The energy of a system as a function of time | vedclass

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(D) Given the equation $E(t) = A^2 e^{-\alpha t}$.Taking the natural logarithm on both sides: $\ln E = 2 \ln A - \alpha t$.Differentiating both sides

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$4$

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(D) Given the equation $E(t) = A^2 e^{-\alpha t}$.Taking the natural logarithm on both sides: $\ln E = 2 \ln A - \alpha t$.Differentiating both sides to find the relative error: $\frac{dE}{E} = 2 \frac{dA}{A} - \alpha dt$.For maximum percentage error,we consider the absolute values of the errors: $\left| \frac{dE}{E} ight| = 2 \left| \frac{dA}{A} ight| + \alpha |dt|$.Given $\frac{dA}{A} = 1.25 \% = 0.0125$ and the error in time $dt = 1.50 \% 	ext{ of } t = 0.015 	imes 5 \ s = 0.075 \ s$.Given $\alpha = 0.2 \ s^{-1}$.Substituting the values: $\frac{dE}{E} 	imes 100 = 2(1.25 \%) + (0.2 \ s^{-1})(0.075 \ s) 	imes 100$.$\frac{dE}{E} 	imes 100 = 2.5 \% + (0.2 	imes 0.075) 	imes 100 \% = 2.5 \% + 1.5 \% = 4 \%$.Thus,the percentage error in $E(t)$ is $4 \%$.

The energy of a system as a function of time $t$ is given as $E(t)=A^2 \exp(-\alpha t)$,where $\alpha=0.2 \ s^{-1}$. 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

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