Functions of several variables

[Arushi :))]

Someone help with these questions..
1.
Show that x^2+y^2 = 6 is a level curve of f(x,y)= [ x^2+y^2 ]^(1/2) -x^2-y^2+2
And that all the level curves of f must be centered at the origin.
2.
Suppose F(x,y) is a function about which all we know is that F(0,0)= 0,
F'1(x,y)>,= 2 for all (x,y) and F'2(x,y)<,= 1 for all (x,y)
What can be said about the relative sizes of F(0,0) , F(1,0) , F(2,0) , F(0,1) and F(1,1)?
Write down the inequalities that have to hold between these numbers.

[Noel]

[Noel]

Now put c = (-4)+ root(6)
in the above equation to get x^2+y^2=6

[Arushi :))]

Thanksssss :)

[Gayyam]

Could you please clarify the second question? Specifically, I'm unable to understand what "F'1(x,y)>,= 2 for all (x,y) and F'2(x,y)<,= 1 for all (x,y)" means.

I assumed this, F'1(x,y)>,= 2 means the partial derivative of F(x,y) with respect to x is greater than or equal to 2; and F'2(x,y)<,= 1 means the partial derivative of F(x,y) with respect to y is less than or equal to 1.

△F = F'1*△x + F'2*△y.
With this, we get that the function increases in x more than (or equal to as) 2x does, and it increases in y less than (or equal to as) y.
F(0,0)=0. Thus, F(1,0)>=2, F(2,0)>=4, F(0,1)<=1.
For F(1,1), it must increase in value by at least 2 (due to a change of 1 in x) and increase at most by 1 (due to a change of 1 in y). Thus F(1,1) is also >=2. If we assume F'2(x,y)>0, then F(1,1)>2.

But I'm not sure about the interpretations.

Also, thank you Noel! :) That was a nice solution.

[kangkan]

Hammond heh?

[Arushi :))]

yes kangkan , hammond