Paint is poured onto a flat surface and a circular patch is formed. The area of the patch increases at a rate of 5 cm² s^-1 . a Find, in terms of π, the radius of the patch after 8 seconds. b Find, in terms of π, the rate of increase of the radius after 8 seconds.

To solve the problem, let's break it into two parts.

---

### Given:
- The **area** of the circular patch increases at a constant rate:  
  \[
  \frac{dA}{dt} = 5 \, \text{cm}^2/\text{s}.
  \]
- The area of a circle is given by \( A = \pi r^2 \).

---

### Part (a): Radius of the patch after 8 seconds

1. **Find the total area of the patch after 8 seconds**:  
   Since the area increases at a constant rate of \( 5 \, \text{cm}^2/\text{s} \), after 8 seconds:
   \[
   A = \frac{dA}{dt} \cdot t = 5 \cdot 8 = 40 \, \text{cm}^2.
   \]

2. **Relate the area to the radius**:  
   Using the formula for the area of a circle:
   \[
   A = \pi r^2.
   \]
   Substitute \( A = 40 \):
   \[
   40 = \pi r^2.
   \]

3. **Solve for \( r \)**:
   \[
   r^2 = \frac{40}{\pi}, \quad r = \sqrt{\frac{40}{\pi}} = \frac{\sqrt{40}}{\sqrt{\pi}}.
   \]
   Simplify:
   \[
   r = \frac{2\sqrt{10}}{\sqrt{\pi}} \, \text{cm}.
   \]

---

### Part (b): Rate of increase of the radius after 8 seconds

1. **Differentiate the area formula**:  
   From \( A = \pi r^2 \), differentiate both sides with respect to \( t \):
   \[
   \frac{dA}{dt} = 2\pi r \frac{dr}{dt}.
   \]

2. **Rearrange to find \( \frac{dr}{dt} \)**:  
   \[
   \frac{dr}{dt} = \frac{\frac{dA}{dt}}{2\pi r}.
   \]

3. **Substitute known values**:  
   - \( \frac{dA}{dt} = 5 \),
   - \( r = \frac{2\sqrt{10}}{\sqrt{\pi}} \) (from part (a)).

   Substitute into the formula:
   \[
   \frac{dr}{dt} = \frac{5}{2\pi \cdot \frac{2\sqrt{10}}{\sqrt{\pi}}}.
   \]

4. **Simplify**:
   \[
   \frac{dr}{dt} = \frac{5}{\frac{4\sqrt{10}\sqrt{\pi}}{\pi}} = \frac{5\pi}{4\sqrt{10}\sqrt{\pi}} = \frac{\sqrt{\pi}}{4\sqrt{10}} \, \text{cm/s}.
   \]

---

### Final Answers:
1. Radius after 8 seconds:
   \[
   r = \frac{2\sqrt{10}}{\sqrt{\pi}} \, \text{cm}.
   \]

2. Rate of increase of the radius after 8 seconds:
   \[
   \frac{dr}{dt} = \frac{\sqrt{\pi}}{4\sqrt{10}} \, \text{cm/s}.
   \]

---

Do you want details or have any questions?  
Here are **5 related questions** to consider:  
1. What happens to \( \frac{dr}{dt} \) as \( t \to \infty \)?  
2. How does the rate of area increase affect the radius' rate of change?  
3. What if the area increases non-linearly? How would the approach differ?  
4. How do units play a role in verifying these results?  
5. How would the result change if \( \pi \) is approximated to 3.14?

**Tip**: When working with related rates, always ensure you differentiate correctly with respect to the same variable!

Paint is poured onto a flat surface and a circular patch is formed. The area of the patch increases at a rate of 5 cm² s^-1. a. Find, in terms of π, the radius of the patch after 8 seconds. b. Find, in terms of π, the rate of increase of the radius after 8 seconds.

Discover how to calculate the radius and rate of radius change for a circular patch whose area increases at 5 cm²/s. Detailed step-by-step solution using geometry and calculus. Perfect for Grades 10-12 students studying related rates and area formulas.

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