Question

# The circumference of a sphere was measured to be 84

Circles
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?

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: $$\displaystyle={\frac{{\triangle{S}}}{{{S}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{r}^{{{2}}}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{\left({\frac{{{C}}}{{{2}\pi}}}\right)}^{{{2}}}}}}={\frac{{{\frac{{{84}}}{{\pi}}}}}{{{4}\pi{\left({\frac{{{84}}}{{{2}\pi}}}\right)}^{{{2}}}}}}={\frac{{{1}}}{{{84}}}}\approx{0.012}$$