where t represents some convenient parameter, say time. The derivative of the
position vector r with respect to the parameter tis defined as
dr
dt = lim∆t→ 0
∆r
∆t
= lim∆t→ 0
r (t+ ∆ t)−r (t)
∆t
(6 .49)
In component form the derivative is represented in a form where one can recognize
the previous definition of a derivative of a scalar function. One finds
dr
dt = lim∆t→ 0
[x(t+ ∆ t)ˆe 1 +y(t+ ∆ t)eˆ 2 +z(t+ ∆ t)ˆe 3 ]−[x(t)eˆ 1 +y(t)ˆe 2 +z(t)ˆe 3 ]
∆t
dr
dt
= lim∆t→ 0
[
x(t+ ∆ t)−x(t)
∆t
ˆe 1 +y(t+ ∆ t)−y(t)
∆t
eˆ 2 +z(t+ ∆ t)−z(t)
∆t
ˆe 3
]
dr
dt
=dx
dt
ˆe 1 +dy
dt
ˆe 2 +dz
dt
ˆe 3 =x′(t)ˆe 1 +y′(t)ˆe 2 +z′(t)ˆe 3
This shows that the derivative of the position vector (6.47) is obtained by differenti-
ating each component of the vector. It will be shown that this derivative represents
a vector tangent to the space curve at the point (x(t), y(t), z (t)) for any fixed value of
the parameter t. Second-order and higher order derivatives are defined as derivatives
of derivatives.
Example 6-17.
The two dimensional curve y=f(x) can be
represented by the position vector
r =r (x) = xˆe 1 +f(x)ˆe 2
with the derivative
dr
dx
=ˆe 1 +df
dx
ˆe 2 =ˆe 1 +dy
dx
ˆe 2
Note that at the point (x, f (x)) on the curve one can draw the derivative vector and
show that it lies along the tangent line to the curve at the point (x, f (x)). This shows
that the derivative dxdr is a tangent vector to the curve y=f(x).
In general, if r =r (t)is the position vector of a three dimensional curve, then the
vector drdt will be a tangent vector to the curve. This can be illustrated by drawing
the secant line through the points r (t)and r (t+ ∆ t)and showing the secant line then
approaches the tangent line as ∆tapproaches zero.
