Exponential Growth and Decay: Fresh Take

Lumen Learning

Exponential Growth and Decay: Fresh Take

Apply the exponential growth formula to real-world cases like increasing populations or investments
Describe how long it takes for quantities to double or reduce by half
Implement the exponential decay formula for scenarios like radioactive substances decaying or objects cooling down

Exponential Growth Model

The Main Idea

Exponential Growth Model:
- General form: $y = y_0 e^{kt}$
- $y_0$ : Initial state of the system
- $k$ : Positive growth constant
Key Feature:
- Rate of growth is proportional to current function value
- $y' = ky$
Applications:
- Population growth
- Compound interest
Compound Interest Formula:
- $\text{Balance} = P e^{rt}$
- $P$ : Principal (initial investment)
- $r$ : Interest rate (as a decimal)
- $t$ : Time (in years)
Doubling Time:
- Time for a quantity to double in exponential growth
- Formula: $\text{Doubling time} = \frac{\ln 2}{k}$

Consider a population of bacteria that grows according to the function $f(t)=500{e}^{0.05t},$ where $t$ is measured in minutes. How many bacteria are present in the population after $4$ hours? When does the population reach $100$ million bacteria?

There are $81,377,396$ bacteria in the population after $4$ hours. The population reaches $100$ million bacteria after $244.12$ minutes.

Suppose instead of investing at age $25\sqrt{{b}^{2}-4ac},$ the student waits until age 35. How much would she have to invest at $5\text{%}?$ At $6\text{%}?$

At $5\%$ interest, she must invest $$223,130.16.$ At $6\%$ interest, she must invest $$165,298.89.$

Suppose it takes $9$ months for the fish population in the last example to reach $1000$ fish. Under these circumstances, how long do the owner’s friends have to wait?

$38.90$ months

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.8 Try It Problems” here (opens in new window).

A bacteria culture starts with $1000$ cells and grows exponentially. After $2$ hours, there are $1500$ cells. How many cells will be present after $5$ hours?

Use the exponential growth model: $y = y_0 e^{kt}$

We know:

$y_0 = 1000$ (initial number of cells)
After $2$ hours, $y = 1500$

Find $k$ using the $2$ -hour data point:

$\begin{array}{rcl} 1500 &=& 1000 e^{k(2)} \\ 1.5 &=& e^{2k} \\ \ln 1.5 &=& 2k \\ k &=& \frac{\ln 1.5}{2} \approx 0.2027 \end{array}$

Now use this $k$ value to find the number of cells after $5$ hours:

$\begin{array}{rcl} y &=& 1000 e^{0.2027(5)} \\ &\approx& 2740 \text{ cells} \end{array}$

Therefore, after $5$ hours, there will be approximately $2740$ bacteria cells in the culture.

Exponential Decay Model

The Main Idea

Exponential Decay Model:
- General form: $y = y_0 e^{-kt}$
- $y_0$ : Initial state of the system
- $k$ : Positive decay constant
Key Feature:
- Rate of decay is proportional to current function value
- $y' = -ky$
Applications:
- Newton’s Law of Cooling
- Radioactive decay
- Population decline
Newton’s Law of Cooling:
- $T = (T_0 - T_a)e^{-kt} + T_a$
- $T$ : Temperature of object
- $T_0$ : Initial temperature
- $T_a$ : Ambient temperature
Half-Life:
- Time for a quantity to reduce by half in exponential decay
- Formula: $\text{Half-life} = \frac{\ln 2}{k}$

Suppose the room is warmer $(75\text{°}\text{F})$ and, after $2$ minutes, the coffee has cooled only to $185\text{°}\text{F}.$ 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.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.8 Try It Problems” here (opens in new window).

If we have $100$ g of $\text{carbon-}14,$ how much is left after $t$ years? If an artifact that originally contained $100$ g of carbon now contains $20g$ 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.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.8 Try It Problems” here (opens in new window).

A radioactive sample initially contains $100$ grams of a certain isotope. After $30$ days, the sample contains $80$ grams of the isotope. What is the half-life of this isotope?

Use the exponential decay model: $y = y_0 e^{-kt}$

We know:

$y_0 = 100$ (initial amount)
After $30$ days, $y = 80$

Find $k$ using the given data:

$\begin{array}{rcl} 80 &=& 100 e^{-k(30)} \\ 0.8 &=& e^{-30k} \\ \ln 0.8 &=& -30k \\ k &=& -\frac{\ln 0.8}{30} \approx 0.00743 \end{array}$

Use the half-life formula:

$\begin{array}{rcl} \text{Half-life} &=& \frac{\ln 2}{k} \\ &=& \frac{\ln 2}{0.00743} \\ &\approx& 93.3 \text{ days} \end{array}$

Therefore, the half-life of this isotope is approximately $93.3$ days.