Proportional rates, half-life and model limits
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| proportional rate/prəˈpɔːʃənl reɪt/ | tỷ lệ biến đổi tỷ lệ |
Why does one fixed half-life not mean a fixed subtraction?
- A decaying amount loses more per hour when more remains.
- This lesson studies proportional rate 比例变化率: A rate of change equal to a constant multiplied by the current amount.
The method
- For y=Ae^(kt), dy/dt=ke^(kt)A=ky.
- A>0 is the initial amount; k is a constant proportional rate in inverse time units.
- Positive k gives growth, negative k decay, zero k a constant.
- Over elapsed time Δt the multiplier is e^(kΔt).
- Doubling time is ln2/k for k>0; half-life is ln2/|k| for k<0.
For Q=80e^(-λt) with half-life 4 hours, why are the hourly losses NOT equal?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A constant PROPORTIONAL rate means the same fraction per unit time, not the same amount. The absolute loss in each interval falls as the amount itself falls.
A worked case
Use an illustrative decay Q=80e^(−λt), t in hours and λ=ln2/4. Then Q₄=40, Q₈=20 and Q₁₂=10: each four-hour step halves the amount. At t=4 the rate is Q′=−40λ=−10ln2 amount units per hour. Solving Q≤10 gives t≥12; the strict condition Q<10 instead needs t>12. For an illustrative mathematical account P=100e^(0.1t), t in years, P₁≈110.517 and the one-year effective growth is about 10.517%, not 10%. If the given effective yearly growth is 10%, use k=ln1.1 instead. Continuous compounding and a stated annual multiplier must not be confused.
Proportional rates, half-life and model limits
For y=Ae^(kt), dy/dt=ke^(kt)A=ky
State the valid input, plotted variable or time unit before using the logarithmic/exponential relationship.
Find Q at t=4 hours.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
80e^(−ln2)=40.
Where it goes wrong
- The exponent must be dimensionless: changing hours to minutes changes k by a factor 1/60.
- A half-life is not the time to reach zero; positive A times an exponential stays positive for finite time.
- Absolute losses vary with the amount although the proportional rate is fixed.
- A first whole-period threshold needs adjacent integer checks rather than only a continuous crossing time.
Not true: A constant exponential proportional rate means the same absolute amount is lost in each equal time interval. A constant proportional rate loses the same percentage each interval, so the absolute amount lost falls.
Find Q at t=8 hours.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
80e^(−2ln2)=20.
A constant exponential proportional rate means the same absolute amount is lost in each equal time interval.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A constant proportional rate loses the same percentage each interval, so the absolute amount lost falls.
Applying it
- State time/amount units, initial conditions and the constant-rate assumption.
- Compare observations with predictions before extrapolating.
- Limited resources, changing environmental conditions, fees or withdrawals can invalidate a simple growth model; changing decay conditions can require a varying-rate or multi-stage model.
- These are illustrative mathematical models, not product terms or treatment instructions.
Find the first time in hours when Q≤10.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
ln(80/10)/(ln2/4)=12; monotone decay gives Q≤10 from that time onward.
Match each exponential-model quantity to its formula.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The exponent must be dimensionless, so k carries inverse-time units tied to the stated time unit; changing hours to minutes rescales k.
In your course
- AQA A-level Mathematics (7357), specification section F.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Proportional rate: A rate of change equal to a constant multiplied by the current amount.
- For y=Ae^(kt), dy/dt=ke^(kt)A=ky.
- Watch for: the exponent must be dimensionless: changing hours to minutes changes k by a factor 1/60.