Exact vector fieldsIf U is an open subset of Rn, then a vector field F : U → Rn is exact if F is thegradient of some C1 scalar function f : U → R.
Theorem (Fundamental Theorem of Calculus for vector fields)If F : U → Rn is exact (ie F = ∇f ) then
ˆC
F · dx = f (x1)− f (x0)
if C is any oriented C1 curve that starts at x0 and ends at x1.
Conservative vector fieldsIf U is an open subset of Rn, then a vector field F : U → Rn is conservative if
ˆC
F · dx = 0 for any closed curve C in U.
March 13, 2017 1 / 50
TheoremAssume that F is a vector field on a subset U of Rn
Then the following are equivalent:
F is conservative, i.e.ˆ
CF · dx = 0 for any closed curve C in U.
F satisfies: ˆC1
F · dx =
ˆC2
F · dx
whenever C1 and C2 are two oriented curves in U with the sameendpoints.
F is exact.
March 13, 2017 2 / 50
Closed vector fieldsIf U is an open subset of Rn, then a C1 vector field F : U → Rn is closed
∂iFj = ∂jFi for i , j = 1, . . . ,n
TheoremEvery exact vector field is closed.
On some domains U, there are vector fields that are closed but notexact.
However, if U is star-shaped, then every closed vector field on U is alsoexact.
Above we have used the
DefinitionA set U ⊂ Rn is star-shaped if there exists a point a ∈ U such that for everyx ∈ U, the line segment connecting x and a is contained in U. That is:
for all x ∈ U and t ∈ [0,1], ta + (1− t)x ∈ U
March 13, 2017 3 / 50
Theorem (Green’s Theorem)
Asssume that S ⊂ R2 is a regular region with piecewise smoothboundary ∂S, and that F : R2 → R2 is a C1 vector field. Then
ˆ∂S
F · dx =
¨S
(∂1F2 − ∂2F1) dA
where ∂S is positively oriented.
Remark. We often write F = (P,Q). With this notation, Green’sTheorem can be written
ˆ∂S
P dx + Q dy =
¨S
∂Q∂x− ∂P∂y
dA
March 13, 2017 4 / 50
An example of spherical coordinates.
Problemfind the volume of the set S ⊂ R3 bounded by the cone z = 5
√x2 + y2 and
the sphere x2 + y2 + z2 = 1.
solution: We will write the integral in spherical coordinates
g(ρ, θ, φ) = (ρ sin θ cosφ, ρ sin θ sinφ, ρ cos θ),
where ρ is nonnegative, 0 ≤ φ < 2π and 0 < θ < π. This seems reasonablebecause of the symmetry of the problem. But it is a challenge to write the setS in spherical coordinates.To start, we will rewrite the equations for the sphere and the cone.a. It is easy to check that x2 + y2 + z2 = 1 if and only if ρ = 1.b. Also, z = 5
√x2 + y2 if and only if
ρ cos θ = 5√
(ρ sin θ cosφ)2 + (ρ sin θ sinφ)2 = 5ρ| sin θ|.
Since θ ∈ [0, π], we can forget the absolute values and just write sin θ. Thusz = 5
√x2 + y2 if and only if θ = arctan(1/5).
March 13, 2017 5 / 50
solution, continued
c. With the above formulas and a bit of geometric reasoning (try to draw apicture of S) we can figure out that in term of polar coordinates, S is definedby the inequalities
0 ≤ ρ ≤ 1, 0 ≤ θ ≤ arctan(1/5), 0 ≤ φ < 2π.
In other words,
g−1(S) := {(ρ, θ, φ) : 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ arctan(1/5), 0 ≤ φ < 2π.}
Thus (using a formula for the Jacobian determinant of g from the onlinenotes, see p. 133)
volume(S) =
˙S
1dV =
˙g−1(S)
|det Dg|(ρ, θ, φ)dV
=
ˆ 1
0
ˆ arctan(1/5)
0
ˆ 2π
0ρ2 sin θ dφdθ dρ
March 13, 2017 6 / 50
solution, continuedWe evaluate the integral to find
volume(S) = −23π[cos(arctan(1/5))− cos 0] =
23π[1− cos(arctan(1/5))].
