Question

find the directional derivative of A=4xz^3-3x^2y^2z at (2,-1,2) along 2i-3j+6k​

Answer 1

Directional derivative

Solution. The given function is

$\quad\quad A=4xz^{3}-3x^{2}y^{2}z$

The unit vector along $(2\hat{i}-3\hat{j}+6\hat{k})$ is

$\quad\quad\hat{a}=\frac{2\hat{i}-3\hat{j}+6\hat{k}}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}$

$\quad\quad\quad=\frac{2\hat{i}-3\hat{j}+6\hat{k}}{\sqrt{4+9+36}}$

$\quad\quad\quad=\frac{2\hat{i}-3\hat{j}+6\hat{k}}{\sqrt{49}}$

$\quad\quad\quad=\frac{1}{7}(2\hat{i}-3\hat{j}+6\hat{k})$

$\therefore grad(A)=\hat{i}\frac{\partial}{\partial x}(4xz^{3}-3x^{2}y^{2}z)\\ \quad+\hat{j}\frac{\partial}{\partial y}(4xz^{3}-3x^{2}y^{2}z)\\ \quad+\hat{k}\frac{\partial}{\partial z}(4xz^{3}-3x^{2}y^{2}z)$

$=(4z^{3}-6xy^{2}z)\hat{i}-6x^{2}yz\hat{j}\\ \quad+(12xz^{2}-3x^{2}y^{2})\hat{k}$

Now, $\frac{dA}{ds}$

$=grad(A).\hat{a}$

$=\{(4z^{3}-6xy^{2}z)\hat{i}-6x^{2}yz\hat{j}\\ \quad+(12xz^{2}-3x^{2}y^{2})\hat{k}\}.\frac{1}{7}(2\hat{i}-3\hat{j}+6\hat{k})$

$=\frac{1}{7}(8y^{3}-12xy^{2}z+18x^{2}yz+72xz^{2}-18x^{2}y^{2})$ at the point $(x,\:y,\:z)$

$=\frac{1}{7}(304)$ at the point $(2,\:-1,\:2)$

Answer:

Required directional derivative is $\frac{304}{7}$.

Answer 2

Answer:

Step-by-step explanation:

we know directional derivative (DD)=delphi.n^

so by solving Delphi value and n vector we substitute in the formula. Good luck