Construct rational models from rates

Try

Two fictional pipes fill the same tank. One takes 4 hours alone and the other takes 6 hours alone. Adding the times would predict 10 hours together, slower than either pipe. Instead, define a common quantity: tanks filled per hour. Their rates are 1/4 and 1/6 tank per hour. Under constant simultaneous inflow, the combined rate is 5/12 tank per hour, so one tank takes 12/5 hours.

This is a modeling step before it is algebra. The reciprocal turns a completion time into a rate with different units.

Predict

Predict whether doubling both individual completion times doubles the joint time. Predict whether the result remains valid if the pipes share a supply line that limits their combined flow. The algebra follows from the constant additive-rate assumptions; it does not establish those assumptions for a real system.

Experiment

Track contributions over the proposed joint time. Each contribution is rate multiplied by elapsed time.

Process	Rate, tank/hour	Contribution in 12/5 hours
First pipe	1/4	3/5 tank
Second pipe	1/6	2/5 tank
Combined	5/12	1 tank

Change the second pipe into a drain with rate 1/6 tank per hour. The net rate becomes 1/4−1/6=1/12 tank per hour, predicting 12 hours to fill under the same idealized assumptions. If net rate were zero, no reciprocal fill time would exist; if negative, a negative computed time would not describe filling an empty tank.

Understand

For two positive solo completion times T₁,T₂ and a joint time t, each process contributes t/T₁ or t/T₂ of a job. Completing exactly one divisible job gives t/T₁+t/T₂=1. Hence t=1/(1/T₁+1/T₂). This equation is meaningful only when the tasks really can be worked on simultaneously and the rates add. Two machines doing different indivisible stages of one product may not fit it.

Now construct an unknown-time model. Suppose the slower pipe takes 2 hours longer than the faster one and together they fill in 4/3 hours. Let x>0 be the faster time, making the slower time x+2. Their combined rate is 3/4, so 1/x+1/(x+2)=3/4. Multiply by 4x(x+2), legal on x>0, to get 4(x+2)+4x=3x(x+2). Thus 3x²−2x−8=0=(3x+4)(x−2). The candidates are 2 and −4/3. Only x=2 satisfies the physical domain; the other solo time is 4. Original rates 1/2+1/4=3/4 verify the result.

Travel uses the same reciprocal structure: time equals distance divided by positive speed. For equal distances at different speeds, add the travel times, not the speeds. Whole-trip average speed is total distance divided by total time. For 20 km at 10 km/hour and 20 km at 20 km/hour, the times are 2 and 1 hours, giving average 40/3 km/hour rather than 15. The slower segment occupies more time.

Apply

Define the unknown and its units first, then derive a rational equation from contributions or elapsed times. Record all positive-speed or positive-time restrictions before clearing denominators. After solving, check each candidate in the original model and explain any rejection in context.

A negative candidate can be algebraically valid for the cleared polynomial but physically meaningless. An excluded denominator value is invalid even without a contextual objection. These are different reasons for rejection and should be named separately. In an actual application, observed rate variability would also limit precision; the exact fractions here belong to stipulated synthetic constant-rate models.

A cumulative rate-model repair

A fill pipe takes x hours and a slower drain takes x+3 hours to move one tank. If they operate together and fill in six hours, start with rates, not a difference of completion times: 1/x−1/(x+3)=1/6, with x>0. The left side reduces to 3/[x(x+3)]. Clearing denominators on this domain gives x²+3x−18=0, or (x−3)(x+6)=0. Only x=3 survives. Check the original statement: the drain takes six hours, and the net rate 1/3−1/6=1/6 fills one tank in six hours.

The discarded root −6 is not an excluded-denominator root: the original rational expression is defined there. It is rejected because negative solo times contradict the model. A complete solution identifies which restriction does the work rather than saying every rejected root is “extraneous.”

Use the input control to vary the positive solo fill time while keeping the three-hour difference fixed. Predict whether a larger x increases or decreases the joint fill time, then compare the computed rates. The derived joint time x(x+3)/3 increases on x>0 because its derivative (2x+3)/3 is positive; alternatively, compare both positive factors without calculus. Do not treat the exact computed numbers as measured data.

For the changed retry, alter both the time difference and net filling time. Retain four pieces of written evidence: a unit-labeled rate equation, the domain, an original-equation check and one real-world assumption that would need measurement. A shared supply line can invalidate additivity even when every algebraic step is correct.

Rates, domains and model validation

Review criterion: Write rates in matching units, retain the physical domain, and check the original equation.

Predict, then change one input

Fill-pipe time x (hours):

3

Fill rate (tank/hour)

0.333333

Drain rate (tank/hour)

0.166667

Net rate (tank/hour)

0.166667

Joint fill time (hours)

6

The drain time is x+3. Rates are stipulated constant and additive; this is a model prediction, not a measured installation.

Explain the changed case: A slower drain

A fill pipe takes x hours and a drain takes x+3 hours for one tank. Together they fill an empty tank in 6 hours under constant rates. Which conclusion is valid?

• 1/x − 1/(x+3) = 1/6
• Physical domain: x > 0

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

x = 3 hours; the drain takes 6 hours.x = −6 hours; the drain takes −3 hours.x = 9 hours because 6+3=9.

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

The following lessons connect equations to complete sets of coordinate pairs, then to intersections and feasible regions. Keep both algebraic conditions and contextual domains visible. A plotted point is useful evidence only when it satisfies the original relationships in the same units.

Continue to An equation describes a set of points.