By using the results tan θ=dy
dxand ds^2 =dx^2 +dy^2 ,one can calculate the derivatives
dθ
dx =d^2 y
dx^2
1 +(dy
dx) 2 and
ds
dx =±√
1 +(
dy
dx) 2
.The chain rule for differentiation can be employed to calculate the curvature
κ=∣∣
∣∣dθ
ds∣∣
∣∣=∣∣
∣∣dθ
dxdx
ds∣∣
∣∣=∣∣
∣d(^2) y
dx^2
∣∣
∣
[
1 +
(dy
dx
) 2 ]^32. (6 .67)
The unit tangent vector ˆet satisfies ˆet·ˆet= 1.Differentiating this relation with
respect to arc length sand simplifying produces
ˆet·dˆet
ds +dˆet
ds ·ˆet= 2 ˆet·dˆet
ds = 0. (6 .68)When the dot product of two nonzero vectors is zero, the two vectors are perpendic-
ular to one another. Hence, the vector d
ˆet
dsis perpendicular to the tangent vector ˆet
when evaluated at a common point on the curve. It is known that the vector ˆenis
perpendicular to the tangent vector. The vectors ˆenand dˆet
dsare therefore colinear.
Consequently there exists a suitable constant csuch that
dˆet
ds =cˆen.It is now demonstrated that c=κ, the curvature associated with the curve. To
solve for the constant cdifferentiate ˆet with respect to the arc length s. From the
expression
dˆet
dsds
dx =dˆet
dx =√
1 + [f′(x)]^2 f′′(x)ˆe 2 −[ˆe 1 +f′(x)ˆe 2 ][1 + [f′(x)]^2 ]−^12 f′(x)f′′(x)
1 + [f′(x)]^2 ,the derivative of the unit tangent vector with respect to arc length is given by
dˆet
ds =f′′(x)
[1 + [f′(x)]^2 ]^32[
−√f′(x)eˆ 1 +ˆe 2
1 + [f′(x)]^2]
=f′′(x)
[1 + [f′(x)]^2 ]^32ˆen.Taking the absolute value of both sides of this equation shows that the scalar cur-
vature κis a function of position and is given by
κ= |f′′(x)|
[1 + [f′(x)]^2 ]^32.