Confusing differentials problem - Electrical resistance $ R = \frac{k}{r^2} $ with $ dr = 5\% $

Question

I am struggling with a confusing differentials' problem. It seems like there is a key piece of information missing:

The problem:

The electrical resistance $ R $ of a copper wire is given by $ R = \frac{k}{r^2} $ where $ k $ is a constant and $ r $ is the radius of the wire. Suppose that the radius has an error of $ \pm 5\% $, find the $\%$ error of $ R $.

My solution:

\begin{align*} R &= \frac{k}{r^2}\\ \frac{dR}{dr} &= k \cdot (-2) \cdot r^{-3} \quad \therefore \quad dR = \frac{-2k \cdot 0.05}{r^3} = \frac{-0.1k}{r^3}\\ \end{align*}

So the percentage error is given by

\begin{align*} E_\% = \frac{\frac{-0.1k}{r^3}}{\frac{k}{r^2}} = - \frac{0.1}{r} \end{align*}

My question: Am I missing something? Should I have arrived in a real value (not a function of $ r $ )? Is there information missing on the problem?

Thank you.

Answer 1

When we say the radius has an error of $5\%$, we mean that as a relative error, so $dr=0.05r$

Answer 2

Taking logs we have $ln (R)=ln (k)-ln (r^2) $ thus differentiating we have $\frac {dR}{R}=-2\frac {dr}{r} $ now multiplying by $100$ we have $\text {percent error in R}=-2\text {percent error in r}$ thus $\text {percent error in R}=- (\pm 10)$(as radius increases resistance decreases and vice-versa.)

Answer 3

Just doing this by brute force:

$$R=\frac{k}{r^2}\quad\text{percentage error in $r$ is $\pm 5$ percent}$$

So $R_{min} =\frac{k}{(r+0.05r)^2} =\frac{k}{1.05^2r^2}=\frac{400}{441}\cdot\frac{k}{r^2}$

$R_{max} = \frac{k}{(0.95r)^2} =\frac{400}{361}\cdot \frac{k}{r^2}$

Then the percentage error is given by:

$$\frac{R_{max}-R_{min}}{2R}\cdot 100 =\frac{\frac{k}{r^2}\left(\frac{400}{361}-\frac{400}{441}\right)}{2\cdot \frac{k}{r^2}}\cdot 100 \approx 10.1\%\quad\text{3s.f.}$$