Finally, to simplify, let α = arctan(1/5), so that tanα = 15 . By rewriting this we
find that sinα = 15 cosα. Solving the system of equations
sin2 α + cos2 α = 1, sinα =15
cosα,
we find that cosα =√
25/26. (Recall that we already know from above thatcosα > 0.) So we conclude that
volume(S) =23π[1−
√25/26].
March 13, 2017 7 / 50
DefinitionA set S ⊂ R has Jordan measure 0 if, for any ε > 0, there exists a finitenumber of intervals I1, . . . , Ik such that
S ⊂ ∪kj=1Ij
k∑j=1
`(Ij ) < ε
where `(I) is the length of interval I. (So for example `((a,b)) = b − a.)
The “middle thirds" Cantor set is defined to be C := ∩∞k=0Ck , where
C0 = [0,1]
C1 = [0, 13 ] ∪ [ 2
3 ,1] = C0 with the “middle third" removed
C2 = [0, 19 ] ∪ [ 2
9 ,13 ] ∪ [ 2
3 ,79 ] ∪ [ 8
9 ,1]
= C1 with the “middle third" removed from each sub-interval· · · · · · · · ·
Ck+1 = Ck with the “middle third" removed from each sub-interval · · · .
This is an interesting set with m(C) = 0.March 13, 2017 8 / 50
DefinitionA partition of [a,b] is a set of points P = {a = x0 < x1 < . . . < xn = b}.Also define: |P| := “the order of P" := n (for the example above),`(P) := “the length of P" := max |xi=1,...,|P||xi − xi−1|.
DefinitionGiven f : [a,b]→ R and a partition P of [a,b], a Riemann sum of f withrespect to P is a sum of the form
S(f ,P) =n∑
i=1
f (ti )(xi − xi−1), for some choice of ti ∈ [xi−1, xi ].
Also define
U(f ,P) =n∑
i=1
Mi (xi − xi−1), Mi := supx∈[xi−1,xi ]
f (x),
u(f ,P) =n∑
i=1
mi (xi − xi−1), mi := infx∈[xi−1,xi ]
f (x).
March 13, 2017 9 / 50
Definitions1. A set S ⊂ Rn is disconnected if there are nonemepty sets S1,S2 such that
S = S1 ∪ S2, and
S̄1 ∩ S2 = S1 ∩ S̄2 = ∅.
The sets S1,S2 are sometimes called a disconnection of S.
2. S ⊂ Rn is connected if it is not disconnected.
3. S ⊂ Rn is path-connected if, for every a,b ∈ S, there is a continuousfunction γ : [0,1]→ S (that is, a path in S) such that γ(0) = a and γ(1) = b.
Theorems1. If S ⊂ Rn is connected and f : Rn → Rm is continuous, then f (S) isconnected. Same for path-connected.
2. In 1 above, when m = 1, we get the Intermediate Value Theorem in Rn.
3. Every path-connected set is connected.A set that is connected and open is also path-connected.
March 13, 2017 10 / 50
DefinitionA function f : R→ R is differentiable at a point a ∈ R if
limh→0
f (a + h)− f (a)
hexists.
If it exists, the limit is defined to be the derivative of f at a, which isdenoted f ′(a) or df
dx (a).
Alternate DefinitionA function f : R→ R is differentiable at a point a ∈ R if there exists anumber m such that
limh→0
f (a + h)− f (a)−mhh
= 0.
If it exists, the number m is defined to be the derivative of f at a. whichis denoted f ′(a) or df
dx (a).
March 13, 2017 11 / 50
DefinitionA function f : R→ Rm is differentiable at a point a ∈ R if
limh→0
f(a + h)− f(a)
hexists. (note, this limit is a vector in Rm)
If it exists, the above limit is defined to be the derivative of f at a, whichis denoted f′(a) or df
dx (a).
