The energy of a system as a function of time `t` is given as `E(t) = A^(2)exp(-alphat)`, `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

Explanation

Given energy of the system can be written as follows :
`E= A^(2)e^(-alpha t)`
We can differentiate the above to get the following :
`dE= 2A(dA)(e^(-alpha t))+A^(2)(-alpha e^(-alpha t))(dt)`
`implies (dE)/(E )= (2A(dA)(e^(-alpha t))+A^(2)(-alpha e^(-alpha t))(dt))/(A^(2)e^(-alpha t))`
`implies (dE)/(E )=(2(dA))/(A)+(-alpha)(dt)`
In the calculation of error, all terms are added so that we may get maximum possible error. Hence using just magnitudes of each term we can write the above expression as follows :
`implies (triangleE)/(E )=2(triangleA)/(A)+alpha trianglet`
`implies (tringleE)/(E )xx100=2(triangleA)/(A)xx100+100 alpha triangle t" "".........."(i)`
Given : `(triangleA)/(A)xx100= 1.25" "".........."(ii)`
Given `(triangle t)/(t)xx100= 1.5`
Hence at t = 5s we can write the following :
`triangle t= (1.5xx5)/(100)" ""........."(iii)`
Given `alpha = 0.2s^(-1)" ""............"(iv)`
Substituting values from equations (ii), (iii) and (iv) in equation (i) we get the following :
`implies (triangleE)/(E )xx100=2xx1.25+100xx0.2xx(1.5xx5)/(100)`
`implies (triangle E)/(E )xx100=4`
Hence percentage error in the value of energy at t = 5s is 4.