University Physics V · topic UPV-15.06
Decay chains as a linear system
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Decay chains as a linear system: predict a finite change
A decay chain is the constant-coefficient linear system dN/dt = A N, with A carrying −λ_i on the diagonal and the feeding term b_i λ_i one row below, whose solution is the matrix exponential acting on the starting inventory, N(t) = e^(At) N(0). Use the supplied topic model Y_out = (sqrt((r-1)*(r-1)+b*b)+b)/max(r-1,0.02), where Y_out is EIGENVECTOR CONDITION (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. r is Rate ratio λ₂/λ₁, dimensionless. Keep b (Feeding branch b) = 0.9; f (Initial daughter N₂(0)/N₁(0)) = 0 fixed. For this exercise use only 1.76162 ≤ r ≤ 7.25837 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at r = 2.97091 and 3.74045. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EIGENVECTOR CONDITION against r
Predictions from the supplied Decay chains as a linear system model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): EIGENVECTOR CONDITION
Horizontal axis (x): r: Rate ratio λ₂/λ₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Rate ratio λ₂/λ₁ | EIGENVECTOR CONDITION |
|---|---|
| 1.761625 | 2.729707426016311 |
| 3.1358124999999997 | 1.5065422396513257 |
| 4.51 | 1.2887601097024306 |
| 5.8841874999999995 | 1.2011037587671267 |
| 7.258374999999999 | 1.1540946528979767 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 2.97091, substitute into Y_out = (sqrt((r-1)*(r-1)+b*b)+b)/max(r-1,0.02): Y₀ = 1.55597.
- At r = 3.74045, the same fixed-parameter model gives Y₁ = 1.38096.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.17501; the input change is Δr = 0.769545.
- Average sensitivity is ΔY_out/Δr = -0.22742 1 per unit input. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 1.55597, Y₁ = 1.38096; ΔY_out = -0.17501; average sensitivity = -0.22742 1 per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Decay chains as a linear system: infer an input with uncertainty
A decay chain is the constant-coefficient linear system dN/dt = A N, with A carrying −λ_i on the diagonal and the feeding term b_i λ_i one row below, whose solution is the matrix exponential acting on the starting inventory, N(t) = e^(At) N(0). Use the supplied topic model Y_out = (sqrt((r-1)*(r-1)+b*b)+b)/max(r-1,0.02), where Y_out is EIGENVECTOR CONDITION (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. r is Rate ratio λ₂/λ₁, dimensionless. Keep b (Feeding branch b) = 0.9; f (Initial daughter N₂(0)/N₁(0)) = 0 fixed. For this exercise use only 1.76162 ≤ r ≤ 7.25837 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.2423732, with a bounded reading uncertainty ±0.013242. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000549675, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
EIGENVECTOR CONDITION against r
Predictions from the supplied Decay chains as a linear system model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): EIGENVECTOR CONDITION
Horizontal axis (x): r: Rate ratio λ₂/λ₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Rate ratio λ₂/λ₁ | EIGENVECTOR CONDITION |
|---|---|
| 1.761625 | 2.729707426016311 |
| 3.1358124999999997 | 1.5065422396513257 |
| 4.51 | 1.2887601097024306 |
| 5.8841874999999995 | 1.2011037587671267 |
| 7.258374999999999 | 1.1540946528979767 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.76162) = 2.72971 and Y_out(7.25837) = 1.15409. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 1.76162 + (1.24237 − 2.72971)(7.25837 − 1.76162)/(1.15409 − 2.72971) = 6.9504. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 5.11464; substituting back gives Y_out = 1.2423732.
- The interpolation error in the input is |r_line − r| = 1.83576. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2291312 and 1.2556152. Sort the roots even if the response decreases: 4.91992 ≤ r ≤ 5.33163.
- The lower and upper input deviations are 0.194721 and 0.216983. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 5.11464; compatible interval [4.91992, 5.33163]. Endpoint interpolation gives 6.9504.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.