Answer to Question #111585 in Electricity and Magnetism for Ram

Question #111585
Obtain the directional derivative for a scalar field ϕ(x, y, z)=(3x^2y-y^3z^2) at the point (1, -2, -1) in the direction Î+ ĵ +k̂
1
Expert's answer
2020-04-23T12:44:17-0400

scalar field ϕ(x,y,z)=3x2yy3z2\phi (x,y,z)=3x^2y-y^3z^2

point M(1;2;1)M(1;-2;-1)

direction i+j+k\vec{i}+\vec{j}+\vec{k} with normalized vector n(13;13;13)\vec{n}({\frac 1 {\sqrt3}};{\frac 1 {\sqrt3}};{\frac 1 {\sqrt3}})

To find directional derivative we need to get scalar product grad(ϕ)ngrad(\phi )\cdot \vec{n}

grad(ϕ)=(ϕx;ϕy;ϕz)grad(\phi)=({\frac {\partial\phi} {\partial x}};{\frac {\partial\phi} {\partial y}};{\frac {\partial\phi} {\partial z}}) at point MM

ϕx=6xy=12{\frac {\partial\phi} {\partial x}}=6xy=-12

ϕy=3x23y2z2=9{\frac {\partial\phi} {\partial y}}=3x^2-3y^2z^2=-9

ϕz=2y3z=16{\frac {\partial\phi} {\partial z}}=-2y^3z=-16

Then grad(ϕ)=(12;9;16)grad(\phi)=(-12;-9;16)

Directional derivative equal grad(ϕ)n=12139131613=373grad(\phi )\cdot \vec{n}=-12\cdot{\frac 1 {\sqrt3}}-9\cdot{\frac 1 {\sqrt3}}-16\cdot{\frac 1 {\sqrt3}}=-{\frac {37} {\sqrt3}}


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