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Please help check my solution to gradient question.

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A wildebeest is charging across a plain. His path takes him to location (x,y) where x is his distance (in miles) east of his starting point and y is his distance in miles north of his starting point at time t. So x and y are functions of t. The air temperature is a function of both location and time.
    Right now he is moving at a velocity of 12 miles per hour in the northward direction and 7 miles per hour westward. In the northward direction the temperature changes at a rate of -0.10 degrees per mile, and in the eastward direction the temperature changes at a rate of -0.05 degrees per mile. Also, overall the temperature (irrespective of location) is changing at a rate of -1.6 degrees per hour.
    What is the rate of change in air temperature that the wildebeest is experiencing right now?



    3. The attempt at a solution

    I just want to make sure I set this up right. I first created a function called r(t). We then have

    [itex] ∇r = -7 i + 12 j + 1 k [/itex]

    Next I made a unit vector for the 3 respective changes in temperature
    [itex] u = -.05/1.603)i - .1/1.603) j + 1.6/1.63 k [/itex]

    Next I just summed the dot product [itex] ∇r * u [/itex]

    I just wanted to make sure there are no errors in this solution, and that I set it up in an efficient manner. I have trouble when I have to start mixing (x,y) with t.

    Thank you.
     
  2. jcsd
  3. Jun 8, 2012 #2

    HallsofIvy

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    I really don't understand at all. if the wildebeast can only move in a two dimensional plane (North-South and East-West), what does "k" represent? And how is the wildebeast moving in that direction at 1 mile per hour?

    But then I don't understand the question itself! If the temperature is changing at "-0.10 degrees per mile" in the northern direction and at -0.05 degrees per mile" in the eastern direction, then how can it be changing at -1.6 degrees per hour irrespective of location?
     
  4. Jun 8, 2012 #3
    Lol that is why I posted it here! A teacher at my college put this on a calc III test, and I want to make sure that I am prepared for an upcoming exam. I chose to set z = 1 in the gradient because I wanted to consider this problem as a 3 dimensional function where the z value would be a shift up and down a 3 dimensional axis perhaps.

    However I'm not sure if this would yield a correct answer.

    edit : the change in global temperature is what I considered to be our z value.
     
  5. Jun 8, 2012 #4
    You could call it r(t)=(x,y,t). Then r'(t)=(-7,12,1)

    ∇T=(-.05,-.1,-1.6)

    Then dT/dt=∇T[itex]\cdot[/itex]r'(t)

    Comments:

    I'm not sure you need a unit vector.

    Rate of change with respect to what, time?

    Notice that dT/dt≠∂T/∂t=-1.6
     
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