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Question 1060474: A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 13s?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! we use the formula for the area of a circle and use the implicit derivative
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A = pi * r^2, where A is the area, r is the radius and pi = 22/7
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we now take the implicit derivative with respect to time
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dA/dt = 2 * pi * r * (dr/dt)
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we are given that (dr/dt) = 2 and asked to evaluate the Area's rate of change after 13 seconds
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we use the circumference of the circle to get r after 13 seconds
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C = 2 * pi * r, where C is the circumference, then
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dC/dt = 2 * pi * (dr/dt) = 2 * pi * 2 = 4 * pi per second
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after 13 seconds C = 4 * pi * 13 = 52 * pi
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52 * pi = 2 * pi * r
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r = 26
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dA/dt = 2 * pi * 26 * 2 = 326.7256 square feet / second
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