Question

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

Answer 1

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

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

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

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

$grad(\phi)=({\frac {\partial\phi} {\partial x}};{\frac {\partial\phi} {\partial y}};{\frac {\partial\phi} {\partial z}})$ at point $M$

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

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

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

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

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