Why the Sign Matters
A decay model is meant to describe an amount that decreases as time passes. In a differential equation, the sign of the rate tells us whether the amount is increasing or decreasing. In an exponential formula, the sign in the exponent controls the same behavior. If these signs are reversed or mixed between conventions, a supposed decay model can instead predict growth.
The most direct check is simple: if the modeled amount is positive, its derivative should be negative while it is decaying. This check can catch errors before you calculate a half-life or use the model to make a prediction.
Two Consistent Sign Conventions
There are two common ways to write an exponential decay model. The conventions differ in how the rate parameter is defined, not in the behavior they describe.
In both definitions, \(A_0>0\), and the rate parameter has units of inverse time. The two conventions describe the same model when \(k=-r\). The parameter called \(k\) is not positive in every convention: its sign depends on the equation in which it appears.
The equation and the exponential formula must use matching signs. For example, \(A'=kA\) pairs with \(A_0e^{kt}\); \(A'=-rA\), for \(r>0\), pairs with \(A_0e^{-rt}\). Taking the rate from one convention and the formula from the other can reverse decay into growth.
Why the Exponential Formula Has That Sign
Consider the signed-rate initial-value problem \(A'=kA\), \(A(0)=A_0\), where \(k\) is constant. The exponential expression \(A_0e^{kt}\) has derivative \(kA_0e^{kt}\), which is \(kA(t)\), and at \(t=0\) it equals \(A_0\). Thus it satisfies both the differential equation and initial condition.
We can also see why no different differentiable solution can satisfy those conditions. If \(A\) is any solution, differentiate the product \(A(t)e^{-kt}\):
A function whose derivative is zero is constant on its interval of definition. At \(t=0\), this product equals \(A_0\), so \(A(t)e^{-kt}=A_0\) and therefore \(A(t)=A_0e^{kt}\). For the positive-decay-rate convention, substitute \(k=-r\) to obtain \(A(t)=A_0e^{-rt}\). The sign in the exponent is therefore fixed by the sign in the differential equation.
Worked Example: Diagnose a Sign That Predicts Growth
Worked Example: Diagnose a Sign That Predicts Growth
A substance has initial amount \(120\) mg. A proposed signed-rate model is \(A'=0.15A\), with \(t\) measured in days. Does this model describe decay? If not, correct the sign and find the amount after \(2\) days using a decay rate of \(0.15\) per day.
First check the proposed derivative at the initial time:
The positive derivative says the amount is increasing initially. The proposed equation is a growth model, not a decay model. Its solution would be \(A(t)=120e^{0.15t}\). At \(t=2\), it predicts
The prediction is greater than the initial \(120\) mg, consistent with growth and inconsistent with decay.
For decay at \(0.15\) per day, use the positive-decay-rate convention \(A'=-0.15A\). Its solution is \(A(t)=120e^{-0.15t}\), so
This is less than \(120\) mg. The derivative check also gives \(A'(0)=-0.15(120)=-18\text{ mg/day}\), so the amount is decreasing initially. Finally, differentiating the corrected formula gives \(A'(t)=-0.15(120e^{-0.15t})=-0.15A(t)\), and substituting \(t=0\) gives \(A(0)=120\) mg. Both the equation and initial condition check.
Half-Life and the Sign of the Rate
For a decaying amount with initial value \(A_0>0\), the half-life \(T_{1/2}\) is the positive time at which the amount has fallen to \(A_0/2\). Under the signed-rate convention, set \(A(T_{1/2})=A_0/2\) in \(A(t)=A_0e^{kt}\). Under the positive-decay-rate convention, use \(A(t)=A_0e^{-rt}\). These give the corresponding formulas below.
For instance, in the signed-rate convention, \(e^{kT_{1/2}}=1/2\). Taking natural logarithms gives \(kT_{1/2}=\ln(1/2)=-\ln 2\), hence \(T_{1/2}=-\ln 2/k\). Since both \(k\) and \(-\ln 2\) are negative, this time is positive. In the positive-rate convention, \(e^{-rT_{1/2}}=1/2\) gives \(-rT_{1/2}=-\ln 2\), so \(T_{1/2}=\ln 2/r>0\).
Worked Example: Keep Half-Life Positive with a Negative Signed Rate
A sample follows \(A'=kA\), where \(k=-\ln 2/8\) per hour, and starts with \(64\) mg. Find its half-life and amount after \(12\) hours.
The signed rate is negative, as required for decay. Its decimal value is
Apply the signed-rate half-life formula. Keeping the exact value first avoids a rounding error:
For \(12\) hours, the model gives
To check the half-life, substitute \(t=8\): \(A(8)=64e^{-(\ln 2/8)(8)}=64e^{-\ln 2}=32\) mg, exactly half the initial amount. To check the differential equation, differentiate \(A(t)=64e^{kt}\) to get \(A'(t)=64ke^{kt}=kA(t)\), and \(A(0)=64\) mg. The negative \(k\) produces a positive half-life and a decreasing amount.
