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31 July, 14:30

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 6 inches per second. When the radius is 4 feet, at what rate (in ft2/sec) is the total area A of the disturbed water changing?

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  1. 31 July, 15:58
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    dA/dt = 32π ft²/s

    Step-by-step explanation:

    r = the radius of the circle (in ft)

    A = the area of the circle (in ft²)

    t = time since the pebble hit the water (in s)

    Rates of change;

    dr/dt = 6 in/s = 6 x 0.0833 ft/s = 0.5 ft/s

    Now, we are to find dA/dt when r = 4 ft

    Area (A) = πr²

    Let's Differentiate with respect to

    t

    Thus,

    dA/dt = (d/dt) (πr²)

    Using chain rule,

    dA/dt = (2πr) (dr/dt)

    Plugging in the relevant values to get;

    dA/dt = (2 x π x 4) / 0.5 = 32π ft²/s
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