The surface area of a sphere `A = 4pir^2`

The rate of increase of radius is 3 per minute

`(dr)/(dt) = 3`

`(dA)/(dt) = 4pixx2rxx(dr)/(dt)`

`(dA)/(dt) = 8pir(dr)/(dt)`

At `r = 10,`

`(dA)/(dt) = 8pixx10xx3`

`(dA)/(dt) = 240pi `

Therefore at `r = 10` , the rate of area...

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The surface area of a sphere `A = 4pir^2`

The rate of increase of radius is 3 per minute

`(dr)/(dt) = 3`

`(dA)/(dt) = 4pixx2rxx(dr)/(dt)`

`(dA)/(dt) = 8pir(dr)/(dt)`

At `r = 10,`

`(dA)/(dt) = 8pixx10xx3`

`(dA)/(dt) = 240pi `

Therefore at `r = 10` , the rate of area increase is `240pi m^2` per minute.