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Finding arbitrary rate of change.

  1. Jul 31, 2006 #1
    I graphed the function and found the critical points, saddle points, local/absolute max/mins etc.., then i graphed the course parametrically and it resembled a four-leafed rose.

    rate of change makes me think of taking the derivative, but course makes me think of integration. which direction should i take? vague replies only please (as always) :)
     
  2. jcsd
  3. Jul 31, 2006 #2
    I assume that [itex]z[/itex] is the elevation? The question is asking for a rate of change, so you want to differentiate (and my first sentence should be enough of a hint, if the assumption is correct). I am not sure why "course" made you think of integration :rolleyes:
     
  4. Jul 31, 2006 #3

    Office_Shredder

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    The word course is referring to the parametrization of the curve
     
  5. Jul 31, 2006 #4
    because course = path, thought of doing a double integral setting the limits to the position functions of x and y.
     
  6. Jul 31, 2006 #5

    Office_Shredder

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    That doesn't help at all.

    Here's what the problem actually is. You have a mountain, with the height at (x,y) defined in the problem. There is a trail, that a person is going up, defined by the parametrization. So you want to find the rate of change in the trail's height based on the parametrization variable, t. How do we start?

    First question:
    Can you find the height of the trail at time t?

    Second question:
    Can you find the rate of change of that height?
     
  7. Aug 1, 2006 #6

    HallsofIvy

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    Find [tex]\frac{dz}{dt}[/tex]. This is an exercise in using the chain rule.
     
    Last edited: Aug 1, 2006
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