Alternate DefinitionA function f : R→ Rm is differentiable at a point a ∈ R if there exists avector v ∈ Rm such that
limh→0
f(a + h)− f(a)− vhh
= 0.
I it exists, the vector v is defined to be the derivative of f at a. which isdenoted f′(a) or df
dx (a).
March 13, 2017 12 / 50
DefinitionA function f : Rn → R is differentiable at a point a ∈ Rn if there exists avector c ∈ Rn such that
lim‖h‖→0
f (a + h)− f (a)− c · h‖h‖
= 0.
If it exists, the vector c is called the gradient of f at a. It is oftendenoted ∇f (a) or occasionally grad f (a).
Note: the “other definition” does not work in this setting: even if f is alinear function f (x) = v · x + c, as long as v 6= 0,
lim‖h‖→0
f (a + h)− f (a)
‖h‖does not exist for any a ∈ Rn !!
March 13, 2017 13 / 50
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March 13, 2017 15 / 50
March 13, 2017 16 / 50
DefinitionA function f : Rn → R is differentiable at a point a ∈ Rn if there exists avector c ∈ Rn such that
lim‖h‖→0
f (a + h)− f (a)− c · h‖h‖
= 0. (1)
If it exists, the vector c is called the gradient of f at a. It is oftendenoted ∇f (a) or occasionally grad f (a).
RemarkSuppose that c1 and c2 are two vectors in Rn and that
lim‖h‖→0
f (a + h)− f (a)− cj · h‖h‖
= 0 for j = 1 and 2.
Then it is a fact that c1 = c2. (Exercise!)This implies that ∇f (a) is well-defined.
March 13, 2017 17 / 50
TheoremIf all partial derivatives of f exist in an open ball around a and arecontinuous at a, then f is differentiable at a and∇f (a) = ( ∂f
∂x1(a), . . . , ∂f
∂xn(a)).
March 13, 2017 18 / 50
optional ! sketch of proof when n = 2Let a = (a1,a2) and h = (h1,h2). Then
f (a + h)− f (a)
=[f (a1 + h1,a2 + h2)− f (a1,a2 + h2)
]+[f (a1,a2 + h2)− f (a1,a2)
]=
∂f∂x1
(a1 + θ1h1,a2 + h2)h1 +∂f∂x2
(a1,a2 + θ2h2)h2
by the mean value theorem for functions of a single variable, for someθ1, θ2 ∈ [0,1]. So
1‖h‖
[f (a + h)− f (a)− h · ( ∂f
∂x1,∂f∂x2
)(a)
]=
1‖h‖
[h1
(∂f∂x1
(a1 + θ1h1,a2 + h2)− ∂f∂x1
(a1,a2)
)+ h2
(∂f∂x1
(a1,a2 + θ2h2)− ∂f∂x2
(a1,a2)
)]Then you can use the continuity of ∂f/∂x1 and ∂f/∂x2 to show that thisconverges to 0 as h→ 0. (exercise, very optional !!)
March 13, 2017 19 / 50
DefinitionA function f : Rn → R is differentiable at a point a ∈ Rn if there exists avector c ∈ Rn such that
lim‖h‖→0
f (a + h)− f (a)− c · h‖h‖
= 0. (2)
If it exists, the vector c is called the gradient of f at a, denoted ∇f (a).
DefinitionA function f : Rn → Rm is differentiable at a point a ∈ Rn if there existsa m × n matrix A such that
lim‖h‖→0
‖f(a + h)− f(a)− Ah‖Rm
‖h‖Rn= 0. (3)
(Here we think of f, a,h as a column vector.) If it exists, the matrix A is called theJacobian matrix of f at a. It is often denoted Df(a). When it exists, it is unique.
