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using (S=4(pie)R^2) the radius of a ballon increasing at a rate of 3 cm per minute. How...
using (S=4(pie)R^2) the radius of a ballon increasing at a rate of 3 cm per minute. How fast is the surface are increasing, radius is 10
The radius of a spherical balloon is increasing at the rate of 3 cm per minute. How fast is the surface area of the sphere increasing, when the radius is 10 cm? Round to 1 decimal place. (S = 4πR2)
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You should use the sphere surface area formula such that:
`S = 4pi*r^2`
You need to differentiate this formula with respect to t to evaluate the rate of change of area of sphere such that:
`(dS)/(dt) = 8pi*r*(dr)/(dt)`
The problem provides the informations that the radius is of 10 cm and the rate of change of radius is `(dr)/(dt) = 3` (cm)/(min),hence, you should substitute these values in formula `(dS)/(dt)` such that:
`(dS)/(dt) = 8pi*10*3`
`(dS)/(dt) = 240 pi`
Hence, evaluating the rate of change of sphere surface area, under given conditions, yields `(dS)/(dt) = 240 pi (cm^2)` /(min).
Posted by sciencesolve on June 21, 2012 at 6:20 AM (Answer #1)
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