domain of two-variable, natural log function.

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November 2nd 2009, 12:06 PM
adkinsjr

domain of two-variable, natural log function.

I'm having trouble graphing the domain of the function $f(x,y)=ln(9-x^2-9y^2)$ . I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

The domain should be all $x,y$ such that $9-x^2-9y^2 \geq 0$

$-9y^2 \geq x^2-9$

$y^2 \leq 1-\frac{1}{9}x^2$

$y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

So do I just need to sketch the region below the curves $y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

? It seems like there must be an easier way.
November 2nd 2009, 12:26 PM
Defunkt

Quote:

Originally Posted by adkinsjr

I'm having trouble graphing the domain of the function $f(x,y)=ln(9-x^2-9y^2)$ . I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

The domain should be all $x,y$ such that $9-x^2-9y^2 \geq 0$

$-9y^2 \geq x^2-9$

$y^2 \leq 1-\frac{1}{9}x^2$

$y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

So do I just need to sketch the region below the curves $y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

? It seems like there must be an easier way.

Do you know what the ellipse $\frac{x^2}{9} + y^2 \leq 1$ looks like?
November 2nd 2009, 12:33 PM
adkinsjr

Quote:

Originally Posted by Defunkt

Do you know what the ellipse $\frac{x^2}{9} + y^2 \leq 1$ looks like?

yeah, I see it now. I need to shade in the ellipse with center at origin, vertices at $(0,\pm 1)$ and $(\pm 3,0)$ .

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