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How to determine the isotherms of given temperature field?

  1. Oct 11, 2013 #1
    How can I determine the isotherms(curves of constant temperature) of given Temperature field like
    T = x^2 -y^2 + 8y.
    Also how can I draw at least five isotherms of the above temperature field?
     
  2. jcsd
  3. Oct 11, 2013 #2

    mfb

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    Pick at least five different values for T, draw solutions to the equation.
    That's the way to get all isotherms. Consider T as constant, find a relation between x and y. Your equation is such a relation, but you can express it in a nicer way.

    I don't see a differential equation here.
     
  4. Oct 11, 2013 #3

    HallsofIvy

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    Use the definition of "isotherm". For a given temperature, C, you are points where the temperature is C: [itex]x^2- y^2+ 8y= C[/itex]. Now, if you are asking "how can I write that as a function I know how to graph quickly?", the answer is, as so often for things like this, complete the square.

    8/2= 4 and [itex]4^2= 16[/tex] so [tex]y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16[/tex]

    We can write [itex]x^2- y^2+ 8y= x^2- (y^2+ 8y)= x^2- ((y- 4)^2- 16)= x^2- (y- 4)^2- 16[/itex] so that [itex]x^2- y^2+ 8y= C[/itex] becomes [itex]x^2- (y- 4)^2- 16= C[/itex] or [itex]x^2- (y- 4)^2= C+ 16[/itex]. To get that to "standard form", divide through by C+ 16:
    [tex]\frac{x^2}{C+ 16}- \frac{(y- 4)^2}{C+ 16}= 1[/tex]

    That is an equilateral hyperbola with center at (0, 4), vertices at [tex](\sqrt{C+ 16}, 4)[/tex] and [tex](-\sqrt{C+ 16}, 4), and asymptotes y= x+ 4 and y= -x+ 4.
     
  5. Oct 11, 2013 #4

    mfb

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    It is easier to find a function x=f(y).
     
  6. Oct 12, 2013 #5

    HallsofIvy

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    Not when there is NO such function, as here.
     
  7. Oct 13, 2013 #6

    mfb

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    ... or two functions.
     
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