# Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm^2?

Hence,

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Consider the side of the square as a function of time

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To find the rate at which the area of the square is increasing when the area is 16 cm², we use the chain rule of differentiation.

Let ( A ) represent the area of the square and ( s ) represent the length of a side of the square.

Given: ( \frac{ds}{dt} = 6 ) cm/s (rate at which the side length is increasing) ( A = s^2 )

Differentiating both sides of the equation ( A = s^2 ) with respect to time ( t ), we get: ( \frac{dA}{dt} = 2s \frac{ds}{dt} )

Substitute the given values: ( A = 16 ) cm² and ( \frac{ds}{dt} = 6 ) cm/s: ( 16 = s^2 ) ( s = 4 ) cm

( \frac{dA}{dt} = 2(4)(6) ) ( \frac{dA}{dt} = 48 ) cm²/s

Therefore, when the area of the square is 16 cm², the rate at which the area is increasing is ( 48 ) cm²/s.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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