## Task 2 - V10932

The volume of a wooden pole for telegraph lines is calculated using the following formula:

$$
V=\frac{\pi h}{12}\left(d_{1}^{2}+d_{1} d_{2}+d_{2}^{2}\right)
$$

Here, $h$ is the height, $d_{1}$ is the lower diameter, and $d_{2}$ is the upper diameter. In practice, however, the following approximate formula is often used:

$$
V^{\prime}=\frac{\pi h}{4} d^{2}
$$

where $d$ is the average diameter of the wooden pole:

$$
d=\frac{d_{1}+d_{2}}{2}
$$

a) Calculate the volume of a wooden pole for which the following values are given, using both the exact and the approximate formula:

$h=10 \mathrm{~m}, d_{1}=20 \mathrm{~cm}, d_{2}=14 \mathrm{~cm}!$

b) What percentage is the error when using the approximate formula?

c) Derive a formula for $\frac{V-V^{\prime}}{V}$ by setting $d_{1}=d+\delta$ and $d_{2}=d-\delta$!

What value does this expression yield when using the values given in part a)?
a) The volumes are $V=229.336 \mathrm{~cm}^{3}$ and $V^{\prime}=226.980 \mathrm{~cm}^{3}$.

b) The error is $1.03 \%$.

c) If we substitute the formulas for $d_{1}$ and $d_{2}$ into the equation for $V$, we get

$$
\begin{gathered}
V=\frac{\pi h}{12}\left((d+\delta)^{2}+(d+\delta)(d-\delta)+(d-\delta)^{2}\right) \\
V=\frac{\pi h}{12}\left(d^{2}+2 d \delta+\delta^{2}+d^{2}-\delta^{2}+d^{2}-2 d \delta+\delta^{2}\right) \\
V=\frac{\pi h}{12}\left(3 d^{2}+\delta^{2}\right)=\frac{\pi h}{4} d^{2}+\frac{\pi h}{12} \delta^{2}
\end{gathered}
$$

The relative error is then

$$
\frac{V-V^{\prime}}{V}=\frac{\frac{\pi h}{12} \delta^{2}}{\frac{\pi h}{12}\left(3 d^{2}+\delta^{2}\right)}=\frac{\delta^{2}}{3 d^{2}+\delta^{2}}
$$

This results in $\frac{3}{292}$, which corresponds to the 1.03\% from above.