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Differentiation of
Vectors
Introduction
The area of mathematics known as vector calculus is used to model mathematically a vast range of
engineering phenomena including electrostatics, electromagnetic fields, air flow around aircraft and
heat flow in nuclear reactors. In this Section we introduce briefly the differential calculus of vectors.
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Prerequisites
Before starting this Section you should...
- have a knowledge of vectors, in Cartesian
form
- be able to calculate the scalar and vector
products of two vectors
- be able to differentiate and integrate scalar
functions
Learning Outcomes
On completion you should be able to...
54 HELM (2008):
Workbook 12: Applications of Differentiation
®
1. Differentiation of vectors
Consider Figure 31.
P
C
r
x
y
Figure 31
If r represents the position vector of an object which is moving along a curve C, then the position
vector will be dependent upon the time, t. We write r = r(t) to show the dependence upon time.
Suppose that the object is at the point P , with position vector r at time t and at the point Q, with
position vector r(t + δt), at the later time t + δt, as shown in Figure 32.
P
x
y
Q
r(t)
r(t + δt)
P Q
Figure 32
Then
P Q represents the displacement vector of the object during the interval of time δt. The length
of the displacement vector represents the distance travelled, and its direction gives the direction of
motion. The average velocity during the time from t to t + δt is defined as the displacement vector
divided by the time interval δt, that is,
average velocity =
P Q
δt
r(t + δt) − r(t)
δt
If we now take the limit as the interval of time δt tends to zero then the expression on the right
hand side is the derivative of r with respect to t. Not surprisingly we refer to this derivative as
the instantaneous velocity, v. By its very construction we see that the velocity vector is always
tangential to the curve as the object moves along it. We have:
v = lim δt→ 0
r(t + δt) − r(t)
δt
dr
dt
HELM (2008):
Section 12.5: Differentiation of Vectors
®
Example 6
If w = 3t
2 i + cos 2tj, find
(a)
dw
dt
(b)
dw
dt
(c)
d
2 w
dt
2
Solution
(a) If w = 3t
2 i + cos 2tj, then differentiation with respect to t yields:
dw
dt
= 6ti − 2 sin 2tj
(b)
dw
dt
(6t)^2 + (−2 sin 2t)^2 =
36 t^2 + 4 sin
2 2 t
(c)
d
2 w
dt^2
= 6i − 4 cos 2tj
It is possible to differentiate more complicated expressions involving vectors provided certain rules
are adhered to as summarized in the following Key Point.
Key Point 9
If w and z are vectors and c is a scalar, all these being functions of time t, then:
d
dt
(w + z) =
dw
dt
dz
dt
d
dt
(cw) = c
dw
dt
dc
dt
w
d
dt
(w · z) = w ·
dz
dt
dw
dt
· z
d
dt
(w × z) = w ×
dz
dt
dw
dt
× z
HELM (2008):
Section 12.5: Differentiation of Vectors
Example 7
If w = 3ti − t
2 j and z = 2t
2 i + 3j, verify the result
d
dt
(w · z) = w ·
dz
dt
dw
dt
· z
Solution
w · z = (3ti − t
2 j) · (2t
2 i + 3j) = 6t
3 − 3 t
2 .
Therefore
d
dt
(w · z) = 18t
2 − 6 t (1)
Also
dw
dt
= 3i − 2 tj and
dz
dt
= 4ti
so w ·
dz
dt
dw
dt
= (3ti − t
2 j) · (4ti) + (2t
2 i + 3j) · (3i − 2 tj)
= 12 t
2
2 − 6 t
= 18 t
2 − 6 t (2)
We have verified
d
dt
(w · z) = w ·
dz
dt
dw
dt
· z since (1) is the same as (2).
Example 8
If w = 3ti − t
2 j and z = 2t
2 i + 3j, verify the result
d
dt
(w × z) = w ×
dz
dt
dw
dt
× z
Solution
w × z =
i j k
3 t −t
2 0
2 t
2 3 0
= (9t + 2t
4 )k implying
d
dt
(w × z) = (9 + 8t
3 )k (1)
w ×
dz
dt
i j k
3 t −t
2 0
4 t 0 0
= 4t
3 k (2)
dw
dt
× z =
i j k
3 − 2 t 0
2 t
2 3 0
= (9 + 4t
3 )k (3)
We can see that (1) is the same as (2) + (3) as required.
58 HELM (2008):
Workbook 12: Applications of Differentiation
