Re-formulate changing models and interpret their results

Use the completed model foundations

This investigation follows differential-equation formulation, separation, logistic solutions and signed motion. The task is to assemble these methods and explain their domain

Unequal flows change the denominator

For inflow q_in and outflow q_out, volume satisfies V′=Δ=q_in−q_out. Perfect mixing gives M′=q_in c_in−q_out M/V(t). The outgoing concentration is the current M/V, not the inlet concentration and not M/V₀ when the volume changes

Use concentration to separate the equation

Write M=VC. Then VC′+CV′=q_in c_in−q_out C, so C′=q_in(c_in−C)/V. For Δ≠0 and V>0, integration yields C=c_in+(C₀−c_in)(V/V₀)^(−q_in/Δ). Multiplication by V gives M=c_in V+(M₀−c_in V₀)(V/V₀)^(−q_out/Δ)

Work a shrinking-tank example

Start with50L at1g/L. Inflow is3L/min at2g/L and outflow is4L/min. Then V=50−t, M′=6−4M/(50−t), C=2−[(50−t)/50]³ and M=2(50−t)−50[(50−t)/50]⁴. Use0≤t<50; the empty-tank endpoint is excluded

For an inflow of 3 L/min and an outflow of 4 L/min, give the volume rate Δ in L/min

For inflow 3 L/min and outflow 4 L/min, Δ is inflow minus outflow and the concentration power is −q_in/Δ; give that power

If the outflow changes again, which action is valid?

Differentiate the logistic rate as a composition

For y′=r y(1−y/K), differentiate with respect to time: y″=r(1−2y/K)y′=r²y(1−y/K)(1−2y/K). The middle factor is positive for0<y<K. Hence curvature changes at K/2 even though y′ stays positive throughout that interval

Connect a target time and its curvature

Take K=60,r=1/2,y(0)=15. Then y=60/(1+3e^(−t/2)). The target40 gives3e^(−t/2)=1/2, so t=2ln6. At y=40, y″=(1/4)×40×(1/3)×(−1/3)=−10/9. Population is increasing but its growth rate is decreasing

For y′ = (1/2)y(1 − y/60) with y(0) = 15, write the time when y = 40 as s ln R; give s

For y′ = (1/2)y(1 − y/60) with y(0) = 15, write the time when y = 40 as 2 ln R; give R

For y′ = (1/2)y(1 − y/60), give the exact value of y″ when y = 40

Integrate each velocity with its own initial position

On one oriented line, v_A=at+k and x_A(0)=0 give x_A=at²/2+kt. A second particle with v_B=b and x_B(0)=h has x_B=h+bt. The condition x_A(T)−x_B(T)=L gives k=b+(h+L)/T−aT/2 for T>0.

Work a signed-separation example

Let v_A=2t+k, x_A(0)=0, v_B=1 and x_B(0)=3. Require x_A(2)−x_B(2)=−2. Then k=−1/2, x_A(2)=3 and x_B(2)=5. The relative velocity is(4−1/2)−1=5/2m/s. A is behind B by2m at this instant; that is not a collision

Given v_A = 2t + k, x_A(0) = 0, v_B = 1, x_B(0) = 3 and x_A(2) − x_B(2) = −2, give k

For v_A = 2t − 1/2 and x_A(0) = 0, give x_A(2) in metres

What does the endpoint table establish at t=2?

Explain and solve the changed mixing model, ungraded

Derive the variable-volume concentration and mass solutions from the rate balance. State assumptions, units and validity limits. Explain the equal-flow case and what to do at a later flow-rate change. Verify the example with initial volume 50 L, initial concentration 1 g/L, inflow 3 L/min at 2 g/L and outflow 4 L/min, and the constant-concentration special case

Model: preserve volume, mass and the initial condition

Assume a perfectly mixed nutrient with no reaction or unmodelled loss, constant inlet concentration/flows and positive volume below capacity. V=V₀+Δt and M′=q_in c_in−q_out M/V. With M=VC, separate C′=q_in(c_in−C)/V to obtain the displayed power solution for Δ≠0; multiplying by V gives the mass formula. At t=0 both reproduce C₀ and M₀. If C₀=c_in, then C stays c_in while M=c_in V varies with volume. For equal flows q, use V=V₀ and C=c_in+(C₀−c_in)e^(−qt/V₀), not division by Δ=0. In the shrinking example, differentiating M=2V−50(V/50)⁴ with V′=−1 gives M′=−2+4(V/50)³, identical to6−4M/V. Stop before V=0. At a later flow change, retain the current M,V as new initial data and derive the next interval equation

Derive and interpret logistic curvature and time, ungraded

Starting from y′=r y(1−y/K), derive y″ and its sign for0<y<K. For y = K/(1 + Be^(−rt)), with B > 0, r > 0 and target y = 2K/(2 + j) where 0 < j < 2B, derive the target-time formula and justify positivity and uniqueness. Explain why negative curvature does not imply declining population and why K is not reached at finite time from below

Model: distinguish the two derivatives

The chain rule gives y″=r(1−2y/K)y′=r²y−3r²y²/K+2r²y³/K². Since r>0 and0<y<K, y′>0; y″ is positive below K/2, zero at K/2 and negative above. With B = K/y(0) − 1 > 0, r > 0 and 0 < j < 2B, from y=K/(1+B e^(−rt)), a target2K/(2+j) gives B e^(−rt)=j/2 and t=(1/r)ln(2B/j). The generated restriction0<j<2B gives a strictly larger target than the initial population, a positive finite time and a unique crossing by monotonicity. Negative y″ means the positive growth rate is falling. For any finite t, B e^(−rt)>0, so y<K; the capacity is approached asymptotically

Recover and interpret a relative-position condition, ungraded

Given v_A = 2t + k, x_A(0) = 0, v_B = 1, x_B(0) = 3 and x_A(2) − x_B(2) = −2, integrate the velocities, recover k from the signed separation, and verify the result by differentiation and substitution. Explain the distinction between position, displacement, distance and relative velocity. State precisely which collision claim the data justify

Model: the requested instant is not a collision

For v_A = 2t + k, x_A(0) = 0, v_B = 1 and x_B(0) = 3, integration gives x_A=t²+kt and x_B=3+t. At t=2, x_A−x_B=−1+2k=−2, so k=−1/2. Differentiation returns2t−1/2 and1. At t=2 the positions are3 and5, separation−2 and physical distance2. Relative velocity5/2 is the rate of change of separation; it is not the separation itself. Position includes the initial coordinate, displacement subtracts it, and total travelled distance integrates speed. The data show no collision at t=2; determining other collision times would require solving the position equality on the allowed time domain

Assessment boundary

Full balance explanations, separated derivations, sign arguments, validation against data and written interpretation remain ungraded