real s.
(b) We may parameterize X by choosing any point on C1 and any point on C2 and
calculating their midpoint. That is, X is parameterized by
 t 
γ1 (t) + γ2 (s)  t22+s 
.
γ(s, t) =
=
2
2
s2
2
MATH 31BH HOMEWORK 8 SOLUTIONS
3
 
x
2
2
(c) If y  is a point in X, then x = 2t , y = t 2+s , and z = s2 for some real t and s.
z
The first equation tells us that t = 2x. Using this information in the second equation,
we see that s = 2y − t2 = 2y − 4x2 . Therefore, from the last equation, we find that
points in X satisfy
2
1
2y − 4x2 ,
z=
2
or
z = 2y 2 − 8x2 y + 8x4 .
(d) Let f (x, y) = 2y 2 − 8x2 y + 8x4 . Then f is a C 1 mapping, and X is defined by the
surface z = f (x, y). Therefore, X is a smooth surface.
Problem 3.2.10 (a)


cos(t)
Let C be the helicoid parameterized by γ(t) =  sin(t) .
t
The tangent line to C at the point γ(t) is given by
r(s) = γ 0 (t)s + γ(t),
so



 

− sin(t)
cos(t)
−s sin(t) + cos(t)
r(s) =  cos(t)  s +  sin(t)  =  s cos(t) + sin(t)  .
1
t
s+t
If we allow t to vary, we get the parameterization for the union X of all tangent lines to
C:


−s sin(t) + cos(t)
τ (s, t) =  s cos(t) + sin(t)  .
s+t
Problem 3.2.6 Let X ⊆ R4 be the set of points such that
x21 + x22 − x23 − x24 = 0 and x1 + 2x2 + 3x3 + 4x4 = 4.
 
1
0

(a) Let p = 
1. Let
0
x21 + x22 − x23 − x24
F(x1 , x2 , x3 , x4 ) =
.
x1 + 2x2 + 3x3 + 4x4 − 4
Then F(x) = 0 for every x ∈ X, and
2x1 2x2 −2x3 −2x4
[DF] =
.
1
2
3
4
4
MATH 31BH HOMEWORK 8 SOLUTIONS
Now we have
2 0 −2 0
[DF(p)] =
,
1 2 3 4
which is onto. So by the implicit function theorem, X is a smooth manifold in some
neighborhood of p.
(b) The tangent space to X at p is spanned by the null space of [DF(p)]. We have
2 0 −2 0
1 0 −1 0
[DF(p)] =
→
,
1 2 3 4
0 1 2 2
   
1
0 


   

−2
−2
,  .
so the tangent space to X at p is span 
 1   0 





0
1
(c) We notice that every pair of columns of [DF(p)] except for columns 2 and 4 are
linearly independent. Therefore, by the implicit function theorem, each pair of variables
besides x2 and x4 may be written as a function of the other two in some neighborhood
of p.
(d) We claim that X is a smooth manifold. Indeed, recall
2x1 2x2 −2x3 −2x4
1
2
3
4
[DF] =
→
.
1
2
3
4
0 2x2 − 4x1 −2x3 − 6x1 −2x4 − 8x1
The only way this matrix is not onto is if 2x2 = 4x1 , −2x3 = 6x1 , and −2x4 = 8x1 .
But plugging these values into the first equation that defines X tells us that x1 = 0.
Therefore, x1 = x2 = x3 = x4 = 0. But this point does not satisfy the second equation
defining X, and therefore the point does not belong to X.
It follows that [DF] is onto for every point in X, and thus by Theorem 3.1.10 X is a
smooth manifold.
Problem 3.1
(a) The set X ⊆ R3 of equation x3 + xy 2 + yz 2 + z 3 = 4 is a smooth surace.
Proof. Let F (x, y, z) = x3 + xy 2 + yz 2 + z 3 − 4. Then F (x, y, z) = 0 if and only if
(x, y, z) ∈ X.
Moreover,
[DF ] = 3x2 + y 2 2xy + z 2 2yz + 3z 2 .
Now this matrix is not onto if and only if all of its entries are 0. Notice that the first
entry is 0 if and only if x = y = 0.
In this case, the second and third entries will simultaneously be 0 if and only if z = 0.
Therefore this matrix is onto for all points except (0, 0, 0). But notice that (0, 0, 0) 6∈ X,
because it does not satisfy the equation defining X.
Thus, by Theorem 3.1.10, X is a smooth surface.
MATH 31BH HOMEWORK 8 SOLUTIONS
5
(b) We have
[DF (1, 1, 1)] = 4 3 5 .
It follows that the tangent plane to X at (1, 1, 1) has equation
4(x − 1) + 3(y − 1) + 5(z − 1) = 0
and the tangent space to X at (1, 1, 1) has the form
4ẋ + 3ẏ + 5ż = 0.
Problem 3.4
 
