If area of circle increases at a rate of `2cm^(2)//sec` then find the rate at which area of the inscribed square increases.

Explanation

Area of circle `,A, =pir^(2) .` Area of square `,A_(2) =2r^(2)` (see figure)

`(dA)/(dt) , =2pir .(dr)/(dt) " "(dA_(2))/(dt) =4r . (dr)/(dt)`
`:' " "2=2 pi r .(dr)/(dt) " "rArr " " r(dr)/(dt)=(1)/(pi)`
`:. (dA_(2))/(dt) =4 .(1)/(pi) =(4)/(pi) . cm^(2)//sec`
Area of square increases at the rate .`(4)/(pi) cm^(2)//sec.`