7.2 Exponential Growth, Decay, and Continuous Compounding

7.2 Exponential Growth, Decay, and Continuous Compounding

Quick Answer: The foundational exponential function is $y = a \cdot b^x = a(1 \pm r)^x$, where $a$ is the non-zero initial value, $b$ is the base factor, and $r$ is the fractional growth/decay rate. Growth occurs when $b > 1$ ($b = 1 + r$), while decay occurs when $0 < b < 1$ ($b = 1 - r$). Discrete periodic compounding follows $A = P\left(1 + \frac{r}{n}\right)^{nt}$, where $n$ represents compounding intervals per year. Continuous compounding is modeled by Euler's constant: $A = P e^{rt}$. Radioactive decay and doubling phenomena are modeled by $A = A_0(0.5)^{t/h}$ and $A = A_0(2)^{t/d}$. Equivalent periodic rates are revealed algebraically by rewriting $b^t = (b^{1/k})^{kt}$.

---

The Exponential Model Architecture (F-LE.2, F-LE.5)

An exponential function is characterized by a constant multiplicative rate of change. Whereas a linear function adds a constant amount over equal intervals (constant first difference $\Delta y$), an exponential function multiplies by a constant factor over equal intervals (constant ratio $\frac{y_{k+1}}{y_k} = b$).

The standard form of an exponential model is:

$$y = a \cdot b^x \quad \text{or} \quad f(t) = a(1 \pm r)^t$$

• Initial Value ($a$): The output when the independent variable is zero ($f(0) = a$). On a graph, $(0, a)$ is the $y$-intercept ($a \neq 0$).
• Base Factor ($b$): The constant growth or decay multiplier per unit change in $x$. The base must be positive and non-equal to one ($b > 0, b \neq 1$).
• Percentage Rate ($r$): The fractional change per unit of time, expressed as a decimal ($r = |b - 1|$).
• Asymptote & Domain/Range: For the parent function $y = b^x$ where $a > 0$, the domain is $(-\infty, \infty)$, the range is $(0, \infty)$, and the $x$-axis ($y = 0$) serves as a horizontal asymptote.

---

Growth Factors vs. Percentage Rates (F-LE.5)

A frequent source of error on Regents examinations is conflating the growth rate ($r$) with the growth factor ($b$):

1. Exponential Growth ($b > 1$)

When a quantity grows by a percentage rate $r > 0$ each period, the new amount equals $100\%$ of the previous quantity plus $r$. The growth factor is:

$$b = 1 + r$$

• Example: A town population growing by $4.5\%$ annually has $r = 0.045$. The growth factor is $b = 1 + 0.045 = 1.045$. The model is $P(t) = P_0(1.045)^t$.

2. Exponential Decay ($0 < b < 1$)

When a quantity decreases by a percentage rate $r > 0$ each period, the new amount equals $100\%$ of the previous quantity minus $r$. The decay factor is:

$$b = 1 - r$$

• Example: A new vehicle depreciating by $14\%$ annually has $r = 0.14$. The decay factor is $b = 1 - 0.14 = 0.86$. The model is $V(t) = V_0(0.86)^t$.

Extracting Rates from Factors

Given an exponential equation, identify whether $b$ is greater than or less than 1:

• If $b = 1.135$, then $b > 1 \implies$ growth with rate $r = 1.135 - 1 = 0.135 = 13.5\%$.
• If $b = 0.72$, then $0 < b < 1 \implies$ decay with rate $r = 1 - 0.72 = 0.28 = 28\%$.

---

Discrete Periodic Compound Interest (A-CED.1)

When interest is compounded periodically (such as monthly or quarterly), the annual interest rate is partitioned into smaller increments, and earned interest begins earning interest of its own during subsequent periods.

The official Regents Reference Sheet provides the formula for periodic compounding:

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

• $A$: Final accumulated balance (future value).
• $P$: Principal sum invested (present value).
• $r$: Nominal annual interest rate in decimal form (e.g., $6.2\% = 0.062$).
• $n$: Number of compounding periods per year.
• $t$: Time elapsed in years.
• $\frac{r}{n}$: Periodic interest rate applied at the end of each compounding interval.
• $nt$: Total number of compounding events over $t$ years.

Standard Compounding Frequencies

As $n$ increases, interest compounds more frequently, generating slightly higher yields on identical principal and interest rates.

---

Continuous Compounding and Euler's Number $e$

What happens if interest compounds every second, every millisecond, or continuously? As the compounding frequency $n$ approaches infinity ($n \to \infty$), the expression $\left(1 + \frac{r}{n}\right)^{nt}$ approaches a mathematical limit governed by Euler's constant $e \approx 2.7182818\dots$:

$$\lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt} = P e^{rt}$$

The continuous compounding formula (also provided on the reference sheet) is:

$$A = P e^{rt}$$

This continuous model applies not only to financial instruments, but also to continuous biological population growth, unconstrained bacterial proliferation, and chemical reactions.

---

Half-Life and Doubling Time Models

Certain exponential models are naturally calibrated to the time interval required for a quantity to double or halve:

1. Doubling Time ($b = 2$)

If a population doubles every $d$ units of time, the amount after time $t$ is:

$$A(t) = A_0 (2)^{\frac{t}{d}}$$

When $t = d$, the exponent becomes $\frac{d}{d} = 1$, yielding $A_0(2)^1 = 2A_0$. When $t = 2d$, the exponent becomes $2$, yielding $A_0(2)^2 = 4A_0$.

2. Half-Life Decay ($b = 0.5$)

The half-life ($h$) is the duration required for half of a radioactive substance to decay:

$$A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}} = A_0 (0.5)^{\frac{t}{h}}$$

Because $(0.5)^{\frac{t}{h}} = (2^{-1})^{\frac{t}{h}} = 2^{-\frac{t}{h}}$, this model can also be written with negative exponents in base 2.

