# Chapter 17 - Exercises and Problems - Page 314: 63

$d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$

#### Work Step by Step

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: $d = \frac{L_0}{2}\sqrt{2\alpha T +\alpha^2\Delta T^2}$

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