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

850 Chapter 26 Vector calculus
Partial differentiation w.r.t.yandzis defined in a similar way, as are higher derivatives.
For example,
∂ 2 v
∂x 2 = ∂2 v x
∂x 2 i + ∂2 v y
∂x 2 j + ∂2 v z
∂x 2 k
Example26.1 Ifv = 3x 2 yi +2xyzj −3x 4 y 2 k, find ∂v
∂x , ∂v
∂y , ∂v
∂z . Further, find ∂2 v
∂x 2 and ∂2 v
∂x∂z .
Solution We find
∂v
∂x = 6xyi +2yzj −12x3 y 2 k
∂v
∂y = 3x2 i +2xzj −6x 4 yk
∂v
∂z = 2xyj
∂ 2 v
∂x 2 = 6yi −36x2 y 2 k
∂ 2 v
∂x∂z = 2yj
EXERCISES26.2
1 Givenv = 2xi +3yzj +5xz 2 kfind
(a)
∂v
∂x
(b)
∂v
∂y
(c)
2 Iff =2i−xyzj+3x 2 zkfind
(a)
(d)
∂f
∂x
∂ 2 f
∂x 2
(b)
∂f
∂y
(e) ∂2 f
∂y 2
(c)
(f)
∂v
∂z
∂f
∂z
∂ 2 f
∂z 2
3 GivenE = (x 2 +y)i + (1 −z)j + (x +2z)kfind
(a)
(d)
∂E
∂x
∂ 2 E
∂x 2
(b)
(e)
∂E
∂y
∂ 2 E
∂y 2
(c)
(f)
∂E
∂z
∂ 2 E
∂z 2
4 Ifv=3xyzi+(x 2 −y 2 +z 2 )j+(x+y 2 )kfind
∂v
∂x , ∂v
∂y , ∂v
∂z , ∂2 v ∂2 v ∂2 v
∂x 2, ∂y 2,and ∂z 2.
5 Ifv = sin(xyz)i +ze xy j −2xykfind
∂v
∂x , ∂v
∂y , ∂v
∂z .
6 Ifv =xi+x 2 yj−3x 3 k,and φ =xyz,find
φv, ∂ ∂φ
(φv),
∂x ∂x , ∂v .Deduce that
∂x
∂
(φv) = φ∂v
∂x ∂x + ∂φ
∂x v
7 Ifv = ln(xy)i +2xycoszj −x 4 yzk,find
∂v
∂x , ∂v
∂y , ∂v
∂z , ∂2 v
∂x 2 , ∂2 v
∂y 2 , and ∂2 v
∂z 2.
Solutions
1 (a) 2i +5z 2 k (b) 3zj
(c) 3yj +10xzk
2 (a) −yzj +6xzk (b) −xzj
(c) −xyj +3x 2 k
(d) 6zk
(e) 0 (f) 0