From a Known Net Change to an Unknown Rate Parameter
A rate function describes how quickly a quantity changes, but its formula may contain an unknown constant. If the net change over a time interval is known, integration converts that information into an equation for the constant.
The key is to integrate the rate over the stated interval. The value of the rate at one instant is not the same as the total change over an interval: accumulated change depends on the rate throughout the interval.
Set Up the Equation Carefully
Suppose \(r(t;k)\) is a rate function with unknown parameter \(k\), and the net change from \(t=a\) to \(t=b\) is \(D\). The parameter must satisfy
This equation uses net change, not necessarily total distance or total amount of activity. If the rate is negative over part of the interval, that part contributes negatively to the integral. The bounds also matter: the integral is accumulated from the stated starting time to the stated ending time.
For a continuous rate function, the Fundamental Theorem of Calculus justifies this relationship. If \(Q'(t)=r(t;k)\), then the theorem gives
Thus, the information about a change in \(Q\) becomes an equation involving \(k\). The equation may be linear or nonlinear in the parameter. It can have one solution, several solutions, or no solution consistent with the model's constraints.
Worked Example: A Linear Parameter in the Rate
Worked Example: A Linear Parameter in the Rate
A tank's volume changes at rate \(r(t)=kt+2\) liters per minute, where \(t\) is measured in minutes and \(k\) is an unknown constant. The volume increases by \(15\) liters from \(t=0\) to \(t=3\). Find \(k\).
The stated increase is the net change, so set the integral of the rate equal to \(15\):
Integrate each term and evaluate at both bounds:
Now solve the equation:
Check by substituting \(k=2\) into the accumulated-change expression:
The parameter is \(k=2\). Since \(kt\) and \(2\) are both rates in liters per minute, \(k\) has units of liters per minute squared. The checked integral gives the required \(15\)-liter net increase.
Why the Integral Produces the Parameter Equation
The calculation can be understood directly from the Fundamental Theorem of Calculus. Assume the rate is continuous on \([a,b]\), and let \(Q\) be a quantity whose derivative is that rate. Then \(Q(b)-Q(a)\) is the difference between the ending and starting values. The theorem says that this difference equals the integral of the rate.
For example, suppose a family of rate functions has the form \(r(t;k)=g(t)+k h(t)\), where \(g\) and \(h\) are continuous functions and \(k\) is a constant. By the linearity of the definite integral,
Define \(G=\int_a^b g(t)\,dt\) and \(H=\int_a^b h(t)\,dt\). If the known net change is \(D\), the condition becomes \(G+kH=D\). When \(H\ne 0\), it has the unique solution \(k=(D-G)/H\). When \(H=0\), the accumulated change is always \(G\): there is no solution if \(D\ne G\), and every allowed \(k\) satisfies the condition if \(D=G\).
This explains why a net-change condition often determines a parameter directly: integration turns the rate model into a scalar equation. If the parameter occurs nonlinearly, integration still gives an equation, but solving it may require considering multiple candidate values.
Worked Example: A Parameter in a Quadratic Rate
Worked Example: A Parameter in a Quadratic Rate
A quantity changes at rate \(r(t)=a(t-2)^2-1\) units per minute, where \(t\) is measured in minutes and \(a\) is unknown. Its net change from \(t=0\) to \(t=4\) is \(4\) units. Find \(a\).
Use the stated interval and set the accumulated rate equal to the net change:
An antiderivative is \(\frac{a(t-2)^3}{3}-t\). Evaluate it at the upper and lower bounds:
Solve for \(a\):
Substitute \(a=\frac{3}{2}\) back into the integral to check:
The answer satisfies the accumulated-change condition. In this model, \(a\) has units of units per minute cubed, so that \(a(t-2)^2\) has units of units per minute, matching the rate.
Nonlinear Equations Can Have Multiple Answers
A parameter does not have to appear linearly in the rate. After integration, the equation may be quadratic or otherwise nonlinear. Every candidate must be checked, and any restrictions on the parameter must be applied.
Worked Example: A Nonlinear Parameter with Two Solutions
A rate is \(r(t)=\bigl((k-1)^2+1\bigr)\) units per hour for \(0\le t\le2\), where \(k\) is dimensionless. The net change over the two-hour interval is \(6\) units. Find all possible values of \(k\).
The rate is constant with respect to \(t\), so integrate it over the given interval:
Set this equal to \(6\) and solve:
Check both candidates in the rate. For either value, \(k-1\) is either \(\sqrt{2}\) or \(-\sqrt{2}\), so \((k-1)^2=2\). The rate is therefore \(2+1=3\) units per hour for either candidate. Its integral over two hours is
Both values, \(k=1+\sqrt{2}\) and \(k=1-\sqrt{2}\), satisfy the condition. If the model additionally required \(k\ge 0\), then \(1-\sqrt{2}<0\) would be excluded, leaving only \(k=1+\sqrt{2}\). Without such a restriction, both are valid.
Common Pitfall: Using a Rate Difference Instead of an Integral
A rate has units of quantity per unit of time. Subtracting its values at two endpoints gives a change in the rate, not the accumulated change in the original quantity. The net change comes from integrating across the interval.
For the first example, an incorrect setup might use the endpoint rates and assert
Since \(r(3)=3k+2\) and \(r(0)=2\), this mistake gives \(3k=15\), or \(k=5\). But that value does not produce the stated net change. The correct check is
The correction is to integrate the rate over the full interval before solving. The integral accounts for the rate at every time, with its sign, and has units of quantity rather than quantity per time.
A Reliable Procedure
When a rate model contains an unknown parameter and a net change is given, organize the work in this order:
Record the rate, the starting and ending times, and the stated net change. Keep the direction of the change and the units clear.
Set the definite integral of the rate from the starting time to the ending time equal to the known net change.
Keep the parameter in the expression until the definite integral has been evaluated, then solve the resulting equation.
Substitute each candidate into the accumulated-change equation and apply any stated restrictions on the parameter.
The procedure also helps catch inconsistent information. For example, a nonlinear equation may have no real solutions, or its solutions may violate a stated parameter restriction. In that case, the condition and the model do not yield an allowable parameter.
Check Your Understanding
For each question, begin by writing the definite-integral equation. Then solve and check the result by substitution.
- A rate is \(r(t)=ct+1\) units per minute. The net change from \(t=0\) to \(t=2\) is \(6\) units. Find \(c\).
- A rate is \(r(t)=b(t-1)^2\) units per hour. The net change from \(t=0\) to \(t=2\) is \(4\) units. Find \(b\).
- If \(r(t;k)=g(t)+kh(t)\), what condition on \(\int_a^b h(t)\,dt\) ensures that a specified net change determines exactly one value of \(k\)?
- Why is \(r(b)-r(a)\) generally not the net change in the quantity whose rate is \(r(t)\)?
- If a nonlinear parameter equation produces two values, what two checks should be made before reporting the answers?
