Why this maximization approach fails?

[friendbobbiny]

Homework Statement

Find all points at which the direction of fastest change of the function f(x,y) = x^2 + y^2 -2x - 2yis in the direction of <1,1>.

Homework Equations

&lt;\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}&gt;

The Attempt at a Solution

\frac{\nabla f}{|\nabla f|} = <1,1>

This doesn't work but

\frac{\delta f}{\delta x} = \frac{\delta f}{\delta y}

does. This latter approach makes sense. The partial derivatives at x and y have the same value. The former approach should work, however. In general, \nabla f gives the direction of maximum change.

 

[friendbobbiny]

NVM

set \frac{\nabla f}{|\nabla f|} to <root(2)/2, root(2)/2>

Resolved!

 

[Mark44]

Homework Statement

Find all points at which the direction of fastest change of the function f(x,y) = x^2 + y^2 -2x - 2yis in the direction of <1,1>.

Homework Equations

&lt;\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}&gt;

Your function is one with two variables, so the gradient should be a vector with two components, not three.
$$\nabla f = <\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>$$

The Attempt at a Solution

\frac{\nabla f}{|\nabla f|} = <1,1>

This doesn't work but

\frac{\delta f}{\delta x} = \frac{\delta f}{\delta y}

does. This latter approach makes sense. The partial derivatives at x and y have the same value. The former approach should work, however. In general, \nabla f gives the direction of maximum change.

Do you have a feel for what the surface for this function looks like? The graph of the surface is a paraboloid that opens upward.

 

[HallsofIvy]

Homework Statement

Find all points at which the direction of fastest change of the function f(x,y) = x^2 + y^2 -2x - 2<b>y </b>is in the direction of <1,1>.

Homework Equations

&lt;\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}&gt;

Just a note on Latex. First, your \nabla F belongs outside the < > denoting the vector. Second you can use "\partial" to indicate partial derivatives rather than "\delta":
\nabla f = &lt;\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}&gt;

3. The Attempt at a Solution

\frac{\nabla f}{|\nabla f|} = <1,1>

This doesn't work but

\frac{\delta f}{\delta x} = \frac{\delta f}{\delta y}

does. This latter approach makes sense. The partial derivatives at x and y have the same value. The former approach should work, however. In general, \nabla f gives the direction of maximum change.