Multivariable-calculus, two variable function, gradient

Question

$$T(x, y)= \frac{x^2-2y^2}{L^2}T_0\ \ \ \ \ \ T_0, L>0$$ I have this two variable equation that describes the temperature of each point in a given plane.$$ \ $$ I want to find the equation of the curve along which an object must move so that its temperature decreases in the fastest way possible.

Answer 1

You want integral curves of the gradient of $T$. That is, you're looking for a curve $(x(t),y(t))$ such that $$x'(t) = \frac{\partial T}{\partial x}(x(t),y(t)) = \frac{2T_0}{L^2}x(t)\quad \mbox{and} \quad y'(t) = \frac{\partial T}{\partial y}(x(t),y(t)) = \frac{-4T_0}{L^2}y(t).$$This system is easy, you can solve each differential equation separately, and obtain $$(x(t), y(t)) = \left(x_0 \exp\left(\frac{2T_0}{L^2}(t-t_0)\right), y_0 \exp \left(\frac{-4T_0}{L^2}(t-t_0)\right)\right).$$Note that $(x(t_0),y(t_0)) = (x_0,y_0)$, so this is the integral curve passing through $(x_0,y_0)$ at time $t_0$.