Mathematics

22/3848–12FE Reference Handbook 10.6 · Mathematics · p. 53 solution y = C e^{−at} (half-life is derived)

Handbook formula

Exponential Growth and Decay

N = N_{0} e^{k t}, t_{1/2} = \frac{ln 2}{∣ k ∣}

p. 53 prints y′ + a y = 0 with y = C e^{−at}. Growth or decay N = N0 e^{kt} is that solution with a = −k and C = N0. Half-life ln 2 / |k| is obtained by setting N/N0 = 1/2; it is not a separate printed line.

$N 0$: Initial value
$k$ (1/time): Growth (or −decay) rate
$t 1/2$: Half-life

Step-by-step solved example

BOD decays with k=0.23 /d. What fraction remains after 3 days?

Exponential remaining fraction vs time — same decay shape as a sag/BOD curve.

1. Evaluate
N/N0=e^{−0.23×3}=e^{−0.69}≈0.50.
$e^{- 0.69} a pp r o x 0.50$
2. Read
About one half-life (ln2/0.23≈3.0 d).

Answer: ≈ 50% remains

10 practice questions

0/10 correct

1.If k>0 the quantity

2.Half-life for k=0.0693 /yr is

3.Doubling time is

4.e^{−0.693} is nearest

5.Continuous compounding F=Pe^{rt} matches this model with k=

6.After 2 half-lives the remainder is

7.k has units of

8.N=100 e^{−0.1 t} at t=0 is

9.Linearize by plotting

10.If remainder is 37% , kt is nearest