Example 7-5. Determine how to calculate the torsion τ of a space curve.

Solution The derivative

d
r

ds =

ˆet is a unit tangent vector to the space curve and

d^2 
r

ds^2

=dˆet

ds

=κˆen. Calculating the third derivative of the position vector with respect

to the arc length parameter gives

d^3 
r

ds^3

=kd

ˆen

ds

+dκ

ds

ˆen=κ(τˆeb−κˆet) + dκ

ds

ˆen=κτ ˆeb−κ^2 ˆet+dκ

ds

ˆen

Use the properties of the triple scalar product with

d
r

ds ·

(

d^2 
r

ds^2 ×

d^3 
r

ds^3

)

=eˆt·

(

κˆen×[κτ ˆeb−κ^2 ˆet+dκds ˆen]

)

(7 .23)

together with the cross products

ˆen׈eb=ˆet, ˆen׈et=−ˆeb, ˆen׈en=
 0

and show equation (7.23) simplifies to

d
r

ds

·

(

d^2 
r

ds^2

×

d^3 
r

ds^3

)

=eˆt·[k^2 τˆet+κ^3 ˆeb] = κ^2 τ

Using the result from the previous example that κ^2 = [x′′(s)]^2 + [y′′(s)]^2 + [z′′(s)]^2 one

can write

τ=

d
r

ds ·

(

d^2 
r

ds^2 ×

d^3 
r

ds^3

)

[x′′(s)]^2 + [y′′(s)]^2 + [z′′(s)]^2

which can also be expressed as the determinant

τ=^1

[x′′(s)]^2 + [y′′(s)]^2 + [z′′(s)]^2

∣∣

∣∣

∣∣

x′(s) y′(s) z′(s)

x′′(s) y′′(s) z′′(s)

x′′′(s) y′′′(s) z′′′(s)

∣∣

∣∣

∣∣

Example 7-6. Velocity and Acceleration.

A physical example illustrating the use of the unit tangent and normal vectors

is found in determining the normal and tangential components of the velocity and

acceleration vectors as a particle moves along a curve. If
r denotes the position

vector of the particle,
v its velocity, and
a its acceleration, then

v =d
r

dt

, 
a =d
v

dt

=d

(^2)
r
dt^2
, v^2 =
v ·
v =d
r
dt
·d
r
dt

(
ds
dt
) 2
, (7 .24)

where sis the arc length along the curve. Using chain rule differentiation gives

v =

d
r

dt =

d
r

ds

ds

dt =

ˆetv.