Each side of a square is increasing at a rate of 6 cm/s . At what rate is the area of the square increasing when the area of the square is 16 cm² ?

Solution:

Given each side of the square is increasing at a rate of 6cm/s

Area, A = s²

s is the side length

=> (dA)/(dt)

= (dA/ds)*(ds/dt)

= 2s*(ds)/(dt) {chain rule of differentiation}

(ds)/(dt) = 6 cm/s

When A = 16 cm²

=> 16 = s²

=> s = √(16)

=> s = 4 cm

Hence (dA)/(dt) = 2 × 4 × 6 = 48 cm²s^-1

Each side of a square is increasing at a rate of 6 cm/s . At what rate is the area of the square increasing when the area of the square is 16 cm² ?

Summary:

Each side of a square is increasing at a rate of 6 cm/s . At 48 cm²s^-1 rate, the area of the square is increasing when the area of the square is 16 cm².