If A=2 x i+3 y j+4 z k and u=x^{2}+y^{2}+z^{2}, then div(u A) at (1,1,1) is

GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 37 of 65

Marks: +1, -0

A=2 x i+3 y j+4 z k

and

u=x^{2}+y^{2}+z^{2}

, then

div(u A)

(1,1,1)

Solution:

Given the vector

\ov{ {A}}=2 {x} { {i}↖{∧}+3 {y} { {j}↖{∧}+4 {z} { {k}↖{∧}

Scalar

u=x^{2}+y^{2}+z^{2}

we know

∇ ⋅(\table {u} , \ov{ {A}} ; ↓ , ↓; \;\text " salar "\; , \;\text " vectar "\;)=(∇ {u}) ⋅ \ov{ {A}}+ {u}(∇ ⋅ \ov{ {A}})

∇ {u}=\;{∂ {u}}/{∂ {x}} { {i}↖{∧}+\;{∂ {u}}/{∂ {y}} { {j}↖{∧}+\;{∂ {u}}/{∂ {z}} { {k}↖{∧}

(∇ u)=(2 x {i}↖{∧}+2 y {j}↖{∧}+2 z {k}↖{∧})

(∇ u) ⋅ \ov{A}=(2 x {i}↖{∧}+2 y {j}↖{∧}+2 z {k}↖{∧}) ⋅(2 x {i}↖{∧}+3 y {j}↖{∧}+4 z {k}↖{∧})

=4 x^{2}+6 y^{2}+8 z^{2}

∇ ⋅ \ov{ {A}}=\;{∂ {Ax}}/{∂ {x}} {i}+\;{∂ {Ay}}/{∂ {y}} {j}+\;{∂ {Az}}/{∂ {z}} {k}

∇ ⋅ \ov{ {A}}=2+3+4=9

∇ ⋅( {u} \ov{ {A}})=(∇ {u}) ⋅ \ov{ {A}}+ {u}(∇ ⋅ \ov{ {A}})

=4 {x}^{2}+6 {y}^{2}+8 {z}^{2}+( {x}^{2}+ {y}^{2}+ {z}^{2}) 9

∇ ⋅( {u} \ov{ {A}}) ∣=13 {x}^{2}+15 {y}^{2}+17 {z}^{2}

∇ ⋅( {u} \ov{ {A}})|_{(1,1,1)}=13+15+17

\;\; =45

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