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 *= 4*πR*2)

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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).**

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