# At what rate, in cm/s, is the radius of the circle increasing when the radius is 5 cm if oil is poured on a flat surface, and it spreads out forming a circle and the area of this circle is increasing at a constant rate of 5 cm2/s?

deriving you get that your area changes and your radius changes as:

But:

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To find the rate at which the radius of the circle is increasing when the radius is 5 cm, given that the area of the circle is increasing at a constant rate of 5 cm²/s, we can use the formula for the rate of change of the area of a circle with respect to its radius.

Let ( A ) be the area of the circle and ( r ) be its radius.

Given: [ \frac{dA}{dt} = 5 , \text{cm}^2/\text{s} ]

Using the formula for the area of a circle: [ A = \pi r^2 ]

Differentiating both sides with respect to time: [ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} ]

We can solve for ( \frac{dr}{dt} ): [ 5 = 2\pi \times 5 \times \frac{dr}{dt} ] [ \frac{dr}{dt} = \frac{5}{10\pi} ] [ \frac{dr}{dt} = \frac{1}{2\pi} ]

So, the rate at which the radius of the circle is increasing when the radius is 5 cm is ( \frac{1}{2\pi} ) cm/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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