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

Step-by-step explanation:

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

Answer 2

Answer:

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

Step-by-step 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,

$\frac{dl}{dt} = 5 cm/s$

Now, to determine the rate at which the area of the square is increasing, That is $\frac{dA}{dt}$

To find $\frac{dA}{dt}$, differentiate $A = l^{2}$ with respect to $t$

From, $A = l^{2}$

$\frac{dA}{dt} = \frac{d(l^{2}) }{dl}$ ×$\frac{dl}{dt}$

$\frac{dA}{dt} = 2l$ ×$\frac{dl}{dt}$

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, $A = l^{2}$

$36 = l^{2}\\ \sqrt{36} = l\\6 = l\\l = 6cm$

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,  $\frac{dl}{dt} = 5 cm/s$

Hence,

$\frac{dA}{dt} = 2l$ ×$\frac{dl}{dt}$

$\frac{dA}{dt} = 2(6)$ × $5$

$\frac{dA}{dt} = 60 cm^{2}/s$

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