Page 140 - Intro to Tensor Calculus

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Observe that by differentiating the relations in equation (1.5.30), with respect to both u and v, one can
derive the results
2
∂ ~r ∂~ r ∂bn
e = · bn = − · = b 11
∂u 2 ∂u ∂u
2
r
∂ ~r ∂~ ∂bn ∂bn ∂~ r
f = · bn = − · = − · = b 21 = b 12 (1.5.31)
∂u∂v ∂u ∂v ∂u ∂v
2
∂ ~r ∂~ ∂bn
r
g = · bn = − · = b 22
∂v 2 ∂v ∂v
and consequently the curvature tensor can be expressed as
∂~ r ∂bn
b αβ = − · . (1.5.32)
∂u α ∂u β
The quadratic forms from equations (1.5.20) and (1.5.28) enable us to represent the normal curvature
in the form of a ratio of quadratic forms. We find from equation (1.5.26) that the normal curvature in the
direction du is
dv
2
B e(du) +2fdu dv + g(dv) 2
κ (n) = = . (1.5.33)
2
A E(du) +2Fdu dv + G(dv) 2
r
If we write the unit tangent vector to the curve in the form T = d~ r = ∂~ du α and express the derivative
b
ds ∂u α ds
n
n
of the unit surface normal with respect to arc length as db = ∂b du β , then the normal curvature can be
ds ∂u β ds
expressed in the form
α β
r
dbn ∂~ ∂bn du du
κ (n) = −T · = − α · β
b
ds ∂u ∂u ds ds
(1.5.34)
α
α
b αβ du du β b αβ du du β
= = .
α
ds 2 a αβ du du β
Observe that the curvature tensor is a second order symmetric tensor.
In the previous discussions, the plane containing the unit normal vector was arbitrary. Let us now
consider all such planes that pass through this unit surface normal. As we vary the plane containing the unit
surface normal bn at P we get different curves of intersection with the surface. Each curve has a curvature
associated with it. By examining all such planes we can find the maximum and minimum normal curvatures
associated with the surface. We write equation (1.5.33) in the form
e +2fλ + gλ 2
κ (n) = (1.5.35)
E +2Fλ + Gλ 2
where λ = dv . From the theory of proportions we can also write this equation in the form
du
(e + fλ)+ λ(f + gλ) f + gλ e + fλ
κ (n) = = = . (1.5.36)
(E + Fλ)+ λ(F + Gλ) F + Gλ E + Fλ
Consequently, the curvature κ will satisfy the differential equations

(e − κE)du +(f − κF)dv =0 and (f − κF)du +(g − κG)dv =0. (1.5.37)

dκ (n)
The maximum and minimum curvatures occur in those directions λ where =0. Calculating the deriva-
dλ
tive of κ (n) with respect to λ and setting the derivative to zero we obtain a quadratic equation in λ
2
(Fg − Gf)λ +(Eg − Ge)λ +(Ef − Fe)= 0, (Fg − Gf) 6=0.