The temp. in a neighborhood of the origin is given by a function of the form T(x,y) = T(not) + (e^y) * sinx.?

Question

Find the path followed by a heat fleeing particle that originates at the origin.

Answer 1

I assume by T(not) you mean some constant (sure it's not supposed to be T_0?).
∂T/∂x = e^y cos x
∂T/∂y = e^y sin x
A heat fleeing particle will have (dx/dt, dy/dt) = -k(∂T/∂x, ∂T/∂y) for some positive constant k. So we get
dx/dt = -k e^y cos x
dy/dt = -k e^y sin x
Hence dy/dx = sin x / cos x, from which we get
y = ∫(sin x dx / cos x) = -ln |cos x| + c
We start at the origin, so y(0) = -ln (1) + c = 0 => c = 0.
So y = -ln |cos x|.
Now initially (at time t=0, at the origin), dx/dt = -k and dy/dt = 0, so we start off moving in the negative x direction. At any time for -π/2 < x < 0 we have dx/dt < 0 and dy/dt > 0, so we keep moving towards the asymptote at x = -π/2 but never reach it. In this region we also have cos x > 0, so we don't need the absolute value around it.
Hence the path followed by the particle is given by
y = -ln (cos x), -π/2 < x ≤ 0.