Let A(t)A(t) be the area of a circle with radius r(t)r(t), at time tt in min. Suppose the radius is changing at the rate of drdt=3drdt=3 ft/

Question

Let A(t)A(t) be the area of a circle with radius r(t)r(t), at time tt in min. Suppose the radius is changing at the rate of drdt=3drdt=3 ft/min. Find the rate of change of the area at the moment in time when r=6r=6 ft.

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Autumn 1 week 2021-11-19T10:06:56+00:00 1 Answer 0 views 0

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    2021-11-19T10:08:06+00:00

    Answer:

    At a point when r = 6 ft

    The rate of change of Area with time is given as

    (dA/dt) = 36π = 113.1 ft²/min

    Step-by-step explanation:

    Area of a circle, A = πr²

    A = πr²

    (dA/dt) = (dA/dr) × (dr/dt)

    (dA/dr) = 2πr, (dr/dt) = 3

    (dA/dt) = 2πr × 3 = 6πr

    At a point when r = 6 ft

    (dA/dt) = 6πr = 6π × 6 = 36π = 113.1 ft²/min

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