• Mar 24th 2008, 10:04 PM
caelestis
I'm having so much trouble with this question and would really appreciate if someone could give me hints or suggestions on how to solve it??

The electrical resistance R of a wire is given by

k / r^2 k=constant, r=radius of wire

Use differentials to estimate the percentage error in the measured value of r if we want the the percentage error in R to be within +/- 3%

Thank you so much for your time. :)
• Mar 25th 2008, 05:56 AM
TKHunny
The Equation

$\displaystyle R(r) = \frac{k}{r^{2}}$

The Differential Equation

$\displaystyle dR = -2\frac{k}{r^{3}}dr$

The Constraint

$\displaystyle \frac{|dR|}{R} \leq 0.03$

Substitute and Solve
• Mar 25th 2008, 06:04 AM
CaptainBlack
Quote:

Originally Posted by caelestis
I'm having so much trouble with this question and would really appreciate if someone could give me hints or suggestions on how to solve it??

The electrical resistance R of a wire is given by

k / r^2 k=constant, r=radius of wire

Use differentials to estimate the percentage error in the measured value of r if we want the the percentage error in R to be within +/- 3%

Thank you so much for your time. :)

What couses is this for? If its calculus you will get one answer, if physics
analysis of errors you will get a different answer.

$\displaystyle R=\frac{k}{r^2}$

if we in fact have $\displaystyle r+\epsilon$, we get:

$\displaystyle R+\rho=\frac{k}{(r+\epsilon)^2}\approx \frac{k}{r^2} \left(1-2 \left(\frac{\epsilon}{r}\right)\right)$

so:

$\displaystyle \rho \approx -2R\frac{\epsilon}{r}$

Therefore:

$\displaystyle \frac{\rho}{R} \approx -2\frac{\epsilon}{r}$

which relates the fractional error in $\displaystyle r$ to that in $\displaystyle R$ and vice versa, which is all you need to answer the question.

RonL