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Recovering a Quantity from Its Rate

Use a rate function and one known value to reconstruct the quantity at any time in its interval.

Intermediate 10 min read

What You'll Learn

  • Express a known rate as the derivative of the quantity being reconstructed
  • Set up a definite integral from the time with known quantity to the desired time
  • Use the Fundamental Theorem of Calculus to evaluate the integral
  • Recover a position, amount, or other quantity from an initial value
  • Check a reconstructed function by differentiating it and applying its initial condition
  • Identify why a rate alone does not determine the quantity’s absolute value

From a Rate to the Quantity

A rate describes how quickly a quantity changes, but it does not by itself tell us the quantity’s value. For example, knowing a moving object’s velocity tells us how its position changes; it does not reveal where the object started. To recover the position, we also need one known position at a particular time.

The same idea applies to many changing quantities. If we know a rate function and the value of the quantity at a reference time, integration lets us determine the quantity at other times. The initial value supplies the information that the rate alone is missing.

Key idea: The quantity at a target time equals its known value at a reference time plus the integral of its rate from the reference time to the target time.

Rate, Initial Value, and Reconstructed Quantity

Let \(q(t)\) be a quantity that is differentiable on an interval, and let \(r(t)\) be its rate of change. By definition, \(q'(t)=r(t)\). Suppose we know \(q(a)=q_a\), where \(a\) is a time in the interval. If \(r\) is continuous between \(a\) and a target time \(t\), the Fundamental Theorem of Calculus gives:

$$ q(t)=q(a)+\int_a^t r(u)\,du. $$

The variable \(u\) is a dummy variable of integration. It represents the time over which the rate is accumulated; \(t\) remains the target time. The integral measures the change in \(q\) between \(a\) and \(t\), while \(q(a)\) sets the starting value.

Definition: Given a continuous rate \(r\) and an initial condition \(q(a)=q_a\), the quantity reconstructed from the rate is \(q(t)=q_a+\int_a^t r(u)\,du\), for target times \(t\) in the interval where the rate is defined and continuous between \(a\) and \(t\).

The reference time does not have to be \(0\). It is the time at which the quantity’s value is known. The target time may be greater or less than \(a\); the definite integral uses its usual signed bounds. This formula is a way to recover the quantity, not merely a way to find an antiderivative of the rate.

Why the Formula Works

The Fundamental Theorem of Calculus connects an accumulation function to its integrand. Define \(Q(t)=q_a+\int_a^t r(u)\,du\). If \(r\) is continuous, the theorem says that the derivative of the integral with respect to its upper bound is \(r(t)\). Therefore \(Q'(t)=r(t)\), and at the reference time the integral is zero, so \(Q(a)=q_a\). Thus \(Q\) has the required rate and the required initial value.

We can also show that these conditions determine only one quantity. Suppose \(q_1\) and \(q_2\) both have derivative \(r\) and both equal \(q_a\) at \(a\). Their difference \(d(t)=q_1(t)-q_2(t)\) has derivative \(d'(t)=r(t)-r(t)=0\). A function with zero derivative on an interval is constant there, and \(d(a)=q_a-q_a=0\). Hence that constant is zero, so \(q_1(t)=q_2(t)\) throughout the interval. The initial value is what fixes the constant that an antiderivative alone cannot determine.

Theorem: If \(r\) is continuous on an interval containing \(a\) and \(t\), then \(q(t)=q_a+\int_a^t r(u)\,du\) has derivative \(r(t)\) and satisfies \(q(a)=q_a\). It is the unique differentiable function on that interval with those properties.

Applying the Formula

A reliable setup keeps the reference time and target time visible. First identify the known value \(q(a)\). Next write the rate \(r(u)\) inside the integral, with lower bound \(a\) and upper bound \(t\). Evaluate the integral, add the known value, and check both the derivative and the initial condition.

1
Identify the quantities.
Write the rate function \(r(t)\), the reference time \(a\), and the known value \(q(a)=q_a\).
2
Set up the accumulation.
Use \(q(t)=q_a+\int_a^t r(u)\,du\), keeping the bounds in the order reference time to target time.
3
Evaluate and verify.
Find the integral, simplify the expression for \(q(t)\), then check that its derivative is \(r(t)\) and that its value at \(a\) is \(q_a\).

Worked Examples

Worked Example: Recovering Position from Velocity

An object has velocity \(v(t)=6t-2\) meters per second for \(t\geq 0\). Its position at \(t=0\) is \(s(0)=3\) meters. Find its position function and its position at \(t=4\) seconds.

