Expand the quadratic term for resistance equation

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When a cylindrical wire's length is increased by x percent while maintaining a fixed volume and resistivity, the new cross-sectional area can be expressed as A_new = A / (1 + x/100). The resistance is proportional to the ratio of length to area, leading to R_old proportional to ℓ/A and R_new proportional to ℓ(1 + x/100)²/A. By expanding the quadratic term and noting that the squared term is negligible, the resistance change can be approximated. The conclusion suggests that the resistance will increase by roughly 2x percent.
amcavoy
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A cylindrical wire of length \ell and cross-sectional area \textrm{A} has a fixed volume \textrm{V}. If \ell is increased by +x percent and the volume and resistivity stay the same, by what percentage (in terms of x) will the resistance change?
I said that A\ell=A_{\textrm{new}}\ell\left(1+\frac{x}{100}\right)\implies A_{\textrm{new}}=\frac{A}{1+\frac{x}{100}}. Also, we know that R\propto\frac{\ell}{A}.. Therefore,

R_{\textrm{old}}\propto\frac{\ell}{A}

R_{\textrm{new}}\propto\frac{\ell\left(1+\frac{x}{100}\right)^{2}}{A}

Somehow I'm not getting as nice of an answer as I expected.
 
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Expand the quadradic term, you can claim that the {(\frac x {100})}^2 term is very small and ignore it. Can you see where to go from there?
 
So roughly 2x percent?
 
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