Question

the sides of a square are increasing at a rate of 10cm/sec. how fast is the area enclosed by the square increasing when the area is 150 cm^2

Answer 1

A = s²

dA/dt = 2s(ds/dt)

= (2√150)(10) = 20√150

= 100√6 cm²/sec

Answer 2

The area of the square is increasing at a rate of 244.94 cm^2/s when the area is 150 cm^2.

Step-by-step explanation:

In this scenario, we are dealing with the sides of a square increasing at a rate of 10 cm/sec, and we are asked to determine how fast the area of the square is increasing when the area is 150 cm^2.

Let's denote the side of the square as 's' and the area of the square as 'A'.

Since A = s^2, we can use differentiation with respect to time (t) to find the rate of change of the area.

We know that A = s^2

dA/dt = 2s · ds/dt.

We are given ds/dt = 10 cm/s.

To find dA/dt when A = 150 cm^2, we first have to find the side length of the square when the area is 150 cm^2.

So, s = √(150 cm2)

= 12.247 cm (approximately).

Now, plug this value into the differentiated formula:

dA/dt = 2 · 12.247 cm · 10 cm/s

= 244.94 cm^2/s.

Therefore, the area of the square is increasing at a rate of 244.94 cm^2/s when the area is 150 cm^2.