STEM Problem Detail

Problem Statement

Prompt

Provide me the solution for question 3.5.

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$$
\begin{aligned}

\textbf{Given velocity field:} \\
\vec{V} &= u \mathbf{i} + v \mathbf{j} \\
u &= 1 + 3t, \quad v = y \\[6pt]

\textbf{1. Streamlines} \\
\text{Equation of streamline at an instant } t=\text{constant:} \\
\frac{dy}{dx} &= \frac{v}{u} \\
&= \frac{y}{1+3t} \\

\frac{1}{y}\,dy &= \frac{1}{1+3t}\,dx \\
\int \frac{1}{y}\,dy &= \int \frac{1}{1+3t}\,dx \\
\ln y &= \frac{x}{1+3t} + C \\

y &= C\, e^{\frac{x}{1+3t}} \\

\text{For streamline passing through } (x_1, y_1): \\
C &= y_1 e^{-\frac{x_1}{1+3t}} \\

\boxed{
y = y_1 \exp\left(\frac{x - x_1}{1+3t}\right)
} \\[10pt]

\textbf{2. Pathline (trajectory of a fluid particle)} \\

\frac{dx}{dt} &= 1 + 3t, \quad \frac{dy}{dt} = y \\

\textbf{(a) Equation for } x(t): \\
\frac{dx}{dt} &= 1 + 3t \\
x &= \int (1 + 3t)\,dt \\
x &= t + \frac{3}{2}t^2 + C_1 \\

\text{If particle passes through } (x_1, y_1) \text{ at } t=0: \\
C_1 &= x_1 \\

\boxed{
x = x_1 + t + \frac{3}{2}t^2
} \\[10pt]

\textbf{(b) Equation for } y(t): \\
\frac{dy}{dt} &= y \\
\frac{1}{y}\,dy &= dt \\
\ln y &= t + C_2 \\

\text{At } t=0,\ y=y_1: \\
C_2 &= \ln y_1 \\

\boxed{
y = y_1 e^{t}
} \\[10pt]

\textbf{3. Relation between pathline and streamline} \\

\text{From } y = y_1 e^{t}: \quad
t = \ln\left(\frac{y}{y_1}\right) \\

\text{Substitute into } x\text{-equation:} \\

x - x_1 &=
\ln\left(\frac{y}{y_1}\right)
+ \frac{3}{2}
\left[\ln\left(\frac{y}{y_1}\right)\right]^2 \\

\text{This is the pathline equation in } x\text{–}y \text{ plane.} \\[10pt]

\textbf{4. Comparison and conclusion} \\

\text{Since velocity field contains } t,\ \text{flow is unsteady.} \\
\text{Hence: Streamlines } \neq \text{ Pathlines} \\

\text{If flow were steady } (u,v \text{ independent of } t): \\
\text{Streamlines and pathlines would coincide.} \\

\boxed{
\text{For steady flow, streamline and pathline patterns coincide}
}

\end{aligned}
$$

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Problem #36