The temperature $T(x,y)$ at points in the $xy$-plane is given by $T(x,y)= x^2 -2 y^2$. An ant wishes to cool off as quickly as possible. Along what curve through $(2,1)$ should the ant move in order to continue to experience maximum rate of cooling?
$$∇T = 2x \hat i-4y \hat j$$
I think the tangent of the ant's path has to be parallel to the negative of the gradient of $T$ at all times, but I don't know how to go on.
The tangent to the ant path at time $t$ is directed along the vector $(x'(t),y'(t))$ and the gradient of $T$ at every point $(x(t),y(t))$ is directed along the vector $(x(t),-2y(t))$ hence one asks that the vectors $(x'(t),y'(t))$ and $(x(t),-2y(t))$ are colinear at all time, that is, that $$\left|\begin{array}{cc}x'(t)&x(t)\\ y'(t)&-2y(t)\end{array}\right|=0$$ or, equivalently, that $$2x'(t)y(t)+y'(t)x(t)=0$$ Multiplying the LHS by $x(t)$ yields $$2x'(t)x(t)y(t)+y'(t)x(t)^2=(x(t)^2y(t))'$$ hence the ant moves along a path of equation $$x(t)^2y(t)=c$$ If $(x(0),y(0))=(2,1)$, one gets $c=$ [to be completed by the OP] hence the curve has equation [to be completed by the OP].