Respuesta :
Answer:
The area of the square is increasing at a rate of 60 cm²/s when the area of the square is 36 cm²
Step-by-step explanation:
From the formula for area of a square
[tex]A = l^{2}[/tex]
Where [tex]A[/tex] is the area
and [tex]l[/tex] is the length of a side
Differentiate the equation with respect to time t
[tex]\frac{d}{dt} A = \frac{d}{dt} (l^{2})[/tex]
Let [tex]u = l^{2}[/tex]
Then,
[tex]\frac{dA}{dt} = \frac{du}{dt}[/tex]
Using Chain rule
[tex]\frac{du}{dt} = \frac{du}{dl} \times \frac{dl}{dt}[/tex]
Since, [tex]u = l^{2}[/tex]
[tex]\frac{du}{dl} = 2l[/tex]
∴ [tex]\frac{dA}{dt} = \frac{du}{dt}[/tex] becomes
[tex]\frac{dA}{dt} = \frac{du}{dl} \times \frac{dl}{dt}[/tex]
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt}[/tex]
To determine the rate at which the area of the square is increasing when the area of the square is 36 cm². We will find the length [tex]l[/tex] at this instant.
From
[tex]A = l^{2}[/tex]
[tex]36cm^{2} = l^{2} \\[/tex]
∴ [tex]l = \sqrt{36cm^{2} }[/tex]
[tex]l = 6cm[/tex]
From the question,
Each side of a square is increasing at a rate of 5 cm/s, that is
[tex]\frac{dl}{dt} = 5cm/s[/tex]
Putting the parameters into the equation,
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt}[/tex]
[tex]\frac{dA}{dt} = 2(6)(5)[/tex]
[tex]\frac{dA}{dt} = 60 cm^{2}/s[/tex]
Hence, the area of the square is increasing at a rate of 60 cm²/s when the area of the square is 36 cm².
