Question

the circumference of a sphere was measured to be 89 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: Find the radius formula and its derivative

The circumference formula is \(C = 2\pi r\), so \(r=\frac{C}{2\pi}\). The derivative \(dr=\frac{dC}{2\pi}\).
The surface - area formula of a sphere is \(S = 4\pi r^{2}\). Substitute \(r=\frac{C}{2\pi}\) into it, we get \(S=\frac{C^{2}}{\pi}\). Then \(dS=\frac{2C}{\pi}dC\).

Step2: Calculate the maximum error in surface area

Given \(C = 89\) cm and \(dC=0.5\) cm.
Substitute into \(dS=\frac{2C}{\pi}dC\), we have \(dS=\frac{2\times89}{\pi}\times0.5=\frac{89}{\pi}\approx28.33\) \(cm^{2}\).

Step3: Calculate the relative error

The relative error formula is \(\frac{dS}{S}\). Since \(S=\frac{C^{2}}{\pi}\) and \(dS=\frac{2C}{\pi}dC\), then \(\frac{dS}{S}=\frac{\frac{2C}{\pi}dC}{\frac{C^{2}}{\pi}}=\frac{2dC}{C}\).
Substitute \(C = 89\) cm and \(dC = 0.5\) cm, we get \(\frac{dS}{S}=\frac{2\times0.5}{89}=\frac{1}{89}\approx0.0112\).

Answer:

The maximum error in the calculated surface area is approximately \(28.33\) \(cm^{2}\).
The relative error in the calculated surface area is approximately \(0.0112\).