Question

$\orange{\rm GATE\ 2019\ EE}$
If A = 2xi + 3yj + 4zk and u = x² + y² + z² , then div(uA) at (1,1,1) is ? ( 2 m )

Answer 1

Given A = 2xi + 3yj + 4zk and u = x²+y²+z²

Note : div(A) = ∇.A and grad(u) = ∇A

We have a vector identity ,

∇.(fB) = ∇f.B + f∇.B

where ,

f is a scalar and B is a vector quantity

Here we have a scalar quantity , u = x²+y²+z² and vector quantity A = 2xi+3yj+4zk

So , ∇.(uA) = ∇u.A + u∇.A

Lets calculate ∇u ,

$\longrightarrow \sf \qquad grad(u)$

$\longrightarrow \sf \dfrac{\partial u}{\partial x} i + \dfrac{\partial u}{\partial y} j + \dfrac{\partial u}{\partial z} k$

$\longrightarrow \sf 2x i+2yj+2zk$

Lets calculate ∇.A

$\longrightarrow \sf \qquad div(A)$

$\longrightarrow \sf \dfrac{\partial }{\partial x}(2x) + \dfrac{\partial }{\partial x}(3y)+\dfrac{\partial }{\partial x}(4z)$

$\longrightarrow \quad 2+3 + 4$

$\longrightarrow \quad 9$

So ∇.(uA) ,

$\longrightarrow$ ∇u.A + u∇.A

$\longrightarrow \sf (2xi+2yj+2zk).(2xi+3yj+4zk) + (x^2+y^2+z^2)(9)$

$\longrightarrow \sf (4x^2+6y^2+8z^2) + (9x^2+9y^2+9z^2)$

$\longrightarrow \sf 13x^2+15y^2+17z^2$

At (1,1,1)

$\longrightarrow \sf 13(1)^2+15(1)^2+17(1)^2$

$\longrightarrow \sf 45$

So , div(uA) at (1,1,1) is 45