Here the rate of position is velocity, so \(s'(t)=v(t)=6t-2\). Apply the reconstruction formula with reference time \(0\):

$$ s(t)=3+\int_0^t(6u-2)\,du. $$

An antiderivative of \(6u-2\) is \(3u^2-2u\). Evaluating at the bounds gives:

$$ \begin{aligned} s(t) &=3+\left[3u^2-2u\right]_0^t\\ &=3+\bigl(3t^2-2t\bigr)-\bigl(3(0)^2-2(0)\bigr)\\ &=3+3t^2-2t. \end{aligned} $$

At \(t=4\), the position is:

$$ s(4)=3+3(4)^2-2(4)=3+48-8=43\text{ meters}. $$

Check the reconstruction: \(s'(t)=6t-2=v(t)\), and \(s(0)=3+3(0)^2-2(0)=3\) meters. Both the velocity and the initial position agree with the information given.

Worked Example: Using an Initial Value at a Nonzero Time

A tank contains \(10\) liters of water at \(t=2\) minutes. Its volume changes at the rate \(r(t)=4-\frac{1}{2}t\) liters per minute. Find the volume at \(t=6\) minutes, assuming the rate applies throughout the interval from \(2\) to \(6\).

Let \(V(t)\) be the volume. The known value is \(V(2)=10\), so integrate from \(2\), not from \(0\):

$$ V(6)=10+\int_2^6\left(4-\frac{1}{2}u\right)\,du. $$

An antiderivative of \(4-\frac{1}{2}u\) is \(4u-\frac{u^2}{4}\). Evaluate at both endpoints:

$$ \begin{aligned} \int_2^6\left(4-\frac{1}{2}u\right)\,du &=\left[4u-\frac{u^2}{4}\right]_2^6\\ &=\left(4(6)-\frac{6^2}{4}\right)-\left(4(2)-\frac{2^2}{4}\right)\\ &=(24-9)-(8-1)\\ &=15-7\\ &=8\text{ liters}. \end{aligned} $$

Therefore \(V(6)=10+8=18\) liters. For a function on the interval, the same calculation gives \(V(t)=10+\left[4u-\frac{u^2}{4}\right]_2^t\), which simplifies to \(V(t)=3+4t-\frac{t^2}{4}\). Check: \(V'(t)=4-\frac{1}{2}t=r(t)\), and \(V(2)=3+8-1=10\) liters. The reconstructed function satisfies both conditions.

Worked Example: Finding a Value Before the Reference Time

A quantity \(q(t)\) has rate \(q'(t)=\cos t\), and \(q\left(\frac{\pi}{2}\right)=5\). Find \(q(0)\) and a formula for \(q(t)\).

The known time is \(a=\frac{\pi}{2}\). Keep it as the lower bound even though the target time \(0\) is earlier:

$$ q(t)=5+\int_{\pi/2}^t\cos u\,du. $$

An antiderivative of \(\cos u\) is \(\sin u\). Thus:

$$ \begin{aligned} q(t) &=5+\left[\sin u\right]_{\pi/2}^t\\ &=5+\sin t-\sin\left(\frac{\pi}{2}\right)\\ &=5+\sin t-1\\ &=4+\sin t. \end{aligned} $$

At \(t=0\), \(q(0)=4+\sin(0)=4\). Check: \(q'(t)=\cos t\), and \(q\left(\frac{\pi}{2}\right)=4+\sin\left(\frac{\pi}{2}\right)=5\). The function therefore has the stated rate and reference value.

A Common Mistake: Forgetting the Initial Value

An antiderivative of the rate gives a family of possible quantities, not necessarily the quantity that matches the stated initial condition. For instance, suppose \(s'(t)=2t\) and \(s(0)=5\). A common incorrect step is to write \(s(t)=t^2\), treating the antiderivative as the complete answer. This produces \(s(0)=0\), not \(5\), and gives the incorrect position \(s(3)=9\).

The correct reconstruction includes the initial value:

$$ s(t)=5+\int_0^t 2u\,du =5+\left[u^2\right]_0^t =5+t^2. $$

Now \(s'(t)=2t\), \(s(0)=5\), and \(s(3)=5+9=14\). The derivative check confirms the rate; the initial-value check confirms the constant.

Takeaway: Integrating the rate determines how the quantity changes. The known value at the reference time determines its starting level. Use both pieces of information.

Check Your Understanding

For each question, use the rate and initial value together, and check your result where appropriate.

  1. If \(q'(t)=3t^2\) and \(q(1)=4\), write a formula for \(q(t)\).
  2. A position has velocity \(v(t)=5\) meters per second and \(s(2)=7\) meters. Use an integral to find \(s(5)\).
  3. If \(q'(t)=\sin t\) and \(q(0)=2\), find \(q(t)\) and verify the initial condition.
  4. Why does an antiderivative of a rate, without an initial value, fail to determine one unique quantity?
  5. In the formula \(q(t)=q(a)+\int_a^t r(u)\,du\), what does the lower bound \(a\) represent?

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