Approximating error in area of a circle from its radius

Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?

Given that $$A=\pi r^2$$ $$A'=2\pi r$$ it can be found that the radius when A = 15 must be $$r=\sqrt{\frac{15}{\pi}}$$ so using differentials $$A(r+\Delta r)\approx A(r)+A'(r)\Delta r$$ $$15\approx\frac{\sqrt{15}}{\pi}+2\pi\sqrt{\frac{15}{\pi}}\Delta r$$

so $$\Delta r\approx\frac{15-\sqrt{\frac{15}{\pi}}}{2\pi \sqrt{\frac{15}{\pi}}}$$

Is this correct?

• $A=\pi r^2 \implies r=\sqrt{\frac{A}{\pi}}$. Did you mean the radius is $\sqrt{\frac{15}{\pi}}$? – Dando18 Dec 11 '18 at 23:47
• Yes, edited to show that. – ovil101 Dec 11 '18 at 23:51
• Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $\frac{\sqrt{15}}\pi$? What is the justification for both? Where does the $\pm0.2$ enter the formula? – Lutz Lehmann Dec 12 '18 at 8:56

Your approach is correct but likely an overkill. The final result is also not really pretty.

If you have a known area $$A$$ with error $$\delta$$, then $$r = \sqrt{\frac{A \pm \delta}{\pi}}$$. Then its error is given by $$\delta_r = r_\max - r_\min = \sqrt{\frac{A + \delta}{\pi}} - \sqrt{\frac{A - \delta}{\pi}}$$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $$\delta_r = \sqrt{\frac{A}{\pi}}((1+\delta/A)^{1/2} - (1-\delta/A)^{1/2}) \approx \sqrt{\frac{A}{\pi}}(2*\frac{\delta}{2A}) \approx \frac{\delta}{\sqrt{\pi A}}$$.

The errors are related by

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

Therefore

$$\Delta r = \frac{\Delta A}{2\pi r} = \frac{0.2}{2\pi \sqrt{\frac{15}{\pi}}} = \frac{0.1}{\sqrt{15\pi}}$$

Your formula is incorrect because you had $$\Delta A = A - r$$