math

Find
dy/dx and d2y/dx2.
x = et, y = te−t

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  1. x = e^t
    y = te^-t

    dy/dx = y'/x'
    = (1-t)e^-t/e^t
    = (1-t)e^-2t

    d^2/dx^2 = (x'y"-x"y')/x'^3
    = ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/e^3t
    = (2t-3)e^-3t

    or, do it directly

    x = e^t, so
    y = lnx/x

    y' = (1-lnx)/x^2 = (1-t)e^-2t
    y" = (2lnx-3)/x^3 = (2t-3)e^-3t

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  2. thanks. so how woudl i find For which values of t is the curve concave upward?

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  3. just as usual. Where is y" positive?

    e^-3t is always positive, so all you need is

    2t-3 > 0
    t > 3/2

    e^3/2 = 4.48, and you can see from the graph that it is concave upward for x > 4.48

    http://www.wolframalpha.com/input/?i=plot+x+%3D+e^t%2C+y+%3D+te^-t+for+1%3C%3Dt%3C%3D2

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