Question

Let f(x, y, z) be a differentiable function. At the point (1, 1, 2), the directional derivative is 4,3,2 in the direction i + j, j + k and i + k, respectively.

a) Find the directional derivative at the point (1, 1, 2) in the direction 3i + 3j + 3k.

b) Compute ∇f(1, 1, 2).

c) In which direction does the function f increase most rapidly? In which direction does the

function f decreases most rapidly?

Answer 1

Let be

d₁=$\frac{\partial f}{\partial x}(1,1,2);$

$d_2=\frac{\partial f}{\partial y}(1,1,2);\\ d_3=\frac{\partial f}{\partial z}(1,1,2);\\$

Then $\frac{\partial f}{\partial e}(1,1,2)=\frac{d_1\cdot e_1+d_2\cdot e_2+d_3\cdot e_3}{\sqrt{e_1^2+e_2^2+e_3^2}}$ - derivative along vector e

Therefore we have

$\frac{\partial f}{\partial (i+j)}(1,1,2)=\frac{d_1+d_2}{\sqrt{2}}=4;\\ \frac{\partial f}{\partial (j+k)}(1,1,2)=\frac{d_2+d_3}{\sqrt{2}}=3;\\ \frac{\partial f}{\partial (i+k)}(1,1,2)=\frac{d_1+d_3}{\sqrt{2}}=2;\\$

This is the system to solve.

For let we add all eqations:

$\frac{2}{\sqrt 2}\cdot(d_1+d_2+d_3)=9;$

From it we have $d_1+d_2+d_3=\frac{9\cdot \sqrt 2}{2}$

First equation from the system can be written as $d_1+d_2=\frac{8\cdot \sqrt 2}{2}$

Forming difference of two kast equation we have $d_3=\frac{\sqrt 2}{2}$ ;

Analogically we do next

$d_2+d_3=\frac{6\cdot \sqrt 2}{2}$ and $d_1=\frac{3\cdot \sqrt 2}{2}$ .

$d_1+d_3=\frac{4\cdot \sqrt 2}{2}$ , $d_2=\frac{5\cdot \sqrt 2}{2}$ .

As a consequence we have

$\frac{\partial f}{\partial (3i+3j+3k)}(1,1,2)=\frac{\partial f}{\partial (i+j+k)}(1,1,2)= \frac{d_1+d_2+d_3}{\sqrt{3}}=\frac{\frac{9\cdot \sqrt 2}{2}}{\sqrt 3}=\frac{3\cdot \sqrt 6}{2}$

thus a) is done.

b) Gradient $\nabla f(1,1,2)=\begin{pmatrix} \frac{3\cdot \sqrt 2}{2} \\ \frac{5\cdot \sqrt 2}{2} \\ \frac{\sqrt 2}{2} \end{pmatrix}$

c) Function f in point (1,1,2) increase most rapidly in direction of $\nabla f(1,1,2)$ or along vector (3,5,1), correspondingly it decreases most rapidly in inverse direction or along vector (-3,-5,-1).