## Essential University Physics: Volume 1 (3rd Edition)

$d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$
We know the following two equations: $\alpha = \frac{\frac{\Delta L}{L}}{\Delta T}$ And $L=L_0(1+\alpha \Delta T)$ Using trigonometry to solve for d, it follows: $d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$