How can I find the domain and radius of the level curves for a given function?

[jeffreylze]

Homework Statement

z=f(x,y)= -[tex]\sqrt{9-2x^2-y^2}[/tex]

Sketch the level curves for f(x,y)

Homework Equations

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[tex]^{2}[/tex]=2x[tex]^{2}[/tex]+y[tex]^{2}[/tex]

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

 

[tiny-tim]

Hi jeffreylze!

Sketch the level curves for f(x,y)
…
I let z = c (a constant) . Substituting c into the equation and rearranging it, I got this 9-c[tex]^{2}[/tex]=2x[tex]^{2}[/tex]+y[tex]^{2}[/tex]

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?

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.

 

[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 don't know, I am quite confused.

 

[HallsofIvy]

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

 

[jeffreylze]

Hence, 9-c^2 >= 0 ?

 

[tiny-tim]

Hence, 9-c^2 >= 0 ?

Nooo … 9 - c² = a,

and the only restriction is that c² ≥ 0 …

so a … ?

 

[jeffreylze]

so a[tex]\leq[/tex]9 ? Am i right this time?

 

[tiny-tim]

so a[tex]\leq[/tex]9 ? Am i right this time?

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.

 

[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?

 

[tiny-tim]

I let z = c (a constant) . Substituting c into the equation and rearranging it, I got this 9-c[tex]^{2}[/tex]=2x[tex]^{2}[/tex]+y[tex]^{2}[/tex]

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?

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 )