Exponential Decay

**Exponential decay** happens when a quantity decreases by the same **percentage** in every equal time period. The amount lost gets smaller each time, because each new loss is a percentage of what is left. $$y = a(1 - r)^t$$ Where: - $a$ = initial value (the amount at $t = 0$) - $r$ = **decay rate**, the percent lost per period, written as a decimal ($15\% \to r = 0.15$) - $1 - r$ = **decay factor** $b$, the part that is kept each period ($1 - 0.15 = 0.85$) - $t$ = number of time periods In the general form $y = a \cdot b^t$ you can tell the type from the base: - $b > 1$: exponential **growth** - $0 < b < 1$: exponential **decay** After $t$ periods, the fraction that remains is $(1 - r)^t$, no matter how big the starting amount is.

Growth or Decay?

For each function, decide whether it shows growth or decay and find the percent change per period: $f(t) = 300 \cdot 0.92^t$ and $g(t) = 50 \cdot 1.03^t$.

Look at the base of $f$: $b = 0.92$ and $0 < 0.92 < 1$ = $f$ shows decay

Find the decay rate of $f$: $r = 1 - b = 1 - 0.92 = 0.08$ = $f$ loses $8\%$ per period

Look at the base of $g$: $b = 1.03 > 1$, so $r = 1.03 - 1 = 0.03$ = $g$ grows $3\%$ per period

Read the initial values: $f(0) = 300 \cdot 0.92^0 = 300$ and $g(0) = 50$ = Starting values $300$ and $50$

Answer: $f$ is exponential **decay** at $8\%$ per period (starting at $300$); $g$ is exponential **growth** at $3\%$ per period (starting at $50$).

Car Depreciation

A new car costs 25000 dollars and loses $15\%$ of its value every year. What is it worth after 4 years? Round to the nearest cent.

Identify the values: $a = 25000$, $r = 0.15$, $t = 4$ = Values identified

Find the decay factor: $1 - r = 1 - 0.15 = 0.85$ = $b = 0.85$

Write the formula: $V(4) = 25000 \cdot 0.85^4$ = Formula set up

Calculate the power: $0.85^4 = 0.52200625$ = About $52.2\%$ of the value is left

Multiply and round: $25000 \cdot 0.52200625 = 13050.15625 \approx 13050.16$ = 13050.16 dollars

Answer: After 4 years the car is worth about **13050.16 dollars**, a little more than half of its price.

Medicine in the Body

A patient takes a 400 mg dose of a medicine. Each hour the body removes $20\%$ of the medicine that is still present. How much is left after 5 hours, and what percent of the dose is that? Round to one decimal place.

Find the decay factor: $1 - 0.20 = 0.8$ = $b = 0.8$

Write the model: $M(t) = 400 \cdot 0.8^t$ = Model written

Calculate the power: $0.8^5 = 0.32768$ = $0.32768$ of the dose remains

Find the amount: $400 \cdot 0.32768 = 131.072 \approx 131.1$ = $131.1$ mg

Turn the factor into a percent: $0.32768 \times 100\% = 32.768\% \approx 32.8\%$ = $32.8\%$ remains

Answer: After 5 hours about **131.1 mg** is left, which is about **32.8%** of the dose.

Cooling Coffee (Simple Model)

A cup of coffee is $90°C$ in a room at $20°C$. In a simple model, the difference between the coffee and the room shrinks by $10\%$ every minute. What is the coffee's temperature after 5 minutes? Round to one decimal place.

Find the starting difference: $90 - 20 = 70$ = $D_0 = 70$ degrees

Model the difference: $D(t) = 70 \cdot 0.9^t$ = Decay factor $0.9$

Calculate the power: $0.9^5 = 0.59049$ = $0.59049$

Find the difference after 5 minutes: $70 \cdot 0.59049 = 41.3343$ = $D(5) \approx 41.33$ degrees

Add the room temperature back: $20 + 41.3343 = 61.3343 \approx 61.3$ = $61.3°C$

Answer: After 5 minutes the coffee is about **61.3°C**. Only the difference to the room decays, so the coffee approaches $20°C$ but never drops below it.

Linear vs Exponential Decrease

A phone costs 800 dollars. Model A: it loses 100 dollars per year. Model B: it loses $20\%$ per year. Compare the values after 3, 6 and 8 years (round to the nearest cent).

Write both models: A: $V = 800 - 100t$; B: $V = 800 \cdot 0.8^t$ = Linear vs exponential

Compare after 3 years: A: $800 - 300 = 500$; B: $800 \cdot 0.512 = 409.6$ = B is lower

Compare after 6 years: A: $800 - 600 = 200$; B: $800 \cdot 0.262144 = 209.7152 \approx 209.72$ = Now A is lower

Compare after 8 years: A: $800 - 800 = 0$; B: $800 \cdot 0.16777216 \approx 134.22$ = A reaches zero, B does not

Explain the pattern: A subtracts the same amount; B subtracts $20\%$ of a smaller and smaller value = Exponential losses shrink over time

Answer: The exponential model loses more at first (409.60 vs 500 dollars after 3 years), but its yearly losses keep shrinking: after 6 years it is higher (209.72 vs 200), and after 8 years the linear value is 0 while the exponential value is still about 134.22 dollars.

