Question

Each side of a square is increasing at a rate of 5 cm/s. At what rate is the area of the square increasing when the area of the square is 36 cm2

Answer 1

Answer: 60

Explanation:

`A = s^2\\\\frac{dA}{dt}=2s (ds)/(dt)\\36=s^2 \\ s=6\\(dA)/(dt) = 2(6)(5)=60 (cm^2)/(s)`

Answer 2

The area of the square is increasing at a rate of 60 cm²/s

Explanation:

Let
`l` be the length of the side of the square

Hence, The area of the square is given by

`A = l^(2)`

Where
`A` is the area of the square

From the question, each side of a square is increasing at a rate of 5 cm/s, that is,

`(dl)/(dt) = 5 cm/s`

Now, to determine the rate at which the area of the square is increasing, That is
`(dA)/(dt)`

To find
`(dA)/(dt)`, differentiate
`A = l^(2)` with respect to
`t`

From,
`A = l^(2)`

`(dA)/(dt) = (d(l^(2)) )/(dl)` ×
`(dl)/(dt)`

`(dA)/(dt) = 2l` ×
`(dl)/(dt)`

Now, to determine the rate at which the area of the square is increasing when the area of the square is 36 cm²

First, we will determine the length at this instance,

From,
`A = l^(2)`

`36 = l^(2)\\ √(36) = l\\6 = l\\l = 6cm`

The length at this instance is 6cm

To determine the rate at which the area of the square is increasing at this instance

`l = 6cm\\`

From the question,
`(dl)/(dt) = 5 cm/s`

Hence,

`(dA)/(dt) = 2l` ×
`(dl)/(dt)`

`(dA)/(dt) = 2(6)` ×
`5`

`(dA)/(dt) = 60 cm^(2)/s`

Hence. the area of the square is increasing at a rate of 60 cm²/s