vector

Vectors 413
Now we have to find such that F
=
We know that
d = dx
dy
dz
x y z
[Total differential coefficient]
=
i j k .( i dx jdy kdz)
x y z
=
i j k .( i dx jdy kdz)
= .( idx jdy kdz)
x y z
= F.( i dx jdy kdz)
= [( x 2y 4 z) i (2x3 y z) j (4x y 2 z) k)].( i dx jdy kdz)
= (x + 2y + 4z) dx + (2x – 3y – z) dy + (4x – y + 2z) dz
= x dx – 3y dy + 2z dz + (2y dx + 2x dy) + (4z dx + 4x dz) + (–z dy – y dz)
( zdy
ydz)
= xdx3 ydy 2 zdz (2ydx 2 xdy) (4zdx 4 xdz)
=
2 2
x 3y
+ z 2 + 2xy + 4zx – yz + c Ans.
2 2
Example 55. Let V (x, y, z) be a differentiable vector function and (x, y, z) be a scalar
function. Derive an expression for div ( V
) in terms of .V , div V and .
(U.P. I Semester, Winter 2003)
Solution. Let V = V 1 i V 2 j V 3 k
div ( V
) = .( F)
= i j k .[ V1 i V2 j V3
k]
= ( V1) ( V2) ( V3)
x y z
x y z
V1 V2
V3
= V1 V2 V3
x x y y
z z
V1 V2
V3
= V1 V2 V3
x y z
x y z
= i j k .( V1 i V2 j V3
k)
i j k .( V1 i V2 j V3
k)
x y z
x y z
= ( . V) ( ). V (div V) (grad ).
V
Ans.
Example 56. If A is a constant vector and R = x î + y ĵ + z ˆk , then prove that
Curl A.
R
A A R
(K. University, Dec. 2009)
Solution. Let A = A1 î + A 2
ĵ + A 3
ˆk , R = xî + y ĵ + z ˆk
A.
R (A 1 î + A 2
ĵ + A 3
ˆk ) . (x î + y ĵ + z ˆk ) = A 1
x + A 2
y + A 3
z
[ A. R]
R = (A1 x + A 2
y + A 3
z) (x î + y ĵ + z ˆk )
= (A 1
x 2 + A 2
xy + A 3
zx) î + (A 1
xy + A 2
y 2 + A 3
yz) ĵ + (A 1 xz + A 2 yz + A 3 z2 ) ˆk