# Rate of cooling given a direction and speed

1. Oct 2, 2012

### explorer58

1. The problem statement, all variables and given/known data
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=e^(-x²-2y²-3z²). He is currently at (1,1,1).
a)In which direction should Ralph go to cool the fastest?

b) If the ship can travel at a speed of e^8, what will be the rate of cooling?

2. Relevant equations

3. The attempt at a solution
a) The direction he should go would be the negative of the gradient, which I got to be (2e^-6, 4e^-6, 5e^-6). Easy.

b) Here's where I got lost. the answers online (source http://www.cds.caltech.edu/~marsden/wiki/uploads/math1c-08/assignments/homework_sol3.pdf) say just take the gradient and multiply it by the speed. My question is, shouldn't we need to take a unit vector in the direction of the gradient?

2. Oct 2, 2012

### LCKurtz

Given a curve $\vec R(t) = \langle x(t),y(t),z(t)\rangle$ and temperature $T(x,y,z)$ you want to use the chain rule to calculate the rate of change of $T$:$$\frac{dT}{dt} = T_x\frac {dx}{dt} +T_y\frac{dy}{dt}+T_z\frac {dz}{dt} =\nabla T\cdot \vec R'(t)$$