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Rate of cooling given a direction and speed

  1. Oct 2, 2012 #1
    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. jcsd
  3. Oct 2, 2012 #2


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    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##:$$
    \frac{dT}{dt} = T_x\frac {dx}{dt} +T_y\frac{dy}{dt}+T_z\frac {dz}{dt}
    =\nabla T\cdot \vec R'(t)$$
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