Why this maximization approach fails?

[friendbobbiny]

Homework Statement

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

Homework Equations

$$<\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}>$$

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 $$f(x,y) = x^2 + y^2 -2x - 2y$$is in the direction of <1,1>.

Homework Equations

$$<\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}>$$

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

$$<\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}>$$

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":
$$\nabla f = <\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}>$$

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.