---
title: "Four-point implicit derivatives of Q"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:four-point-implicit-numerical-m0"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/four-point-implicit-numerical-m0
site: BCT Engineering
---

# Four-point implicit derivatives of Q

Computational Techniques in Civil Engineering · Chapter 5: Finite difference method

- Asked in: 2070 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/four-point-implicit-numerical-m0
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

The value of flow rate $Q$ at four points in the space-time grid are shown in the figure below. $\Delta t = 1$ h, $\Delta x = 1000$ m and $\theta = 0.55$. Calculate the values of $\partial Q/\partial t$ and $\partial Q/\partial x$ by the four-point implicit method. ($\theta$ = weighting factor.)

[Figure: space-time grid with time levels j and j+1 and distance points i and i+1 ($\Delta x = 1000$ m, $\Delta t = 1$ h); $Q$ at (i, j+1) $= 3580$ cms, at (i+1, j+1) $= 3470$ cms, at (i, j) $= 3500$ cms, at (i+1, j) $= 3350$ cms]

## Short answer

Using the four-point implicit formulas with θ = 0.55: ∂Q/∂t = [(3580−3500)+(3470−3350)]/(2×3600) = 2.78×10⁻² m³/s², and ∂Q/∂x = 0.55×(−110/1000) + 0.45×(−150/1000) = −0.128 m²/s.

## Answer

#### Four-point implicit method
$$
\frac{\partial Q}{\partial t}\approx\frac{(Q_i^{j+1}-Q_i^j)+(Q_{i+1}^{j+1}-Q_{i+1}^j)}{2\Delta t}
$$
$$
\frac{\partial Q}{\partial x}\approx\theta\,\frac{Q_{i+1}^{j+1}-Q_i^{j+1}}{\Delta x}+(1-\theta)\,\frac{Q_{i+1}^j-Q_i^j}{\Delta x}
$$
Data: $Q_i^{j+1}=3580$, $Q_{i+1}^{j+1}=3470$, $Q_i^j=3500$, $Q_{i+1}^j=3350$; $\Delta t=3600$ s, $\Delta x=1000$ m, $\theta=0.55$.

**Time derivative**
$$
\frac{\partial Q}{\partial t}=\frac{(3580-3500)+(3470-3350)}{2\times3600}=\frac{200}{7200}=2.7778\times10^{-2}\ \text{m}^3/\text{s}^2
$$
**Space derivative**
$$
\frac{\partial Q}{\partial x}=0.55\,\frac{3470-3580}{1000}+0.45\,\frac{3350-3500}{1000}=-0.06050+(-0.06750)=-0.1280\ \text{m}^3/\text{s per m}
$$
**Answer:** $\partial Q/\partial t=2.7778\times10^{-2}$ m$^3$/s$^2$ and $\partial Q/\partial x=-0.1280$ m$^2$/s.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the four-point method dQ/dt is:
   A. average of the two time differences, divided by 2 dt
   B. a single forward difference
   C. central difference over 2 dx
   D. theta-weighted
   Answer: A. [(Qi^{j+1}-Qi^j)+(Qi+1^{j+1}-Qi+1^j)]/(2 dt). (Key point: Time derivative)

2. The value of dQ/dt is:
   A. 2.78 x 10^-2 m^3/s^2
   B. 5.56 x 10^-2
   C. 0.2
   D. 1.0 x 10^-3
   Answer: A. (80+120)/(2 x 3600) = 0.027778. (Key point: Value of dQ/dt)

3. dQ/dx with theta = 0.55 is:
   A. -0.128
   B. +0.128
   C. -0.0605
   D. -0.1355
   Answer: A. 0.55(-110)/1000 + 0.45(-150)/1000 = -0.0605 - 0.0675 = -0.128. (Key point: Space derivative)

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