Discussion Forum

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. If the error in the measurement  of time  is 1.50%  , the percentage error in the value of E (t) at t=5s is

Answer:

Explanation:

$E(t)= A^{2}e^{-\alpha t},$             ...........(i)

                             $\alpha=$ 0.2 s^-1

                  $\left(\frac{dA}{A}\right)\times 100=$ 1.25%

                      $\left(\frac{dt}{t}\right)\times 100=$   1.50

$\Rightarrow$                    $(dt\times100)=1.5t=1.5\times5=7.5$

$\therefore$       $\left(\frac{dE}{E}\right)\times100=\pm 2\left(\frac{dA}{A}\right)\times100$

                                                                                              $\pm\alpha(dt\times100)$

Taking log on both sides of Eq.(i), we get

                 $\log E=2\log A-\alpha t$

   $\frac{dE}{E}= 2\frac{dA}{A}\pm\alpha dt$

$\therefore$     $\left(\frac{dE}{E}\right)\times 100=\pm 2\left(\frac{dA}{A}\right)\times100\pm\alpha(dt \times 100)$

            $=\pm 2(1.25)\pm 0.2(7.5)$

              $=\pm 2.5\pm1.5=\pm4$%