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.