Definitions, growth vs. decay, compounding, doubling time and half-life with real-world models.
Identify increasing/ decreasing behavior via base a and rate k; analyze end behavior and asymptotes.
Discrete vs. continuous compounding; effective rates; financial applications.
Decay modeling with physical meaning; relate to exponential constants and percent decrease.
Convert multiplicative change to additive logs for linearization and parameter estimation.
Doubling time T_d and half-life T_{1/2}; connect to k through natural logarithms.
An exponential function grows or decays by a constant percentage per equal time step. The two most common forms are base form and natural-base form:
Key features:
A convenient modeling form uses a growth factor per period:
Here is the percentage change per base period . If , we have growth; if , we have decay.
Converting to natural-base form yields , where. For small |r|, the approximation holds.
Financial growth often uses compounding. With nominal annual rate , principal, and compounding periods per year for years:
As , we obtain continuous compounding with base :
Effective Annual Rate (EAR) compares different compounding frequencies on an equal footing:
For with (growth), the doubling time solves , hence:
For decay (), the half-life makes, so:
These relations allow quick reasoning about time scales in biology, physics, and finance.
Exponentials have horizontal asymptote . With vertical shift , the asymptote becomes. Horizontal scaling modifies the time scale; vertical scaling changes initial value.
Invest at nominal annual rate for years. Compute with annual, monthly, and continuous compounding, and compare.
Annual (n=1): ≈ 12763
Monthly (n=12): ≈ 12834
Continuous: ≈ 12840
More frequent compounding yields higher returns; continuous compounding is the theoretical maximum for a given nominal rate.
A substance decays according to with half-life 30 days. Find and compute remaining mass after 20 days when g.
Half-life relation: ⇒
After 20 days: ≈ 126 g
The mass decreases by a constant percentage per equal time interval; the decay is multiplicative, not linear.
A population follows with in years. Find the doubling time.
≈ 8.66 years.
At 8% continuous growth, the quantity doubles about every 8.66 years.
Compare EAR for with monthly compounding vs. continuous compounding.
Monthly: ≈ 12.68%
Continuous: ≈ 12.75%
Continuous compounding gives a slightly higher effective rate for the same nominal rate.
With monthly deposits and monthly compounding, the future value follows an annuity-growth model. For simplicity, if the deposit occurs at period end, the closed form is:
This complements the single-sum formulas and is vital in budgeting and retirement planning.
Drug concentration in blood often decays exponentially after reaching peak level. If the elimination constant is , the half-life is . Repeated dosing superposes exponentials.
Understanding half-life informs safe dosing intervals and steady-state analysis.
Newton's Law of Cooling often yields an exponential approach to ambient temperature. With ambient and initial :
The asymptote shifts from 0 to , demonstrating transformation effects on exponential functions.
Short-run population growth can often be approximated by an exponential with rate . The doubling time provides an intuitive summary of pace.
In the long run, logistic models may be more appropriate, but exponential fits are useful locally.
Classify each function as growth or decay and state its asymptote:
(a) Decay (factor 0.7<1), asymptote y=0.
(b) Growth (k=0.3>0), asymptote y=0.
(c) Decay toward y=10 (vertical shift), asymptote y=10.
Find such that under continuous compounding with rate .
We solve ⇒ ⇒ .
A culture grows from 100 to 135 in 2 hours. Assume . Find and the doubling time.
⇒ ≈ 0.1508 h−1.
≈ 4.596 hours.
A medication has half-life 6 hours. If the initial dose produces concentration 40 units, find concentration after 18 hours.
Three half-lives: units.
For , state growth/decay and its horizontal asymptote.
Decay toward ; asymptote is as .
For data following , take ln to get (linear in t).
Estimate slope k by linear regression on ; intercept gives .
Q: When should I prefer over ?
Use for naturally continuous change (continuous compounding, decay constants). Use for discrete-period change.
Q: How to estimate k robustly?
Prefer multi-point log–linear regression over two-point estimates; inspect residuals and outliers for model adequacy.