Mixing a Brine Tank: Rational Function With a Hole
A chemistry class models the concentration C(t), in grams per litre, of salt in a 200 L tank as brine is added and drained. Their model is C(t) = (5t + 20)/(t^2 - 16) for t ≥ 0, where t is minutes after the experiment begins. A student claims the concentration is undefined at t = 4 because the denominator is zero, and wants to declare the experiment invalid at that moment. The teacher points out that the numerator also vanishes at t = 4 and asks the class to analyse the function's behaviour near t = 4, at large t, and at t = 0, and to decide whether the model can be repaired by defining a single value at t = 4.
1. Factor and simplify C(t), identify the vertical asymptote and the hole, and state the limit of C(t) as t approaches 4.
Read the worked answer
Factor: numerator 5t + 20 = 5(t + 4); denominator t^2 - 16 = (t - 4)(t + 4).
So C(t) = 5(t + 4)/[(t - 4)(t + 4)] = 5/(t - 4), provided t ≠ -4. Since t ≥ 0, the factor (t + 4) never vanishes in the domain except at t = -4, which is outside the domain anyway. The cancellation removes the common factor, leaving C(t) = 5/(t - 4) for t ≥ 0, t ≠ 4.
At t = 4 the simplified form has a vertical asymptote: as t → 4⁻, C → -∞; as t → 4⁺, C → +∞. The original expression had a removable-looking factor, but after cancellation the singularity at t = 4 is a genuine vertical asymptote, not a hole. The 'hole' would occur at t = -4, where both numerator and denominator vanish; that point is outside the physical domain t ≥ 0.
Therefore the limit as t → 4 does not exist (it is unbounded, with opposite signs from each side). No single value at t = 4 can make the function continuous, so the model cannot be repaired there; the concentration genuinely blows up as the tank composition approaches the singular time.
2. Describe the end behaviour as t → ∞ and evaluate C(0). Interpret both in terms of the tank, and explain whether the model is physically reasonable for large t.
Read the worked answer
End behaviour: C(t) = 5/(t - 4). As t → ∞, the denominator grows without bound, so C(t) → 0⁺. The concentration tends to zero from above.
At t = 0: C(0) = 5/(0 - 4) = -5/4 g/L. A negative concentration is physically impossible, so the model already fails at the start: the formula gives -1.25 g/L, which cannot describe salt content.
Interpretation: the model predicts that just before t = 4 the concentration is large and negative (unphysical), and just after t = 4 it is large and positive. The vertical asymptote at t = 4 signals a breakdown of the model, not a real infinite concentration. For large t the model predicts dilution toward zero, which is plausible qualitatively, but the negative values for 0 ≤ t < 4 make the model unreliable over that interval.
Conclusion: the rational model is a poor physical description on [0, 4) and near t = 4. A repaired model would need a numerator and denominator that stay positive on the physical domain, for example by shifting the singularity outside t ≥ 0 or by using a different rate structure. The exercise illustrates that algebraic simplification can reveal a vertical asymptote where a student expected a harmless hole.