# A spherical snowball with diameter 4 inches is removed from the freezer in june and begins melting uniformly such that it is shrinking 2 cubic inches per minute. How fast (in square inches per minute) is its surface area decreasing when the radius is one inch?

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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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.

Approved by eNotes Editorial Team