The circumference of a sphere was measured to be 89 cm with a possible error of 0.5 cm. Use linear approximation to estimate the maximum error in the calculated surface area.

  

Estimate the relative error in the calculated surface area.

  

### Part 1: Maximum Error in the Calculated Surface Area

Given:
- The circumference of a sphere \( C = 89 \, \text{cm} \)
- Possible error in circumference \( dC = 0.5 \, \text{cm} \)
- The surface area \( A \) of a sphere is related to the radius \( r \) by the formula:  
  \[
  A = 4 \pi r^2
  \]

First, let's express the radius in terms of the circumference:
\[
C = 2 \pi r \quad \Rightarrow \quad r = \frac{C}{2 \pi}
\]

Now differentiate the surface area \( A \) with respect to \( r \):
\[
dA = \frac{dA}{dr} \, dr
\]
Since \( A = 4 \pi r^2 \), the derivative of \( A \) with respect to \( r \) is:
\[
\frac{dA}{dr} = 8 \pi r
\]
Thus, the differential of the surface area is:
\[
dA = 8 \pi r \, dr
\]

Next, express \( dr \) in terms of \( dC \). Differentiating the relationship \( C = 2 \pi r \), we get:
\[
dC = 2 \pi \, dr \quad \Rightarrow \quad dr = \frac{dC}{2 \pi}
\]

Substitute \( dr \) into the equation for \( dA \):
\[
dA = 8 \pi r \cdot \frac{dC}{2 \pi} = 4 r \, dC
\]

Now, substitute \( r = \frac{C}{2 \pi} \) into the above equation:
\[
dA = 4 \cdot \frac{C}{2 \pi} \cdot dC = \frac{2 C \, dC}{\pi}
\]

Finally, plug in the values \( C = 89 \, \text{cm} \) and \( dC = 0.5 \, \text{cm} \):
\[
dA = \frac{2 \cdot 89 \cdot 0.5}{\pi} \approx 28.33 \, \text{cm}^2
\]

So, the maximum error in the calculated surface area is approximately **28.33 cm²**.

---

### Part 2: Relative Error in the Calculated Surface Area

The relative error is given by the ratio of the maximum error in the surface area \( dA \) to the actual surface area \( A \):
\[
\text{Relative Error} = \frac{dA}{A}
\]

The actual surface area \( A \) is:
\[
A = 4 \pi r^2 = 4 \pi \left( \frac{C}{2 \pi} \right)^2 = \frac{C^2}{\pi}
\]

Substitute \( C = 89 \, \text{cm} \):
\[
A = \frac{89^2}{\pi} \approx 2523.09 \, \text{cm}^2
\]

Now, calculate the relative error:
\[
\text{Relative Error} = \frac{28.33}{2523.09} \approx 0.0112
\]

Thus, the relative error in the surface area is approximately **0.0112**, or **1.12%**.

---

Would you like further details or clarification on any part of the process?

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Here are 5 questions related to this topic:
1. How does linear approximation work in error estimation for different shapes?
2. How does the error in volume relate to the error in surface area for a sphere?
3. What happens to the error when measuring a sphere with a larger or smaller circumference?
4. How do differentials help in understanding small changes in geometric quantities?
5. How could the error be minimized in practical measurements?

**Tip:** When measuring geometric quantities, always be aware of how errors propagate through the formulas to avoid compounding inaccuracies.

The circumference of a sphere was measured to be 89 cm with a possible error of 0.5 cm. Use linear approximation to estimate the maximum error in the calculated surface area. Estimate the relative error in the calculated surface area.

Learn how to estimate the maximum error and relative error in the surface area of a sphere, given a measured circumference of 89 cm and a possible error of 0.5 cm. This example uses linear approximation and covers important geometric and calculus concepts.

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