Observe that [tex]y=\pm x[/tex] form a level curve for the surface in question. This means that if we substitute [tex]y=x[/tex] or [tex]y=-x[/tex] in to the equation for the surface, then [tex]z=f(x,y)=c[/tex] is a constant.
This is only true for the surface in option A.
- [tex]z = x^2 - y^2 + 6[/tex]
[tex]y = x \implies z = x^2 - x^2 + 6 \implies z = 6[/tex]
[tex]y = -x \implies z = x^2 - (-x)^2 + 6 \implies z = 6[/tex]
[tex]z[/tex] is constant, so both of [tex]y=\pm x[/tex] are level curves.
- [tex]z^2 = x + y + 6[/tex]
[tex]y=x \implies z^2 = x + x + 6 \implies z = \pm\sqrt{2x + 6}[/tex]
[tex]y=-x \implies z^2 = x + (-x) + 6 \implies z = \pm\sqrt6[/tex]
[tex]y=x[/tex] is not a level curve.
- [tex]z = \sqrt{x^2 + y^2 + 6}[/tex]
[tex]y = x \implies z = \sqrt{x^2 + x^2 + 6} = \sqrt{2x^2 + 6}[/tex]
[tex]y = -x \implies z = \sqrt{x^2 + (-x)^2 + 6} = \sqrt{2x^2 + 6}[/tex]
Neither are level curves.
- [tex]z = \sqrt{x^2 + y^2}[/tex]
[tex]y = x \implies z = \sqrt{x^2 + x^2} = \sqrt2 |x|[/tex]
[tex]y=-x \implies z = \sqrt{x^2 + (-x)^2} = \sqrt2 |x|[/tex]
Again, neither are level curves.
