The circumference of a sphere was measured to be 80 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.

To solve the problem using linear approximation, we first recognize that the surface area \( S \) of a sphere is calculated using the formula \( S = 4\pi r^2 \). The radius \( r \) can be related to the circumference \( C \) by the formula \( C = 2\pi r \). If the circumference is measured as \( 80 \) cm, the radius can be expressed as \( r = \frac{C}{2\pi} = \frac{80}{2\pi} \). Now, the maximum error in \( C \) is given as \( \Delta C = 0.5 \) cm. Using the relationship between the circumference and the radius, we can differentiate \( S \) with respect to \( C \) to find how the error in circumference translates to error in surface area. The derivative of surface area with respect to circumference can be given as: \[ \frac{dS}{dC} = \frac{dS}{dr} \cdot \frac{dr}{dC} = 8\pi r \cdot \frac{1}{2\pi} = 4r \] Substituting for \( r \): \[ \Delta S \approx 4r \Delta C \] First, we need to compute \( r \): \[ r = \frac{80}{2\pi} \approx \frac{80}{6.2832} \approx 12.73 \text{ cm} \] Now, substituting \( r \) and \( \Delta C \) into the equation: \[ \Delta S \approx 4 \times 12.73 \times 0.5 \approx 12.73 \text{ cm}^2 \] Therefore, the maximum error in the calculated surface area is approximately \( 12.73 \text{ cm}^2 \). To find the relative error, we first need to find the surface area itself: \[ S = 4\pi r^2 \approx 4\pi (12.73^2) \approx 4\pi (162.32) \approx 2030.95 \text{ cm}^2 \] Then, the relative error \( \text{Relative Error} = \frac{\Delta S}{S} \): \[ \frac{12.73}{2030.95} \approx 0.00626 \text{ or } 0.626\% \] Thus, the relative error in the calculated surface area is approximately \( 0.626\% \).