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We are asked to find the maximum error and the relative error in the calculated surface area of a sphere using linear approximation. Here's how we can approach the problem:

### Step 1: Formulas Involved
1. **Circumference of a sphere**: \( C = 2\pi r \), where \( r \) is the radius.
2. **Surface area of a sphere**: \( A = 4\pi r^2 \).
   
   We are given the circumference \( C = 84.000 \, \text{cm} \), and we know there is an error in the circumference measurement of \( \Delta C = 0.50000 \, \text{cm} \). We need to find the corresponding maximum error in the surface area using linear approximation.

### Step 2: Radius from the Circumference
From the circumference formula \( C = 2\pi r \), we can solve for the radius:
\[
r = \frac{C}{2\pi} = \frac{84.000}{2\pi} \approx 13.37 \, \text{cm}
\]

### Step 3: Error in Surface Area
Using linear approximation, the differential of the surface area is:
\[
dA = \frac{dA}{dr} \cdot dr
\]
The derivative of the surface area \( A = 4\pi r^2 \) with respect to \( r \) is:
\[
\frac{dA}{dr} = 8\pi r
\]
Thus, the change in surface area \( dA \) due to a small change \( dr \) in the radius is:
\[
dA = 8\pi r \cdot dr
\]
Now, \( dr \) can be found using the error in the circumference. Since \( C = 2\pi r \), the error in \( r \) is:
\[
\Delta r = \frac{\Delta C}{2\pi} = \frac{0.50000}{2\pi} \approx 0.0796 \, \text{cm}
\]

### Step 4: Maximum Error in Surface Area
Substitute the values of \( r \) and \( \Delta r \) into the expression for \( dA \):
\[
\Delta A = 8\pi r \cdot \Delta r = 8\pi \times 13.37 \times 0.0796 \approx 33.56 \, \text{cm}^2
\]

Thus, the maximum error in the calculated surface area is approximately \( \Delta A \approx 33.56 \, \text{cm}^2 \).

### Step 5: Relative Error in Surface Area
The relative error in surface area is given by:
\[
\text{Relative error} = \frac{\Delta A}{A}
\]
First, we calculate the surface area \( A \) using the radius \( r = 13.37 \, \text{cm} \):
\[
A = 4\pi r^2 = 4\pi \times (13.37)^2 \approx 2246.5 \, \text{cm}^2
\]
Now, the relative error is:
\[
\text{Relative error} = \frac{33.56}{2246.5} \approx 0.01494 \, (\text{or } 1.494\%)
\]

### Final Answers:
- Maximum error in the calculated surface area: \( \approx 33.56 \, \text{cm}^2 \)
- Relative error in the calculated surface area: \( \approx 0.01494 \) or \( 1.494\% \)

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How is linear approximation applied in other physical problems?
2. What is the difference between absolute and relative error?
3. Can we apply a similar process to estimate the error in the volume of the sphere?
4. How would a larger error in the circumference affect the error in surface area?
5. Why is the relative error often a more useful measure than the absolute error?

**Tip:** When dealing with measurement errors, always remember to distinguish between absolute and relative errors, as they provide different insights into the impact of uncertainties.

The circumference of a sphere was measured to be 84.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area and the relative error.

Learn how to estimate the maximum and relative error in the surface area of a sphere using linear approximation, given a measured circumference of 84 cm with a possible error of 0.5 cm.

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