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.

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 4

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

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

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask

	Social Dances 1 By Marion Cabalfin Start Test
	U.s. history By OpenStax Read Online Course
	Anthropology Marriage Family Household By Richley Crapo Start Assignment
	13 Biology 13 Modern Understandings of Inheritance By OpenStax Start Quiz
	26 AP 26 Fluid, Electrolyte, Acid-Base Balance Essay By OpenStax Start Flashcards
	2 Psychology MCQ 2009 1 Exam By John Gabrieli Start Exam
©flickr: Derek	Treatment Of Psychological Disorders By Michael Nelson Start Quiz
	1 CDL Quiz - Air Brakes Endorsement Part 1 By Jazzycazz Jackson Start Quiz
	4 Neuroanatomy 04 Spinal Cord Brain Stem By Stephen Voron Start Quiz
	Does your crush like you? By Hoy Wen Start Quiz