March 13, 2017 20 / 50
∇f vs Df : two slightly different perspectives
m = 1, original definitionf : Rn → R is differentiable at a ∈ Rn if there exists c ∈ Rn such that
lim‖h‖→0
f (a + h)− f (a)− c · h‖h‖
= 0. (4)
If it exists, c is called the gradient of f at a, denoted ∇f (a).Here we did not specify whether a,h,∇f (a) are row or column vectors.
m = 1 definition as a special case of f : Rn → Rm
f : Rn → R is differentiable at a ∈ Rn if there exists a 1× n matrix A (i.e.
a row vector) such that
lim‖h‖→0
|f (a + h)− f (a)− Ah|‖h‖Rn
= 0. (5)
(Here we think of a,h as column vectors.) If it exists, the row vector A is denotedDf (a).
March 13, 2017 21 / 50
From the last lecture:
Mean Value Theorem for f : Rn → RLet U ⊂ Rn, and let a,b be points in U such that
γ(t) := (1− t)a + tb satisfies γ(t) ∈ U for every t ∈ [0,1].
If f : U → R is a differentiable function on U, then there exists a point con the line segment connecting a to b (that is, c = γ(t) for some t ∈ (0, 1)) suchthat
f (b)− f (a) = (b− a) · ∇f (c).
RemarkIn fact it suffices to assume that f ◦ γ : [0,1]→ R is continuous on [0,1]and differentiable on (0,1).
March 13, 2017 22 / 50
Example of second partial derivatives
Letf (x , y) = xecos(xy)
Then
∂x f = ecos(xy) − xecos(xy)y sin(xy) = ecos(xy)(1− xy sin(xy)).
∂y f = −xecos(xy)x sin(xy) = −ecos(xy)x2 sin(xy).
So
∂xx f = ecos(xy)[−y sin(xy)(1− xy sin(xy))− xy2 cos(xy)
]∂xy f = −ecos(xy)
[−y sin(xy)x2 sin(xy) + 2x sin(xy) + x2y cos(xy)
]∂yx f = ecos(xy)
[−x sin(xy)(1− xy sin(xy))− x sin(xy)− x2y cos(xy)
]∂yy f = −ecos(xy)
[−x sin(xy)x2 sin(xy)−+x3 cos(xy) cos(xy)
]March 13, 2017 23 / 50
Taylor’s theorem in 1-d and n-d1. If f : R→ R is Ck+1, then
f (x) =k∑
j=0
(x − a)j
j!f (j)(a) + rk,a(x),
and ∃θ ∈ (0,1) such that
rk,a(x) =(x − a)k+1
(k + 1)!f (k+1)(c), for c := θx + (1− θ)a.
——————————————————————————————–2. If f : Rn → R is Ck+1, then
f (x) =∑|α|≤k
(x− a)α
α!∂αf (a) + rk,a(x),
and ∃θ ∈ (0,1) such that
rk,a(x) =∑|α|=k+1
(x − a)α
α!∂αf (c) for c := θx + (1− θ)a.
In both cases, limx→a|rk,a(x)||x−a|k = 0.
March 13, 2017 24 / 50
Proof summary for Taylor approximation for f : Rn → RGiven f : Rn → R and x,a in Rn.
1 Define γ(t) := (1− t)a + tx and
g(t) = f (γ(t))
2 Apply 1-d Taylor Theorem to g with a = 0, f = 1. This gives
g(1) =k∑
j=0
1j
j!g(j)(0) + rk ,a(x)
3 use the chain rule to rewrite in terms of f . This gives
f (x) = f (a)+k∑
j=1
1j!
n∑i1,...,ij=1
(x − a)i1 · · · (x − a)ij∂i1···ij f (a)
+rk ,a(x)
( This can also be written in less awful-looking ways)
4 rewrite, if we wish in terms of multi-indices. Doing this carefully involvesan induction argument (omitted).
March 13, 2017 25 / 50
Assume f : Rn → R is Ck+1.Given a,x ∈ Rn, let
γ(t) = a + t(x− a), g(t) = f (γ(t))
Note g(0) = f (a) and g(1) = f (x).1d Taylor Theorem
g(t) = pk ,0(t) +1k !
tk+1g(k+1)(c) for some c ∈ (0, t), if t > 0.
where
pk ,0(t) = g(0) + tg′(0) + · · ·+ tk
k !g(k)(0)
=k∑
j=0
t j
j!g(j)(0).
