Directional derivative and temperature function Suppose the temperature at the point $(x,y)$ is $T(x,y)$. A particle move such that it position after $t$ seconds it's given by $x=\sqrt{1+t}$, $y=2+ \frac{t}{3}$. The $T$ function satisfies: $T_x(2,3)=4$ and $T_y(2,3)=3$. ¿How fast does the temperature of the particle increase in it's trajectory after 3 seconds?
I don't understand what they're asking for, I think they're asking for the directional derivate of the function at $(2,3)$ at the direction $u=(1,1)$, I give this $u$ because at $t=0$, I get the position $(1,2)$. Also, they give the partial derivatives at $(2,3)$.
What do you guys understand by how fast..? How would you approach this problem?
 A: You've described the following situation. Let $T(x,y)$ denote the temperature at the point $(x,y) \in \mathbb{R}^2$. Furthermore, assume that for some time parameter $t>0$ (given in seconds), we follow a particle taking the trajectory described by $x(t) = \sqrt{1+t}, y(t) = 2 + \frac{t}{3}$ for $t>0$. Then, we can describe the temperature of that particle over time via the parametrisation $ T(x(t),y(t)) $ for $t>0$.
The aim is to determine the rate of change of the temperature when $t=3$, i.e. $\frac{d}{dt}T(x(t),y(t)) \big\vert_{t=3}$. One can compute this directly via the chain rule, i.e.
$$
\begin{align}
\frac{d}{dt}T(x(t),y(t)) &= \frac{dx(t)}{dt} T_x(x(t),y(t)) + \frac{dy(t)}{dt} T_y(x(t),y(t)) \\
&= \frac{1}{2\sqrt{1+t}}T_x(x(t),y(t)) + \frac{1}{3}T_y(x(t),y(t)).
\end{align}
$$
Evaluating the above at $t=3$ gives us
$$
\begin{align}
\frac{d}{dt}T(x(t),y(t))\big\vert_{t=3} &= \frac{1}{2\sqrt{1+3}}T_x(x(3),y(3)) + \frac{1}{3}T_y(x(3),y(3)) \\
&= \frac{1}{4}T_x(2,3) + \frac{1}{3}T_y(2,3) \\
&= \frac{4}{4} + \frac{3}{3} = 2.
\end{align}
$$
Therefore the rate of change in temperature after $3$ seconds is $2$ units of temperature per second.
