Let's solve the problem by breaking it down step by step.

Given:

• The circumference of the sphere $C = 86$ cm
• The possible error in measurement of the circumference $dC = 0.5$ cm

(a) Estimating the maximum error in surface area:

1. Formula for the circumference of a sphere: $C = 2\pi r$ Solving for the radius $r$: $r = \frac{C}{2\pi}$

2. Surface area of a sphere: $A = 4\pi r^2$

3. Differential of the surface area: The differential $dA$, representing the maximum error in surface area, is: $dA = \frac{dA}{dr} \cdot dr = 8\pi r \, dr$ Here, $dr$ is the error in the radius. To find $dr$, we differentiate the expression for circumference: $C = 2\pi r \quad \Rightarrow \quad dC = 2\pi \, dr \quad \Rightarrow \quad dr = \frac{dC}{2\pi}$ Now, substitute this into $dA$.

4. Relative error: The relative error for surface area is given by: $\text{Relative Error} = \frac{dA}{A}$

(b) Estimating the maximum error in volume:

1. Volume of a sphere: $V = \frac{4}{3} \pi r^3$

2. Differential of the volume: The differential $dV$, representing the maximum error in volume, is: $dV = \frac{dV}{dr} \cdot dr = 4\pi r^2 \, dr$

3. Relative error: The relative error for volume is given by: $\text{Relative Error} = \frac{dV}{V}$

Let me calculate each part step-by-step for you based on these formulas.Here are the results for the given problem:

(a) Surface Area:

• The maximum error in surface area is $27 \, \text{cm}^2$ (rounded to the nearest integer).
• The relative error in surface area is $0.012$ (rounded to three decimal places).

(b) Volume:

• The maximum error in volume is $187 \, \text{cm}^3$ (rounded to the nearest integer).
• The relative error in volume is $0.017$ (rounded to three decimal places).

Would you like more details on any step or have any questions?

Here are some related questions:

1. How does the radius influence the error in surface area?
2. How would increasing the possible error in circumference affect the maximum error in surface area?
3. What happens to the relative error if the sphere's circumference is smaller?
4. Can you use this method to estimate errors in other geometric shapes?
5. How does the use of differentials help in error estimation?

Tip: Using differentials for error estimation is highly effective for small measurement errors, as it approximates changes in functions for tiny variations.