Respuesta :

Answer:

80cm^2/s

Step-by-step explanation:

This is a related rates problem where we are considering the rate at which the area of a square changes with respect to time.

So lets consider the area of a square:

A = s^2 (where s represents the length of one side of the square)

Related rates problem deal with functions of time so if we take the area and side length as a function of time and then differentiate implicitly we get:

[tex]\frac{dA}{dt} = 2s(\frac{ds}{dt})[/tex]

The problem states that the side of a square is increasing at a rate off 8cm/s so we can conclude that ds/dt = 8cm/s leaving us with:

[tex]\frac{dA}{dt} = 16s[/tex]

Now, to solve for s we have to consider the other value given. If the area of the square is initially 25cm^2 we can plug this into our formula for area to solve for the side length.

25 = s^2

s = +/- 5 (since side lengths are only positive we only consider +5)

s = 5

Now we can plug this back in for s:

[tex]\frac{dA}{dt} = 80[/tex]

Therefore, the rate at which the area of the square is increasing is 80cm^2 per second.

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