5.3 Compound Interest, Doubling Time & Half-Life Applications

Compound Interest, Doubling Time & Half-Life Applications

Real-world exponential applications are heavily represented in the Advanced Math section of the Digital SAT. The three most common contextual frameworks are Compound Interest, Doubling Time, and Half-Life Decay.

Mastering these formulas allows you to dissect complex word problems rapidly, set up exact mathematical models, and solve for final values, initial principals, or elapsed time periods.

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1. Discrete Compound Interest Mechanics

When money earns compound interest, the interest generated in each period is added to the principal, earning interest on top of interest in all subsequent periods.

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

+-----------------------------------------------------------------------------+
|                   COMPOUND INTEREST FORMULA BREAKDOWN                       |
|                                                                             |
|                                  r   (n · t)                                |
|                      A(t) = P ( 1 + --- )                                   |
|                                  n                                          |
|                             |    |   |     |                                |
|      +----------------------+    |   |     +----------------------+         |
|      v                           v   v                            v         |
|  [ PRINCIPAL (P) ]         [ RATE (r) ] [ FREQUENCY (n) ]   [ TOTAL PERIODS]|
|  Initial deposit (\$)       Annual decimal Compounding times Total compound  |
|                            (e.g., 5% = 0.05) per year       events = n · t  |
+-----------------------------------------------------------------------------+

Compounding Frequencies ($n$ Values):

• Annually: $n = 1$
• Semi-annually: $n = 2$
• Quarterly: $n = 4$
• Monthly: $n = 12$
• Daily: $n = 365$

Example Comparison:

Suppose $10,000 is invested at an annual interest rate of $6\%$ ($r = 0.06$) for $5$ years ($t = 5$):

• Annual ($n=1$): $A = 10,000\left(1 + \frac{0.06}{1}\right)^{1 \times 5} = 10,000(1.06)^5 \approx 13,382.26$, yielding $13,382.26.
• Quarterly ($n=4$): $A = 10,000\left(1 + \frac{0.06}{4}\right)^{4 \times 5} = 10,000(1.015)^{20} \approx 13,468.55$, yielding $13,468.55.
• Monthly ($n=12$): $A = 10,000\left(1 + \frac{0.06}{12}\right)^{12 \times 5} = 10,000(1.005)^{60} \approx 13,488.50$, yielding $13,488.50.

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2. Continuous Compounding & Euler's Constant $e$

As compounding frequency $n$ approaches infinity (compounding every microsecond continuously), the discrete compounding expression $\left(1 + \frac{r}{n}\right)^{nt}$ converges to the continuous exponential base $e^{rt}$, where $e \approx 2.71828$ is Euler's number.

$$A(t) = P e^{rt}$$

• $P$: Initial Principal amount deposited.
• $e$: Base of the natural logarithm ($e \approx 2.7182818$).
• $r$: Annual nominal interest rate (in decimal form).
• $t$: Time elapsed in years.

For the same $10,000 at $6\%$ over $5$ years compounded continuously:

$$A = 10,000 e^{0.06 \times 5} = 10,000 e^{0.30} \approx 10,000(1.3498588) \approx 13,498.59$$

This yields an ending balance of $13,498.59.

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3. Half-Life Radioactive Decay Models

The half-life ($t_{1/2}$ or $h$) of a substance is the time required for half ($50\%$) of the existing quantity to decay. Radioactive decay is naturally modeled by a base of $\frac{1}{2}$ (or $0.5$):

$$N(t) = N_0\left(\frac{1}{2}\right)^{\frac{t}{h}} = N_0(0.5)^{\frac{t}{h}}$$

+-----------------------------------------------------------------------------+
|                          HALF-LIFE DECAY PATTERN                            |
|                                                                             |
|   Time Elapsed (t):   0       1 Half-Life (h)   2h         3h         4h    |
|   Remaining Mass:    N_0 ----> (1/2) N_0 ----> (1/4) N_0 -> (1/8) N_0 -> 1/16|
|   Percent Remaining: 100%        50%             25%        12.5%      6.25%|
+-----------------------------------------------------------------------------+

Step-by-Step Half-Life Cycle Counting:

When time $t$ is an integer multiple of $h$, the remaining mass is calculated by halving the initial amount for each cycle:

• Number of cycles: $k = \frac{t}{h}$
• Remaining Amount: $N(t) = N_0 \cdot \left(\frac{1}{2}\right)^k$

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4. Doubling Time Models

In biological systems, bacterial colonies, or data storage expansions, quantities often double at fixed time intervals $d$:

$$N(t) = N_0 \cdot 2^{\frac{t}{d}}$$

• $N_0$: Initial population or count at $t = 0$.
• $d$: Doubling period (time required to multiply by $2$).
• $t$: Total elapsed time in the same time units as $d$.

+-----------------------------------------------------------------------------+
|                         DOUBLING TIME PATTERN                               |
|                                                                             |
|   Time Elapsed (t):   0       1 Doubling (d)    2d         3d         4d    |
|   Population:        N_0 ----> 2 N_0 --------> 4 N_0 ----> 8 N_0 ----> 16 N_0|
+-----------------------------------------------------------------------------+

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5. Step-by-Step Worked SAT Exam Examples

Worked Example 1: Discrete Compound Interest Calculation

Problem: An entrepreneur deposits $8,000 into a high-yield business savings account with an annual interest rate of $4.8\%$ compounded monthly. What is the total balance in the account after $3$ years, rounded to the nearest dollar?

Step-by-Step Solution:

1. Identify parameters: $P = 8,000$, $r = 0.048$, $n = 12$, $t = 3$.
2. Calculate monthly rate: $\frac{r}{n} = \frac{0.048}{12} = 0.004$.
3. Calculate total compounding periods: $nt = 12 \times 3 = 36$.
4. Apply the formula: $$A = 8,000(1 + 0.004)^{36} = 8,000(1.004)^{36}$$
5. Evaluate in Desmos / calculator: $$1.004^{36} \approx 1.1545524 \implies A = 8,000 \times 1.1545524 \approx 9,236.42 \approx 9,236$$

The total balance in the account after $3$ years is $9,236.

Worked Example 2: Solving for Elapsed Time with Desmos

Problem: A researcher starts with $300$ milligrams of a radioactive isotope that has a half-life of $14$ days. After how many days will exactly $37.5$ milligrams of the isotope remain?

Method A: Algebraic Cycle Halving

1. Set up the half-life ratio: $\frac{N(t)}{N_0} = \frac{37.5}{300} = 0.125 = \frac{1}{8}$.
2. Express $\frac{1}{8}$ as a power of $\frac{1}{2}$: $\frac{1}{8} = \left(\frac{1}{2}\right)^3$.
3. Equate exponents: $\left(\frac{1}{2}\right)^{\frac{t}{14}} = \left(\frac{1}{2}\right)^3 \implies \frac{t}{14} = 3$.
4. Multiply: $t = 3 \times 14 = 42$ days.

Method B: Desmos Graphical Solver

1. In Desmos, type Line 1: y = 300 * (0.5)^(x / 14)
2. Type Line 2: y = 37.5
3. Click the intersection point: Desmos displays $(42, 37.5)$, confirming $x = 42$ days immediately.

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6. Desmos Solver Playbook & SAT Pitfalls