Chain Rule with partial derivatives

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Homework Statement



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.



Homework Equations



chain rule with partial derivatives



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?
 

Answers and Replies

  • #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.
 
  • #3
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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?
 
  • #4
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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
 
  • #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.
 
  • #6
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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.
 
  • #7
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Ok, found my problem, it should be -16costsint for the second derivative. do i need the second derivative at all?
 
  • #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?
 
  • #9
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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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