Q.A circular template has a radius of $10$ cm. The measurement of radius has an approximate error of $0.02$ cm. Then the percentage error in calculating area of this template is

When a quantity is approximated instead of computed exactly, the mismatch between the actual and approximate values is measured by three related notions of error, all built on linear approximation (equation $f(x_0+\Delta x)-f(x_0)\approx f'(x_0)\Delta x$).

Definition (Absolute Error).

If $h=x-x_0$ is the change in the input and $E(h)=f(x_0+h)-f(x_0)-f'(x_0)h$, then $E(h)$ is the absolute error made by using the linear approximation in place of the true value; $E(0)=0$ and $\displaystyle\lim_{h\to0}\frac{E(h)}{h}=0$ whenever $f$ is differentiable at $x_0$, i.e. the error vanishes faster than $h$ itself.

Absolute error alone doesn't say whether an approximation is good — an absolute error of $5$ is trivial if the actual value is $10{,}000$ but huge if the actual value is $6$. This is fixed by dividing out the actual value:

Definition (Relative and Percentage Error). If the actual value is nonzero,

Working method for an error problem. Given a quantity $Q=f(x)$ measured via $x$, and a measurement/change $dx$ in $x$:

1. Absolute error in $Q$ $\approx dQ = f'(x)\,dx$ (evaluated at the reference value of $x$).
2. Relative error $\approx \dfrac{dQ}{Q} = \dfrac{f'(x)}{f(x)}\,dx$.
3. Percentage error $= 100\times$ relative error.

Powers scale the percentage error linearly. If $Q=x^n$ (or more generally $Q=cx^n$), logarithmic differentiation gives $\dfrac{dQ}{Q}=n\dfrac{dx}{x}$ — so the percentage error in $x^n$ is $n$ times the percentage error in $x$. The same idea run in reverse (with $n=\tfrac1n$ replaced by a root) shows the percentage error in the $n$th root of a number is $\dfrac1n$ times the percentage error in the number — e.g. an area $A=\pi r^2$ has percentage error $2\times$ that of $r$ (since $n=2$), while a volume $V=\tfrac43\pi r^3$ has percentage error $3\times$ that of $r$.