Exponential Models with Differential Equations
Introduced| English |
|---|
| exponential growth/ˌekspəˈnenʃl ɡrəʊθ/ |
| exponential model/ˌekspəˈnenʃl ˈmɒdl/ |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ |
The equation behind exponential growth
- One differential equation appears everywhere: $\dfrac{dy}{dt}=ky$ — "rate proportional to amount."
- Its solution is the famous exponential model 指数模型 $y=y_0 e^{kt}$.
- Population, radioactive decay, compound interest, cooling — all obey this pattern.
- Solve it once and you own a huge family of real-world models.
Deriving $y=y_0 e^{kt}$
- Separate: $\dfrac{1}{y}\,dy = k\,dt$. Integrate: $\ln|y|=kt+C$.
- Exponentiate: $y = e^{kt+C}=A\,e^{kt}$.
- Apply the initial amount $y(0)=y_0$: $A=y_0$.
- Solution: $y=y_0 e^{kt}$ — the initial value times $e^{kt}$.
The solution of $\dfrac{dy}{dt}=ky$ with $y(0)=y_0$ is...
Separate and integrate: $y=y_0 e^{kt}$.
Growth when $k>0$
- If $k>0$, then $e^{kt}$ increases, so $y$ shows exponential growth 指数增长.
- The quantity keeps multiplying — the bigger it gets, the faster it grows.
- $k$ is the (continuous) growth rate; a larger $k$ means faster growth.
- Examples: unchecked populations, compound interest.
Growth (k>0) vs decay (k<0)
y = a·e^{bx}
$y=y_0 e^{kt}$ rises when $k>0$ and falls toward zero when $k<0$ — the sign of $k$ sets the behaviour.
In $y=y_0 e^{kt}$, exponential growth occurs when...
$k>0$ makes $e^{kt}$ increase → growth.
For $P=500e^{0.3t}$, find $P(0)$ (the initial amount).
$P(0)=500 e^0=500$.
Decay when $k<0$
- If $k<0$, then $e^{kt}$ decreases, so $y$ shows exponential decay 指数衰减.
- The quantity shrinks toward $0$ but never quite reaches it.
- Radioactive decay and cooling differences follow $k<0$.
- The magnitude $|k|$ sets how fast the decay happens (linked to half-life).
Exponential decay occurs when $k$ is ____ (its sign).
$k<0$ makes $e^{kt}$ shrink toward $0$.
The model $y=y_0 e^{kt}$ applies to a situation with a constant rate $\dfrac{dy}{dt}=c$.
Constant rate gives a line; the exponential model needs rate ∝ amount.
Which situations are modeled by $\dfrac{dy}{dt}=ky$?
The first three have rate proportional to the amount; the constant-rate tap does not.
The exponential model comes only from $\frac{dy}{dt}=ky$ (rate proportional to the amount). Don't apply $y=y_0 e^{kt}$ to a constant-rate situation ($\frac{dy}{dt}=c$, which gives a line). And the sign of $k$ is everything: $k>0$ grows, $k<0$ decays — a wrong sign flips the whole behavior.
A colony of $500$ bacteria grows so that $\dfrac{dP}{dt}=0.3P$ ($t$ in hours). Find $P(t)$ and $P(4)$.
- Model: $P=P_0 e^{kt}=500\,e^{0.3t}$.
- At $t=4$: $P(4)=500\,e^{1.2}\approx500(3.32)\approx1660$ bacteria.
- Since $k=0.3>0$, this is exponential growth.
The differential equation $\frac{dy}{dt}=ky$ gives the exponential model $y=y_0 e^{kt}$, where $y_0$ is the initial amount. $k>0$ → exponential growth; $k<0$ → exponential decay. It applies only when the rate is proportional to the current amount.