6.8 Exponential growth and decay (Page 4/9)

Calculus volume 1 Page 4 / 9

y = y_{0} e^{− k t},

and we see that

\begin{matrix} T - T_{a} & = & (T_{0} - T_{a}) e^{− k t} \\ T & = & (T_{0} - T_{a}) e^{− k t} + T_{a} \end{matrix}

where $T_{0}$ represents the initial temperature. Let’s apply this formula in the following example.

Newton’s law of cooling

According to experienced baristas, the optimal temperature to serve coffee is between $155 ° F$ and $175 ° F .$ Suppose coffee is poured at a temperature of $200 ° F,$ and after $2$ minutes in a $70 ° F$ room it has cooled to $180 ° F .$ When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.

We have

\begin{matrix} T & = & (T_{0} - T_{a}) e^{− k t} + T_{a} \\ 180 & = & (200 - 70) e^{− k (2)} + 70 \\ 110 & = & 130 e^{−2 k} \\ \frac{11}{13} & = & e^{−2 k} \\ ln \frac{11}{13} & = & −2 k \\ ln 11 - ln 13 & = & −2 k \\ k & = & \frac{ln 13 - ln 11}{2} . \end{matrix}

Then, the model is

T = 130 e^{(ln 11 - ln 13 / 2) t} + 70 .

The coffee reaches $175 ° F$ when

\begin{matrix} 175 & = & 130 e^{(ln 11 - ln 13 / 2) t} + 70 \\ 105 & = & 130 e^{(ln 11 - ln 13 / 2) t} \\ \frac{21}{26} & = & e^{(ln 11 - ln 13 / 2) t} \\ ln \frac{21}{26} & = & \frac{ln 11 - ln 13}{2} t \\ ln 21 - ln 26 & = & \frac{ln 11 - ln 13}{2} t \\ t & = & \frac{2 (ln 21 - ln 26)}{ln 11 - ln 13} \approx 2.56. \end{matrix}

The coffee can be served about $2.5$ minutes after it is poured. The coffee reaches $155 ° F$ at

\begin{matrix} 155 & = & 130 e^{(ln 11 - ln 13 / 2) t} + 70 \\ 85 & = & 130 e^{(ln 11 - ln 13) t} \\ \frac{17}{26} & = & e^{(ln 11 - ln 13) t} \\ ln 17 - ln 26 & = & (\frac{ln 11 - ln 13}{2}) t \\ t & = & \frac{2 (ln 17 - ln 26)}{ln 11 - ln 13} \approx 5.09. \end{matrix}

The coffee is too cold to be served about $5$ minutes after it is poured.

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Suppose the room is warmer $(75 ° F)$ and, after $2$ minutes, the coffee has cooled only to $185 ° F .$ When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.

The coffee is first cool enough to serve about $3.5$ minutes after it is poured. The coffee is too cold to serve about $7$ minutes after it is poured.

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Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have

\begin{matrix} \frac{y_{0}}{2} & = & y_{0} e^{− k t} \\ \frac{1}{2} & = & e^{− k t} \\ - ln 2 & = & − k t \\ t & = & \frac{ln 2}{k} . \end{matrix}

Note : This is the same expression we came up with for doubling time.

Definition

If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by

Half-life = \frac{ln 2}{k} .

Radiocarbon dating

One of the most common applications of an exponential decay model is carbon dating . $Carbon- 14$ decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of $carbon- 14$ is approximately $5730$ years—meaning, after that many years, half the material has converted from the original $carbon- 14$ to the new nonradioactive $nitrogen- 14 .$ If we have $100$ g $carbon- 14$ today, how much is left in $50$ years? If an artifact that originally contained $100$ g of carbon now contains $10$ g of carbon, how old is it? Round the answer to the nearest hundred years.

We have

\begin{matrix} 5730 & = & \frac{ln 2}{k} \\ k & = & \frac{ln 2}{5730} . \end{matrix}

So, the model says

y = 100 e^{− (ln 2 / 5730) t} .

In $50$ years, we have

\begin{matrix} y & = & 100 e^{− (ln 2 / 5730) (50)} \\ \approx & 99.40 . \end{matrix}

Therefore, in $50$ years, $99.40$ g of $carbon- 14$ remains.

To determine the age of the artifact, we must solve

\begin{matrix} 10 & = & 100 e^{− (ln 2 / 5730) t} \\ \frac{1}{10} & = & e^{− (ln 2 / 5730) t} \\ t & \approx & 19035. \end{matrix}

The artifact is about $19,000$ years old.

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If we have $100$ g of $carbon- 14,$ how much is left after. years? If an artifact that originally contained $100$ g of carbon now contains $20 g$ of carbon, how old is it? Round the answer to the nearest hundred years.

A total of $94.13$ g of carbon remains. The artifact is approximately $13,300$ years old.

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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