Accumulated Rates and Instantaneous Change

From a Rate to an Accumulated Quantity

A rate describes how quickly a quantity changes. For example, a pump’s rate might be measured in litres per minute, while the amount of water in a tank is measured in litres. If the rate varies over time, the total change depends on the rate throughout the interval, not just on its value at the end.

When the rate is continuous, integration gives a precise model for this accumulation. The first part of the Fundamental Theorem of Calculus then says that differentiating the accumulated quantity recovers the rate at that time. This makes the theorem useful both as a mathematical result and as a way to interpret real processes.

The variable \(s\) is a placeholder for time inside the integral; \(t\) is the time at which we evaluate the accumulated quantity. The initial value \(Q(t_0)\) must be included when the model describes the quantity itself. The integral alone represents the change from \(t_0\), not the starting amount.

The Accumulation Derivative

The first part of the Fundamental Theorem of Calculus gives the central relationship. If \(r\) is continuous and \(Q(t)=Q(t_0)+\int_{t_0}^{t}r(s)\,ds\), then at an interior time \(t\), \(Q'(t)=r(t)\). In words, the instantaneous rate of change of the accumulated quantity is the rate being accumulated at that instant.

This result has a direct interpretation in a rate model. The quantity \(Q(t)\) describes how much has accumulated by time \(t\). Its derivative \(Q'(t)\) describes how quickly that amount is changing at that moment. Because the rate is continuous, the nearby values of \(r\) are close to \(r(t)\), so over a sufficiently short time interval the rate is nearly constant at \(r(t)\).

Why Differentiation Recovers the Rate

Here is a derivation of the theorem’s central claim. Write the accumulation function as \(Q(t)=Q(t_0)+\int_{t_0}^{t}r(s)\,ds\). For a nonzero increment \(h\), subtract its value at \(t\) from its value at \(t+h\). The fixed starting amount cancels, leaving the integral over the short interval from \(t\) to \(t+h\).

To see why this average approaches \(r(t)\), subtract \(r(t)\) from the difference quotient. Continuity means that for any \(\varepsilon>0\), there is a \(\delta>0\) such that \(|r(s)-r(t)|<\varepsilon\) whenever \(|s-t|<\delta\). If \(0<|h|<\delta\), every point between \(t\) and \(t+h\) satisfies that bound. Therefore, using the absolute-value bound for an integral,

Thus the difference quotient approaches \(r(t)\) as \(h\) approaches zero. By the definition of the derivative, \(Q'(t)=r(t)\). The argument works for positive or negative \(h\); the integral’s orientation accounts for the direction of the increment.

Worked Examples

Units and Signs in Rate Models

Units provide a useful check on an accumulation model. If \(Q\) is measured in litres and \(t\) in minutes, then \(Q'(t)\) and \(r(t)\) are measured in litres per minute. In the integral, the time element \(ds\) contributes minutes, so \(r(s)\,ds\) has units of litres. The integral therefore has the same units as a change in \(Q\).

The rate can be positive, zero, or negative. A positive rate means the quantity is increasing at that instant; a negative rate means it is decreasing. A zero rate means there is no instantaneous change at that instant. The sign of the rate describes direction, while its magnitude describes how fast the change is occurring.

The integral accumulates net change. If a rate is positive during part of an interval and negative during another part, the contributions have opposite signs. The resulting net change need not equal the total amount of movement in both directions.

A Common Mistake: Adding the Initial Amount to the Rate

A frequent error is to carry the initial quantity into the derivative. Suppose \(Q(t)=20+\int_0^t(5+2s)\,ds\), where \(Q\) is measured in litres and \(t\) in minutes. The initial \(20\) litres belongs to the amount model, but it is constant. Its derivative is zero.

The incorrect calculation is \(Q'(t)=20+(5+2t)=25+2t\). This incorrectly adds a quantity measured in litres to a rate measured in litres per minute. The correct calculation is

At \(t=3\), the correct instantaneous rate is \(5+2(3)=11\) litres per minute, not \(31\). Differentiation removes the fixed initial amount and recovers only the current rate.

Why This Connection Matters

Many quantities are understood through their rates: the amount of water entering a tank, the distance travelled by a moving object, or the amount of material produced by a process. When a continuous rate is known, its integral describes the accumulated change. Conversely, when a quantity is defined by accumulating a continuous rate, its derivative returns the rate at each time.

This is the practical meaning of the first part of the Fundamental Theorem of Calculus. It connects a process measured over an interval with what is happening at a particular instant. When applying it, identify the accumulated quantity, its initial value, the rate, and the units. Then use continuity of the rate to conclude that the derivative of the accumulation is the rate itself.