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7.2 Exponential Growth and Decay

Consider a quantity that grows or decays at a rate that is proportional to its own size.

y: quantity that is growing / decaying

\[ \frac{dy}{dt} = k \cdot y \]

k: constant of proportionality ("proportional to __")

y: size of y

The larger $ y $ is, the faster it grows/decays.

$ k $ is also the relative growth / decay rate.

For example, if something is growing at a rate that is 2% of its size, then:

\[ \frac{dy}{dt} = 0.02y \]

(for example, savings account w/ 2% interest rate)

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\[ \frac{dy}{dt} = ky \]

The right side $ (ky) $ is also called the absolute growth / decay rate.

If $ \frac{dy}{dt} = ky $, what is $ y(t) $?

$ y $ is a type of function whose rate of change is a constant multiple of itself (or whose derivative looks like itself).

Exponential: \[ y = Ce^{kt} \]

$ C, k $ are constants

Check: Is $ \frac{dy}{dt} = ky $ if $ y = Ce^{kt} $?

\[ \frac{dy}{dt} = C \cdot e^{kt} \cdot k \]

\[ = k \cdot (Ce^{kt}) = ky \text{ so yes} \]

So, growth / decay at rate proportional to its size is described by \[ y = Ce^{kt} \rightarrow \text{exponential growth / decay} \]

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If $ k > 0 $, then the rate of change is positive so $ y $ is growing (population, interest, spread of disease)

If $ k < 0 $, then the rate of change is negative so $ y $ is decaying (radioactive decay, metabolism of drugs)

\[ y = Ce^{kt} \]

$ C $: the initial size of $ y $

so we usually use $ y_0 $ for that

\[ y = y_0 e^{kt} \longleftrightarrow \frac{dy}{dt} = ky \]

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Example

The average yearly inflation rate between 2016 and 2019 in the US is 2.1%. If a loaf of bread cost $2 in 2016, assuming the inflation rate stays the same, what will a loaf of bread cost in 2030? When will the price double ($4)?

\[ y(t) = y_0 e^{kt} \]

$ y $: cost of bread
$ t $: number of years since 2016
($ t = 1 \rightarrow 2017 $)

Find: $ y(14) $ and when $ y = 4, t = ? $

2030 is 14 years after 2016

so $ t = 14 $

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Exponential Growth Model: Finding the Growth Constant

k is the key part

Initially, cost is $ \$2 \rightarrow y_0 = 2 $

$ y(t) = 2e^{kt} $

now find $ k $

$ y(0) = 2 $ and $ y(1) = 1.021 y(0) = (1.021)(2) = 2.042 $

2% higher $ \rightarrow 1 + 0.021 $

now we sub $ y(1) = 2.042 $ into $ y(t) = 2e^{kt} $

$ 2.042 = 2e^{k \cdot 1} $ (where $ t=1 $)

$ \frac{2.042}{2} = e^k = 1.021 $

$ k = \ln(1.021) = $$ 0.0208 $

Price Prediction for 2030

In 2030, the price is

$ y(14) = 2e^{0.0208(14)} = $$ 2.68 $ $ \$2.68 $

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When will price double?

$ y = 4, t = ? $

back to $ y(t) = 2e^{0.0208t} $

$ 4 = 2e^{0.0208t} $

$ 2 = e^{0.0208t} $

$ \ln 2 = 0.0208t \rightarrow t = \frac{\ln 2}{0.0208} \approx 33 $ (yrs after 2016)

in year 2049

General Doubling Time Formula

the time to double is the same for all $ y_0 $

$ y(t) = y_0 e^{kt} $

time to double $ y_0 \rightarrow y(t) = 2y_0 $

$ 2y_0 = y_0 e^{kt} $

$ 2 = e^{kt} $

$ \ln 2 = kt $

$ t = \frac{\ln 2}{k} $

time to double whatever $ y_0 $

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Exponential Decay Example: Caffeine Half-Life

Example: The half-life of caffeine in the human body is around 5 hours. If 100 mg of caffeine was ingested by drinking coffee at 6 am, how long before 80% of the caffeine is eliminated? How much remains at 10 pm?

Definition of Half-life:

Time to decay to half of initial size ($y = \frac{1}{2} y_0$, find $t = ?$)

Setting up the Model

Start with the general exponential growth/decay model:

\[y(t) = y_0 e^{kt}\]

$t$: hours after 6 am (e.g., $t = 1 \rightarrow 7 \text{ am}$)
$y$: mg of caffeine in body
$y_0$: initial amount = 100

The specific equation for this problem is:

\[y(t) = 100 e^{kt}\]

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Finding the Decay Constant (k)

Find $k$ from the half-life of 5 hours:

\[y(t) = y_0 e^{kt}\]\[\frac{1}{2} y_0 = y_0 e^{k(5)}\]

Note: $\frac{1}{2} y_0$ represents the amount half left, and $t = 5$ is the half-life.

\[\frac{1}{2} = e^{5k}\]\[\ln \frac{1}{2} = 5k\]\[k = \frac{\ln \frac{1}{2}}{5} \approx -0.139\]

The relative decay rate is approximately 13.9%.

Calculating Time to Lose 80%

Time to lose 80% of initial 100 mg:

If 80% is lost, then 20% remains.

\[y(t) = (0.2)(100) = 20\]

Substitute the known values into the equation:

\[y(t) = 100 e^{-0.139t}\]\[20 = 100 e^{-0.139t}\]\[\frac{1}{5} = e^{-0.139t}\]\[\ln \frac{1}{5} = -0.139t\]\[t = \frac{\ln \frac{1}{5}}{-0.139} \approx 11.6 \text{ hours after 6 am}\]

Result: $t \approx 11.6$ hours

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Exponential Decay Calculation

How much remains at 10 pm?

10 pm $\rightarrow t = 16$ (16 hours after 6 am)

Equation: $y(t) = 100 e^{-0.139 t}$

$y(16) = 100 e^{-0.139(16)} \approx $

10.8

mg