The energy of a system as a function of time t is given as E t = A 2 - α t , where α = 0.2  …

The energy of a system as a function of time t is given as Et=A2-αt , where α=0.2 s-1 . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t=5 s is

Solution

Energy E=A2e-αt
For small % errors, we can, do differentiation
dE=2AdAe-αt+A2-αe-αtdt
Fractional error =dEE=2Ae-αtdA+-αA2e-αtdtA2e-αt=2dAA+-αdttt 
% error =21.25%+0.2×1.5%×5
=4% (errors always add up)
Alternate solution:
E=A2e-αt
Taking natural logarithm on both sides,
lnE=lnA2+-αt
Differentiating
dEE=2dAA+-αdt
For small fractional erros, errors always add up
dEE=2dAA+αdtt×t
=21.25%+0.21.5%5 
=4%

Asked in: JEE Advanced 2015 (Paper 2)

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