Directional derivative The governor Ralph has trouble on the bright side of Mercury. The temperature in the wall of the vessel, when it is in the position $(x, y, z)$ is given by $T(x, y, z)=e^{-x^2-2y^2-3z^2}$, where $x$, $y$ and $z$ are measured in meters. Now it is located at $(1,1,1)$. 


*

*In what direction should it move to lower the temperature as quickly as possible? 

*If the vessel is traveling with velocity $e^8$ per second, how quickly will temperature be decreased if it moves in that direction? 

*Unfortunately the metal of the wall will break if it cooled with rate greater than $\sqrt{14}e^2$ degrees per second. Describe the set of possible directions where it can move to reduce the temperature at a rate not greater than the permissible. 


$$$$ 
I have done the following: 


*

*The direction to that the vessel should move so that the temperature decreases as fast as possible is given by : 
$$-\nabla T(1, 1, 1)$$ 
Is this correct?? 
So, we have the following: 
$$\nabla T(x, y, z)=(-2x e^{-x^2-2y^2-3z^2}, -4ye^{-x^2-2y^2-3z^2}, -6ze^{-x^2-2y^2-3z^2}) \\ \Rightarrow -\nabla T(1, 1, 1)=(2e^{-6}, 4e^{-6}, 6e^{-6})$$ 
Is this correct?? 

*Are we looking for the rate of change at the direction we found at the question $1$ ?? 
So, is it as followed?? $$grad f \cdot \overrightarrow{v}$$ where $\overline{v}$ is the unit vector of the qquestion $1$ ?? But how can we use the fact that the velocity is $e^8$ ?? 

*Could you give me some hints what we could do ?? 
$$$$ 
EDIT: 
At the question $3$ are we looking for the directions $\overrightarrow{v}$ such that $$\nabla T \cdot \overrightarrow{v}\leq \sqrt{14}e^2$$ ??
 A: *

*It is correct.

*Yes it should be $\nabla f \cdot \vec{v}=||\nabla f|| ||\vec{v}||\cos\theta$ where $\theta=0$, and $||\vec{v}||=e^8$.
For 2, think about it this way:
It is asking the rate of change of temperature, 
$$\frac{dT}{dt}=\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}+\frac{\partial T}{\partial z}\frac{dz}{dt}=\nabla T \cdot \frac{d\vec{r}}{dt}=\nabla T \cdot \vec{v}=||\nabla T|| ||\vec{v}||\cos\theta$$
Since it says the velocity is $e^8$, then $||v||=e^8$. The others are obvious.


*We need $||\nabla T|| ||\vec{v}||\cos\theta\leq \sqrt{14}e^2$. You can solve for $\cos\theta$ to find the direction.

A: Hints:


*

*A direction is usually given by a unit vector. The direction to move to get the fastest cooling would be
$$
-\frac{\nabla T(1,1,1)}{\|\nabla T(1,1,1)\|}
$$
which is the direction of $-\nabla T(1,1,1)$; so, mod the magnitude, your thinking is correct. The direction would therefore be
$$
\frac{(1,2,3)}{\sqrt{14}}
$$

*The velocity described by a speed of $e^8$ in the direction given in 1. is
$$
e^8\frac{(1,2,3)}{\sqrt{14}}
$$
The rate of change of $T$ moving at that velocity is
$$
\begin{align}
e^8\frac{(1,2,3)}{\sqrt{14}}\cdot\nabla T(1,1,1)
&=-e^8\frac{\nabla T(1,1,1)}{\|\nabla T(1,1,1)\|}\cdot\nabla T(1,1,1)\\[6pt]
&=-e^8\|\nabla T(1,1,1)\|
\end{align}
$$

*Let the direction of travel be $d$, where $|d|=1$. The rate of change of $T$ achieved by moving at speed $e^8$ in direction $d$ is
$$
e^8d\cdot\nabla T(1,1,1)
$$
Now, since $u\cdot v=|u|\,|v|\cos(\theta)$, where $\theta$ is the angle between the directions of $u$ and $v$, the rate of change of $T$ is
$$
e^8\|\nabla T(1,1,1)\|\cos(\theta)
$$
so we need to find $\theta$ so that
$$
-\sqrt{14}e^2\le e^8\|\nabla T(1,1,1)\|\cos(\theta)\le\sqrt{14}e^2
$$
