University Physics V · topic UPV-15.08
Nuclear dating and isochron regression
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.
Nuclear dating and isochron regression: predict a finite change
An isochron is the straight line traced in daughter/stable against parent/stable coordinates by samples that closed at the same instant with the same initial ratio; its slope is e^(λt) − 1 and its intercept is that initial ratio. Use the supplied topic model Y_out = 1000*(0.0004/max(0.745*w,0.3))/(0.01397*exp(0.01397*t)), where Y_out is AGE 1σ FROM SPREAD (Myr). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. w is Lever arm in ⁸⁷Rb/⁸⁶Sr, dimensionless. Keep t (Age of the suite t) = 3.6 Ga; r0 (Initial ⁸⁷Sr/⁸⁶Sr) = 0.701 fixed. For this exercise use only 0.80375 ≤ w ≤ 3.79625 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 w = 1.4621 and 1.88105. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
AGE 1σ FROM SPREAD against w
Predictions from the supplied Nuclear dating and isochron regression 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): AGE 1σ FROM SPREAD / Myr
Horizontal axis (x): w: Lever arm in ⁸⁷Rb/⁸⁶Sr
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Lever arm in ⁸⁷Rb/⁸⁶Sr | AGE 1σ FROM SPREAD / Myr |
|---|---|
| 0.80375 | 45.472076663915274 |
| 1.551875 | 23.55098291977248 |
| 2.3 | 15.89051374722692 |
| 3.0481249999999998 | 11.990381502931118 |
| 3.7962499999999997 | 9.627443297628426 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 1.4621, substitute into Y_out = 1000*(0.0004/max(0.745*w,0.3))/(0.01397*exp(0.01397*t)): Y₀ = 24.997 Myr.
- At w = 1.88105, the same fixed-parameter model gives Y₁ = 19.4297 Myr.
- Signed change: ΔY_out = Y₁ − Y₀ = -5.56738 Myr; the input change is Δw = 0.41895.
- Average sensitivity is ΔY_out/Δw = -13.2889 Myr 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 = 24.997 Myr, Y₁ = 19.4297 Myr; ΔY_out = -5.56738 Myr; average sensitivity = -13.2889 Myr 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.
Nuclear dating and isochron regression: infer an input with uncertainty
An isochron is the straight line traced in daughter/stable against parent/stable coordinates by samples that closed at the same instant with the same initial ratio; its slope is e^(λt) − 1 and its intercept is that initial ratio. Use the supplied topic model Y_out = 1000*(0.0004/max(0.745*w,0.3))/(0.01397*exp(0.01397*t)), where Y_out is AGE 1σ FROM SPREAD (Myr). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. w is Lever arm in ⁸⁷Rb/⁸⁶Sr, dimensionless. Keep t (Age of the suite t) = 3.6 Ga; r0 (Initial ⁸⁷Sr/⁸⁶Sr) = 0.701 fixed. For this exercise use only 0.80375 ≤ w ≤ 3.79625 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 = 13.901008 Myr, with a bounded reading uncertainty ±0.64103 Myr. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925, 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.
AGE 1σ FROM SPREAD against w
Predictions from the supplied Nuclear dating and isochron regression 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): AGE 1σ FROM SPREAD / Myr
Horizontal axis (x): w: Lever arm in ⁸⁷Rb/⁸⁶Sr
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Lever arm in ⁸⁷Rb/⁸⁶Sr | AGE 1σ FROM SPREAD / Myr |
|---|---|
| 0.80375 | 45.472076663915274 |
| 1.551875 | 23.55098291977248 |
| 2.3 | 15.89051374722692 |
| 3.0481249999999998 | 11.990381502931118 |
| 3.7962499999999997 | 9.627443297628426 |
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(w) = 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(0.80375) = 45.4721 Myr and Y_out(3.79625) = 9.62744 Myr. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.80375 + (13.901 − 45.4721)(3.79625 − 0.80375)/(9.62744 − 45.4721) = 3.43947. 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 w = 2.62917; substituting back gives Y_out = 13.901008 Myr.
- The interpolation error in the input is |w_line − w| = 0.810295. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 13.259978 Myr and 14.542038 Myr. Sort the roots even if the response decreases: 2.51328 ≤ w ≤ 2.75628.
- The lower and upper input deviations are 0.115897 and 0.127103. 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 w ≈ 2.62917; compatible interval [2.51328, 2.75628]. Endpoint interpolation gives 3.43947.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.