From Several Rates to One Change
Many quantities change because several processes act at the same time. A tank may receive liquid through two pipes while a drain removes it. A store may receive deliveries while orders remove items. In each case, the rates do not all push the quantity in the same direction.
The central task is to combine those rates with their correct signs. Add rates that increase the quantity; subtract rates that decrease it. The resulting net rate describes the instantaneous direction and speed of change. Integrating that net rate over time gives the total change, which can then be added to the initial amount.
Every rate in the formula must refer to the same quantity and use compatible time units. For example, a rate in liters per hour cannot be added directly to one in liters per minute until the time units have been converted. When the rates are continuous or piecewise continuous, their definite integrals give the accumulated contributions over the interval.
Why the Net-Rate Formula Works
By definition, the net rate is the rate at which the quantity itself changes. Thus, wherever the rates describe the model, \(Q'(t)=R(t)\). The Fundamental Theorem of Calculus connects that instantaneous change to the total change between two times. Applying it gives the accumulation equation.
At each time, the quantity increases at the combined source rate and decreases at the combined sink rate, so \(Q'(t)=\sum_i S_i(t)-\sum_j L_j(t)\).
Integrating both sides from \(a\) to \(b\) gives \(\int_a^b Q'(t)\,dt=\int_a^b R(t)\,dt\).
The left side is \(Q(b)-Q(a)\). Therefore \(Q(b)-Q(a)=\int_a^b R(t)\,dt\).
By linearity of the definite integral, \(\int_a^b R(t)\,dt=\sum_i\int_a^b S_i(t)\,dt-\sum_j\int_a^b L_j(t)\,dt\).
This derivation explains why it is valid either to combine rates first and integrate once, or to integrate each source and sink separately and then combine the accumulated amounts. Both approaches produce the same net change, provided the rates use consistent units and cover the same time interval.
Worked Example: Two Inflows and a Drain
Worked Example: Two Inflows and a Drain
A tank initially contains \(12\) liters. For \(0\leq t\leq 2\) hours, two pipes add liquid at rates \(3+t\) liters per hour and \(2\) liters per hour. A drain removes liquid at rate \(1+t^2\) liters per hour. Find the net change and the amount in the tank after two hours.
Step 1: Form the net rate. Add the inflow rates and subtract the outflow rate:
Step 2: Integrate from \(0\) to \(2\).
Step 3: Add the change to the initial amount.
Check by integrating the separate rates. The first inflow contributes \(\int_0^2(3+t)\,dt=6+2=8\) liters, and the second contributes \(\int_0^2 2\,dt=4\) liters. The drain removes \(\int_0^2(1+t^2)\,dt=2+\frac{8}{3}=\frac{14}{3}\) liters. The net is \(8+4-\frac{14}{3}=\frac{36}{3}-\frac{14}{3}=\frac{22}{3}\) liters, matching the result. The final amount is therefore \(\frac{58}{3}\) liters.
The Sign of the Change Is Not the Sign of Every Rate
The net rate \(R(t)\) can be positive at some times and negative at others. A positive net rate means the quantity is increasing at that time; a negative net rate means it is decreasing. The sign of the integral over an interval determines the overall change across that interval. It does not require the net rate to have the same sign at every moment.
In particular, a net rate that becomes negative near the end of an interval can still have a positive integral if earlier positive contributions are larger. Conversely, the quantity can increase for part of an interval and still finish below its initial amount if the negative contributions are larger.
Worked Example: A Net Rate That Changes Sign
A warehouse initially has \(20\) kilograms of a material. Over \(0\leq t\leq 5\) minutes, deliveries arrive at \(6+t\) kilograms per minute and \(4\) kilograms per minute. Material is removed at rates \(3t\) kilograms per minute and \(2\) kilograms per minute. Find the final amount and explain the sign of the change.
Step 1: Combine all four rates.
Step 2: Find where the net rate changes sign. Solve \(8-2t=0\):
Thus, the net rate is positive for \(0\leq t<4\), zero at \(t=4\), and negative for \(4<t\leq 5\). The material increases during most of the interval and decreases during the final minute.
Step 3: Integrate to find the total change.
Step 4: Calculate the final amount.
