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Chain Rule with partial derivatives

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Let T= g(x,y) be the temperature at the point (x,y) on the ellipse x=2sqrt2 cos(t) and y= sqrt2 sin(t), t is from 0 to 2pi. suppose that partial derivative of T with respect to x is equal to y and partial derivative of T with respect to y is equal to x. Locate the max and min temperatures by examining dT/dt and the derivative of dT/dt.



    2. Relevant equations

    chain rule with partial derivatives



    3. The attempt at a solution

    I got an answer of 4 for dT/dt. this means that the temperature is increasing at a constant rate of 4 units/sec. meaning that the min is at t=0 and the max at t=2pi. do i need to find the area under the dT/dt graph to find the max and min temps? or am i way off?
     
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  3. Oct 10, 2009 #2

    LCKurtz

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    You haven't shown your work for dT/dt but I'm pretty sure you have a sign mistake. If you think about this problem physically, it represents the temperature in an ellipse shaped wire. The temperature can't be increasing all the way around because when t = 0 and t = 2pi you are at the same point of the wire.
     
  4. Oct 10, 2009 #3
    Ok, this is what i have
    dT/dt= ∂T/∂x (dx/dt) + ∂T/∂y (dy/dt)
    =y(-2sqrt2 sint(t)) + x(sqrt 2 cos(t))
    then plug in y and x from the given and i got...
    (-4sint*sint) + (4 cost*cost)
    when i graphed that, i got a straight line at 4.
    Where did i go wrong?
     
  5. Oct 10, 2009 #4
    Ok nevermind, i just graphed in degrees instead of radians. it's been a long week, obviously :( ok, i think i know what i'm doing now. sorry for the confusion :P
     
  6. Oct 10, 2009 #5

    LCKurtz

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    Your derivative is OK, but you don't get a straight line. Sure, T(t) = 4 at t = 0 and t = 2pi. Did you try any other points? Like t = pi/2? Trig functions don't generally give you straight lines.
     
  7. Oct 10, 2009 #6
    I forgot to graph in radians. i switched modes and got a curve with maximums at t=0, pi, and 2 pi. minimums at pi/2 and 3pi/2. how do i find which is the actual max and min if i'm given just the parametric? do i need the second derivative? for the second derivative i got -8sint*cost + 8cost*sint. but when i try and graph that, i get nothing in my window. i'm so lost.
     
  8. Oct 10, 2009 #7
    Ok, found my problem, it should be -16costsint for the second derivative. do i need the second derivative at all?
     
  9. Oct 10, 2009 #8

    LCKurtz

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    "do i need the second derivative at all? "

    I guess it couldn't hurt. What do you know about the extrema of a continuous function on a closed interval?
     
  10. Oct 10, 2009 #9
    Nevermind, i got it all figured out. the zeros of the first derivative give you the maxs and mins and you just plug back in. apparently i can't remember calc 1 anymore. whoops. but thank you for your help!
     
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