To determine how fast the radius of the circle is changing when the oil is spilling at a rate of 10 cm²/sec and the radius is 5 cm, we can apply calculus and related rates concepts.
1. Understanding the Problem:
- We are given the rate at which the area of the circle is changing, \(\( \frac{dA}{dt} = 10 \)\) cm²/sec.
- We are asked to determine how fast the radius \(\( r \)\) is changing when \(\( r = 5 \)\) cm.
2. Formulating the Problem:
- The area \(\( A \)\) of a circle is given by the formula:
\(\[
A = \pi r^2
\]\)
- To find how the radius is changing over time, we differentiate both sides of this equation with respect to time \(\( t \)\):
\(\[
\frac{dA}{dt} = \frac{d}{dt}(\pi r^2)
\]\)
3. Applying Differentiation:
- Differentiating \(\( A = \pi r^2 \)\) with respect to time \(\( t \)\), using the chain rule, we get:
\(\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
\]\)
- Here, \(\( \frac{dA}{dt} \)\) is the rate of change of the area (given as 10 cm²/sec), \(\( r \)\) is the radius (given as 5 cm), and \(\( \frac{dr}{dt} \)\) is the rate at which the radius is changing, which we need to determine.
4. Solving for \(\( \frac{dr}{dt} \)\):
- Rearrange the formula to solve for \(\( \frac{dr}{dt} \)\):
\(\[
\frac{dr}{dt} = \frac{\frac{dA}{dt}}{2\pi r}
\]\)
5. Substituting Known Values:
- Substitute the known values \(\( \frac{dA}{dt} = 10 \)\) cm²/sec and \(\( r = 5 \)\) cm into the equation:
\(\[
\frac{dr}{dt} = \frac{10}{2\pi \cdot 5}
\]\)
\(\[
\frac{dr}{dt} = \frac{10}{10\pi}
\]\)
\(\[
\frac{dr}{dt} = \frac{1}{\pi}
\]\)
6. Interpreting the Result:
- The rate of change of the radius when the radius is 5 cm is:
\(\[
\frac{dr}{dt} = \frac{1}{\pi} \text{ cm/sec}
\]\)
- Numerically, \(\( \frac{1}{\pi} \approx 0.318 \text{ cm/sec} \)\).
Therefore, the radius of the circle is increasing at a rate of approximately 0.318 cm/sec when the radius is 5 cm.
