From a Verbal Rate to an Equation
A differential equation can express a rule for how a quantity changes without giving an explicit formula for the quantity itself. In a rate model, the key translation is from a sentence about change to an equation involving a derivative.
For example, “the volume increases by 4 liters per minute” describes the rate at which volume changes over time. If \(V\) is volume and \(t\) is time, the sentence can be written as \(\frac{dV}{dt}=4\). The derivative identifies the changing quantity’s instantaneous rate with respect to the stated independent variable.
The goal here is to translate the words accurately, not to solve the resulting equation. Before writing one, identify the quantity that changes, what it changes with respect to, and what the rate depends on.
Identify the Variables First
In this topic, \(t\) will often represent time and \(y\) a quantity that changes over time. The dependent variable is the quantity whose value changes; the independent variable is the quantity with respect to which its rate is measured. Thus \(\frac{dy}{dt}\) means the rate of change of \(y\) with respect to \(t\).
Use symbols that make the context clear when possible: \(P\) for a population, \(V\) for a volume, or \(A\) for an amount of a substance. The derivative must match those choices. If \(P\) is a population measured over time, for instance, its rate is \(\frac{dP}{dt}\), not \(\frac{dt}{dP}\).
A rate is signed. A positive derivative means the dependent quantity is increasing as the independent variable increases; a negative derivative means it is decreasing. The words “increases” and “decreases” therefore matter when assigning a sign.
Constant Rates
“Increases at a constant rate” means that the derivative has a fixed positive value, provided the sentence refers to the rate with respect to the chosen independent variable. “Decreases at a constant rate” means that the derivative has a fixed negative value. A constant rate does not depend on the current amount or on time.
Worked Example: A Constant Increase in Volume
A container’s volume \(V\) increases at a constant rate of 12 liters per hour. Let \(t\) be time in hours. Translate the statement into a differential equation.
Step 1: Identify the variables. The changing quantity is \(V\), measured in liters. The independent variable is \(t\), measured in hours. The rate is therefore \(\frac{dV}{dt}\).
Step 2: Translate the rate. The volume increases, so the rate is positive. Its constant value is 12 liters per hour:
Check: The equation says that the volume’s rate of change is 12 liters per hour, exactly the stated rate. Its units are liters per hour, as required.
Worked Example: A Constant Decrease
A supply of liquid decreases at a constant rate of 3 milliliters per second. Let \(A\) be the amount in milliliters and \(t\) be time in seconds.
Step 1: Write the rate using the chosen variables. It is \(\frac{dA}{dt}\).
Step 2: Apply the sign and magnitude. The amount decreases, so the derivative is negative. The magnitude of the rate is 3 milliliters per second:
Check: The negative sign represents a decrease, and the magnitude 3 represents 3 milliliters per second. Thus the equation matches both parts of the sentence.
Rates Proportional to the Amount
“The rate is proportional to the amount” describes a different relationship from a constant rate. The rate changes when the amount changes: doubling the amount doubles the rate, assuming the same constant of proportionality.
By definition, if a quantity \(R\) is proportional to an amount \(y\), then \(R=ky\) for a constant \(k\). In a rate model, \(R\) is the derivative \(\frac{dy}{dt}\), so this proportionality gives a differential equation.
The constant \(k\) is called a constant of proportionality. Its units must make the equation consistent: if \(y\) is measured in people and \(t\) in hours, then \(\frac{dy}{dt}\) has units of people per hour, so \(k\) has units of inverse hours.
Worked Example: Population Growth Proportional to Population
A population \(P\), measured in bacteria, increases at a rate proportional to its current size. The rate is 0.3 times the current population per hour. Write a differential equation.
Step 1: Identify the rate. Since \(P\) changes over time \(t\), the rate is \(\frac{dP}{dt}\).
Step 2: Translate “proportional to its current size.” The current size is \(P\), and the proportionality constant is \(0.3\) per hour. Because the population increases, the sign is positive:
Step 3: Check the units and meaning. The right-hand side has units \((0.3\text{ per hour})(P\text{ bacteria})\), or bacteria per hour. If \(P=10\) bacteria at a particular instant, the equation gives the rate \(0.3(10)=3\) bacteria per hour at that instant. This is positive and proportional to the population, as stated.
Worked Example: A Decrease Proportional to the Amount
A substance’s amount \(A\) decreases at a rate proportional to the amount present. The constant of proportionality has magnitude 0.08 per minute. Write the equation using \(t\) in minutes.
