The surface area of a sphere with radius r is given by `4*pi*r^2` and the volume of the sphere is `(4/3)*pi*r^3` .

It is given that the surface area of the sphere is increasing at the rate 18 m^2/s.

`SA = 4*pi*r^2`

`(dSA)/(dt) = 8*pi*r*(dr)/(dt) = 18`

=> `(dr)/(dt) =...

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The surface area of a sphere with radius r is given by `4*pi*r^2` and the volume of the sphere is `(4/3)*pi*r^3` .

It is given that the surface area of the sphere is increasing at the rate 18 m^2/s.

`SA = 4*pi*r^2`

`(dSA)/(dt) = 8*pi*r*(dr)/(dt) = 18`

=> `(dr)/(dt) = 18/(8*pi*r)`

The volume V = `(4/3)*pi*r^3`

`(dV)/(dt) = (4/3)*pi*3*r^2*(dr)/(dt)`

substitute `(dr)/(dt) = 18/(8*pi*r)`

=> `(dV)/(dt) = (4/3)*pi*3*r^2*18/(8*pi*r)`

=> `(dV)/(dt) = r*9`

**The rate of increase of the sphere is given by `(dV)/(dt) = 9*r` **