Find particular path on a plane I'm working on the following question:
Let $T(x,y) = 1-x^2-2y^2$ be the temperature at each point $P(x,y)$ in the plane. A heat-loving bug is placed in the plane at the point $P(-1,1)$. Find the path that the bug should take to stay as warm as possible.
So, from $T(x,y)$ we know that the warmest possible temperature will be 1, since increasing $x$ and $y$ will just subtract from 1. That means that the warmest point is $Q(0,0)$ on the plane. Thus, the bug will want to head in the direction of this point from $P(-1,1)$. From here, I believe we'll create $PQ = (1,-1)$. Then, I know that we can find a unit-vector $U$ in the direction of $PQ$, which would be $<1,-1>$ again. But this still doesn't answer the question?
Any help would be greatly appreciated!
 A: I believe what you are describing and what the question is asking are two different things, although I can understand why you came up with what you did because this question is worded poorly in my opinion.
My take on re-wording the question for clarity: Let $T(x,y) = 1-x^2-2y^2$ be the temperature at each point $(x,y)$ in the plane. A heat-loving bug is placed in the plane at the point $(-1,1)$. Find the path that the bug should take to always be heading in the direction of most warmth. 
This is a standard calculus problem. The way to approach it is recognize that at every point, the gradient of $T$ = $\nabla{T}$ points in the direction of maximum increase of your function $T$. In this case, that means that at every point, heading in the direction $\nabla{T}$ will be the best plan for the bug in order to be most warm.
$$\nabla{T} = (-2x,-4y)$$
If you can think about following this direction starting from $(-1,1)$, it would be a inward spiraling path that ends as you noted at $(0,0)$.
A: While it is true that the point $(0,0)$ is the warmest, the bug doesn't know this. The direction that the bug takes will always be in the direction of steepest ascent (in the direction that temperature increases the most). This is the direction of the gradient:
$$\nabla T=\langle -2x,-4y\rangle.$$ This means that $$\frac{dx}{dt}=-2x$$ and $$\frac{dy}{dt}=-4y,$$ where $t$ is just a variable that paramaterizes the path. The solutions to these differential equations are $$x=C_1e^{-2t}$$ and $$y=C_2e^{-4t},$$ where $C_1$ and $C_2$ are constants we use to make sure the bug starts at $\langle -1,1\rangle$. Putting in $t=0$ into these equations gives us $C_1=-1$ and $C_2=1$. This leaves us with the path $$p(t)=\langle -e^{-2t},e^{-4t}\rangle.$$
A change of coordinates $t\mapsto t'=e^{-2t}$ gives us $p(t)=\langle -t,t^2\rangle$, where $t$ now starts at $1$ and goes to $0$.
A: I don't believe any of the above solutions are complete.  The one from Alex S gives the path as a function of time t, and the one from NoseKnowsAll gives just the gradient and implies there will be a spiral--by which I assume he means wrapping around the origin.  Moreover, the comment from Cardinal stating that the shortest route is along the gradient direction is clearly false:  The shortest route is a straight line to the origin, $(0,0)$.
Alex's intermediate solution shows:  ${\bf p}(t) = (-e^{-2t}, e^{-4 t})$.  But the path cannot depend upon $t$.  Think of the path as a road:  Where would you place and pave the road?  One must therefore eliminate $t$ from the description of the path.  Simple algebra gives the path:  $y = x^2$, and is shown on this stream plot.  Note that it does not wrap around the origin.

