Chapter 4 - Differentiation - 4.1 Linear Approximation and Applications - Exercises - Page 173: 31

$-0.00171$ $cm$

$\frac{dL}{dt}$ = $kL$ = $(1.9\times{10}^{-5})\times18$ = $3.42\times{10}^{-4}$ $cm/°C$ $\frac{ΔL}{Δt}$ = $\frac{dL}{dt}$ $ΔL$ = $\frac{dL}{dt}(Δt)$ $ΔL$ = $(3.42\times{10}^{-4})\times(-5)$ = $-0.00171$ $cm$ at $T = 25 °C$ wire length is about $18-0.00171 = 17.99829$ $cm$