# In Trouble by the Sunny Side of Mercury

I have the following homework problem, and have been able to work out parts a) and b), but not part c).

Captain Ralph is in trouble near the sunny side of Mercury. 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?

I got: $\left(2e^{-6},4e^{-6},6e^{-6}\right)$

b) If the ship travels at $e^9$ meters per second, how fast (in degrees per second) will the temperature decrease if he proceeds in that direction?

I got: $\sqrt{56}e^3$

c) Unfortunately, the metal of the hull will crack if cooled at a rate greater than $\sqrt{{{17}}}e^3$ 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.

I'm lost here.

I found this other math.slackexchange post, but I'm lost following him. I've tried recreating his steps to no avail for my own problem.

I also found this Caltech homework. The problem I'm doing is exactly the same as their last problem (just different numbers), I don't understand how they got to their answer.

Also, I am submitting to a computer, so even though the answer might be correct, it could be that the answer isn't the one they're looking for. I submitted to my instructors for more verification on this problem last night, but have still not heard back from them.

Any clarification/help would be much appreciated!

Edit: I also tried $\sqrt{56}e^3\cos \left(\theta \right)$, which wasn't accepted either. I still haven't heard back from either of the course professors.

Edit 2: I found this website where they go through and solve for theta.

Additionally, my answer needs to be something in the form of this (whatever it means, please explain it to me if you understand it). $$\{ai+bj+ck \mid a^2+b^2+c^2=1, 0<[answer]\leq \sqrt{17}e^3\}$$

• tried to edit the mathjax, had to guess where you wanted the braces. For reference, \{ and \} will insert left and right braces---{ and } on their own are used for grouping commands, i.e. \sqrt{1-x^2} or \frac{a}{b} – erfink May 12 '17 at 2:40
• @erfink I didn't know that! Thanks for teaching me! Is there a website with all this stuff? – TobyTobyo May 12 '17 at 2:52
• math.meta.stackexchange.com/questions/5020/… gives a good introduction – erfink May 12 '17 at 2:55

The directional derivative of a function in the direction of $\hat n$ is given by $\vec \nabla f \cdot \hat n$

In this case, at the point $(1,1,1)$ with the ship travelling a distance $s$ in the direction $(a,b,c)$ with $a^2+b^2+c^2=1$ we have ...

$$\frac{dT}{ds}=-2 e^{-6}(1,2,3)\cdot (a,b,c)$$

so $$\frac{dT}{dt} = e^9\frac{dT}{ds} = -2 e^3 (a+2b+3c)$$

You require $-\sqrt{17}e^3 < \frac{dT}{dt} <0$

$$0< a+2b+3c < \frac{ \sqrt{17}}2$$
• Thank you! The answer that ended up working for me was $-2e^3(a+2b+3c)$, since the condition on the right hand side was preset to $\sqrt{17}e^3$. – TobyTobyo May 12 '17 at 16:32