# Sketching Level Curves

1. Dec 28, 2008

### jeffreylze

1. The problem statement, all variables and given/known data
z=f(x,y)= -$$\sqrt{9-2x^2-y^2}$$

Sketch the level curves for f(x,y)

2. Relevant equations

3. The attempt at a solution

I am really poor at this. I let z = c (a constant) . Substituting c into the equation and rearranging it, I got this 9-c$$^{2}$$=2x$$^{2}$$+y$$^{2}$$

From this, I know i will get a ellipse. But I am stuck here. How do I find the domain for c and the radius? Some hints will be helpful. Thanks

2. Dec 28, 2008

### tiny-tim

Hi jeffreylze!
Hint: first, put 9 - c2 = a, so it's 2x2 + y2 = a …

i] what values can a have?

ii] yes, it's an ellipse, and obviously its centre is at the origin, so just find the points where it intersects the x and y axes.

3. Dec 28, 2008

### jeffreylze

i) Since ellipse has a general equation, (x-h)^2/a^2 + (y-k)^2/b^2 = 1, hence the a you were referring to would have the values >= 0 ? I dont know, I am quite confused.

4. Dec 28, 2008

### HallsofIvy

Staff Emeritus
No, a2 must be non-negative but a itself can be any (non-zero) number. Bacause of the symmetry, it really doesn't matter.

5. Dec 28, 2008

### jeffreylze

Hence, 9-c^2 >= 0 ?

6. Dec 28, 2008

### tiny-tim

Nooo … 9 - c2 = a,

and the only restriction is that c2 ≥ 0 …

so a … ?

7. Dec 28, 2008

### jeffreylze

so a$$\leq$$9 ? Am i right this time?

8. Dec 28, 2008

### tiny-tim

Yup!

oops … sorry … i forgot what this was all about … i forgot about the √(9 - 2x2 + y2), so your first answer, a ≥ 0, is also correct: 0 ≤ a ≤ 9.

And now … find the points where it intersects the x and y axes.

9. Dec 28, 2008

### jeffreylze

but the i still couldn't get the answer -3 ≤ c ≤ 0. This is what i did, for 0 ≤ a ≤ 9, I subbed a= 9 - c^2 into it to get 0 ≤ 9 - c^2 ≤ 9 , rearranged and i got 0≤|c|≤3 . Where did i go wrong?

10. Dec 29, 2008

### tiny-tim

Well, basically, you're right,

but you dropped a condition when you started to use c2 instead of c …

when you go back from c2 to c, you have to remember that c to c2 is single-valued, but c2 to c is double-valued, and you have to select the right square-root (of c2) …

in this case, the original equation for f specified the negative square-root …

so you combine the condition c ≤ 0 with your result 0≤|c|≤3 to give …

da-daa! … -3 ≤ c ≤ 0.

(actually, it would have been a lot quicker, and just as accurate, to say, just by looking, that the square-root of nine-minus-something-positive must be between 0 and 3 )