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Homework Help: Chain Rule Problem

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    [12] 3.Thetemperature T(x, y) at a point of the xy-plane is given by
    T(x,y)= ye^(x^2).
    A bug travels from left to right along the curve y = x^2
    at a speed of 0.01m/sec. The bug
    monitors T(x, y) continuously. What is the rate of change of T as the bug passes through
    the point (1, 1)?

    2. Relevant equations
    Parameterizing x and y in terms of t, taking into account the velocity given (and assuming x and y are in meters, t is in seconds):

    x = 0.01t
    y = (0.01t)^2
    t = 100 s

    Chain rule
    dT/dt = Tx dx/dt + Ty dy/dt

    3. The attempt at a solution

    dT/dt = 2xye^(x^2) * 0.01 + e^(x^2) * 0.02(0.01t) = 2*1*1*e * 0.01 + 0.02*e * 1 =
    0.04*e degrees/sec

    Right? Lol.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 3, 2011 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For that parametrization, the bug's speed is 0.01m/sec only at the origin.

    So the answer is likely incorrect.
  4. Dec 4, 2011 #3


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    Science Advisor

    You have the horizontal component of the bug's speed equal to 0.01 m/s, not its speed. If the bugs horizontal position is given by u(t), and its vertical position by v(t), then its speed is given by [itex]\sqrt{(u')^2+ (v')^2}= 0.01[/tex]
  5. Dec 4, 2011 #4
    Hey thanks for the replies... the horizontal component eh, so if it's traveling at 0.01m/s from left to right then x = u(t) = 0.01t, and then if y = x^2, so wouldn't y = (0.01t)^2 then? How do you arrive at sqrt{(u')^2+ (v')^2}= 0.01? Wouldn't the bug be at (1,1) at 100 seconds given the parametrizations?
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