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    Topics || Problems

    If \(s\) is measured in feet and \(t\) in seconds, what is the rate at which \(s\) is changing at the end of 2 seconds when \(s = \frac{1}{1-t^2}\).

    \(s = \frac{1}{1-t^2}\)

    \( \frac{ds}{dt} = \frac{0 -1 (1-t^2)^2}{(1-t^2)}\)

    \(\frac{ds}{dt} = v = \frac{-(1-t^2)^2}{1-t^2}\)

    At \(t = 2 ~sec\)

    \(v = \frac{-(1-4)^2}{1-4}\)

    \(v = \frac{9}{4} ~fps\)