Exponential decay describes many things that shrink over time: - **Depreciation**: A new car or phone loses a percentage of its value every year, so it loses the most money at the start - **Medicine**: The body removes a fixed percentage of a drug each hour, which decides how often a dose is needed - **Cooling**: A hot drink cools quickly at first and then more and more slowly as it nears room temperature - **Light and sound**: Each meter of water or wall absorbs a percentage of the light or sound passing through Knowing the decay formula lets you predict values, compare offers (a fixed amount off per year versus a percentage off) and understand why some things never quite reach zero.

Mistake: Using the decay rate as the base: writing $25000 \cdot 0.15^t$ for a $15\%$ loss

Why: $0.15^t$ keeps only $15\%$ each year, which means losing $85\%$ per year. The base must be the part that is **kept**.

Correct: Decay factor $= 1 - r$. A $15\%$ decrease gives $25000 \cdot 0.85^t$.

Mistake: Subtracting the same amount every period: '20% of 1000 is 200, so after 5 years the value is 0'

Why: That is linear thinking. Each year's loss is $20\%$ of the **current** value, not of the original value, so the losses shrink.

Correct: Multiply by $0.8$ each year: $1000 \to 800 \to 640 \to 512 \to 409.6 \to 327.68$. After 5 years $327.68$ is left.

Mistake: Computing $a \cdot 1 - r^t$ instead of $a(1 - r)^t$

Why: The exponent belongs to the whole decay factor. $1 - 0.15^2 = 0.9775$, but $(1 - 0.15)^2 = 0.7225$.

Correct: Find $1 - r$ first, then raise it to the power $t$, then multiply by $a$.

Mistake: Believing a decaying quantity eventually reaches exactly zero

Why: Multiplying a positive number by a factor like $0.8$ always gives a positive number.

Correct: In the model $y = a(1 - r)^t$ the values get closer and closer to $0$ (the horizontal asymptote $y = 0$) but never reach it.

Depreciation of Phones and Laptops

Electronics lose a percentage of their value each year. Resale websites and insurers use decay models to estimate what a used device is worth.

A 1000-dollar laptop that loses $30\%$ per year is worth $1000 \cdot 0.7^2 = 490$ dollars after 2 years.

Medicine Dosing

Doctors and pharmacists use exponential decay to know how much of a drug is still in the body and when the next dose should be taken.

If the body removes $15\%$ of a drug each hour, then after 4 hours $0.85^4 \approx 0.522$, so about $52.2\%$ of the dose remains.

Cooling Drinks

Engineers and food scientists model cooling with decay: the temperature difference between an object and the room shrinks by a percentage each minute.

Coffee at $90°C$ in a $20°C$ room: if the difference of $70$ shrinks $10\%$ per minute, after 1 minute it is $63$, so the coffee is $83°C$.

Exponential decay follows $y = a(1 - r)^t$: the quantity loses the same percentage every period

The decay rate $r$ is the percent lost; the decay factor $1 - r$ is the part kept (a $15\%$ decrease gives $0.85$)

In $y = a \cdot b^t$, a base $0 < b < 1$ means decay and $b > 1$ means growth

After $t$ periods the fraction remaining is $(1 - r)^t$; multiply by $100\%$ for the percent remaining

Linear decrease subtracts the same amount and can reach zero; exponential decrease loses less each period and only approaches zero

Round only at the end and state the rounding (nearest cent, one decimal place)

Q: What is the difference between the decay rate and the decay factor?

A: The decay rate $r$ is what you lose each period (for example $15\% = 0.15$). The decay factor is what you keep: $1 - r = 0.85$. The decay factor is the number that goes into the base of the power.

Q: How can I tell growth from decay just by looking at the formula?

A: Look at the base $b$ in $y = a \cdot b^t$. If $b > 1$, the quantity grows (for example $1.04$ means $+4\%$). If $0 < b < 1$, it decays (for example $0.96$ means $-4\%$).

Q: Does a decaying quantity ever reach zero?

A: Not in the model. Each period keeps a fixed fraction of a positive number, so the result stays positive. The graph gets closer and closer to the line $y = 0$, its horizontal asymptote. In real life we stop counting once the amount is too small to matter.

Q: What is half-life?

A: The half-life is the time it takes for exactly half of the quantity to be left. It is a special case of exponential decay with its own formula, covered in the lesson Half-Life Problems.

Exit Presentation

Exponential Decay

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Pre-Calculus / Exponential Functions

Exponential Decay

Learn how quantities shrink by a constant percentage, use the decay formula y = a(1 - r)^t and solve depreciation, medicine and cooling problems.

Advanced25 min

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