Question

Find all the points at which the direction of fastest change of the function f(x, y) = x2 + y2 − 4x − 2y is i + j.

Answer 1

Answer:

All the points that lie on the line:

$y(x)=x-1$

Step-by-step explanation:

In order to find the maximum rate of change, we need to find the gradient of the function. The gradient of a function of two variables is defined to be:

$\nabla f =\frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j$

So:

$\frac{\partial f}{\partial x} = 2x-4=2(x-2)\\\\\frac{\partial f}{\partial y} = 2y-2=2(y-1)$

Hence:

$\nabla f(x,y)=2(x-2)i+2(y-1)j$

Since we need to find all the points at which the direction of fastest change of the function $f(x,y)=x^{2} +y^{2} -4x-2$ is $i+j$. Then:

$2(x-2)i+2(y-1)=ai+aj\\\\a>0$

Therefore:

$2(x-2)=2(y-1)\\\\Solving\hspace{3}for\hspace{3}y\\\\y=x-1$

So, we can conclude, that all the points where the direction of fastest change of $f(x,y)$ lie on the line:

$y(x)=x-1$