3000

1. Exercise -1(A) : If r = sinti = costj = tk, find
(i)
𝑑𝑟
𝑑𝑡
(ii)
𝑑2
𝑟
𝑑𝑡2 (iii)|
𝑑𝑟
𝑑𝑡
| (iv)|
𝑑2
𝑟
𝑑𝑡2 |
Solution : (i)
𝑑𝑟
𝑑𝑡
=
𝑑
𝑑𝑡
(sinti = costj = tk)
= costi-sinti+tk
(ii)
𝑑2
𝑟
𝑑𝑡2 =
𝑑𝑟
𝑑𝑡
{
𝑑𝑟
𝑑𝑡
(sinti + costj + tk)}
𝑑𝑟
𝑑𝑡
(cosi - sintj + tk)
=-sinti-costj
(iii)
𝑑𝑟
𝑑𝑡
= cosi + sintj + tk
|
𝑑𝑟
𝑑𝑡
| = √𝑐𝑜𝑠2 𝑡 + 𝑠𝑖𝑛𝑡 𝑡 + 12 = √1 + 1 = √2
(iv) |
𝑑2
𝑟
𝑑𝑡2 |=√𝑠𝑖𝑛2 𝑡 + 𝑐𝑜𝑠2 𝑡 = √1 = 1
Solution : Let –r,be the position vector of the particle at any time t.
R=xi+yj+zk
=𝑒−𝑡
𝑖 + 2𝑐𝑜𝑠3𝑡𝑗 + 2𝑠𝑖𝑛3𝑡𝑘
V=
𝑑𝑟
𝑑𝑡
=
𝑑
𝑑𝑡
(𝑒−𝑡
+ 2𝑐𝑜𝑠3𝑡𝑗 + 2𝑠𝑖𝑛3𝑡𝑘)
=𝑒−𝑡
𝑖 − 6𝑠𝑖𝑛3𝑡𝑗 + 6𝑐𝑜𝑠3𝑡𝑘
𝑎 =
𝑑2
𝑟
𝑑𝑡2
=
𝑑
𝑑𝑡
(−𝑒−𝑡
𝑖 − 6𝑐𝑜𝑠3𝑡𝑗 + 6𝑐𝑜𝑠3𝑡𝑘)
=𝑒−𝑡
𝑖 − 18𝑐𝑜𝑠3𝑡𝑗 − 18𝑠𝑖𝑛3𝑡𝑘 t=0,
At t=0,
V=-i+6k
| 𝑣| = √−12 + 62 = √37
At t= 0, a= i-18 cos 0.j-18 sin 3.0k
=i-0-18
| 𝑎|= √(1)2 + (−18)2 = √1 + 324 = √325