Continuous model:
N(t)=N0ekt.
N0 = initial value, k = continuous rate. k>0: growth. k<0: decay.
Half-life (for decay): T=∣k∣ln2.
Doubling time (for growth): T=kln2.
Linearisation. Taking ln of N=N0ekt gives lnN=lnN0+kt — LINEAR in t. This is the basis of fitting exponential models to data (linear regression on lnN vs t).
Worked example (decay). A radioactive isotope has half-life 12 years. Find k, then how long until 90% has decayed.
T=12⇒k=−12ln2≈−0.0578.
90% decayed ⇒ 10% remains: N0ekt=0.1N0⇒kt=ln0.1=−ln10.
t=−ln2/12−ln10=ln212ln10≈39.86 years.
Worked example (compound interest comparison). $5000 at 4.5%. Compare end-of-10-year value if compounded (a) annually, (b) continuously.
(a) A = 5000(1.045)^{10} \approx \7765$.
(b) A = 5000 e^{0.045 \cdot 10} \approx 5000 e^{0.45} \approx \7841$.
Continuous compounding gives slightly more — the limiting case as compounding frequency goes to infinity.
Newton's law of cooling. T(t)−Troom=(T0−Troom)e−kt — applied physics problem on Paper 3.