Question

The circumference of a sphere was measured to be 84

Circles
ANSWERED
asked 2021-08-06
The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm.
Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?

Answers (1)

2021-08-07
Circumference is
\(\displaystyle{C}={2}\pi{r}\)
Differentiate both sides with respect to r
\(\displaystyle{\frac{{{d}{C}}}{{{d}{r}}}}={2}\pi\)
When \(\displaystyle\triangle{r}\) is small, we can write
\(\displaystyle{\frac{{\triangle{C}}}{{\triangle{r}}}}={2}\pi\)
\(\displaystyle{\frac{{\triangle{C}}}{{{2}\pi}}}=\triangle{r}\)
\(\displaystyle\triangle{r}={\frac{{\triangle{C}}}{{{2}\pi}}}\)
It is given that \(\displaystyle\triangle{C}={0.5}\)
\(\displaystyle\triangle{r}={\frac{{{0.5}}}{{{2}\pi}}}\)
\(\displaystyle\triangle{r}={\frac{{{1}}}{{{4}\pi}}}\)
\(\displaystyle{S}{\left({r}\right)}-{4}\pi{r}^{{{2}}}\)
Differentiate both sides with respect to r
\(\displaystyle{\frac{{{d}{S}}}{{{d}{r}}}}={8}\pi{r}\)
if \(\displaystyle\triangle{r}\) is small, we can say
\(\displaystyle{\frac{{\triangle{S}}}{{\triangle{r}}}}={8}\pi{r}\)
\(\displaystyle\triangle{S}={8}\pi{r}\triangle{r}\)
\(\displaystyle\triangle{S}={4}\cdot{2}\pi{r}\triangle{r}\)
\(\displaystyle\triangle{S}={4}{C}\triangle{r}\)
Where Cis the circumference.
Note that \(\displaystyle\triangle{S}\) will be maximum when \(\displaystyle\triangle{r}\) is maximum. That means, the maximum error in S is:
\(\displaystyle\triangle{S}={4}\cdot{84}\cdot{\frac{{{1}}}{{{4}\pi}}}={\frac{{{84}}}{{\pi}}}\approx{27}\ \text{cm}^{{{2}}}\)
Relative Error: \(\displaystyle={\frac{{\triangle{S}}}{{{S}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{r}^{{{2}}}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{\left({\frac{{{C}}}{{{2}\pi}}}\right)}^{{{2}}}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{\left({\frac{{{84}}}{{{2}\pi}}}\right)}^{{{2}}}}}}={\frac{{{1}}}{{{84}}}}\approx{0.012}\)
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