Approximate Concentration by Taylor Extrapolation of an Error-Function Profile: 23-Chem-B3 Simulation, Modelling, and Optimization, December 2014, Q3

Question 3: Approximate Concentration by Taylor Extrapolation of an Error-Function Profile (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $c/c_0 = 1 - \operatorname{erf}\!\big(x/[2\sqrt{{Dt}}]\big)$ with $D = 3\ \text{{m}}^2/\text{{month}}$, $t = 3\ \text{{month}}$ (so $2\sqrt{{Dt}} = 6\ \text{{m}}$), and the known anchor value $c/c_0\big|_{{x=30}} = 0.5230$.

Find. The approximate $c/c_0$ at $x = 31$ m (same $t$), using the known value at $x = 30$ m rather than evaluating the error function afresh.

Figure 2 — Profile c/c₀ = 1 − erf(x/6). By x ≈ 30 m the front's tail is extremely flat, so a 1 m step barely changes c/c₀.

Approach. Expand $c/c_0$ to first order (a Taylor / finite-difference step) about the known point $x=30$ m: the correction is the analytic slope times $\Delta x = 1$ m.

Analytic slope. Differentiating the profile ($\tfrac{d}{du}\operatorname{erf}(u) = \tfrac{2}{\sqrt\pi}e^{-u^2}$, $u = x/2\sqrt{{Dt}}$): $$\frac{d(c/c_0)}{dx} = -\frac{1}{\sqrt{\pi Dt}}\,\exp\!\left(-\frac{x^2}{4Dt}\right).$$
Evaluate at $x = 30$ m. Here $\dfrac{x^2}{4Dt} = \dfrac{900}{36} = 25$ and $\dfrac{1}{\sqrt{\pi Dt}} = \dfrac{1}{\sqrt{9\pi}} = 0.1881$, so $$\left.\frac{d(c/c_0)}{dx}\right|_{{30}} = -0.1881\,e^{-25} = -2.6\times10^{-12}\ \text{{m}}^{-1}.$$
First-order step to $x = 31$ m. The change over $\Delta x = 1$ m is $$\Delta(c/c_0) \approx \left.\frac{d(c/c_0)}{dx}\right|_{{30}}\!\!\cdot(1\ \text{{m}}) = -2.6\times10^{-12},$$ which is utterly negligible, so $$\left.\frac{c}{c_0}\right|_{{x=31}} \approx 0.5230 - 2.6\times10^{-12} = \boxed{0.5230}.$$

Check. As printed, the anchor $c/c_0 = 0.5230$ is not numerically consistent with the nominal data (the profile itself gives $1-\operatorname{erf}(5) \approx 1.5\times10^{-12}$ at $x=30$ m). This does not affect the requested answer: the point of the question is the sensitivity, and the first-order slope at $x=30$ m is $\sim10^{-12}$ per metre, so to every reported figure $c/c_0$ is unchanged over a 1 m step. The concentration at $x = 31$ m equals the given value, $0.5230$.

Quantity	Result
Slope $d(c/c_0)/dx$ at 30 m	$-2.6\times10^{-12}\ \text{{m}}^{-1}$
Change over $\Delta x = 1$ m	$\approx -2.6\times10^{-12}$ (negligible)
$c/c_0$ at $x = 31$ m	$\approx 0.5230$

Concept & Theory

Topic. First-order Taylor (finite-difference) extrapolation and the sensitivity of a diffusion/dispersion front.

Key relations. $f(x+\Delta x) \approx f(x) + f'(x)\Delta x$; error-function derivative $\tfrac{d}{du}\operatorname{erf}(u) = \tfrac{2}{\sqrt\pi}e^{-u^2}$; similarity variable $u = x/2\sqrt{{Dt}}$.

Why this works. The complementary error function decays like a Gaussian, so its derivative at $u=5$ carries the factor $e^{-25}\sim10^{-11}$. Physically the dispersion front is essentially flat five "diffusion lengths" out, so moving the probe 1 m produces no measurable change — the correct engineering conclusion is "no change," obtained without a single erf-table lookup.

Common pitfall. Trying to evaluate $\operatorname{erf}(5)$ by series or by a table in a closed-book exam. The question deliberately supplies the anchor value so you extrapolate with the derivative instead.

Source. Chapra & Canale, Numerical Methods for Engineers §4.1 (Taylor-series approximation and error); Kreyszig, Advanced Engineering Mathematics (error function).

Question 3 of 7: Approximate Concentration by Taylor Extrapolation of an Error-Function Profile

Question 3: Approximate Concentration by Taylor Extrapolation of an Error-Function Profile (20 marks)