From a Verbal Description to a Differential Equation
A model often begins with a sentence rather than an equation. For example, a population may be described as “increasing at a rate proportional to its current size.” To write a mathematical model, we must decide what quantity changes, what variable measures time, and what “proportional” means for the rate of change.
The equation \( \frac{dP}{dt}=kP \) records a precise relationship: the instantaneous rate at which \(P\) changes is a constant multiple of the current value of \(P\). This tutorial focuses on translating that relationship from words into symbols. It does not require finding a formula for \(P\).
What “Proportional to the Current Amount” Means
Two quantities are proportional when one is a fixed constant multiple of the other. In this setting, the rate \( \frac{dP}{dt} \) is the first quantity and the current amount \(P\) is the second. Thus, proportional change is not a claim that the amount \(P\) is constant, nor that the rate itself is constant. It says that the rate changes in proportion to the amount.
For example, if a population is large, a fixed percentage growth model predicts a larger increase per unit time than it does for a small population. The rate per unit of current population remains the same. When \(P\ne 0\), this relative rate can be written as \( \frac{1}{P}\frac{dP}{dt} \), so the proportional-change equation says that this relative rate is the constant \(k\).
The equation \( \frac{dP}{dt}=kP \) itself remains meaningful at \(P=0\); only the quotient by \(P\) is unavailable there. The constant \(k\) must be fixed over the time interval being modeled. If the proportionality factor varies with time, the equation would instead involve a time-dependent factor, such as \(k(t)P\), which is a different assumption.
Why the Translation Gives \( \frac{dP}{dt}=kP \)
Let \(R(P)\) denote the rate of change associated with a current amount \(P\). Saying this rate is proportional to the amount means there is one fixed number \(k\) such that \(R(P)=kP\) throughout the modeled range. But the rate of change of \(P\) with respect to time is \(R(P)=\frac{dP}{dt}\). Substituting this definition of rate gives the differential equation.
Let \(P(t)\) denote the changing quantity at time \(t\).
The rate of change with respect to time is \( \frac{dP}{dt} \).
A rate proportional to \(P\) is a constant multiple \(kP\).
Therefore \( \frac{dP}{dt}=kP \).
This argument also explains why the model uses the current amount, not a fixed starting amount: the stated proportionality is to the amount present at each time \(t\). The constant \(k\) describes the rate per unit of current amount. Its sign must match the direction of change: a positive signed rate of increase gives \(k>0\), while a negative signed rate of decrease gives \(k<0\).
Worked Examples
Worked Example: A Population Grows by a Fixed Percentage Rate
A population \(P(t)\), measured in organisms, increases at an instantaneous rate of \(3\%\) of its current size per day. Translate the statement into a differential equation. Also find the instantaneous rate when the population is \(1200\) organisms.
First, identify the variables: \(P(t)\) is the population and \(t\) is time in days. The phrase “\(3\%\) of its current size per day” means that the rate is \(0.03\) times the current population. Convert the percentage to a decimal:
Because this is an increase, the rate is positive. The model is therefore
At \(P=1200\), substitute into the right-hand side:
Thus, at that moment the population increases at \(36\) organisms per day. Check the translation by comparing the rate with the current amount:
which is \(3\%\) per day as stated.
Worked Example: A Substance Decays at a Fixed Percentage Rate
An amount \(P(t)\), measured in grams, decreases at a rate whose magnitude is \(8\%\) of the amount currently present per hour. Write the differential equation. Then determine the rate of change when \(P=250\) grams.
The time variable \(t\) is measured in hours, and \(P(t)\) is the amount remaining. The magnitude of the rate of decrease is \(0.08P\). Since \(P\) is decreasing, its signed derivative is negative the magnitude of that decrease:
At \(P=250\) grams, substitution gives
So the amount is changing at \(-20\) grams per hour at that instant, meaning it is decreasing by \(20\) grams per hour. Check the relative rate:
The negative sign confirms a decrease, and the magnitude \(0.08\) is \(8\%\) per hour.
Worked Example: Find the Proportionality Constant from a Rate and Amount
At a particular time, a culture contains \(400\) cells and is increasing at \(60\) cells per hour. Suppose its rate of increase is proportional to its current amount, with the same proportionality constant throughout the model. Find \(k\) and write the differential equation.
