From a Changing Quantity to a Model
A differential-equation model begins with a quantity that changes and a description of how quickly it changes. The first task is to make that description precise: define the quantity, define the independent variable, include units, and express the stated rate as a derivative.
For example, if \(V\) is a volume measured in liters and \(t\) is time measured in minutes, then \(\frac{dV}{dt}\) has units of liters per minute. A statement about the volume’s rate of change is a statement about this derivative.
Once the equation and initial condition are set, the initial value identifies the point \((t_0,y_0)\) where the solution path begins. A supplied slope field gives the local direction of that path: at each point, its short line segment has slope \(F(t,y)\). Sketching means following those local directions from the initial point, not guessing a formula for the entire solution.
A Reliable Modeling Process
Separate the scenario into two questions: What quantity is changing? and What determines its rate of change? If several processes add to or remove from a quantity, account for each one with the correct sign.
Choose the independent variable, such as time \(t\), and name the changing quantity, such as volume \(V(t)\). State what the units measure.
Write the derivative of the changing quantity. Express the stated rate using the variables and parameters in the scenario.
Add rates that increase the quantity and subtract rates that decrease it. Check that every term has the same rate units.
Mark the initial point, then sketch a curve whose tangent direction agrees with the field along its path.
Why the Slope Field Gives the Path’s Direction
Suppose a model is written as \(\frac{dy}{dt}=F(t,y)\), and a differentiable solution passes through \((t_0,y_0)\). By the definition of derivative, the slopes of secant lines from that point approach the tangent slope:
Therefore, the field’s segment at \((t_0,y_0)\) has exactly the slope of the solution’s tangent there. The same reasoning applies at every point on the solution. This is why a solution curve follows the field. A sketch is an approximation: it should match the direction segments, but it need not identify an exact formula.
Worked Example: Inventory with a Changing Order Rate
Worked Example: Inventory with a Changing Order Rate
A shop begins with 20 boxes in stock. It receives deliveries at a constant rate of 9 boxes per hour. Customer orders are being filled at a rate of \(2+t\) boxes per hour, where \(t\) is the number of hours since tracking began. Build the model and use the supplied field to describe the inventory path.
Let \(I(t)\) be the inventory in boxes and \(t\) the time in hours. The derivative \(\frac{dI}{dt}\) has units of boxes per hour. Deliveries increase stock; filled orders decrease it. Thus,
The initial condition places the path at \((0,20)\). To read its direction, compute the field slopes at several times:
These slopes are in boxes per hour. Since the slope is positive before \(t=7\), the path rises from \((0,20)\). Its positive slope gets smaller, so it becomes less steep as it rises. At \(t=7\), its slope is zero. After that, the field points downward, so the path begins to fall. The sketch should therefore rise, flatten at its turning point, and then descend.
Check: Substituting the initial time into the model gives \(I'(0)=7-0=7\) boxes per hour, so the path leaves \((0,20)\) with positive slope, as described. The delivery and order rates are both in boxes per hour, so their difference has the correct units for \(\frac{dI}{dt}\).
Worked Example: Water Volume with a Proportional Leak
Worked Example: Water Volume with a Proportional Leak
A tank receives water at 12 liters per minute. At the current volume \(V\) liters, the leak removes water at \(0.25V\) liters per minute. Initially, the tank contains 32 liters. Build the differential equation and describe the path indicated by the supplied field.
Let \(V(t)\) be the volume in liters and \(t\) be time in minutes. Inflow adds water and leakage removes it, so the net rate is
The coefficient \(0.25\) has units of inverse minutes: multiplying it by a volume in liters gives a rate in liters per minute. The initial point is \((0,32)\). At that volume, the leak rate is \(0.25(32)=8\) liters per minute, so the initial net rate is \(12-8=4\) liters per minute.
