If the length of the diagonal of a square is increasing at the rate of 0.2 cm/sec, then the rate of increase of its area when its side is `30//sqrt(2)` cm, is

Options

A. `3 cm^(2)//sec`
B. `(6)/(sqrt(2))cm^(2)//sec`
C. `3sqrt(2)cm^(2)//sec`
D. `6cm^(2)//sec`

Answer

D

Explanation

Let at any time t the length of the diagonal be x cm. Then, each side = `(x)/(sqrt(2))` cm.
We have,
`(dx)/(dt) = 0.2 cm//sec`
Let A be the area of the squar. Then,
`A=((x)/(sqrt(2)))^(2)=(1)/(2)x^(2)`
`implies (dA)/(dt)=x(dx)/(dt)`
`implies ((dA)/(dt))_((x)/(sqrt(2))=(30)/(sqrt(2)))=30xx0.2 = 6cm^(2)//sec`