To find the rate of increase over time for the radius of a disk, we’re given that:

1. The area $A$ of a disk with radius $r$ is given by: $A(r) = \pi r^2$

2. The rate at which the area is increasing with respect to time is: $\frac{dA}{dt} = 2 \, \text{cm}^2/\text{s}$

We are asked to find $\frac{dr}{dt}$, the rate of change of the radius over time.

Solution

1. Differentiate the area formula with respect to time $t$: Since $A = \pi r^2$, we can differentiate both sides with respect to $t$: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$

2. Substitute the given values: We know that $\frac{dA}{dt} = 2$ cm²/s, so we can substitute this into the equation: $2 = 2\pi r \frac{dr}{dt}$

3. Solve for $\frac{dr}{dt}$: Divide both sides by $2\pi r$ to isolate $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{2}{2\pi r} = \frac{1}{\pi r}$

Thus, the rate of change of the radius with respect to time is: $\frac{dr}{dt} = \frac{1}{\pi r} \quad \text{cm/s}$

Final Answer

The rate of increase of the radius $r$ of the disk over time is: $\frac{dr}{dt} = \frac{1}{\pi r} \quad \text{cm/s}$

Would you like more details or have any questions about this solution?

---

Here are some related questions you might consider:

1. How would the rate $\frac{dr}{dt}$ change if the area increased at a different rate, such as 5 cm²/s?
2. How does the radius change over time if $r$ is initially large or small?
3. What happens to $\frac{dr}{dt}$ as $r$ approaches infinity?
4. How could we integrate $\frac{dr}{dt} = \frac{1}{\pi r}$ to find $r$ as a function of $t$?
5. What is the relationship between $\frac{dA}{dt}$ and $\frac{dA}{dr}$ in this problem?

Tip: When solving related rates problems, carefully identify the variables that change over time and use implicit differentiation.