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Functions of Several Variables, Temperature?

  1. Aug 2, 2008 #1
    Hello, my name is Calleigh and i am new to the forum! I am in Calculus II and have a few questions on some problems. I am using the textbook Calculus 8th edition by Larson, Hostetler and Edwards. Could someone please help me?

    The problem is on pg 942 in chapter 13.6 in the text, number 76. It reads:

    The temperature at point (x,y) on a metal plate is modeled by:
    T(x,y)=400e^-((x^2+y)/2) where x>=0 and y>=0.

    It asks to find the directions of no change in heat on the plate from the point (3,5).
    It also asks to find the direction of greatest increase in heat from the point (3,5).


    Does anyone know what this problem is talking about? Usually it helps if i can picture it in my head but i'm lost... My professor suggested drawing a picture, but i haven't the slightest clue even where to begin.

    Any help would be greatly appreciated! Thanks guyssss ;)
     
  2. jcsd
  3. Aug 3, 2008 #2

    Hootenanny

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    If you are looking for the direction in which there is no change in temperature*, what can you say about the directional derivative in that direction?

    For the second part, what does the gradient of a vector field represent?

    (*)From a physics point of view, I will mention that temperature and heat are not the same thing, but since we're in Maths, I'll let you off :wink:.
     
  4. Aug 3, 2008 #3
    thanks for the reply hootenanny!

    So you're saying i should take the derivative of T(x,y)? How do i do that with 2 variables in the parenthesis?
     
  5. Aug 3, 2008 #4
    Someone suggested a different method and i gave it a shot, could someone tell me if this is right?

    -400xe^-((x^2+y)/2), p=-1200e^-7
    400(-.5)e^-((x^2+y)/2), p=200e^-7
    -200e^-7 (6,1)

    Now, i have no idea what this means, i just followed the example and did the same thing. Can anyone understand this and tell me if its right? lol Thanks ;)
     
  6. Aug 3, 2008 #5

    Hootenanny

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    You're studying multi variable calculus, you must know how to take the gradient and determine the directional derivative of a function.
    I honestly have no idea what you have done here.
     
  7. Aug 3, 2008 #6
    Then can you start me off in the right direction? My friends and are are tying these out, we honestly don't know what to do. We havent covered this stuff in class yet our professors just giving us a peak at what w will encounter in next semester, i just want to impress him.
     
  8. Aug 3, 2008 #7

    HallsofIvy

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    Then impress him by looking up "gradient" or "gradient vector" in your text book.
     
  9. Aug 3, 2008 #8

    Hootenanny

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    If you click on the words gradient and directional derivative you'll be taken to the appropriate articles in our Library. However, these only offer a brief overview. For a fuller treatment I suggest that you consult your text.

    Have you done any partial differentiation yet?
     
  10. Aug 4, 2008 #9
    Thanks for those links they help a little. I actually found an example that seemed to ask for the same things, so i tried the method on this prob and this is what i got...

    T(x,y)= 400e^-((x^2+y)/2) [-xi-(1/2)j]
    T(3,5)=400e^(-7[-3i-(1/2)j)]

    So there will not be change in directions perpendicular to the gradient +or - (i-6j)

    and the largest increase will be in the direction of the gradient -3i-(1/2)j

    does this seems about right? thanks ;)
     
  11. Aug 4, 2008 #10

    Hootenanny

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    Looks good to me :approve:.
     
  12. Aug 4, 2008 #11
    thanks! :)
     
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