Question

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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Explanation:

Step1: Recall formulas for sphere

The circumference of a sphere (a circle, since we consider the great circle) is \( C = 2\pi r \), and the surface area is \( S = 4\pi r^2 \). We can express \( S \) in terms of \( C \): from \( C = 2\pi r \), we get \( r=\frac{C}{2\pi} \). Substitute into \( S \): \( S = 4\pi(\frac{C}{2\pi})^2=\frac{C^2}{\pi} \).

Step2: Find derivative of \( S \) with respect to \( C \)

To use linear approximation (differentials), we find \( dS \). The derivative \( S^\prime(C)=\frac{2C}{\pi} \). So, \( dS = S^\prime(C)dC=\frac{2C}{\pi}dC \).

Step3: Substitute values for maximum error in \( S \)

We know \( C = 80 \) cm and \( dC = 0.5 \) cm (the possible error in \( C \)). Plugging in: \( dS=\frac{2\times80}{\pi}\times0.5=\frac{80}{\pi}\approx25.46 \) cm² (this is the maximum error in surface area).

Step4: Find relative error in \( S \)

Relative error is \( \frac{dS}{S} \). We have \( S=\frac{C^2}{\pi} \) and \( dS=\frac{2C}{\pi}dC \). So, \( \frac{dS}{S}=\frac{\frac{2C}{\pi}dC}{\frac{C^2}{\pi}}=\frac{2dC}{C} \). Substitute \( C = 80 \), \( dC = 0.5 \): \( \frac{2\times0.5}{80}=\frac{1}{80}=0.0125 \) or 1.25%.

Answer:

Maximum error in surface area: \(\frac{80}{\pi}\) (or approximately \(25.46\)) cm²; Relative error: \(\frac{1}{80}\) (or \(0.0125\) or \(1.25\%\))