Question #dad87

1 Answer
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}) #
#(dT)/T = [cancel{(\frac{2\pi}{\sqrt{g}})}(dl)/(2sqrt{l})]\times[cancel{(\sqrt{g}/(2\pi))}1/\sqrt{l}]=1/2(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#
#dT = C/(2\sqrt{l})dl#

#(dT)/T = (cancel{C}/(2\sqrt{l})dl)\times(1/(cancel{C}\sqrt{l}))=1/2(dl)/l#