Question

The radius of a circular disc is given as $24cm$ with a maximum error in measurement of $0.02cm$. Estimate the maximum error in the calculated area of the disc and compute the relative error by using differentials.

Answer 1

Area of the circular disc is given by $A=\pi r^2$.

Consider $A=\pi r^2$

Differentiate both sides with respect to $r$, we get

$\frac{dA}{dr}=2\pi r$

$\Rightarrow dA=2\pi rdr$

Calculated area: $A=\pi (24{)}^2=576\pi c{m}^2\approx 1809.6c{m}^2$

Because of an error in measurement, the radius might actually be as big as $24+0.02$

If $r$ is increased from $24$ by an amount $\Delta r=dr=0.02$.

Then the actual change in the calculated area would be $\Delta A=A(24+\Delta r)-A\left(24\right)=A\left(24.02\right)-A\left(24\right)$.

$\therefore \Delta A\approx dA=A\left(24.02\right)-A\left(24\right)=\pi (24.02{)}^2-576\pi =576.96\pi -576\pi =0.96\pi c{m}^2$.

So we estimated the maximum error in calculated area as $0.96\pi c{m}^2\approx 3.01718c{m}^2\approx 3.02c{m}^2$

The maximum relative error in calculated area is $\frac{\Delta A}{A}\approx \frac{dA}{A}=\frac{0.96\pi }{576\pi }=\frac{0.96}{576}\approx 0.0017$

The maximum $\%$ relative error in calculated area is about $0.17\%$