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I Second derivative of a curve defined by parametric equations

  1. Mar 1, 2017 #1
    Quick question. I know that if we have a curve defined by ##x=f(t)## and ##y=g(t)##, then the slope of the tangent line is ##\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##. I am trying to find the second derivative, which would be ##\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{\frac{d}{dx}\frac{dy}{dt}}{\frac{dx}{dt}}##. Now, the final correct formula is ##\displaystyle \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}##. My question is, why is it valid that ##\displaystyle \frac{d}{dx} \frac{dy}{dt} = \frac{d}{dt} \frac{dy}{dx}##?
     
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  3. Mar 1, 2017 #2

    andrewkirk

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    The formula can be calculated by the chain rule, treating ##x## as the intermediate function (of ##t##), as:
    $$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
    \left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$
    And that is equal to
    $$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}
    $$
    which is the same as what you describe as the final correct formula.

    How did you get your own formula? It doesn't look correct to me. Doing it your way I get:
    $$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
    \frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right)
    =\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}-
    \frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt}
    \right]\ /\ \left(\frac{dx}{dt}\right)^2
    $$
    (using the quotient rule for the second step), which is not like what you wrote.
     
  4. Mar 1, 2017 #3
    Okay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if ##\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##, then ##\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}##?
     
  5. Mar 1, 2017 #4

    andrewkirk

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    It is correct. But what you wrote here is different from what you wrote in the OP. Perhaps your LaTeX got a bit muddled back there?
     
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