Worked Example: Do Not Mix the Two Rate Conventions
Worked Example: Do Not Mix the Two Rate Conventions
A model uses a positive decay rate \(r=0.04\) per day and starts at \(500\) mg. Find the amount after \(5\) days, and identify the result of mistakenly using a positive exponent.
Because \(r\) is defined as positive in the equation \(A'=-rA\), the matching formula is \(A(t)=A_0e^{-rt}\). Thus
The result is below \(500\) mg, as decay requires. The derivative of this formula is \(A'(t)=-0.04A(t)\). At the initial time, \(A'(0)=-0.04(500)=-20\text{ mg/day}\), confirming the decrease.
A common mismatch is to keep the positive number \(0.04\) but write \(500e^{0.04t}\). That gives
This answer exceeds the initial amount. Differentiating the mistaken formula gives \(A'(t)=0.04A(t)>0\) for a positive amount, so it describes growth. The correction is not to change the half-life formula independently; it is to pair the chosen convention consistently. Either use \(A'=-rA\), \(r=0.04>0\), and \(A=500e^{-0.04t}\), or use \(A'=kA\), \(k=-0.04\), and the same decreasing exponential.
A Reliable Sign-Check Procedure
When checking a decay model, verify its behavior before trusting its numerical predictions. The following sequence separates the physical meaning from the notation choice.
Determine whether the equation is \(A'=kA\) with a signed \(k\), or \(A'=-rA\) with a positive \(r\).
For \(A>0\), a decay model must give \(A'<0\). If the derivative is positive, the proposed equation predicts growth.
Use \(A_0e^{kt}\) for \(A'=kA\), or \(A_0e^{-rt}\) for \(A'=-rA\) with \(r>0\).
At \(t=0\), the formula should give \(A_0\); at positive time, a decay prediction should be below \(A_0\).
The computed half-life must be positive. A negative result often signals that the signed-rate formula was used with the wrong sign.
Common Mistakes and Their Corrections
Several errors arise from treating \(k\) as if it had one universal sign. The sign is meaningful only together with the equation and formula where the parameter is used.
| Mistake | Why it fails | Correction |
|---|---|---|
| Using \(A'=kA\) with \(k>0\) and calling it decay | For \(A>0\), \(A'=kA>0\), so the amount increases. | Use \(k<0\) in the signed-rate convention, or write \(A'=-rA\) with \(r>0\). |
| Using \(A_0e^{rt}\) when \(r>0\) is called a decay rate | The positive exponent makes the amount grow. | For a positive decay rate \(r\), use \(A_0e^{-rt}\). |
| Using \(T_{1/2}=\ln 2/k\) when \(k<0\) | The result is negative, although a half-life is a positive elapsed time. | For \(A'=kA\) with \(k<0\), use \(T_{1/2}=-\ln 2/k\). |
| Changing the sign of \(k\) in the half-life formula but not in the model | The equation, solution, and half-life no longer describe one consistent rate convention. | Write down the differential equation first, then use the matching exponential and half-life formula. |
A negative half-life is a useful warning, not a meaningful prediction that the sample reaches half its amount in negative elapsed time. Return to the original equation and check the sign of its derivative for a positive amount. This directly tests whether the model describes decay.
Why This Check Is Useful
Exponential formulas can produce plausible-looking numerical values even when a sign is wrong. Checking only the arithmetic may not reveal that the model has turned decay into growth. The derivative check is more fundamental: it tests the direction of change encoded by the equation itself. The initial-value check confirms that the model starts at the stated amount, and a later-value check confirms that its predicted behavior agrees with the intended process.
Keep the roles of the two parameters clear. A signed rate constant \(k\) is negative for decay in \(A'=kA\). A positive decay-rate parameter \(r\) is positive in \(A'=-rA\). The same physical model can be written either way, but its equation, solution, and half-life formula must remain consistent throughout the calculation.
Check Your Understanding
For each question, use the derivative sign and the stated convention to justify your answer.
- A positive amount satisfies \(A'=0.07A\). Does it increase or decrease? What sign change would make this a decay equation?
- In \(A'=-rA\), with \(r>0\), which exponential formula matches the equation: \(A_0e^{rt}\) or \(A_0e^{-rt}\)?
- For \(A'=kA\) with \(k<0\), explain why \(-\ln 2/k\) is positive.
- A proposed decay formula gives \(A(0)=A_0\) but \(A(3)>A_0\). What does that tell you to check?
- Write the positive half-life formula for each convention: signed \(k<0\), and positive decay rate \(r>0\).
