Question

The energy of a system as a function of time tis
given as E (t) = Aexp (-at), where a = 0-2 s-!
The measurement of A has an error of 1.25%. If
the error in the measurement of time is 1.50%, the
percentage error is the value of E (t) at t = 5 sec is​

Answer 1

Answer:

E(t)=A

2

e

−αt

Taking natural logarithm on both sides,

ln(E)=2ln(A)+(−αt)

Differentiating both sides

E

dE

=2

A

dA

+(αdt)

Errors always add up for maximum error.

∴

E

dE

=2

A

dA

+α(

t

dt

)×t

Here,

A

dA

=1.25 %,

t

dt

=1.5%, t=5s, α=0.2s

−1

∴

E

dE

=(2×1.25%)+(0.2)×(1.5%)×5=4%

I think this is it.

Answer 2

ANSWER

E(t)=A2e−αt
Taking natural logarithm on both sides,
ln(E)=2ln(A)+(−αt)
Differentiating both sides
EdE​=2AdA​+(αdt)
Errors always add up for maximum error.
∴EdE​=2AdA​+α(tdt​)×t
Here, AdA​=1.25 %, tdt​=1.5%, t=5s, α=0.2s−1
∴EdE​=(2×1.25%)+(0.2)×(1.5%)×5=4%

Answer By Jannatk0218