Let the radius of the snow ball at a given time, t be r. Then,

Volume, `V = 4/3pir^3`

Area, `A = 4pir^2`

The rate of volume shrinkage is `(dV)/(dt)` .

`(dV)/(dt) = 4pir^2(dr)/(dt) `

The rate of volume shrinkage is 3 cubic inches per minute. Therefore when ** radius is...**

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Let the radius of the snow ball at a given time, t be r. Then,

Volume, `V = 4/3pir^3`

Area, `A = 4pir^2`

The rate of volume shrinkage is `(dV)/(dt)` .

`(dV)/(dt) = 4pir^2(dr)/(dt) `

The rate of volume shrinkage is 3 cubic inches per minute. Therefore when **radius is one inch**,

`3 = 4pi xx 1^2 xx (dr)/(dt)`

`(dr)/(dt) = 3/(4pi)` **(This is the rate of radius decrease at r=1)**

The rate of area shrinkage, `(dA)/(dt)` ,

`(dA)/(dt) =4pi xx 2r xx (dr)/(dt)`

Therefore when `r =1` ,

`(dA)/(dt) =4pi xx 2 xx 1 xx 3/(4pi)`

`(dA)/(dt) = 6`

**Therefore when r = 1 inch, the rate of surface area decrease is,**

**6 square inches per minute.**