---
title: "Pathline and streamlines for velocity field"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:pathline-streamlines-bx-t0"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pathline-streamlines-bx-t0
site: BCT Engineering
---

# Pathline and streamlines for velocity field

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2075 Chaitra, 4+4 marks
- Video: [Lines 2 - Example Part 1: Stream and Path Line](https://www.youtube.com/watch?v=QfFkRN_na38) — Till Wagner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pathline-streamlines-bx-t0
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

Velocity field $\vec v = Bx(1+At)\hat i + Cy\hat j$ with A = 0.5 s⁻¹, B = C = 1 s⁻¹. The coordinates are measured in meters. (i) Plot the pathline of the particle that passed through the point (1,1,0) at time t = 0. (ii) Plot the streamlines through the same point (1,1,0) at instants t = 0, 1 and 2 s.

## Short answer

Pathline: x=e^{t+0.25t²}, y=e^t (or ln x = ln y + 0.25(ln y)²). Streamlines: y=x (t=0), y=x^{2/3} (t=1 s), y=x^{1/2} (t=2 s).

## Answer

**Velocity field:** $u = x(1+0.5t)$, $v = y$ (A = 0.5 s⁻¹, B = C = 1 s⁻¹).

#### (i) Pathline of the particle at (1,1) at t = 0

$$
\frac{dy}{dt} = y\Rightarrow y = e^{t}
$$

$$
\frac{dx}{dt} = x(1+0.5t)\Rightarrow \ln x = t + 0.25t^2\Rightarrow x = e^{t+0.25t^2}
$$

Eliminating $t = \ln y$:

$$
\ln x = \ln y + 0.25(\ln y)^2
$$

| t (s) | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|
| x (m) | 1.00 | 1.76 | 3.49 | 7.87 | 20.09 | 58.12 | 190.57 |
| y (m) | 1.00 | 1.65 | 2.72 | 4.48 | 7.39 | 12.18 | 20.09 |

#### (ii) Streamlines through (1,1,0)

$$
\frac{dy}{dx} = \frac{v}{u} = \frac{y}{x(1+0.5t)}\Rightarrow y = x^{1/(1+0.5t)}
$$

| Instant | Equation | y at x = 2 | y at x = 4 |
|---|---|---|---|
| t = 0 | $y = x$ | 2.00 | 4.00 |
| t = 1 s | $y = x^{2/3}$ | 1.59 | 2.52 |
| t = 2 s | $y = x^{1/2}$ | 1.41 | 2.00 |

```text
 y
 4 |    /  <- t=0 (y = x)
   |   / .-' t=1
 2 |  / .'  t=2
   | /.'
 1 |*-----------------------> x
   1    2    3    4
```

The pathline starts along the $t=0$ streamline (slope 1 at $t=0$) but bends below it: as time passes the $x$-velocity grows, so the particle crosses successive streamlines that become flatter. The pathline and streamlines differ because the flow is unsteady.

**Answer: pathline $\ln x = \ln y + 0.25(\ln y)^2$ (parametric $x = e^{t+0.25t^2}$, $y = e^t$); streamlines $y = x$, $y = x^{2/3}$, $y = x^{1/2}$ at $t = 0, 1, 2$ s.**

## Check yourself

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

1. What is the pathline of the particle at (1,1) at t = 0?
   A. x = e^(t+0.25t^2), y = e^t
   B. x = e^t, y = e^(t+0.25t^2)
   C. x = 1+t, y = 1+t
   D. x = e^(0.5t), y = e^t
   Answer: A. dy/dt=y gives y=e^t; dx/dt=x(1+0.5t) gives ln x = t + 0.25t^2. (Key point: Pathline equations)

2. How does the pathline relation in x and y come out?
   A. ln x = ln y
   B. ln x = ln y + 0.25 (ln y)^2
   C. x = y^2
   D. x = y + 0.25y^2
   Answer: B. Since t = ln y, ln x = t + 0.25t^2 = ln y + 0.25(ln y)^2. (Key point: Eliminating t)

3. Which streamline passes through (1,1) at t = 2 s?
   A. y = x^2
   B. y = x
   C. y = x^(1/2)
   D. y = x^(2/3)
   Answer: C. Exponent 1/(1+0.5*2) = 1/2. (Key point: Streamline form)

4. Why do the pathline and streamlines differ?
   A. The particle is faster than the fluid
   B. The streamlines are only valid at x=1
   C. Viscosity bends the pathline
   D. The flow is unsteady, so the particle crosses successive streamlines
   Answer: D. As u grows with time the particle moves across changing streamline patterns; they would coincide only in steady flow. (Key point: Pathline vs streamline)

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