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.

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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%.

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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.