The energy of a system is given as E(t) = alp | vedclass

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(C) Given the energy equation: $E = \alpha^3 e^{-\beta t}$.Taking the natural logarithm on both sides: $\ln E = 3 \ln \alpha - \beta t$.Differentiatin

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$6$

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(C) Given the energy equation: $E = \alpha^3 e^{-\beta t}$.Taking the natural logarithm on both sides: $\ln E = 3 \ln \alpha - \beta t$.Differentiating to find the relative error: $\frac{dE}{E} = 3 \frac{d\alpha}{\alpha} - \beta dt$.For maximum percentage error,we consider the absolute values of the errors: $\left( \frac{dE}{E} ight)_{\max} = 3 \left( \frac{d\alpha}{\alpha} ight) + \beta |dt|$.We are given $\frac{d\alpha}{\alpha} = 1.2 \%$ and the percentage error in $t$ is $\frac{dt}{t} = 1.6 \%$,which means $dt = 0.016 	imes t$.At $t = 5 \ s$,$dt = 0.016 	imes 5 = 0.08 \ s$.Substituting the values: $\left( \frac{dE}{E} ight)_{\max} = 3(1.2 \%) + (0.3 \ s^{-1})(0.08 \ s) 	imes 100 \%$.$\left( \frac{dE}{E} ight)_{\max} = 3.6 \% + (0.024) 	imes 100 \% = 3.6 \% + 2.4 \% = 6 \%$.

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

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