# A hypothetical square grows so that the length of its diagonals are increasing at a rate of 4 m/min. How fast is the area of the square increasing when the diagonals are 14 m each?

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To find the rate at which the area of the square is increasing when the diagonals are 14 meters each, we can use the fact that the diagonal of a square is related to its side length by the formula:

[ d = s\sqrt{2} ]

Differentiating both sides with respect to time, we get:

[ \frac{{dd}}{{dt}} = \frac{{ds}}{{dt}}\sqrt{2} ]

Given that ( \frac{{dd}}{{dt}} = 4 ) m/min, we can rearrange the formula to solve for ( \frac{{ds}}{{dt}} ):

[ \frac{{ds}}{{dt}} = \frac{{4}}{{\sqrt{2}}} ]

Now, when the diagonals are 14 meters each, we can find the side length of the square using the formula:

[ s = \frac{{d}}{{\sqrt{2}}} ]

[ s = \frac{{14}}{{\sqrt{2}}} ]

[ s = 7\sqrt{2} ]

Then, substituting ( s = 7\sqrt{2} ) and ( \frac{{ds}}{{dt}} = \frac{{4}}{{\sqrt{2}}} ) into the formula for the rate of change of the area of a square with respect to time, we get:

[ \frac{{dA}}{{dt}} = 2s \frac{{ds}}{{dt}} ]

[ \frac{{dA}}{{dt}} = 2(7\sqrt{2})\left(\frac{{4}}{{\sqrt{2}}}\right) ]

[ \frac{{dA}}{{dt}} = 56 \text{ m}^2/\text{min} ]

So, when the diagonals are 14 meters each, the area of the square is increasing at a rate of 56 square meters per minute.

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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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