Now we express g(j) in terms of f ....
March 13, 2017 26 / 50
from above: f : Rn → R is Ck+1, g(t) = f (γ(t)) for γ(t) := a + t(x− a).Then by the Chain Rule,
g′(t) = ∇f (γ(t)) · γ′(t) =n∑
i=1
(xi − ai)∂i f (γ(t)).
Differentiating again,
g′′(t) =n∑
i=1
n∑j=1
(xi1 − ai1)(xi2 − ai2)∂i1i2 f (γ(t)).
And in the same way, if we differentiate j times,
g(j)(t) =n∑
i1=1
· · ·n∑
ij=1
(xi1 − ai1) · · · (xij − aij )∂i1···ij f (γ(t))
March 13, 2017 27 / 50
Let’s look at g(4) in n = 2 dimensions:
g′(0) = (x1 − a1)∂1f (a) + (x2 − a2)∂2f (a)
g′′(0) = (x1 − a1)(x1 − a1)∂11f (a) + (x2 − a2)(x1 − a1)∂21f (a)
+ (x1 − a1)(x2 − a2)∂12f (a) + (x2 − a2)(x2 − a2)∂22f (a)
g′′′(0) = (x1 − a1)(x1 − a1)(x1 − a1)∂111f (a) + (x2 − a2)(x1 − a1)(x1 − a1)∂211f (a)+ (x1 − a1)(x2 − a2)(x1 − a1)∂121f (a) + (x2 − a2)(x2 − a2)(x1 − a1)∂221f (a)+ (x1 − a1)(x1 − a1)(x2 − a2)∂112f (a) + (x2 − a2)(x1 − a1)(x2 − a2)∂212f (a)+ (x1 − a1)(x2 − a2)(x2 − a2)∂122f (a) + (x2 − a2)(x2 − a2)(x2 − a2)∂222f (a)
March 13, 2017 28 / 50
g(4)(0)
= (x1 − a1)(x1 − a1)(x1 − a1)(x1 − a1)∂1111f (a) + (x2 − a2)(x1 − a1)(x1 − a1)(x1 − a1)∂2111f (a)+ (x1 − a1)(x2 − a2)(x1 − a1)(x1 − a1)∂1211f (a) + (x2 − a2)(x2 − a2)(x1 − a1)(x1 − a1)∂2211f (a)+ (x1 − a1)(x1 − a1)(x2 − a2)(x1 − a1)∂1121f (a) + (x2 − a2)(x1 − a1)(x2 − a2)(x1 − a1)∂2121f (a)+ (x1 − a1)(x2 − a2)(x2 − a2)(x1 − a1)∂1221f (a) + (x2 − a2)(x2 − a2)(x2 − a2)(x1 − a1)∂2221f (a)+ (x1 − a1)(x1 − a1)(x1 − a1)(x2 − a2)∂1112f (a) + (x2 − a2)(x1 − a1)(x1 − a1)(x2 − a2)∂2112f (a)+ (x1 − a1)(x2 − a2)(x1 − a1)(x2 − a2)∂1212f (a) + (x2 − a2)(x2 − a2)(x1 − a1)(x2 − a2)∂2212f (a)+ (x1 − a1)(x1 − a1)(x2 − a2)(x2 − a2)∂1122f (a) + (x2 − a2)(x1 − a1)(x2 − a2)(x2 − a2)∂2122f (a)+ (x1 − a1)(x2 − a2)(x2 − a2)(x2 − a2)∂1222f (a) + (x2 − a2)(x2 − a2)(x2 − a2)(x2 − a2)∂2222f (a)
This is the same as
(x− a)α∂αf (0) for α = (4, 0)
+4(x− a)α∂αf (0) for α = (3, 1)
+6(x− a)α∂αf (0) for α = (2, 2)
+4(x− a)α∂αf (0) for α = (1, 3)
+(x− a)α∂αf (0) for α = (0, 4)
For every α, the coefficient is |α|!/α! (which is just a binomial coefficient when n = 2).
