Rate of cooling given a direction and speed

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SUMMARY

Captain Ralph must navigate to cool his ship on Mercury, where the temperature is defined by T=e^(-x²-2y²-3z²). To achieve the fastest cooling, he should move in the direction of the negative gradient, calculated as (2e^-6, 4e^-6, 5e^-6). The ship's speed is e^8, and the rate of cooling can be determined by multiplying the gradient by this speed. The correct approach involves using the chain rule to find the rate of change of temperature over time.

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  • Knowledge of exponential functions and their properties
  • Basic concepts of vector calculus
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Homework Statement


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?


Homework Equations





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?
 
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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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