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Math Help - rates of change

  1. #1
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    rates of change

    The time t, in seconds, taken by an object dropped from a height of s metres to reach the ground is given by the formula t = squareroot(s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

    so i set up the solution and everything, but the squareroots are driving me mad! please help me solve this?

    thank you.
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  2. #2
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    Rates of change

    Hello checkmarks
    Quote Originally Posted by checkmarks View Post
    The time t, in seconds, taken by an object dropped from a height of s metres to reach the ground is given by the formula t = squareroot(s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

    so i set up the solution and everything, but the squareroots are driving me mad! please help me solve this?

    thank you.
    t = \sqrt{\frac{s}{5}}

    =\frac{\sqrt{s}}{\sqrt{5}}

    = \frac{s^{\frac{1}{2}}}{\sqrt{5}}

    \Rightarrow \frac{dt}{ds} = \frac{\tfrac{1}{2}s^{-\frac{1}{2}}}{\sqrt{5}}

    =\frac{1}{2\sqrt{5s}}

    When s = 125, \frac{dt}{ds}=\frac{1}{2\sqrt{5\times 125}}

    =\frac{1}{50}

    Grandad
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