Step 1: Represent the proportional rate. A rate proportional to \(A\) has magnitude \(0.08A\) amount-units per minute.
Step 2: Apply the direction of change. Because the amount decreases, the derivative must be negative:
Step 3: Check with a sample amount. If \(A=25\) amount-units at an instant, then the equation gives \(\frac{dA}{dt}=-0.08(25)=-2\) amount-units per minute. The negative value indicates a decrease, and its magnitude is proportional to the current amount.
Rates That Depend on Time and Amount
Some descriptions say that the rate depends on both the time and the current amount. In that case, represent the rate with a function of both variables: \(\frac{dy}{dt}=F(t,y)\). The function may combine contributions in different ways; it need not be a product of \(t\) and \(y\).
For example, a container might receive liquid at a rate that changes with time while also losing liquid at a rate proportional to its current volume. The net rate is the inflow minus the outflow. Adding rates with their signs is a useful way to translate such descriptions.
Worked Example: Time-Varying Inflow and Amount-Dependent Leakage
A tank contains \(V\) liters after \(t\) minutes. Liquid enters at \(4+0.5t\) liters per minute, and liquid leaks out at a rate of \(0.08V\) liters per minute. Write the differential equation for the tank’s volume.
Step 1: Identify the rate being modeled. The changing quantity is \(V\), so the net rate is \(\frac{dV}{dt}\).
Step 2: Record the two contributions. Inflow is positive and equals \(4+0.5t\). Outflow reduces the tank’s volume, so its contribution is \(-0.08V\).
Step 3: Add the signed rates.
Step 4: Check units and signs. The term \(4\) has units of liters per minute. The coefficient \(0.5\) has units of liters per minute squared, so \(0.5t\) has units of liters per minute when \(t\) is in minutes. The coefficient \(0.08\) has units of inverse minutes, so \(0.08V\) has units of liters per minute. For example, at \(t=6\) minutes and \(V=20\) liters, the inflow is \(4+0.5(6)=4+3=7\) liters per minute, and the leak is \(0.08(20)=1.6\) liters per minute. The net rate is \(7-1.6=5.4\) liters per minute. This agrees with the equation, which gives \(4+0.5(6)-0.08(20)=4+3-1.6=5.4\).
A Reliable Translation Process
When a verbal model is more involved, translate it in a fixed order. This helps keep the derivative, dependencies, signs, and units consistent.
Choose a symbol, such as \(y\), \(P\), or \(V\), and state what it measures.
If the rate is measured over time, choose \(t\) and write the rate as \(\frac{dy}{dt}\).
A fixed rate is constant; a rate proportional to the amount includes the amount as a factor; a rate depending on time and amount uses both variables.
Increases and inflows contribute positively; decreases and outflows contribute negatively.
Every term on the right must have the units of the derivative, and the equation should say the same thing as the original sentence.
Common Mistakes and Why the Checks Matter
A common error is to omit the negative sign when translating a proportional decrease. For example, the statement “\(A\) decreases at a rate proportional to \(A\), with constant \(k>0\)” is sometimes incorrectly written as \(\frac{dA}{dt}=kA\). That equation gives a positive rate whenever \(A>0\), so it describes an increase, not a decrease. The correction is:
Another error is to treat every stated rate as constant. “The rate is proportional to the amount” does not mean \(\frac{dy}{dt}=c\) for a fixed \(c\); it means \(\frac{dy}{dt}=ky\). Conversely, a constant rate does not acquire a factor of \(y\) unless the description says that the rate depends on the amount.
Initial information, such as “the tank contains 20 liters at time \(t=0\),” is separate from the rate rule. It can be written as \(V(0)=20\), but it does not change how the verbal rate is translated into a differential equation. The equation describes the rate rule; an initial condition states a particular starting value.
Check Your Understanding
For each item, identify the dependent and independent variables, then write the differential equation. Include signs and units where they are specified.
- A population \(P\) decreases at a constant rate of 50 people per year. Let \(t\) be measured in years.
- An amount \(A\) increases at a rate proportional to its current amount, with positive constant of proportionality \(k\). Write the equation.
- A quantity \(y\) decreases at a rate proportional to its current amount, with constant of proportionality magnitude \(c>0\). Write the equation.
- A tank’s volume \(V\) receives water at 6 liters per minute and loses water at a rate of \(0.2V\) liters per minute. Write the equation for net change.
- In your own words, why does a model whose rate depends on both time \(t\) and amount \(y\) have the general form \(\frac{dy}{dt}=F(t,y)\)?
