Let's solve the problem step-by-step:
Given data:
- Circumference of the sphere, \( C = 86 \, \text{cm} \)
- Possible error in the circumference, \( \delta C = 0.5 \, \text{cm} \)
### Part (a)
To estimate the maximum error in the surface area:
1. Determine the radius \( r \) of the sphere:
The circumference \( C \) of a sphere is related to the radius \( r \) by the formula:
[tex]\[
C = 2 \pi r
\][/tex]
Solving for \( r \):
[tex]\[
r = \frac{C}{2 \pi}
\][/tex]
2. Calculate the error in the radius \( \delta r \):
Given the error in the circumference \( \delta C \), the error in the radius \( \delta r \) can be found using:
[tex]\[
\delta r = \frac{\delta C}{2 \pi}
\][/tex]
3. Find the surface area \( A \) of the sphere:
The surface area \( A \) of a sphere is given by:
[tex]\[
A = 4 \pi r^2
\][/tex]
4. Use differentials to estimate the maximum error \( \delta A \) in the surface area:
Using the derivative, the maximum error in \( A \) can be approximated by:
[tex]\[
\delta A \approx \left| \frac{dA}{dr} \right| \delta r
\][/tex]
Since \( \frac{dA}{dr} = 8 \pi r \), we have:
[tex]\[
\delta A \approx 8 \pi r \delta r
\][/tex]
Substituting \( r = \frac{C}{2 \pi} \) and \( \delta r = \frac{\delta C}{2 \pi} \), we get:
[tex]\[
\delta A \approx 8 \pi \left( \frac{C}{2 \pi} \right) \left( \frac{\delta C}{2 \pi} \right)
\][/tex]
Simplifying this expression and substituting the given values, the maximum error \( \delta A \) in the surface area is:
[tex]\[
\delta A \approx 27 \, \text{cm}^2
\][/tex]
5. Calculate the relative error in the surface area:
The relative error is given by:
[tex]\[
\text{Relative error} = \frac{\delta A}{A}
\][/tex]
Substituting the correct values, the relative error is:
[tex]\[
\text{Relative error} \approx 0.012
\][/tex]
### Part (b)
To estimate the maximum error in the volume:
1. Find the volume \( V \) of the sphere:
The volume \( V \) of a sphere is given by:
[tex]\[
V = \frac{4}{3} \pi r^3
\][/tex]
2. Use differentials to estimate the maximum error \( \delta V \) in the volume:
Using the derivative, the maximum error in \( V \) can be approximated by:
[tex]\[
\delta V \approx \left| \frac{dV}{dr} \right| \delta r
\][/tex]
Since \( \frac{dV}{dr} = 4 \pi r^2 \), we have:
[tex]\[
\delta V \approx 4 \pi r^2 \delta r
\][/tex]
Substituting \( r = \frac{C}{2 \pi} \) and \( \delta r = \frac{\delta C}{2 \pi} \), we get:
[tex]\[
\delta V \approx 4 \pi \left( \frac{C}{2 \pi} \right)^2 \left( \frac{\delta C}{2 \pi} \right)
\][/tex]
Simplifying this expression and substituting the given values, the maximum error \( \delta V \) in the volume is:
[tex]\[
\delta V \approx 187 \, \text{cm}^3
\][/tex]
3. Calculate the relative error in the volume:
The relative error is given by:
[tex]\[
\text{Relative error} = \frac{\delta V}{V}
\][/tex]
Substituting the correct values, the relative error is:
[tex]\[
\text{Relative error} \approx 0.017
\][/tex]
Final answers:
- Maximum error in the calculated surface area: \( 27 \, \text{cm}^2 \)
- Relative error in the surface area: \( 0.012 \)
- Maximum error in the calculated volume: \( 187 \, \text{cm}^3 \)
- Relative error in the volume: [tex]\( 0.017 \)[/tex]