Check by comparing accumulated deliveries and removals. The deliveries add \(\int_0^5(6+t)\,dt=30+\frac{25}{2}=\frac{85}{2}\) kilograms and \(\int_0^5 4\,dt=20\) kilograms. Total deliveries are \(\frac{85}{2}+\frac{40}{2}=\frac{125}{2}\) kilograms. The first removal takes \(\int_0^5 3t\,dt=\frac{75}{2}\) kilograms, and the second takes \(\int_0^5 2\,dt=10=\frac{20}{2}\) kilograms. Total removals are \(\frac{95}{2}\) kilograms. Their difference is \(\frac{125}{2}-\frac{95}{2}=\frac{30}{2}=15\) kilograms, as calculated. Although the net rate is negative after minute \(4\), the total change remains positive.
Keep the Initial Amount Separate from the Change
The definite integral of the net rate gives a change, not the final amount by itself. The initial quantity is a separate piece of information. Once the change has been calculated, use \(Q(b)=Q(a)+\Delta Q\). If the integral is negative, this addition reduces the initial quantity; it does not mean that the amount added or removed is negative.
Worked Example: A Positive Net Change from Four Rates
A container initially holds \(7\) gallons. For \(0\leq t\leq 2\) minutes, one source adds liquid at \(2t\) gallons per minute and another adds liquid at \(6\) gallons per minute. Two drains remove liquid at \(4\) gallons per minute and \(t^2\) gallons per minute. Find the change and the final amount.
Step 1: Form the net rate.
Step 2: Integrate over the stated interval.
Step 3: Add the initial amount.
Check by accumulating each stream. The two sources contribute \(\int_0^2 2t\,dt=4\) gallons and \(\int_0^2 6\,dt=12\) gallons, for a total of \(16\) gallons. The drains remove \(\int_0^2 4\,dt=8\) gallons and \(\int_0^2 t^2\,dt=\frac{8}{3}\) gallons, totaling \(\frac{32}{3}\) gallons. The net change is \(16-\frac{32}{3}=\frac{48}{3}-\frac{32}{3}=\frac{16}{3}\) gallons. Adding the initial \(7\) gallons gives \(\frac{37}{3}\) gallons.
A Common Sign and Arithmetic Pitfall
A frequent error is to add every rate, including rates that remove material. Another is to subtract the sink rates correctly in the formula but then mishandle fractions when combining their accumulated amounts. Labeling the source and sink totals separately helps expose both errors.
Worked Example: Correcting a Source–Sink Sign Error
Suppose total sources add \(24\) kilograms during an interval and total sinks remove \(16\) kilograms. The correct net change is \(24-16=8\) kilograms. A mistaken calculation might write \(24+16=40\) kilograms, treating removed material as if it were added. Correct the sign by identifying the direction of each contribution: sources are positive and sinks are negative. Thus, \(\Delta Q=24+(-16)=8\) kilograms.
When contributions are written with fractions, make a common denominator carefully. For instance, if sources add \(24\) kilograms and \(16\) kilograms while a sink removes \(\frac{64}{3}\) kilograms, then
The final quantity would be the initial quantity plus \(\frac{56}{3}\) kilograms. Do not reuse a numerator or denominator conversion from one term for another; first express each term independently with the common denominator, then combine.
Interpretation and Model Checks
There are several useful checks after calculating an accumulated amount:
- Units: Integrating a rate in kilograms per minute with respect to time in minutes gives kilograms. A result with rate units rather than quantity units signals a unit mismatch.
- Direction: If total source contributions exceed total sink contributions, the net change must be positive. If sinks exceed sources, it must be negative.
- Initial amount: The final amount must equal the initial amount plus the signed net change.
- Physical feasibility: If the model describes a nonnegative physical amount, a calculated negative final quantity may indicate that the assumed rates cannot continue over the entire interval or that the model needs an additional constraint.
These checks do not replace the calculation, but they can reveal a sign, arithmetic, unit, or modeling error. In applications, source and sink rates are usually given as nonnegative magnitudes; the signs enter when constructing the net rate. If a problem instead supplies signed rates, follow its stated sign convention and do not subtract an already negative rate a second time.
Check Your Understanding
For each question, keep the signs, interval, units, and initial amount distinct.
- A reservoir receives water from two inlets at \(5\) and \(t\) liters per minute, while a drain removes \(2+t^2\) liters per minute. Write the net-rate function.
- If a net rate \(R(t)\) is positive during the first part of an interval and negative during the last part, what determines whether the final amount exceeds the initial amount?
- A quantity begins at \(30\) kilograms and has a net change of \(-7\) kilograms. What is its final amount, and what does the negative sign describe?
- Why can the integrals of the source rates and sink rates be computed separately and then combined?
- A rate is given in liters per minute, but the interval is measured in hours. What must be checked or converted before integrating?
