Magnitude of directional and gradient vectors?

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If given only:
f(5,2) = 80
fx(5,2) = 8
fy(5,2) = -6
Suppose 80 is measured in degrees Fahrenheit. Find the direction where the temperature would get cooler.

I just did 8a - 6b = 0 (since using the dot product, <8,-6> * <a,b> = 0.
Then I solved for a, b, and this was the vector equation.

Then the question asked, How many steps would you take to get from 80 to 78 degrees?

I have no idea. Is it asking for the magnitude? :confused:
 
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Do us all a favor and copy the problem exactly as it is given. What you have written is non-sense. First there is no single direction in which "the temperature gets cooler" (That itself makes no sense- temperature is a number and numbers do not get "cooler" or "hotter". I assume you mean "temperature decreases" but what does the problem say?) There exist a range of directions in which the temperature decreases, another in which it gets cooler. It is true that the rate of change of function f, in the direction of unit vector \vec{v}, is \nabla v\cdot\vec{v}. What you have calculated is a vector pointing in a direction such that the temperature does not change (the negative of this vector also gives such a direction). Those two vectors separate directions in which the temperature is increasing from the directions in which the temperature is decreasing.

As far as "How may steps would you take" is concerned, how long is a step??
 
That's exactly what the problem said. It wasn't a textbook question but created by the professor. I assume he meant direction in which the temperature decreases the fastest.

By steps he meant the shortest distance a person would travel in the direction where the temperature decreases from 80 to 78.
 
So I would interpret "steps" as unit lengths then. The direction in which the temperature reduces fastest is directly opposite to the gradient vector. And that gradient is, of course, 8i- 6j which has length \sqrt{64+36}= 10. That is, as long as we stay on the vector -8i+ 6, the temperature, to a linear approximation, decreases by 10 degrees "per step". Notice that I said "linear approximation". From the given information, we cannot be certain what happens as soon as you start moving. Frankly, I don't like this problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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