Understanding Engineering Mathematics

Letf(t)=cbe a constant vector. Thenf(t+h)=c, and so from the above definition

df

dt

=lim

h→ 0

(

f(t+h)−f(t)

h

)

=lim

h→ 0

(

c−c

h

)

= 0

The properties of differentiation of a vector look very much like those of ordinary
differentiation, although the notation may be a bit unnerving at first. We have three types
of product to contend with of course – multiplication by a scalar, scalar product and vector
product. On the other hand we don’t have to worry about a quotient rule, since we have
not defined division of vectors. The rules are as follows.
Ifs(t)is a scalar function andf(t)a vector function, then

(i)

d

dt

(s(t)f(t))=

ds(t)

dt

f(t)+s(t)

df(t)

dt

Ifg(t)is also a vector function, then

(ii)

d

dt

(f(t)+g(t))=

df(t)

dt

+

dg(t)

dt

(iii)

d

dt

(f·g)=f·

dg

dt

+

df

dt

·g

(iv)

d

dt

(f×g)=f×

dg

dt

+

df

dt

×g

where the order of vectors

must be preserved

(i) and (ii) show that differentiation of the component form is simple:

d

dt

(f 1 (t)i+f 2 (t)j+f 3 (t)k)=

df 1 (t)

dt

i+

df 2 (t)

dt

j+

df 3 (t)

dt

k

Problem 11.11

Differentiate the following vector functions with respect tot

(i) cos 4tiY3sin4tjYetk

(ii) tcostiYe−tsintjYt^2 k

(i)

d

dt

(cos 4ti+3sin4tj+etk)

=

d

dt

(cos 4t)i+

d

dt

(3sin4t)j+

d

dt

(et)k

=−4sin4ti+12 cos 4tj+etk

(ii)

d

dt

(tcosti+e−tsintj+t^2 k

=

d

dt

(tcost)i+

d

dt

(e−tsint)j+

d

dt

(t^2 )k

=(cost−tsint)i+(e−tcost−e−tsint)j+ 2 tk