March 13, 2017 29 / 50
Taylor series summaryIf f : Rn → R is Ck+1, then
f (x) =∑|α|≤k
(x− a)α
α!∂αf (a) + rk,a(x),
and
|rk,a(x)|‖x− a‖k+1 is bounded for x near a, hence lim
x→a
|rk,a(x)|‖x− a‖k = 0.
——————————————————————————————–If k = 2, we can write this in the simpler form:
f (x) = f (a) + (x− a) · ∇f (a) +12
(x− a)T H(a)(x− a) + r2,a(x).
where H is the Hessian matrix. We can also write
f (x) = f (a) + (x− a) · ∇(a) +12
(x− a)T H(c)(x− a)
where c = (1− θ)a + θx for some θ ∈ (0,1).March 13, 2017 30 / 50
about local maxima and minimaAssume that U is an subset of Rn and that a ∈ U int .1. If f : U → R is differentiable, then
if a is a local max or local min for f , then ∇f (a) = 0.
2. If f is C2, and a is a local min for f , then in addition,
all eigenvalues of H(a) ≥ 0,
and if a is a local max for f , then
all eigenvalues of H(a) ≤ 0.
Here H(a) denotes the Hessian matrix of f , evaluated at the point a.3. If f is C2, then
if{
∇f (a) = 0 andall eigenvalues of H(a) > 0, then a is a local min for f
if{
∇f (a) = 0 andall eigenvalues of H(a) < 0, then a is a local max for f
March 13, 2017 31 / 50
A curve in R2 can generally be represented in two different ways:1. as the image of a function γ : I → R2, where I ⊂ R is an interval; or
2. as a level set of a function f : R2 → R. This means, as a set of the form
{x ∈ R2 : f (x) = c}
for some constant c. (or equivalently, the set of solutions of an equation f (x) = c).
A curve that is described in the first way is sometimes called a parametrizedcurve, and the function γ is called a parametrization.
exampleConsider the function γ : R→ R2 defined by
γ(t) = (3 + cos t ,2 sin t).
(If we like, we can restrict the variable t to some interval of length 2π such as [0, 2π).) Theimage of γ is exactly the set
{(x , y) ∈ R2 : f (x , y) = 1}, for f (x , y) = (x − 3)2 +14
y2.
March 13, 2017 32 / 50
-5 -4 -3 -2 -1 0 1 2 3 4 5
-3
-2
-1
1
2
3
Figure: the curve from the previous page.
March 13, 2017 33 / 50
1. If γ : I → R2 is a parametrized curve, and if γ′(t0) 6= 0, then the linearTaylor approximation to γ at t0
p1,t0 (t) = γ(t0) + (t − t0)γ′(t0)
gives a parametric form of the tangent line to γ at t0. This is the line thatpasses through γ(t0) and is parallel to γ′(t0).(If γ′(t0) = 0, then the image of the linear Taylor approximation is a point, not a line.)
2. If a curve is represented as a level set
{x ∈ R2 : f (x) = c}
and if ∇f (x0) 6= 0, then the tangent line to the curve at a point x0 is the levelset (for the same value c) of the linear Taylor approximation of f at x0, that is,the set
{x ∈ R2 : p1,x0 (x) = c}This can be rewritten (using the assumption that f (x0) = c and the fact thatp1,x0 (x) = f (x0) +∇f (x0) · (x− x0) as
{x ∈ R2 : ∇f (x0) · (x− x0) = 0}.
This is the line that passes through x0 and is orthogonal to ∇f (x0).(If ∇f (x0) = 0, then this set is not a line; rather it is all of R2.)
March 13, 2017 34 / 50
ExampleConsider the parametric curve
γ(t) = (t3, t2), t ∈ R
One can check that the image of this curve is exactly the set
{x ∈ R2 : f (x) = 0} for f (x) = f (x , y) = x2 − y3.
