From a Rate History to a Threshold Time
A threshold question asks when an accumulated quantity first attains a specified target. Examples include the time when an inventory reaches a required level, when production totals a given amount, or when a queue grows to a chosen size. The target is not reached merely because the rate has been positive for some time: what matters is the net accumulation since the starting time.
To find the time, first express the quantity using its initial value and its rate. Then set that expression equal to the target and identify the earliest time satisfying the equation. This last step matters: an equation may have more than one solution, and only the earliest solution is the first threshold time.
The Accumulation Model
Let \(A(t)\) be an accumulated quantity at elapsed time \(t\), measured from \(t=0\). Let \(A_0\) denote its starting value, so \(A(0)=A_0\), and let \(r(t)\) denote its signed rate of change. A positive rate increases the quantity; a negative rate decreases it. If \(r\) is continuous, or piecewise continuous, on the time interval under consideration, the accumulation model is
The integral gives the net change from time \(0\) to time \(t\). It accounts for increases and decreases; it is not necessarily the total amount of all inflows or all production if the rate can be negative. The letter \(s\) is a variable of integration, while \(t\) is the time at which we evaluate the accumulated quantity.
If \(B=A_0\), the first threshold time is \(t=0\). In the common case \(B>A_0\), the quantity must gain a net amount \(B-A_0\) before reaching the target. Thus the threshold equation is
This equation makes the role of the initial value explicit. The target \(B\) is the desired final amount; the required accumulation is only the gap between the target and the starting amount.
Why the Integral Gives the Change
The accumulation formula follows from the Fundamental Theorem of Calculus. If \(r\) is continuous and \(A\) is differentiable with \(A'(t)=r(t)\), then the theorem gives
Since \(A'(s)=r(s)\) and \(A(0)=A_0\), this becomes
So the threshold equation is not a separate rule: it is the accumulation model evaluated at a time when \(A(t)=B\). If the rate is piecewise continuous, the same formula applies by integrating over each piece and adding the results.
Existence and Uniqueness of a First Threshold Time
Solving the equation algebraically is only part of the task. A target may not be reached within the model’s time interval, and there may be more than one time at which it is reached. A useful simple case is a strictly positive rate. Suppose \(r\) is continuous and \(r(t)>0\) for every \(t\in[0,T]\). Then \(A\) is continuous and strictly increasing: for \(0\leq t_1<t_2\leq T\),
Consequently, a target \(B>A_0\) is reached on \([0,T]\) exactly when \(B\leq A(T)\), and if it is reached, it is reached only once. The existence part follows from continuity and the Intermediate Value Theorem: \(A(0)=A_0<B\), and \(A(T)\geq B\), so \(A(t)=B\) for some \(t\in[0,T]\). Strict increase gives uniqueness.
If the rate can be zero or negative, the accumulation need not be strictly increasing. The integral equation still describes every time the target is attained, but it may have no solution or multiple solutions. In that case, restrict attention to the model’s time interval, solve for all valid candidate times, and select the smallest. A candidate outside the interval, such as a negative elapsed time, is not a valid threshold time for a model that begins at \(t=0\).
Worked Examples
Worked Example: Constant Rate
A tank contains \(120\) litres at the start. Water enters at a constant rate of \(8\) litres per minute. When does the tank reach \(200\) litres, assuming the model remains valid?
The initial amount is \(A_0=120\), the rate is \(r(t)=8\), and the target is \(B=200\). The accumulated amount after \(t\) minutes is
Set this equal to the target and solve:
The rate is positive, so the amount is strictly increasing and this solution is the first threshold time. Check by substitution:
The tank first reaches \(200\) litres after \(10\) minutes.
Worked Example: A Rate That Changes with Time
A workshop has completed \(30\) items at time \(0\). Its production rate is \(r(t)=4+2t\) items per hour for \(t\geq0\). Find when cumulative production reaches \(90\) items.
