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Differentials of Composite Functions

  1. Jan 18, 2012 #1
    I cannot figure out how to do this problem completely:

    If U =x3y, find [itex]\frac{dU}{dt}[/itex] if x5 + y = t and x2 + y3 = t2.

    I know that I am using the chain rule here and I have the partial derivates of U:
    [itex]\frac{∂U}{∂x}[/itex] = 3x2y

    [itex]\frac{∂U}{∂y}[/itex] = x3

    So far I have the equation given below.
    [itex]\frac{dU}{dt}[/itex] = 3x2y [itex]\frac{dx}{dt}[/itex] + x3 [itex]\frac{dy}{dt}[/itex]

    However, I do not know how to calculate [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex]. I tried to calculate them implicitly but I am still working with three variables x, y, and t. Could you please help me with this? Any insight would be greatly appreciated! Thank you!
     
  2. jcsd
  3. Jan 18, 2012 #2
    Well, you haven't used either of the other conditions: [itex]x^5 + y = t[/itex] and [itex]x^2 + y^3 = t^2[/itex]. Try differentiating with respect to [itex]t[/itex]. Then you can at least get [itex]\frac{dx}{dt}[/itex] in terms of [itex]\frac{dy}{dt}[/itex] (or vice versa).
     
  4. Jan 23, 2012 #3
    Thank you for your suggestion! I really appreciate it! :)

    Okay so I calculated [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex].

    [itex]\frac{dx}{dt}[/itex] = -[itex]\frac{\frac{∂F}{∂t}}{\frac{∂F}{∂x}}[/itex] and [itex]\frac{dy}{dt}[/itex] = -[itex]\frac{\frac{∂F}{∂t}}{\frac{∂F}{∂y}}[/itex]

    I substituted x and y into the two different t equations getting x2 + (t - x5)3 - t2 = 0 and (t2 - y3)5/2 + y - t = 0. From there I was able to get [itex]\frac{dx}{dt}[/itex] = - [-2t + 3(t - x5)2 / 2x - 15x4(t - x5)2] and [itex]\frac{dy}{dt}[/itex] = - [5t(t2 - y3)[itex]\frac{3}{2}[/itex] - 1 / -[itex]\frac{15}{2}[/itex]y2(t2 - y3)[itex]\frac{3}{2}[/itex] + 1].

    From there I plugged [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex] into the original equation, [itex]\frac{du}{dt}[/itex] = [itex]\frac{∂u}{∂x}[/itex] ([itex]\frac{dx}{dt}[/itex]) + [itex]\frac{∂u}{∂y}[/itex] ([itex]\frac{dy}{dt}[/itex]).

    The final answer for [itex]\frac{du}{dt}[/itex] looks pretty ugly and cannot be simplified much. Does this seem correct? Is this how I ultimately calculate [itex]\frac{du}{dt}[/itex]?
     
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