Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook