# Water leaking onto a floor forms a circular pool. The radius of the pool increases at a rate of 4 cm/min. How fast is the area of the pool increasing when the radius is 5 cm?

First, we should begin with an equation we know relating the area of a circle, the pool, and its radius:

However, we want to see how fast the area of the pool is increasing, which sounds a lot like rate... which sounds a lot like a derivative.

To put this into words, we say that:

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To find how fast the area of the pool is increasing, you can use the formula for the rate of change of area with respect to time:

( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} )

Given: ( \frac{dr}{dt} = 4 ) cm/min ( r = 5 ) cm

Plug in the values: ( \frac{dA}{dt} = 2\pi \times 5 \times 4 ) ( \frac{dA}{dt} = 40\pi ) square cm/min

So, when the radius is 5 cm, the area of the pool is increasing at a rate of ( 40\pi ) square cm/min.

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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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