Understanding Engineering Mathematics

Second and higher derivatives may be obtained in the obvious way by repeated differ-
entiation. Thus, for example:

d^2 f(t)

dt^2

=

d^2 f 1

dt^2

i+

d^2 f 2

dt^2

j+

d^2 f 3

dt^2

k

Problem 11.12

Evaluate

d^2 f

dt^2

and

d^3 f

dt^3

for f=tiYe−tjYcostk

df

dt

=i−e−tj−sintk

d^2 f

dt^2

=e−t−costk

d^3 f

dt^3

=−e−tj+sintk

When we introduced ordinary differentiation we referred it to the gradient or slope of a
curve. We can do the same for differentiation of vector functions – but it is now a little more
complicated. It is perhaps best to appeal to the example considered earlier of a vector function
r(t)describing the position of a particle at timet. Thus, letr(t)be the position vector of a
moving particleP.Astvaries the particle moves along a curve in space – see Figure 11.19.

O

P

P′

Figure 11.19Definition of derivative of a vector.

Now for two points on the curver(t),r(t+h), close to each other we have:

r(t+h)−r(t)

h

=

−→

OP′−

−→

OP

h

=

−→

PP′

h

Ash→0,P′→Pand the vector

−→

PP′becomes tangential to the curve. Also the magni-

tude

−∣−→

∣PP′

∣

∣

h

is the average speed of the particle over the intervalPP′and so ash→0this

becomes the velocity of the particle. Thus:

lim

h→ 0

(

r(t+h)−r(t)

h

)

=

dr

dt