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

Step-by-step solution

Energy E = A 2 e - α t For small % errors, we can, do differentiation d E = 2 A d A e - α t + A 2 - α e - α t d t Fractional error = d E E = 2 A e - α t d A + - α A 2 e - α t d t A 2 e - α t = 2 d A A + - α d t t t % error = 2 1.25 % + 0.2 × 1.5 % × 5 = 4 % (errors always add up) Alternate solution: E = A 2 e - α t Taking natural logarithm on both sides, l n E = l n A 2 + - α t Differentiating d E E = 2 d A A + - α d t For small fractional erros, errors always add up d E E = 2 d A A + α d t t × t = 2 1.25 % + 0.2 1