To find how fast the radius is changing when the radius is 2 cm, we can use the relationship between the area of a circle and its radius. The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle.
Given that the area is increasing at a rate of \( \frac{dA}{dt} = 1 , \text{cm}^2/\text{s} \), we want to find the rate of change of the radius \( \frac{dr}{dt} \) when the radius \( r = 2 , \text{cm} \).
Using the chain rule to differentiate the area with respect to time \( t \), we have:
\[ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \]
First, we calculate \( \frac{dA}{dr} \):
\[ \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r \]
Now we can substitute \( r = 2 , \text{cm} \):
\[ \frac{dA}{dr} = 2\pi(2) = 4\pi \]
Now, we substitute \( \frac{dA}{dt} \) and \( \frac{dA}{dr} \) into the chain rule equation:
\[ 1 = 4\pi \cdot \frac{dr}{dt} \]
Now we can solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{1}{4\pi} \]
Thus, the rate at which the radius is changing when the radius is 2 cm is:
\[ \frac{dr}{dt} \approx \frac{1}{4 \times 3.14} \approx \frac{1}{12.56} \approx 0.0796 , \text{cm/s} \]
So, the radius is changing at approximately \( 0.0796 , \text{cm/s} \) when the radius is \( 2 , \text{cm} \).
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