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3 September, 03:29

A man 6ft tall walks at a rate of 6ft/s away from a lamppost that is 23 ft high. At what rate is the length of his shadow changing when he is 65 ft away from the lamppost?

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  1. 3 September, 06:11
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    Suppose,

    x = distance of the man

    s = length of the shadow

    Using the idea of similar triangles

    6/s = 23 / (x + s)

    Simplifying:

    we get,

    6 (x + s) = 23s

    6x + 6s = 23s

    6x = 17s

    Differentiating with respect to time,

    6 (dx/dt) = 17 (ds/dt)

    Manipulating the above equation for ds/dt,

    ds/dt = (17/6) (dx/dt)

    ds/dt = (17/6) (6)

    ds/dt = 17 ft/sec.
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