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एक गोलीय गुब्बारे का आयतन `20` `"सेमी"^(3)//"सेकण्ड"` की दर से बढ़ रहा है । बताइये कि जब त्रिज्या `8` सेमी है, तो इसका पृष्ठीय क्षेत्रफल किस दर से बढ़ रहा है?

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माना `r` त्रिज्या के गोलीय गुब्बारे का आयतन `V` तथा पृष्ठीय (सतहों) क्षेत्रफल `S` है। तब
`S=4pir^(2)` और `V=(4)/(3)pir^(3)`
दिया है: `(dV)/(dt)=20"सेमी"^(2)//"सेकण्ड"`, `r=8` सेमी
अब, `V=(4)/(3)pir^(3)`
`implies(dV)/(dt)=(4)/(3)pixx3r^(2)(dr)/(dt)`
`implies20=4pir^(2)(dr)/(dt)`
`implies(dr)/(dt=(5)/(pir^(2))`........`(1)`
और `S=4pir^(2)`
`implies(dS)/(dt)=4pixx2r(dr)/(dt)`
`implies(dS)/(dt)=8pirxx(5)/(pir^(2))` [समी. `(1)` से ]
`implies(dS)/(dt)=(40)/(r )`
`r=8` सेमी पर ,
`[(dS)/(dt)]_(r=8)=((40)/(8))"सेमी"^(2)//"सेकण्ड"`|
अतः पृष्ठीय क्षेत्रफल `5"सेमी"^(2)//"सेकण्ड"` की दर से बढ़ रहा है।

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