Post on 13-Mar-2021
Copyright Β© 2014-2016 by Harold Toomey, WyzAnt Tutor 1
Haroldβs Exponential Growth and Decay Cheat Sheet
16 May 2016
Discrete Continuous
π΄ = π (1 +π
π)
ππ‘
π΄ = ππππ‘
Simple Interest: π΄ = π + πΌ = π + πππ‘ = π(1 + ππ‘)
A = Amount after time t P = Original amount, such as principle e = The natural number (~2.718) r = Rate of growth/loss, e.g. interest rate (15% = 0.15) t = Elapsed time n = Divides time into periods per time unit
e β 2.71828 18284 59045 23536 β¦
π = limπββ
(1 +1
π)
π
π΄ = limπββ
π (1 +π
π)
ππ‘= ππππ‘
e = βπ₯π
π !βπ=0 = 1 +
1
1!+
1
2!+
1
3!+
1
4!+
1
5!+ β―
Savings Account Example:
P = $100 r = 8% = 0.08 t = 1 year n = 4 (quarterly)
π΄ = $100 (1 +0.08
4)
4(1)= $108.24
Savings Account Example:
π΄ = $100 π0.08(1) = $108.33
If n = 1, A = $108.00 (+0Β’) Annually If n = 4, A = $108.24 (+24Β’) Quarterly If n = 12, A = $108.29 (+5Β’) Monthly If n = 365, A = $108.33 (+4Β’) Daily If n = β, A = $108.33 (+0Β’) Continuously
Compounded interest after 3 years: π΄(3) = π (1 + 8%) (1 + 8%) (1 + 8%)
π΄(3) = π(1 + 0.08)3 = 1.26 * P
(See calculus derivation on page 2)
π΄ = π (1 +π
π)
ππ‘
π = π΄
(1 +ππ
)ππ‘
π = π [(π΄
π)
1ππ‘β
β 1]
π‘ = 1
π
ln(π΄π)
ln (1 +ππ)
n = ?
π΄ = ππππ‘
π = π΄
πππ‘
π =1
π‘ln (
π΄
π)
π‘ =1
πln (
π΄
π)
n = β
Copyright Β© 2014-2016 by Harold A. Toomey, WyzAnt Tutor 2
Calculus Derivation Graphs Assume the rate of growth or decay is proportional to the amount of substance (P).
ππ
ππ‘ β π
ππ
ππ‘= π π
Separate variables and integrate:
ππ
π= π ππ‘
β«ππ
π = β« π ππ‘
ln|π| = ππ‘ + π
Solve for P(t):
πln|π| = πππ‘+π
|π| = ππ πππ‘ = πΆπππ‘ At t=0 (initial condition):
π(0) = πΆππβ0 = πΆ β 1 = π0 Therefore,
π(π‘) = π0 πππ‘ or
π΄ = ππππ‘
Left: Exponential Growth (k or r positive) Right: Exponential Decay (k or r negative)
Chemistry Half-Life π΄ = π΄πππ’ππ‘ πππππππππ πππ‘ππ π‘πππ π‘ π΄0 = π΄πππ’ππ‘ π π‘πππ‘πππ π€ππ‘β ππ‘ π‘πππ π‘ = 0 π = π»πππ πΏπππ (π‘πππ π’πππ‘π ) π‘ = π‘πππ (π‘πππ π’πππ‘π )
π΄ = π΄0 (1
2)
π‘π