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The volume of a spherical balloon is increasing at a rate of ` 25 cm^(3)//sec`. Find the rate of increase of its curved surface when the radius of balloon is 5 cm. |
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Answer» Correct Answer - `(dS)/(dt) = 10 cm^(2)//s` `V = (4)/(3) pir^(3) rArr (dV)/(dt) = (4pi r^(2)).(dr)/(dt) .(dr)/(dt) rArr (dr)/(dt) = (25)/(4pir^(2))` `S = 4 pi r^(2) rArr (dS)/(dt) = 8pi r.(dr)/(dt) = (8pi r.(25)/(4pir^(2))) = (50)/(r)` `rArr [(dS)/(dt)]_(r =5) = ((50)/(5)) cm^(2)//s` |
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