Answer to Question #240781 in Calculus for Sakshi

Question #240781

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?


1
Expert's answer
2021-09-23T00:30:51-0400

Let be

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

"d_2=\\frac{\\partial f}{\\partial y}(1,1,2);\\\\\nd_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;\\\\\n\\frac{\\partial f}{\\partial (j+k)}(1,1,2)=\\frac{d_2+d_3}{\\sqrt{2}}=3;\\\\\n\\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)=\n\\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}\n \\frac{3\\cdot \\sqrt 2}{2} \\\\\n \\frac{5\\cdot \\sqrt 2}{2} \\\\\n\\frac{\\sqrt 2}{2}\n\\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).


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