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Abby's question at Yahoo! Answers involving exponential decay

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MarkFL

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Feb 24, 2012
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Here is Abby's question:

Differential Equations time constant problem? The time constant T is the amount of time that an exponentially decaying quantity takes to decay by a factor of 1/e. Given an exponentially decaying quantity r(t) with the initial value r0 = r(0), show that its time constant is the time at which the tangent line to the graph of r(t)/r0 at (0,1) crosses the t-axis.

Please help! I can't figure this out. Please explain your steps so I can understand for future questions. Thank you!
Here is a link to the original question:

Differential Equations time constant problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
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Hello Abby,

Let's begin with:

$\displaystyle r(t)=r_0e^{-kt}$ where $\displaystyle 0<k$

Now, to determine the time constant $\displaystyle T$ as defined, we may set:

$\displaystyle \frac{r_0}{e}=r_0e^{-kT}$

We now solve for $\displaystyle T$:

$\displaystyle e=e^{kT}$

Equating exponents, we find:

$\displaystyle kT=1\,\therefore\,T=\frac{1}{k}$

Next, let's determine the root of the described tangent line.

$\displaystyle \frac{r(t)}{r_0}=e^{-kt}$

Hence, the slope of the tangent line is:

$\displaystyle \frac{d}{dt}\left( e^{-kt} \right)=-ke^{-kt}$

At $\displaystyle t=0$ this gives us a slope of $\displaystyle m=-k$

Then, using the point-slope formula, we find the equation of the tangent line to be:

$\displaystyle y-1=-k(t-0)$

$\displaystyle y=1-kt$

And so the root is:

$\displaystyle t=\frac{1}{k}=T$

Shown as desired.