How do I find small percentage changes (differentiation)?

Question

the time period T for a simple pendulum of length l is given by #T=2pi*sqrt(l/g)# where g is a constant.
find the percentage change in T when l changes by 6%

5176 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-22T06:25:13

#"Approx. "3%#.

#T=2pisqrt(l/g)=(2pi)/sqrtg*sqrtl#.

#:. (dT)/(dl)=(2pi)/sqrtg*1/(2sqrtl)=pi/sqrt(gl)#.

Recall that, #deltaT~~(dT)/(dl)*deltal#.

#:. (deltaT)/T~~(dT)/(dl)*(deltal)/T#,

#=pi/sqrt(gl)*(deltal)/(2pisqrt(l/g))#,

#rArr (deltaT)/T~~=1/2*(deltal)/l#.

#:."The % Change in "T=(deltaT)/T*100~~1/2*(deltal)/l*100#,

#=1/2("the % Change in "l)#,

#=1/2(6%)#.

#:."The % Change in "T~~3%#.