Problem on first order expansions and the chain rule

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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
    $T(x+\Delta x,y + \Delta y) \approx T(x, y) + \partial_x T(x, y) \Delta x + \partial_y T(x, y) \Delta y $
    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 $$\frac{d}{dt}T(x(1), y(1))= - 6 e^{-2}.$$

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