Qand the vector ∆r representing the direction of the secant line through the points
P and Q.
Letting the point Qapproach the point P one finds the direction of the secant
line vector ∆r approaches the direction of the tangent to the curve at the point P.
In this limiting process one can write
dr
ds
= lim∆s→ 0 ∆r
∆s
=dx
ds
ˆe 1 +dy
ds
ˆe 2 +dz
ds
ˆe 3 =ˆet
where ˆetrepresents a unit tangent vector to the curve. Note that this tangent vector
is a unit vector since the magnitude of this derivative is
∣∣
∣∣dr
ds
∣∣
∣∣=
√
dr
ds
·dr
ds
=
√(
dx
ds
) 2
+
(
dy
ds
) 2
+
(
dz
ds
) 2
=
√
(dx )^2 + (dy)^2 + (dz)^2
(ds)^2
= 1
since an element of arc length is given by ds^2 =dx^2 +dy^2 +dz^2. This shows the vector
dr
ds
is a unit vector which is tangent to the space curve r =r (s).
By using chain rule differentiation one can assign a geometric interpretation
to the derivative of a space curve r =r (t) which is expressed in terms of a time
parameter t. Using the chain rule one finds
dr
dt
=dr
ds
ds
dt
=vdr
ds
=vˆet=v
Here v=dsdt is a scalar called speed and represents the change in distance with respect
to time. The above equation shows the velocity vector is also tangent to the curve
at any instant of time.
Differentiation Formulas
The derivative of any vector v =v (t) is defined ∆limt→ 0 v (t+ ∆∆t)t−v (t) =dvdt .Note
the derivative of a constant vector is zero. Using the property that the limit of a
sum is the sum of the limits, the above differentiation formula indicates that each
of the components of a vector must be differentiated. Here it is assumed the unit
vectors ˆe 1 ,ˆe 1 ,ˆe 3 are fixed constants and so their derivatives are zero.
For vector functions of the parameter t
u =u(t) = u 1 (t)ˆe 1 +u 2 (t)ˆe 2 +u 3 (t)eˆ 3
v =v (t) = v 1 (t)ˆe 1 +v 2 (t)ˆe 2 +v 3 (t)ˆe 3 ,
w=w(t) = w 1 (t)ˆe 1 +w 2 (t)ˆe 2 +w 3 (t)ˆe 3
