Understand the idea For a constant proportional rate, the solution is exponential. Determine the constant from data using logarithms and interpret its sign and units. Half-life and doubling time describe multiplicative changes, not fixed additive changes. A model in which a temperature difference decays uses that difference as the dependent quantity, producing a shifted exponential.
y ( t ) = y 0 e k t , T double = ln 2 k ( k > 0 ) , T half = ln 2 ∣ k ∣ ( k < 0 ) y(t)=y_0e^{kt},\qquad T_{\text{double}}=\frac{\ln2}{k}\ (k>0),\quad T_{\text{half}}=\frac{\ln2}{|k|}\ (k<0) y ( t ) = y 0 e k t , T double = k ln 2 ( k > 0 ) , T half = ∣ k ∣ ln 2 ( k < 0 ) Worked example A population satisfies P ′ = k P \displaystyle P'=kP P ′ = k P P ( 0 ) = 100 \displaystyle P(0)=100 P ( 0 ) = 100 P ( 4 ) = 200 \displaystyle P(4)=200 P ( 4 ) = 200 P ( t ) \displaystyle P(t) P ( t )
The model is
100 e k t \displaystyle 100e^{kt} 100 e k t . Since
2 = e 4 k \displaystyle 2=e^{4k} 2 = e 4 k ,
k = ln 2 / 4 \displaystyle k=\ln2/4 k = ln 2/4 . Therefore
P ( t ) = 100 e ( ln 2 ) t / 4 \displaystyle P(t)=100e^{(\ln2)t/4} P ( t ) = 100 e ( l n 2 ) t /4 . The population doubles every four time units under this model.
Check this before moving on The proportional rate constant k \displaystyle k k y ′ \displaystyle y' y ′
Try it yourself A substance has half-life 6 \displaystyle 6 6 80 \displaystyle 80 80 18 \displaystyle 18 18
Show solution and reasoning Three half-lives pass, so the amount is 80 ( 1 / 2 ) 3 = 10 \displaystyle 80(1/2)^3=10 80 ( 1/2 ) 3 = 10 80 e − ( ln 2 ) t / 6 \displaystyle 80e^{-(\ln2)t/6} 80 e − ( l n 2 ) t /6
Work through AP Calculus AB with a tutor Detailed notes are discussed during the course with worked examples, guided practice, and feedback on mathematical reasoning.
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