For $P=\dfrac{F}{\pi R^2}$ with percentage errors of $\pm2$ in $F$ and $\pm1$ in $R$, the percentage error in $P$ is:

1. Take logarithms: $\ln P=\ln F-\ln\pi-2\ln R$. 2. Differentiating gives $\dfrac{\Delta P}{P}=\dfrac{\Delta F}{F}+2\dfrac{\Delta R}{R}$ — the terms ADD when errors are maximised. 3. Multiply through by 100%: $\pm(2\%+2\times1\%)$. 4. That gives $\pm4\%$. 5. Subtracting instead of adding gives $-1\%$, which was the recorded error and shows a misunderstanding of maximising errors. _Source: NECTA ACSEE 2023 Physics 131/1, Question 1: Measurement_