Let \(P(t)\) be the cell count and \(t\) be measured in hours. The model has the form \( \frac{dP}{dt}=kP \). At the given moment, the rate is \(60\) cells per hour and the amount is \(400\) cells, so substitute these values:
Divide both sides by \(400\):
Therefore, the differential equation is
Check using the given amount: when \(P=400\), the equation gives \( \frac{dP}{dt}=0.15(400)=60 \) cells per hour, exactly the stated rate. The rate-to-amount ratio is \(60/400=0.15\) per hour, so the proportionality constant is consistent with the data.
Use the Direction of Change to Choose the Sign
The wording of a problem may describe a positive “rate of decrease.” That phrase often gives the magnitude of the decrease, not the signed derivative. The derivative \( \frac{dP}{dt} \) records signed change: it is negative when \(P\) is decreasing. Keep these two ideas separate.
| Verbal description | Signed model |
|---|---|
| Increases at a rate proportional to its current amount | \( \frac{dP}{dt}=kP \), with \(k>0\) |
| Decreases at a rate proportional to its current amount | \( \frac{dP}{dt}=-cP \), with \(c>0\) |
| Changes at a signed rate proportional to its current amount | \( \frac{dP}{dt}=kP \), with the sign of \(k\) determined by the direction |
In the second row, \(c\) is written as a positive magnitude, so the minus sign is displayed explicitly. Equivalently, the equation can be written \( \frac{dP}{dt}=kP \) with \(k=-c<0\). These are two notational choices for the same decrease model.
A Common Sign Mistake
Suppose an amount decreases at \(5\%\) of its current value per hour. A common incorrect translation is
This equation describes an increase, because if \(P=400\), it gives \( \frac{dP}{dt}=0.05(400)=20 \), a positive rate. The correction is to represent the signed derivative as negative:
Now, when \(P=400\), the equation gives \( \frac{dP}{dt}=-0.05(400)=-20 \). This means a decrease of \(20\) units per hour, and \(20/400=0.05\), the required \(5\%\) magnitude. The check should always compare both the sign and the rate magnitude with the verbal statement.
Proportional Change Is Not Constant Absolute Change
A constant absolute rate is a different assumption. “The amount decreases by \(12\) units per hour” means the derivative is \(-12\), regardless of the current amount:
It does not mean \( \frac{dP}{dt}=kP \) for one fixed constant \(k\). For instance, if the amount were \(100\), the rate-to-amount ratio would be \(-12/100=-0.12\); if the amount were \(200\), the ratio would be \(-12/200=-0.06\). Since those ratios differ, there is no single constant \(k\) that makes the rate proportional to the amount at both levels.
Likewise, a statement that a population grows by a fixed number of organisms per day describes a constant absolute rate, while a statement that it grows by a fixed percentage of its current size per day describes proportional change. Reading whether the rate is tied to the current amount is essential before choosing the equation.
Why This Model Is Useful
The equation \( \frac{dP}{dt}=kP \) is a compact way to express a broad modeling assumption used for populations and decaying quantities. It specifies how the rate responds to the amount currently present: doubling the amount doubles the instantaneous rate, while halving the amount halves that rate, with the same proportionality constant. Whether this assumption is appropriate depends on the situation and the time range being modeled; it is not automatically true of every population or substance.
When translating a new statement, pause to identify the amount and time variables, decide whether the described rate is proportional to the current amount, and determine whether the change is an increase or a decrease. If a percentage is given, convert it to a decimal before using it as a rate constant. The result of this step is the differential equation itself; finding a formula that satisfies it is a separate task.
Check Your Understanding
For each question, identify the changing quantity and time variable, then write or interpret the appropriate rate statement.
- A population increases at \(2.5\%\) of its current size per week. Write the differential equation and state the sign of its rate constant.
- A quantity decreases at a rate whose magnitude is \(6\%\) of its current amount per minute. Write the signed differential equation.
- At a certain moment, a culture has \(500\) organisms and increases at \(40\) organisms per hour. Under a proportional-change assumption, find \(k\) and write the model.
- Explain why “decreases by \(9\) grams per hour” does not, by itself, describe a rate proportional to the current amount.
- If \( \frac{dP}{dt}=-0.04P \) and \(P=300\) at an instant, what is the signed rate at that instant, and is \(P\) increasing or decreasing?
