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

426 Chapter 12 Applications of differentiation
Also,
a × db
dt
i j k
=
3t −t 2 0
∣4t 0 0∣
= 4t 3 k
∣ ∣∣∣∣∣ i j k
da
dt ×b = 3 −2t 0
2t 2 3 0∣
=(9+4t 3 )k
and so
a × db
dt + da
dt ×b=4t3 k +(9+4t 3 )k=(9+8t 3 )k = d dt (a×b)
as required.
EXERCISES12.5
1 Ifr =3ti+2t 2 j+t 3 k,find
dr d 2 r
(a) (b)
dt dt 2
2 GivenB =te −t i+costjfind
dB d 2 B
(a) (b)
dt dt 2
3 Ifr =4t 2 i+2tj−7kevaluaterand dr
dt
4 Ifa=t 3 i−7tk,andb=(2+t)i+t 2 j−2k,
(a) finda ·b
(b) find da
dt
whent = 1.
(c) find db
dt
(d) showthat d dt (a·b)=a·db
dt + da
dt ·b.
5 Givenr =sinti+costj,find
(a)r (b)r (c) |r|
Show thatthe position vectorandvelocityvector are
perpendicular.
6 Show r = 3e −t i + (2 +t)jsatisfies
¨r +ṙ =j
7 Givena=t 2 i−(4−t)j,b=i+tjshow
( ) ( )
d
(a)
dt (a×b)= a × db da
+
dt dt ×b
(b)
d
(a·b)=a·db da
+b ·
dt dt dt
Solutions
1 (a) 3i+4tj+3t 2 k
(b) 4j+6tk
2 (a) (−te −t +e −t )i −sintj
(b) e −t (t −2)i −costj
3 4i+2j−7k,8i+2j
4 (a) t(t 3 +2t 2 +14) (b) 3t 2 i−7k
(c) i+2tj
5 (a) costi−sintj (b) −sinti−costj (c) 1