Taking z= C a constant, you have x^2+ 4(y-1)^2= C. What kind of conic section is that?

"Upside down triangle"? Doesn't your text book tell you what the "gradient" is? And I doubt that the problem really says "show where it is on the graph". exists at2) Find/_\f{upside down triangle in front of'f'} and show where it is on graph.everypoint on the graph. I expect the problem asks you to draw at a specific point on the level curves. It is a vectorperpendicularto the level curve at point.

No, it is not. For one thing, that is NOT a vector! Didn't it occur to you that the directional vector should be aso fx = 2x

fy = 8y - 8

3) Determine the directional derivative of'f'at (1, 1) in the direction of (1, 1).

so fx = 2x

fy = 8y - 8

sub in (1, 1) and i get (2, 0)

is directional vector 1/(1^2 + 1^2)= 1/2 ?vector?

Thelengthof the directional vector, in the direction of theunitvector u, is . That's almost what you have except that (1, 1) has length , not 1, so it is not a unit vector. After you have corrected that, multiply by the unit vector in that direction.

[quoteWhere do i go from here?

4) Determine the maximum and minimum values of 'f'in the domain and show were the values occur on the graph.

{(x, y) | -1 <= x <= 1, 0 <= y <= 2}[/quote]

You have one ' outside the italics and one inside. Do you meam max and min values of f?

Think about this geometrically. What does the graph of this function look like? Do you see that f(x,y) is never negative? Where is f(x,y)= 0?

Thank you, working would be much appreciated.