# increasing at the rate of 3 cm per minute. How fast issurface area of the sphere increasing, when the radius is 10 cm?(S = 4πR^2) ***4n is 4pie 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 = 4πR^2)

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.

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