Resolved: "At what points is the direction of fastest..."

High School
Geometry
Trigonometry
Trigonometric Functions

UkusakazaL

Answered question

2021-09-09

At what points is the direction of fastest change of the function $f (x, y) = x^{2} + y^{2} - 4 x - 8 y$ is $i + j$ .

Answer & Explanation

d2saint0

Skilled2021-09-10Added 89 answers

The required course is established by $\nabla f (x, y)$
$\nabla f (x, y) = < \frac{\partial f}{\partial x}, \frac{\partial f}{\partial} y) >$
$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^{2} + y^{2} - 4 x - 8 y)$
$= 2 x - 4$
$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x^{2} + y^{2} - 4 x - 8 y)$
$= 2 y - 8$
Thus,
$\nabla f (x, y) = < \frac{\partial f}{\partial x}, \frac{\partial f}{\partial} y) >$
$= < 2 x - 4, 2 y - 8 >$
gradf(x,y) has to be parallel to i+j
The direction vector of i+j is $< 1, 1 >$
Thus,
$< 2 x - 4, 2 y - 8 >$ has to be parallel to <1, 1>
Since in $< 1, 1 >$ the coordinates are equal, 2x-4=2y-8
$2 x - 4 = 2 y - 8$
$2 (x - 2) = 2 (y - 4)$
$x - 2 = y - 4$
$y = x + 2$

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