# Question dad87

Dec 14, 2017

T(l) = 2\pi\sqrt{l/g} = ((2\pi)/sqrt{g})\sqrt{l}; \qquad \qquad dT =((2\pi)/sqrt{g})(dl)/(2sqrt{l}) 
$\frac{\mathrm{dT}}{T} = \left[\cancel{\left(\setminus \frac{2 \setminus \pi}{\setminus \sqrt{g}}\right)} \frac{\mathrm{dl}}{2 \sqrt{l}}\right] \setminus \times \left[\cancel{\left(\setminus \frac{\sqrt{g}}{2 \setminus \pi}\right)} \frac{1}{\setminus} \sqrt{l}\right] = \frac{1}{2} \frac{\mathrm{dl}}{l}$

#### Explanation:

In your work I see that your ${T}_{0} '$ is incorrect.
Perhaps bunching everything that are constants into one single constant would help you (at the initial stages)

T = 2\pi\sqrt{l/g} = C\sqrt{l};\qquad C\equiv(2\pi)/\sqrt{g}

T = Cl^{1/2}; \qquad dT = C. 1/2l^{1/2-1} dl= C/2.l^{-1/2} = C/(2l^{1/2})dl#
$\mathrm{dT} = \frac{C}{2 \setminus \sqrt{l}} \mathrm{dl}$

$\frac{\mathrm{dT}}{T} = \left(\frac{\cancel{C}}{2 \setminus \sqrt{l}} \mathrm{dl}\right) \setminus \times \left(\frac{1}{\cancel{C} \setminus \sqrt{l}}\right) = \frac{1}{2} \frac{\mathrm{dl}}{l}$