To read the field at different volumes, calculate:
Starting at 32 liters, the positive slopes direct the path upward. As the volume rises toward 48 liters, the slopes become less steep. At 48 liters the field is horizontal. Above 48 liters the slopes are negative. Thus, from this initial point, sketch an increasing path that flattens as it approaches the horizontal direction near 48 liters. Do not extend that conclusion to initial values above 48 liters; there the field points downward.
Check: At \(V=32\), the equation gives \(V'(0)=12-0.25(32)=4\) liters per minute, matching the separate inflow-minus-leak calculation. Setting the net rate to zero gives \(12-0.25V=0\), so \(0.25V=12\) and \(V=48\) liters, consistent with the zero slope in the field.
Worked Example: Snow Accumulating at a Changing Rate
Worked Example: Snow Accumulating at a Changing Rate
At the start of an observation, snow depth is 10 centimeters. During the observation, snow accumulates at a rate of \(2+0.5t\) centimeters per hour, where \(t\) is time in hours. Build the model and describe the solution path in the supplied field.
Let \(S(t)\) be snow depth in centimeters. The stated accumulation rate is already the rate of change of depth, so
The rate is in centimeters per hour: \(2\) has units of centimeters per hour, and the coefficient \(0.5\) has units of centimeters per hour squared, so \(0.5t\) is also in centimeters per hour. At selected times, the field slopes are
These slopes do not depend on \(S\), so at any fixed time the field segments have the same slope across different depths. Starting at \((0,10)\), the solution rises because the slopes are positive. It becomes steeper as time increases because the rate increases from 2 to 3 to 4 centimeters per hour. Sketch an increasing, steepening path through the initial point.
Check: At \(t=0\), the model gives \(S'(0)=2+0.5(0)=2\) centimeters per hour, which agrees with the stated starting accumulation rate. The initial value \(S(0)=10\) places the path at the correct depth.
Common Pitfall: Reversing an Inflow and an Outflow
In the tank example, a tempting but incorrect model is \(\frac{dV}{dt}=12+0.25V\). This adds the leak rate to the inflow, as if leakage increased the volume. At \(V=32\), that calculation gives \(12+0.25(32)=12+8=20\) liters per minute, even though the tank gains 12 liters per minute and loses 8 liters per minute.
The correction is to use net change: inflow minus outflow. Thus \(\frac{dV}{dt}=12-0.25V\), and at \(V=32\) the rate is \(12-8=4\) liters per minute. Before sketching, check whether each process adds to or removes from the quantity.
What to Check Before Sketching
A coherent model and sketch should agree in three ways:
- Units: The derivative and every term on its right-hand side must have units of quantity per unit time.
- Initial point: The initial condition must locate the starting point on the graph.
- Direction: A positive field slope means the quantity increases as time advances; a negative slope means it decreases; a zero slope means the tangent is horizontal at that point.
Also distinguish a statement about the rate from one about the quantity itself. “The quantity increases by 5 units per minute” describes its derivative. “The quantity is 5 units” describes its value. Mixing these meanings produces an equation with the wrong interpretation, and often the wrong units.
Building a model is useful because it turns a verbal description into a mathematical statement that can be checked against a field and an initial value. At this stage, the goal is to write the rate equation correctly and sketch a path consistent with it. The field provides local direction information; it does not by itself give an exact formula for the solution.
Check Your Understanding
For each question, identify the variables, rate relationship, and initial point where appropriate.
- A container holds 15 liters at \(t=0\). Water enters at 6 liters per minute and exits at 2 liters per minute. Write a differential equation with an initial condition.
- A quantity \(Q(t)\) is decreasing at a rate of \(0.3Q\) units per hour. Write the differential equation and state the units required for the coefficient \(0.3\).
- A field has negative slopes along a solution path for a time interval. What does that tell you about the quantity as time advances on that interval?
- A tank receives 10 liters per minute and loses 3 liters per minute. A student writes a net rate of \(10+3=13\) liters per minute. Identify and correct the error.
- For \(\frac{dS}{dt}=1+t\) and \(S(0)=4\), calculate the field slopes at \(t=0\) and \(t=2\), then describe how the sketch should change in steepness.
