Captain Ralph is in trouble near the sunny side of Mecury.

Question

The temperature of the ship's hull when he is at location $(x,y,z)$ will be given by $T(x,y,z)= e^{-x^2-2y^2-3z^2}$, where $x,y,$ and $z$ are measured in meters. He is currently at $(1,1,1)$.

(A) In what direction should he proceed in order to decrease the temperature most rapidly?

(B) If the ship travels at $e^8$ meters per second, how fast will be the temperature decrease if he proceeds in that direction?

(C) Unfortunately, the metal of the hull will crack if cooled at a rate greater than $\sqrt{14}e^2$ degrees per second. Describe the set of possible directions in which he may proceed to bring the temperature down at no more than that rate.

So I tried to the partial derivatives of each variable, but they all end up negative. However, I think for question (A), the answer should be in the direction of the $z$ direction because it has the greatest negative coefficient in the equation. Is that correct?

For (B), do I just plug in $e^8$ into the partial derivative of $z$?

For (C), what do I do?

Answer 1

(A)

Taking the gradient of $T$ yields $\nabla T (x,y,z) = \langle -2 x e^{-x^2-2 y^2-3 z^2},-4 y e^{-x^2-2 y^2-3 z^2},-6 z e^{-x^2-2 y^2-3 z^2} \rangle$. This gives the direction of maximum increase of any point $(x,y,z)$.

Since we're at $(1,1,1)$ and want the maximum decrease, the appropriate vector is

$$-\nabla T (1,1,1) = \langle \frac{2}{e^6},\frac{4}{e^6},\frac{6}{e^6} \rangle \approx \langle 0.0049575, 0.00991501, 0.0148725 \rangle$$

Since we're only after the direction of maximum decrease, we should normalize this result, giving:

$$\vec{e}= \frac{ - \nabla T (1,1,1)}{\| - \nabla T (1,1,1) \|} = \left\langle -\frac{1}{\sqrt{14}},-\sqrt{\frac{2}{7}},-\frac{3}{\sqrt{14}} \right\rangle$$

Note that this is not simply heading in the $z$ direction, but off at an angle.

(B)

Recall that if $\vec{r}(t)$ is a parametrized path, then $\frac{d}{dt} T(\vec{r}(t)) = \nabla T(\vec{r}(t)) \cdot \vec{r}'(t)$---this is really just the chain rule in disguise. Since we want to head in the direction of maximum decrease, it is useful to know that $\vec{u} \cdot \vec{v} = \| \vec{u} \| \|\vec{v} \| \cos \theta$. So in our case $\theta=0$ and

the rate of decrease is $ \| \nabla T(1,1,1) \| \mathrm{e}^8= 2 \sqrt{14}e^2$ or the rate of change is $-2\sqrt{14}e^2$.

(C)

Since the ship's hull will crack if the rate of change is greater than $\sqrt{14}e^2$, we need to head off at an angle $\theta$ relative to the direction of maximum decrease. This means that we want

$\big| \|\nabla T(1,1,1) \| \mathrm{e}^8 *\cos \theta \big| \leq \sqrt{14}e^2$

which means that

$| 2 \sqrt{14}e^2 \cos \theta | \leq \sqrt{14}e^2$ which gives that $|\cos \theta| \leq \frac{1}{2} \implies \theta \in [\pi/3, 2 \pi 3]$