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