GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 65 of 65

Marks: +1, -0

The directional derivative of

f(x, y, z)=x y z

at point

(-1,1,3)

in the direction of vector

{ {i}↖{∧}-2 { {j}↖{∧}+2 { {k}↖{∧}

Solution:

f(x, y, z)=x y z

∇ f=(i \;{∂}/{∂ x}+j \;{∂}/{∂ y}+{k}↖{∧} \;{∂}/{∂_{2}})(x y z)

=y z {i}↖{∧}+x z {j}↖{∧}+x y {k}↖{∧}

(∇ f)_{(-1,1,3)}=3 {i}↖{∧}-3 {j}↖{∧}-{k}↖{∧}

Normal vector

=\;{{i}↖{∧}-2 {j}↖{∧}+2 {k}↖{∧}}/{√{1+(-2)^{2}+(2)^{2}}}=(\;{{i}↖{∧}-2 {j}↖{∧}+2 {k}↖{∧}}/{3})

So directional derivative of

{f}

in direction of

{ {P}↖{→}

=∇ f ⋅ \;{{P}↖{→}}/{|{P}↖{→}|}

=(3 {i}↖{∧}-3 {j}↖{∧}-k) ⋅ \;{({i}↖{∧}-2 {j}↖{∧}+2 {k}↖{∧})}/{3}

=\;{7}/{3}

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