Although γ and f are C∞ functions, the curve has a corner at t0 = 0,corresponding to the point 0 = (0,0). Note also that γ′(0) = 0 and that∇f (0) = 0.
-1.25 -1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1
0.25
0.5
0.75
1
1.25
March 13, 2017 35 / 50
ExampleConsider the unit circle in the plane R2. The standard parametrization is viathe function
γ(t) = (cos t , sin t)
and the standard “level set" representation is
{x ∈ R2 : f (x) = 0} for f (x) = f (x , y) = x2 + y2 − 1.
Then it is not hard to find the tangent line at the point e.g. (1,0) = γ(0).————————————————————————————————But we could also describe the unit circle in different ways. For example, wecan parametrize it by the function
γ(t) = (cos t3, sin t3), −π1/3 ≤ t ≤ π1/3.
or we can write it as the “level set"
{x ∈ R2 : f (x) = 0} for f (x) = f (x , y) = (x2 + y2 − 1)2.
If we do this, then γ′(0) = 0 and ∇f (x0) = 0 for x0 = (1,0) = γ(0). So thesedescriptions of the unit circle make it hard to find the tangent line at (1,0).
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For reasons suggested by the previous example, whenever possiblewe prefer to have a parametrization γ : I → R2 (where I is an interval)such that
γ′(t) 6= 0 for all t ∈ I
Also, if we represent a curve as a level set
{x ∈ R2 : f (x) = c}
then whenever possible we prefer f to satisfy
∇f (x) 6= 0 for all points x in the curve.
But for some curves (such as curves with cusps) it is not possible tosatisfy these conditions.
March 13, 2017 37 / 50
A curve in R3 can generally be represented in two different ways:1. as the image of a function γ : I → R3, where I ⊂ R is an interval; or
2. as a level set of a function f : R3 → R2. This means, as a set of the form
{x ∈ R3 : f(x) = c}
for some vector c ∈ R2.
A curve that is described in the first way is sometimes called a parametrizedcurve, and the function γ is called a parametrization.
exampleConsider the function γ : R→ R3 defined by
γ(t) = (t cos 3t ,2t sin 3t , t).
The image of γ is exactly the set
{x ∈ R3 : f(x) =
(00
)}, f(x) =
(x2 + 1
4 y2 − z2
x − z cos 3z
)See the pictures below.
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Figure: the parametrized curve
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Figure: the parametrized curve and the set {x : f1(x) = 0}
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Figure: the parametrized curve and the set {x : f2(x) = 0}
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Figure: the parametrized curve and the sets {x : fi (x) = 0}, i = 1,2. Theirintersection is the set {x : f(x) = 0}.
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1. If γ : I → R3 is a parametrized curve, and if γ′(t0) 6= 0, then the linearTaylor approximation to γ at t0
p1,t0 (t) = γ(t0) + (t − t0)γ′(t0)
gives a parametric form of the tangent line to γ at t0. This is the line thatpasses through γ(t0) and is parallel to γ′(t0).(If γ′(t0) = 0, then the image of the linear Taylor approximation is a point, not a line.)
2. If a curve is represented as a level set
{x ∈ R3 :
(f1(x)
f2(x)
)=
(c1
c2
)}
and if {∇f1(x0),∇f1(x0)} are linearly independent, then the tangent line to thecurve at a point x0 is the set
{x ∈ R3 : Df(x0)(x− x0) = 0}.
This can also be written as{x ∈ R3 :
(∇f1(x0) · (x− x0)∇f2(x0) · (x− x0)
)=
(00
)}.
(If {∇f1(x0),∇f1(x0)} are linearly dependent, then this set is not a line.)
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For a curve in R3, whenever possible we prefer parametrizations γsuch that γ′ is never 0.
Also, if we represent a curve as a level set
{x ∈ R3 : f(x) = c}
then whenever possible we prefer f to satisfy
{∇f1(x),∇f2(x)} are linearly independent for all points x in the curve,
or equivalently,
rank(Df)(x) = 2 for all points x in the curve.