The net production after \(t\) hours is the integral of the rate:
Therefore \(A(t)=30+4t+t^2\). Set the accumulated amount equal to \(90\):
Factor the quadratic and find its two algebraic solutions:
Elapsed time must satisfy \(t\geq0\), so \(t=-10\) is not a valid time in this model. Also \(r(t)=4+2t>0\) for \(t\geq0\), so the accumulation is strictly increasing and the valid solution is the first threshold time. Check it directly:
The target is reached after \(6\) hours.
Worked Example: A Piecewise Rate
A queue starts with \(10\) people. During the first \(4\) minutes it grows at \(6\) people per minute. After that, it grows at \(4\) people per minute. Find when it first reaches \(50\) people.
The first interval contributes \(6(4)=24\) people, so at \(t=4\) the queue contains \(10+24=34\) people. The target has not yet been reached. For \(t\geq4\), the additional accumulation after minute \(4\) is \(4(t-4)\). Thus the model on the second interval is
Set this expression equal to \(50\) and solve:
On the first interval \(0\leq t\leq4\), the largest queue size is \(10+6(4)=34\), which is below \(50\). Both rates are positive, so the queue keeps growing; the solution at \(t=8\) is therefore the first threshold time. Check the amount at that time:
The queue first reaches \(50\) people after \(8\) minutes.
A Reliable Method
For a model with a starting amount, a possibly changing rate, and a specified target, organize the calculation in this order:
Write down the starting value \(A_0\), the rate \(r(t)\), the target \(B\), and the time interval on which the model applies.
Use \(A(t)=A_0+\int_0^t r(s)\,ds\). For a piecewise rate, split the integral at each change in rate.
Solve \(A(t)=B\), or equivalently \(\int_0^t r(s)\,ds=B-A_0\), for candidate times.
Discard times outside the valid interval. Determine whether the target is reached and, if there are multiple valid solutions, choose the earliest one.
Substitute the selected time into the accumulation formula and confirm that the result is exactly the target.
Common Mistake: Forgetting the Starting Amount
A frequent error is to treat the entire target as the amount that must be accumulated after \(t=0\). Suppose an inventory starts at \(40\) units and increases at \(10\) units per hour. The target is \(100\) units. The incorrect calculation \(10t=100\) gives \(t=10\) hours, but it ignores the \(40\) units already present. Substitution exposes the error: after \(10\) hours the inventory is \(40+10(10)=140\), not \(100\).
The required increase is \(100-40=60\) units. The correct equation and check are
Always distinguish the target amount from the additional net amount needed to reach it.
Why the First-Time Condition Matters
With a strictly positive rate, accumulation is strictly increasing, so a target can be reached at most once. With signed rates, however, the quantity can rise and fall. Then \(A(t)=B\) might have several valid solutions. Each solution represents a time at which the target is attained, but only the smallest valid time answers a question asking when it is reached first.
There is also a difference between finding an algebraic solution and establishing that a threshold is reached during the modelled period. For a strictly positive rate on \([0,T]\), compare the target with the endpoint amount \(A(T)\). If \(B>A(T)\), there is no threshold time in that interval, even though an equation might have a solution under an extrapolation beyond \(T\). The model’s stated interval is part of the problem, not an optional restriction.
Check Your Understanding
Use the starting value, rate history, and valid time interval in each question. For a first-time question, consider whether there could be an earlier solution.
- A quantity starts at \(25\) units and increases at \(5\) units per minute. Find the time it reaches \(60\) units.
- A process starts with \(12\) completed items and has rate \(r(t)=3+t\) items per hour. Write the threshold equation for reaching \(32\) items and find its nonnegative solution.
- A stock starts at \(50\) units and changes at a constant rate of \(-4\) units per day. Does it reach \(30\) units within \(6\) days? If so, when?
- A quantity starts at \(8\) units and grows at \(2\) units per minute for the first \(3\) minutes, then at \(5\) units per minute. Find its first time at \(30\) units.
- In one sentence, explain why a solution to \(A(t)=B\) outside the model’s stated time interval is not a valid threshold time for that model.
