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A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

5) A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

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Given =   \frac{dr}{dt} = 5 \ cm/s

To find =  \frac{dA}{dt}    at  r = 8 cm

Solution:-
 
Area of the circle (A) = \pi r^{2}
\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt}                         (by chain rule)
\frac{dA}{dt} = \frac{d\pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ cm^{2}/s
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is   80\pi \ cm^{2}/s
 

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