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Thread: Parametric curves (derivative)

  1. #1
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    Parametric curves (derivative)

    $\displaystyle x = e - e^t$
    $\displaystyle y = t e^{-t}$

    Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ and state where the graph is concave upward and downward.

    I found dy/dx already: $\displaystyle \frac{1-e^{-t}}{1-e^t}$

    But when I used the quotient rule to find $\displaystyle \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$ I got $\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}$

    I think this is right, but I might have entered it wrong. Is it?

    I know the concavity is based on when the second derivative is positive and negative.
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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    $\displaystyle x = e - e^t$
    $\displaystyle y = t e^{-t}$

    Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ and state where the graph is concave upward and downward.

    I found dy/dx already: $\displaystyle \frac{1-e^{-t}}{1-e^t}$

    But when I used the quotient rule to find $\displaystyle \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$ I got $\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}$
    $\displaystyle \frac{dy}{dx} = \frac{(1-t)e^{-t}}{-e^t} = \frac{t-1}{e^{2t}}$

    $\displaystyle \frac{d^2y}{dx^2} = \frac{e^{2t} - (t-1) \cdot 2e^{2t}}{e^{6t}} = \frac{3-2t}{e^{4t}}$
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  3. #3
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    Er, sorry. Something got lost when I was formatting. Big time.

    The equations should be:

    $\displaystyle x = t - e^t$
    $\displaystyle y = t + e^{-t}$

    My bad. I meant that I had found dy/dx and it had been marked as correct.
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