---

Revealing Equivalent Rates via Exponent Properties (A-SSE.3c)

A major focus of Next Generation Algebra II examinations is manipulating exponential expressions using the power of a power property: $b^t = \left(b^{1/k}\right)^{kt}$. This transformation reveals equivalent periodic rates (such as converting an annual rate into a monthly or daily rate) without altering the model's fundamental values.

Converting Annual Rates to Monthly Rates

Suppose an investment grows according to $V(t) = 5000(1.06)^t$, where $t$ is in years. To find the equivalent monthly rate:

1. Introduce a factor of $12$ into the exponent while preserving equality: $$(1.06)^t = \left(1.06^{\frac{1}{12}}\right)^{12t}$$
2. Evaluate the interior 12th root using a calculator: $$1.06^{\frac{1}{12}} \approx 1.00486755$$
3. Rewrite the function: $$V(t) \approx 5000(1.00487)^{12t}$$
4. Interpret the monthly parameters:
   • Monthly growth factor: $\approx 1.00487$
   • Monthly growth rate: $r_m = 1.00487 - 1 = 0.00487 = 0.487\%$ per month.

---

Summary of Exponential Models

---

Worked Examples

Worked Problem 1: Comparing Quarterly vs. Continuous Compounding

Problem: A principal sum of $\text{\textdollar}5,000$ is invested at an annual interest rate of $4.8\%$ for $6$ years. Calculate the final balance under quarterly compounding and under continuous compounding. Round both values to the nearest cent and determine the difference.

• Step 1: Compute quarterly balance ($n = 4$).

  $$A = P\left(1 + \frac{r}{n}\right)^{nt} = 5000\left(1 + \frac{0.048}{4}\right)^{4 \times 6}$$ $$\frac{0.048}{4} = 0.012 \implies A = 5000(1.012)^{24}$$ $$A = 5000(1.331473) \approx \$6,657.36$$

• Step 2: Compute continuous balance.

  $$A = P e^{rt} = 5000 e^{0.048 \times 6} = 5000 e^{0.288}$$ $$A = 5000(1.333757) \approx \$6,668.79$$

• Step 3: Calculate the difference. Subtract the unrounded balances and round once, at the end:

  $$6668.7863 - 6657.3637 = 11.4226 \approx \$11.42$$

  Continuous compounding earns about $\text{\textdollar}11.42$ more than quarterly compounding over 6 years. (Rounding each balance to the cent first and then subtracting gives $\text{\textdollar}11.43$ - a one-cent discrepancy that shows exactly why NYSED rubrics require rounding only at the final step.)

Worked Problem 2: Half-Life Calculation for Radioactive Decay

Problem: A laboratory prepares an $80\text{-gram}$ sample of Phosphorus-32, which has a half-life of $14.3$ days. Write an exponential decay model and calculate the mass remaining after $40$ days to the nearest hundredth of a gram.

• Step 1: Formulate the half-life model.

  $$A(t) = A_0(0.5)^{\frac{t}{h}} = 80(0.5)^{\frac{t}{14.3}}$$

• Step 2: Substitute $t = 40$ days.

  $$A(40) = 80(0.5)^{\frac{40}{14.3}}$$

• Step 3: Evaluate the exponent and compute.

  $$\frac{40}{14.3} \approx 2.797203$$ $$(0.5)^{2.797203} \approx 0.143867$$ $$A(40) = 80 \times 0.143867 \approx 11.5094$$

• Step 4: Round to the nearest hundredth.

  $$A(40) \approx 11.51 \text{ grams}$$

Worked Problem 3: Rewriting an Annual Model to Reveal Monthly Growth Rate

Problem: The valuation of a technology firm is modeled by $V(t) = 250000(1.24)^t$, where $t$ is measured in years. Rewrite $V(t)$ in the form $V(t) = 250000(1 + r_m)^{12t}$ to reveal the equivalent monthly percentage growth rate, rounded to the nearest tenth of a percent.

• Step 1: Apply the power of a power exponent rule.

  $$(1.24)^t = \left(1.24^{\frac{1}{12}}\right)^{12t}$$

• Step 2: Evaluate the monthly base factor.

  $$1.24^{\frac{1}{12}} \approx 1.0180875$$

• Step 3: Extract the monthly growth rate $r_m$.

  $$1 + r_m = 1.0180875 \implies r_m = 0.0180875$$

  Multiply by $100\%$ to convert to a percentage:

  $$r_m \approx 1.80875\% \approx 1.8\%$$

• Step 4: State the rewritten function and rate.

  $$V(t) \approx 250000(1.018)^{12t}$$

  The equivalent monthly growth rate is approximately $1.8\%$ per month.

---

Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Confusing percentage rate $r$ with base factor $b$. If an investment grows by $7\%$, writing $y = P(0.07)^t$ models a catastrophic loss of $93\%$ per year. Always add 1 for growth ($b = 1.07$) and subtract from 1 for decay ($b = 1 - 0.07 = 0.93$).
• Pitfall 2: Misinterpreting compounding frequency $n$ in the exponent. In $A = P(1 + r/n)^{nt}$, students often calculate $(1 + r/n)$ raised to $t$ instead of $nt$, forgetting to multiply the number of years by the number of periods per year.
• Pitfall 3: Inconsistent time units in half-life or doubling models. If the half-life is given in days ($h = 5\text{ days}$) and the problem asks for the amount remaining after $3\text{ weeks}$, convert weeks to days ($t = 21\text{ days}$) before substituting into $\frac{t}{h}$.