The energy of a system is given as $E(t)=\alpha^3e^{-\beta t}$, where $t$ is the time and $\beta=0.3\ \mathrm s^{-1}$. The errors in the measurement of $\alpha$ and $t$ are 1.2% and 1.6%, respectively. At $t=5$ s, maximum percentage error in the energy is :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\frac{\Delta E}E=3\frac{\Delta\alpha}\alpha+\beta\,\Delta t=3(1.2\%)+\beta t\cdot\frac{\Delta t}t=3.6\%+(0.3	imes5)(1.6\%)=3.6\%+2.4\%=6\%$. Answer (1).

Q. The energy of a system is given as $E(t)=\alpha^3e^{-\beta t}$, where $t$ is the time and $\beta=0.3\ \mathrm s^{-1}$. The errors in the measurement of $\alpha$ and $t$ are 1.2% and 1.6%, respectively. At $t=5$ s, maximum percentage error in the energy is :

Explanation

Reviewed by Dr. Abubacker

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