The radius of a circular disk is given as 24 cm with a maximum error in measurement of 0.2 cm. Use differentials to estimate the maximum error in the calculated area of the disk.

My attempt:

Look at the linear approximation of $A(r) = \pi r^2$ near $r=24$.

We have that

$$A(r) \approx A(24) + A'(24)(x-24)$$

and maximizing the error in measurement of the radius gives

$$A(r) \approx A(24) + A'(24)(24.2-24)$$

and we read off the differential term


which equals $9.6 \pi$.

I'm just double-checking my solutions before showing it to my calculus students tomorrow.


  • 1
    $\begingroup$ This is correct. I would tell them why you plug in $24.2$ instead of $23.8$. You have the interval $I = (24-.02, 24+.02)$ and by taylor's approximation we have $f(x) - f(24) = f'(24)(x-24) + \epsilon(x)$ where $x \in I$. Hence, the error term is approximated by $g(x) = f'(24)(x-24)$ which you want to maximize on $I$. Okay so now we just plug in $24.2$ since this function in monotonically increasing. Technically I think it would be better if you also got the upper bound on $\epsilon(x)$ and so you want to maximize $g_{\epsilon}(x) = f'(24)(x-24) + \epsilon(x)$. $\endgroup$ – Faraad Armwood Oct 19 '17 at 23:21

Being a physicist, even if your solution is very correct, instead of using linear approximation, I suggest you use differentials (as written in the title of your post).

Let us start with a modified version of your example $$A=k\, r^n\implies \log(A)=\log(k)+n\log(r)$$ Differentiate $$\frac{dA}{A}=n\frac{dr}{r}\implies\frac{|\Delta A|}{A}=n\frac{|\Delta r|}{r}\implies|{\Delta A}|=n k r^{n-1} {|\Delta r|}$$


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