A circular piece of metal with radius $a$ has the temperature given by the following relation: for a point $(x,y)$, the temperature $T(x,y)$ is proportional to the square of the distance of this point $(x,y)$ to the center of the piece of metal, with constant of proportionality $k>0$.
So the exercise asks the following:
a) If a particle at $(\frac{a}{2},0)$ moves to the rigth about the $x$ axis, it'll suffer temperature increase or decrease?
b) What's the rate of change in respect to $y$ at the same point? What does that mean?
Here's what I did:
a) $$T(x,y) = k\sqrt{x^2+y^2}\implies \frac{\partial T(x,y)}{\partial x} = \frac{kx}{\sqrt{x^2+y^2}} \implies \\ \frac{\partial T(\frac{a}{2},0)}{\partial x} = \frac{k\frac{a}{2}}{\sqrt{\frac{a}{2}}^2} = k > 0$$ therefore, there is a temperature increase
b) if I do the same for $y$, we have:
$$\frac{\partial T(\frac{a}{2},0)}{\partial x} = \frac{2y}{\sqrt{x^2+y^2}} = 0$$
therefore, there is no increase or decrease in temperature in the $y$ axis for that contains this point
Am I rigth?
I think it is easier since the square of the distance is just $x^2+y^2$. Indeed $$\frac{\partial T}{\partial y }=2k y~~,$$ therefore in $(a/2,0)$ the rate of change of $T$ along $y$ is zero. This is because this point locally minimizes $T$ respect to displacements in the $y$ direction.