082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

Solution (a) Ifa = 3t 2 i +cos2tj, then differentiation with respect tot yields
(b)
(c)
da
dt =6ti−2sin2tj
da
∣dt
∣ = √ (6t) 2 + (−2sin2t) 2 =
d 2 a
dt 2 =6i−4cos2tj
√
36t 2 +4sin 2 2t
12.5 Differentiation of vectors 425
It is possible to differentiate more complicated expressions involving vectors provided
certain rules are adhered to. If a and b are vectors andcis a scalar, all functions
oftimet, then
d
(ca) =cda
dt dt + dc
dt a
d
dt (a+b)=da dt + db
dt
d
(a·b)=a·db
dt dt + da
dt ·b
d
dt (a×b)=a×db dt + da
dt ×b
Example12.10 Ifa = 3ti −t 2 jandb=2t 2 i +3j,verify
d
(a) (a·b)=a·db
dt dt + da
dt ·b (b) d
dt (a×b)=a×db dt + da
dt ×b
Solution (a) a·b = (3ti −t 2 j)· (2t 2 i +3j)=6t 3 −3t 2
Also
So,
d
dt (a·b)=18t2 −6t
da
dt =3i−2tj
a· db
dt
+b· da
dt
db
dt
=4ti
=(3ti−t 2 j)·(4ti) + (2t 2 i+3j)·(3i−2tj)
= 12t 2 +6t 2 −6t=18t 2 −6t
We have verified d (a·b)=a·db
dt dt + da
dt ·b.
i j k
(b) a×b=
3t −t 2 0
∣2t 2 3 0∣
=(9t+2t 4 )k
d
dt (a×b)=(9+8t3 )k