Suppose that the radius of a circular disk is given as 24 cm with a maximum error in measurement of 0.2cm.
a.) Estimate the maximum error in the calculated area of the disk by using differentials
Area of a circle,
$A = \pi r^2$, where $R = 24$cm and $dr = 0.2cm$
$
\begin{equation}
\begin{aligned}
\frac{dA}{dr} &= \frac{d}{dr}(\pi r^2)\\
\\
dA &= 2 \pi r dr\\
\\
dA &= 2 \pi (24\text{cm})(0.2\text{cm})\\
\\
dA &= 9.6 \pi \text{cm}^2 \qquad \text{Maximum possible error of the area of a circular disk}
\end{aligned}
\end{equation}
$
b.) Find the relative error and the percentage error
For relative error,
$
\begin{equation}
\begin{aligned}
\frac{dA}{A} &= \frac{2 \pi r dr}{\pi r^2}\\
\\
\frac{dA}{A} &= \frac{2dr}{r}\\
\\
\frac{dA}{A} &= \frac{2\left(0.2\cancel{\text{cm}}\right)}{24\cancel{\text{cm}}}\\
\\
\frac{dA}{A} &= 0.0167
\end{aligned}
\end{equation}
$
For percentage error,
$\displaystyle \frac{dA}{A} \times 100\\
0.0167 \times 100 = 1.67 \%
$
No comments:
Post a Comment