To solve this problem, let's break it down into parts.

Given:

• Circumference (C) = 66 cm
• Possible error in C = 0.9 cm

The formulas we’ll use:

1. Circumference of a circle: $C = 2\pi r$
2. Area of a circle: $A = \pi r^2$

Part (a): Approximate the percent error in computing the area

1. Find the radius (r) in terms of circumference (C): $r = \frac{C}{2\pi} = \frac{66}{2\pi} \approx 10.5 \text{ cm}$

2. Compute the error in radius (dr): Since $C = 2\pi r$, $dC = 2\pi \, dr \Rightarrow dr = \frac{dC}{2\pi} = \frac{0.9}{2\pi} \approx 0.143 \text{ cm}$

3. Calculate the percent error in radius: $\text{Percent error in } r = \frac{dr}{r} \times 100 \approx \frac{0.143}{10.5} \times 100 \approx 1.36\%$

4. Relate the percent error in radius to the percent error in area (A): Since $A = \pi r^2$, the percent error in $A$ is approximately twice the percent error in $r$: $\text{Percent error in } A \approx 2 \times 1.36\% \approx 2.72\%$

The correct answer for (a) should be 2.72%.

Part (b): Maximum allowable percent error in circumference if the area error must not exceed 3%

1. Since the area error is twice the radius error, we set up the equation: $2 \times \text{Percent error in } r = 3\%$ $\Rightarrow \text{Percent error in } r = 1.5\%$

2. Since the percent error in circumference is equal to the percent error in radius, the maximum allowable percent error in circumference is 1.5%.

The correct answer for (b) is 1.5%.

Would you like further explanation on any steps?

Relative Questions

1. How is the relationship between radius and circumference derived?
2. Why is the area error twice the radius error?
3. How do small changes in radius affect the area more than the circumference?
4. What is the general formula for error propagation in products and powers?
5. How would these errors change if the circle's size increased?

Tip: For small measurement errors, multiplying by the power factor (like 2 for area in terms of radius) gives a quick approximation of the resulting error.