On a still summer afternoon the pavement downtown is far hotter than the outlying neighborhoods. Model the surface air temperature, in degrees Fahrenheit, by
\begin{equation*}
T(x,y) = 74 + 14\,e^{-(x^2 + 3y^2)/40},
\end{equation*}
where \(x\) and \(y\) are measured in miles east and north of downtown. The core reaches \(88^\circ\)F, and the outskirts settle near \(74^\circ\)F. Two runners meet at the point \(P_0(1,1)\text{.}\)
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The first runner is overheating and wants cooler air as quickly as possible. Which direction should she take, and how fast does the temperature fall along it?
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The second runner has settled into the temperature at \(P_0\) and wants a route on which it never changes at all. What path should she run?
Solution.
Differentiating the exponential with the chain rule,
\begin{align*}
\nabla T \amp = -\frac{14}{40}\,e^{-(x^2+3y^2)/40}\left(2x\,\mathbf i + 6y\,\mathbf j\right),\\
(\nabla T)_{(1,1)} \amp = -\frac{14\,e^{-1/10}}{40}\left(2\,\mathbf i + 6\,\mathbf j\right)\\
\amp \approx -0.633\,\mathbf i - 1.900\,\mathbf j .
\end{align*}
At \(P_0\) the temperature itself is \(T(1,1) = 74 + 14e^{-1/10} \approx 86.7^\circ\)F.
Part 1. Temperature falls fastest in the direction of \(-\nabla T\text{,}\) which is a positive multiple of \(\mathbf i + 3\mathbf j\text{.}\) Normalizing, the heading is north-northeast:
\begin{equation*}
\mathbf u = \frac{1}{\sqrt{10}}\,\mathbf i + \frac{3}{\sqrt{10}}\,\mathbf j,
\end{equation*}
and the rate of change along it is
\begin{align*}
D_{\mathbf u} T \amp = -|\nabla T|
= -\frac{14\,e^{-1/10}}{40}\sqrt{4 + 36}\\
\amp \approx -2.0 \ \frac{{}^\circ\text{F}}{\text{mi}} .
\end{align*}
Notice how modest this is. The gradient hands her the best available direction, not a dramatic one: even running the optimal heading, she must cover a full mile to shed about two degrees. Any other heading does worse.
Part 2. The second runner needs \(D_{\mathbf u}T = 0\text{,}\) that is, \(\mathbf u\) orthogonal to \(\nabla T\text{:}\)
\begin{equation*}
\mathbf u = \pm\frac{1}{\sqrt{10}}\left(3\,\mathbf i - \mathbf j\right).
\end{equation*}
Following that condition continuously, rather than for a single step, means never leaving the level curve of \(T\) through \(P_0\text{.}\) Since \(T\) is constant exactly when \(x^2 + 3y^2\) is constant, her route is
\begin{equation*}
x^2 + 3y^2 = 4,
\end{equation*}
an ellipse with semi-axes \(2\) miles east-west and \(2/\sqrt{3} \approx 1.15\) miles north-south — a closed loop of roughly ten miles, every step of it at \(86.7^\circ\)F.
In the animation below, switch between the two runners. On the loop the thermometer holds at \(86.7^\circ\)F while the blue gradient arrow stays stubbornly perpendicular to her heading; on the escape route the same reading falls steadily, and the path cuts across every isotherm at a right angle.
Use the interactive figure to explore the problem further geometrically, switching between the two routes and watching how the readings respond as the runner moves.
Instructions.
Press “Run the isotherm” to send the runner around the level curve \(x^2+3y^2=4\text{,}\) or “Dash for cool air” to send her along the path of steepest cooling. The readout shows her temperature, position, directional derivative \(D_{\mathbf u}T\text{,}\) and distance travelled; the blue arrow is \(\nabla T\) at her position. Use “Pace” to change her speed, and “Pause” or “Reset” to control the animation.