But for some curves (such as curves with corners) it is not possible tosatisfy these conditions.
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A 2-dimensional surface in R3 can be represented in 2 ways:1. as the image of a parametrization
γ : U → R3
where U is a subset of R2; or
2. As the level set of a function f : R3 → R.
ExampleLet
γ(u, v) = ((cos u)(3 + cos v), (sin u)(3 + cos v),2 sin v),
for 0 < u, v ≤ 2π. This is the same as
{x ∈ R3 : f (x) = 1}
forf(x) = ((x2 + y2)1/2 − 3)2 +
14
z2.
What does it look like?March 13, 2017 45 / 50
DefinitionA p × q matrix has full rank if its rank is min{p,q}. Thus:
if p ≤ q, then “full rank" means “rank p", which is the same assaying that the p rows are linearly independent.
if p ≥ q, then “full rank" means “rank q", which is the same assaying that the q columns are linearly independent.
If p = q, that is, for a square matrix,
full rank ⇐⇒ rows are linearly independent⇐⇒ columns are linearly independent⇐⇒ determinant 6= 0.
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SummaryA k -dimensional surface in Rn can generally be represented in 2 ways:1. as the image of a parametrization γ : U → Rn where U is a subset of Rk .If u0 ∈ U and Dγ(u0) has full rank, then the “tangent plane" to the surface atu0 is parametrized by
γ(u0) + Dγ(u0)(u− u0), u ∈ Rk
Wherever possible, we prefer a parametrization such that Dγ has full rankeverywhere. For “irregular" surfaces this is not possible however.————————————————————————————————2. As a level set {x ∈ Rn : f(x) = c} of a function f : Rn → Rn−k , orequivalently, the set of solutions of n − k equations in Rn. If f (x0) = c, andDf(x0) has full rank, then the “tangent plane" to the surface at x0 is the set
{x ∈ Rn : Df(x0)(x− x0) = 0}Wherever possible, we prefer to represent a surface via a function f such thatDf has full rank everywhere. For “irregular" surfaces this is not possible however.————————————————————————————————Warning: Sometimes “tangent space" is defined as the space of all vectors that are tangent to
the surface at a given point. This is different from the above — this is always a subspace that
contains the origin.March 13, 2017 47 / 50
Figure: The 2d surface in R3 defined in a previous example.
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Implicit Function Theorem, a special case
Assume that U is an open subset of Rn+1, and let F : U → R be a C1
function. Assume that (a,b) is a point such that
F (a,b) = 0, ∂yF (a,b) 6= 0.
Then there exist positive numbers r and h, and a C1 functionf : Br (a)→ R such that
|f (x)− b| < h for all x ∈ Br (a), and
if |x− a| < r and |y − b| < h, then F (x, y) = 0 ⇐⇒ y = f (x) .
Moreover, derivatives of f can be found by implicit differentiation. Thus
∂i f (x) = − ∂iF∂yF
(x, f (x)).
In the Theorem, Br (a) denotes the n-dimensional ball (in Rn, ratherthan in Rn+1) of radius r around a.
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Implicit Function Theorem, general case
Assume that U is an open subset of Rn+k , and let F : U → Rk be a C1
function. Assume that (a,b) ∈ U is a point such that
F(a,b) = 0, (∂yi Fj(a,b))i,j=1,...,k is invertible.
Then there exist positive numbers r and h, and a C1 functionf : Br (a)→ Rk such that
‖f (x)− b‖ < h for all x ∈ Br (a), and
if |x− a| < r and |y− b| < h, then F(x,y) = 0 ⇐⇒ y = f(x) .
Moreover, derivatives of f can be found by implicitly differentiating andsolving to find ∇f.
Above, (∂yi Fj (a,b))i,j=1,...,k denotes the k × k matrix consisting of derivatives(with respect to y1, . . . , yk ) of the components F1, . . . ,Fk of F.
March 13, 2017 50 / 50
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