 
 
x
0
0
Consider the space X of triples p =  0 , q = y , r = 0 such that y 6= 0 and
0
0
z
the segments p, q and q, r form an angle of π/4.
(a) We notice that the angle between these segments is the same as the angle between
~ and qr:
~
vectors qp
 
 
x
0
~ = −y  and qr
~ = −y  .
qp
0
z
Now the cosine of this angle is just
since cos(π/4) = √12 we assert that
~ qr
~
qp·
~
~ .
||qp||||
qr||
Since we desire the angle to be π/4 and
y2
1
√ =p
.
2
(x2 + y 2 )(y 2 + z 2 )
Therefore any point in X will satisfy f (x, y, z) = 0 for
f (x, y, z) = p
y2
(x2
+
y 2 )(y 2
+
z2)
1
−√ .
2
(b) Notice that
(1)2
1
f (0, 1, 1) = p
−√
2
(02 + 12 )(12 + 12 )
1
1
=p
−√
2
(1)(2)
= 0.
We also have
[Df ] = [−xy 2 ((y 2 + z 2 )(x2 + y 2 ))−3/2 (y 2 + z 2 ),
1
2y((x2 + y 2 )(y 2 + z 2 ))−1/2 − y 2 ((x2 + y 2 )(y 2 + z 2 ))−3/2 (4y 3 + 2z 2 y + 2x2 y),
2
−zy 2 ((y 2 + x2 )(z 2 + y 2 ))−3/2 (y 2 + x2 )]
so
√
√ #
2
2
[Df (0, 1, 1)] = 0,
,−
.
4
4
"
6
MATH 31BH HOMEWORK 8 SOLUTIONS
Since the derivative is onto at this point, it follows that X is a smooth manifold in
some neighborhood of (0, 1, 1).
(c) It is true that in some neighborhood of (0, 1, 1) that X is locally the graph of
a function expressing z in terms of x and y. This follows from the implicit function
theorem, since the third column of the derivative of f at (0, 1, 1) is nonzero and hence
invertible.
(d) The tangent plane to X at (0, 1, 1) is given by
√
√
2
2
(y − 1) −
(z − 1) = 0
4
4
and the tangent space to X at this point is given by
√
√
2
2
ẏ −
ż = 0.
4
4
Problem 3.7.16
Let f (x, y, z) = x + y + z. We wish to find the critical points of f constrained to the
surface parameterized by


sin(uv) + u
g(u, v) =  u + v  .
uv
Constrained to this surface, our function takes on the values
f (u, v) = (sin(uv) + u) + (u + v) + uv = sin(uv) + 2u + v + uv
for all real u and v. Therefore, we wish to find the critical points of this function of two
variables.
We have
fu = v cos(uv) + 2 + v and fv = u cos(uv) + 1 + u,
so we seek to solve the system of equations
(
v cos(uv) + 2 + v = 0
u cos(uv) + 1 + u = 0.
By setting equal the lefthand sides of these two equations, we find that v = 2u.
Substituting this into, say, the second equation tells us that any critical point (u, v) will
satisfy
u cos(2u2 ) + u + 1 = 0.