Exponential Growth and Decay (1 Viewer)

johnmiltons · Jun 7, 2019

I've kinda been having some trouble with this question- can anyone help me out?
The current C flowing in a conductor dissipates according to the formula dC/dt = -kC. If it dissipates by 40% in 5 seconds, how long will it take to dissipate to 20% of the original current?
(The answer is 15.8 seconds, but I'm not really sure how the textbook got to that...)

Drongoski · Jun 8, 2019

C(t) = current at time t and C(0) = current at time t = 0 (initial current). When t=5, current has lost 40% of value so that C(5) = 0.6 C(0). Let t = T be the time at which current is reduced to 20% of C(0); i.e. C(T) = 0.2 C(0).

$C(t) = C(0) e^{-kt} \\ \\ \therefore C(5) = C(0)\times e^{-5k} = 0.6 C(0) \implies e^{-5k} = 0.6 \implies e^{-k} =0.6^{\frac {1}{5}}\\ \\C(T) = C(0)e^{-kT} = C(0)(e^{-k})^T = 0.2 C(0) \implies (e^{-k})^T = 0.2 \\ \\ \implies T\times ln (e^{-k}) = ln 0.2 \\ \\ \implies T = \frac {ln 0.2}{ln (e^{-k})} = \frac {ln 0.2}{ln 0.6^{\frac {1}{5}}} = \frac {5ln0.2}{ln0.6} = 15.7533 ...$

Note: I have avoided calculating value of "k", which most of you are usually taught to do; so my solution may be confusing to some of you, especially if you are not very strong at indices and logs.

johnmiltons · Jun 11, 2019

Oh! I ended up solving it like this:
C=Ae^(-kt)
60%A=100%Ae^(-k*5)
ln(0.6)=-k*5
So, -k=ln(0.6)/5

20%A=100%Ae^((ln(0.6)/5)*t)
ln(0.2)/(ln(0.6)/5)=t
So, t=15.7533

So I guess it works! Thank you!

Exponential Growth and Decay (1 Viewer)

johnmiltons

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Drongoski

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