The area of a square is increasing the rate of `0.5" cm"^(2)//"sec"`. Find the rate of increase of its perimeter when the side of square is 10 cm long.

Explanation

If x cm is the side of square, A `"cm"^(2)` is the area and P cm is the perimeter, then P = 4x and `A=x^(2)`.
`:.P=4sqrt(A)`
Differentiating w.r.t. t, we get, `("dP")/("dt")=4xx(1)/(2sqrt("A")).("dA")/("dt")=(2)/(sqrt("A")).("dA")/("dt")`
`:.("dP")/("dt")=(2)/(x).("dA")/("dt")" "...(1)`
Now, `("dA")/("dt")=0.5" cm"^(2)//"sec"` and x = 10 cm
`:." (1) gives",("dP")/("dt")=(2)/(10)xx0.5=0.1" cm"//"sec"`
`:.` the perimeter is increasing at the rate of 0.1 cm/sec.