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.
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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 \( 0.626\% \).
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