Question

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the maximum error in the calculated surface area (round your answer to the nearest integer) and evaluate the relative error (round your answer to three decimal places). B) Use differentials to estimate the maximum error in the calculated volume (round your answer to the nearest integer) and evaluate the relative error (round your answer to three decimal places).

Answer 1

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$($y=f(x)$).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$

The calculated volume is:

$V=\frac{C^3}{6\pi^2}\\V=\frac{(80)^3}{6\pi^2}\\V=\frac{512000}{6\pi^2}\\V=8646.0743$

The relative error is:

$\frac{dV}{V}=\frac{162.1139}{8646.0743}=0.0188\approx0.019$