Vector calculus derivatives finding the isothermal curve

Question

Let the temperature $T = 4x^2+16y^2$ in a body be independent of $z$, identify the isotherms $T(x, y) = \text{const}$. Sketch it.

I just studied vector calculus and only know some basic things.

My attempt :

so I should find $T = c = 4x^2 + 16y^2 $

because $T$ is independent of $z$, $dT/dz = 0$

$z^2=x^2+y^2$

$T = 4(x^2+4y^2) = 4(z^2+3y^2) $

so $dT/dz = 8z = 0$ then $z = 0$ which leads to $x^2+y^2=0 $

thus $x = y = z = 0$, so $T = 4(0) = 0 $? but this will cause the curve to be $x^2+4y^2=0$ which has only $1$ solution $(0,0)$

I really appreciate any help towards this problem

Answer 1

Take $z =T(x,y) = 4x^2 + 16y^2$. If $z = 0$ then $x = y = 0$ and so the isotherm is a point. If $z \not = 0$ then divide out by it.

$$ 1 = \left(\frac{x}{\frac{\sqrt{z}}{2}}\right)^2 + \left(\frac{y}{\frac{\sqrt{z}}{4}}\right)^2$$

and so in the plane, the isotherms form ellipses. Convince yourself that the graph of $T$ is just an elliptic paraboloid.