Exponential Graphs in Context [H]
Exponential Graphs in Context
Many real-world processes involve quantities that grow or decay by a constant multiplicative factor per unit time — rather than by a constant additive amount. These processes produce exponential graphs. Identifying exponential behaviour from a graph or table, and using the model to make predictions, are key skills.
The Exponential Model
An exponential function in context takes the form:
\[ Q = Q_0 \times k^t \]where \( Q_0 \) is the initial value (when \( t = 0 \)), \( k \) is the growth/decay factor per unit time, and \( t \) is time. The graph always passes through \( (0, Q_0) \) and has the x-axis as a horizontal asymptote.
| Scenario | Growth or decay? | \(k\) value | Example |
|---|---|---|---|
| Population growth | Growth | \( k > 1 \) | \( k = 1.03 \) (+3% per year) |
| Compound interest | Growth | \( k > 1 \) | \( k = 1.05 \) (5% p.a.) |
| Radioactive decay | Decay | \( 0 < k < 1 \) | \( k = 0.5 \) (half-life) |
| Drug concentration | Decay | \( 0 < k < 1 \) | \( k = 0.8 \) per hour |
Recognising Exponential Growth from a Table
If successive ratios \( Q_{t+1}/Q_t \) are constant, the relationship is exponential. For example:
| \(t\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| \(Q\) | 200 | 400 | 800 | 1600 |
Each value doubles: \( k = 2 \). Model: \( Q = 200 \times 2^t \).
Worked Examples
A bacterial culture starts with 500 cells and doubles every hour. Write an exponential model and find the count after 6 hours.
\[ N = 500 \times 2^t \]After 6 hours: \( N = 500 \times 2^6 = 500 \times 64 = 32{,}000 \) cells.
£3{,}000 is invested at 4% compound interest per annum. Find the value after 5 years.
\[ V = 3000 \times 1.04^5 = 3000 \times 1.2167 \approx £3{,}650 \]The model \( P = 800 \times 1.06^t \) gives a town's population. Find the population after 10 years and the initial population.
Initial (\(t=0\)): \( P = 800 \times 1 = 800 \).
After 10 years: \( P = 800 \times 1.06^{10} = 800 \times 1.7908 \approx 1{,}433 \).
A radioactive substance has a half-life of 3 hours. Starting with 640 g, write an exponential model and find the mass after 12 hours.
Every 3 hours the mass halves. Let \(t\) be in units of 3-hour periods (half-lives).
\[ M = 640 \times \left(\frac{1}{2}\right)^t \]12 hours = 4 half-lives: \( M = 640 \times (0.5)^4 = 640 \times 0.0625 = 40 \) g.
Alternatively, using hours directly: \( M = 640 \times 0.5^{t/3} \). At \(t=12\): same result.
Key Takeaways
- Exponential model: \( Q = Q_0 \times k^t \). The initial value \( Q_0 \) is always the y-intercept (value at \(t=0\)).
- \( k > 1 \): growth. \( 0 < k < 1 \): decay. The closer \(k\) is to 1, the slower the change.
- To identify an exponential relationship from a table: check that successive values have a constant ratio.
- Half-life problems: if the half-life is \(h\), write \( Q = Q_0 \times (0.5)^{t/h} \) (with \(t\) in the same units as \(h\)).
- Percentage increase of \(r\%\) per period gives \(k = 1 + r/100\); percentage decrease gives \(k = 1 - r/100\).