Problem on first order expansions and the chain rule

Suppose that the temperature in a 2d region is given by $T(x,y) = e^{-x^2 -y^2}$. Suppose that you move along the curve $x(t) = t$ and $y(t) = t^2$.
1. At $t = 1$, approximately how much does $x$ change in time $\Delta t$?
2. How much does $y$ change in a time $\Delta t$?
3. Use a first order expansion to find how much $T(x(t), y(t))$ changes in time $\Delta t$ at $t=1$.
4. What is the rate of change of $T(x(t), y(t))$ with respect to $t$ at $t=1$?.
• Solution

Part a

Recall that
The first order expansion of a single-variable function is $x(t+\Delta t) \approx x(t) + x'(t) \Delta t$.
Hence the change in $x$ after a time $\Delta t$ is apprximately $x'(t) \Delta t$.
Computing that $x'(1) = 1$, we get that $\Delta x \approx \Delta t$.

Part b

By the same logic as above $\Delta y \approx y'(t) \Delta t$.
Computing that $y'(1) = 2$, we get that $\Delta x \approx 2 \Delta t$.

Part c

Recall that the first order expansion of $T(x,y)$ is
Thus, the change in $T$ due to a $\Delta x$ and $\Delta y$ is approximately $\partial_x T\ \Delta x + \partial_y T \ \Delta y$.
Plugging in $\Delta x \approx \Delta t$ and $\Delta y \approx 2 \Delta t$, we get $$\Delta T \approx \partial_x T \cdot \Delta t + \partial_y T \cdot 2 \Delta t \tag{1}$$
Now, we compute the partial derivatives of $T$ at $t=1$.
\begin{align}
\partial_x T(x,y) &= -2 x e^{-x^2 - y^2}\\
\partial_y T(x,y) &= -2 y e^{-x^2 - y^2}\\
\end{align}
Now we evaluate these derivatives at point on the curve corresponding to $t=1$. This point is $x(1) = 1$ and $y(1) = 1$.
\begin{align}
\partial_x T(1,1) &= -2 e^{-2}\\
\partial_y T(1,1) &= -2 e^{-2}\\
\end{align}
As per (1), $$\Delta T \approx -6 e^{-2} \Delta t$$

Part d

As shown in (c), a time $\Delta t$ causes a change in $T$ of approximately $-6 e^{-2} \Delta t$. Hence, the rate of change of $T$ with respect to $t$ is $$\frac{d}{dt}T(x(1), y(1))= - 6 e^{-2}.$$