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

March 13, 2017 14 / 50

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.

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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.

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