Page 116 - Vector Analysis

112 CHAPTER 4. Vector Calculus

A third kind of coordinate chart is given as follows. Let U = (0, 2π) ˆ (0, π), and define

ψ(θ, ϕ) = (sin ϕ cos θ, sin ϕ sin θ, cos ϕ).

Then ψ : U Ñ S2zt(x, 0, z) | 0 ď x ď 1, x2 + z2 = 1u is a continuous bijection with a contin-
uous inverse. We note that for any U = (θ0, θ0 + 2π) ˆ (ϕ0, ϕ0 + π), ψ is a homeomorphism
between U and an open subset of S2.

Next, we would like to define the derivative of f when f : Σ Ñ R3 is a vector-valued

function. We first talk about what the directional derivative is. Let Σ Ď R3 be a regular

surface, p P Σ, and v P TpΣ. It is intuitive to define the directional derivative of f at p in

the direction v by

dˇ ˝ x)(t) , (4.6)
ˇ (f

dt ˇt=0

if the derivative exists, where x : (´δ, δ) Ñ Σ is a C 1-parametrization of a curve on Σ such

that x(0) = p and x 1(0) = v. The first question arising naturally is that if the derivative

in (4.6) depends on the choices of x. Suppose that y : (´δ, δ) Ñ Σ is a C 1-parametrization

of another curve on Σ such that y(0) = p and y 1(0) = v (note that the curve x((´δ, δ))

and y((´δ, δ)) in general are different). Let tV, ψu be a parametrization of Σ at p, and

q = ψ´1(p). Then the chain rule (Theorem 2.49) implies that

v = x 1(0) = dˇ ˝ ψ´1 ˝ x)(t) = (dˇ (ψ´1 ˝ )
ˇ (ψ (Dψ)(q) ˇ x)(t)
dt ˇt=0
dt ˇt=0

and similarly, v = ( d ˇ (ψ´1 ) Therefore,
(Dψ)(q) ˇ ˝ y)(t) .

dt ˇt=0

( d ˇ (ψ´1 ˝ ) = ( d ˇ )
(Dψ)(q) ˇ x)(t) (Dψ)(q) ˇ (ψ´1 ˝ y)(t) .

dt ˇt=0 dt ˇt=0

The injectivity of (Dψ)(ψ´1(p)) then shows that

dˇ (ψ´1 ˝ x)(t) = dˇ (ψ´1 ˝ y)(t) .
ˇ ˇ

dt ˇt=0 dt ˇt=0

Using the chain rule again,

dˇ ˝ x)(t) = Dddt(ˇˇˇft=˝0(ψf)˝(ψψ´˝1(ψp)´)1(˝ddxt ˇˇˇ)t(=t0)(ψ=´D1 ˝(fy˝)(ψt)))(ψ=´1dd(tpˇˇˇ)t)=(0(ddft ˇ ˝ )
ˇ (f = ˇ (ψ´1 x)(t)
ˇt=0
dt ˇt=0
˝ y)(t) .

In other words, the derivative in (4.6) is independent of the choice of x as long as x(0) = p

and x 1(0) = v. This observation implies the following