/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A flow field is given by \(\vec{... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 15

A flow field is given by \(\vec{V}=A x i+2 A y\), where \(A=2 s^{-1}\) Verify that the parametric equations for particle motion are given by \(x_{p}=c_{1} e^{A t}\) and \(y_{p}=c_{2} e^{2 u_{l}} .\) Obtain the equation for the pathline of the particle located at the point \((x, y)=(2,2)\) at the instant \(t=0 .\) Compare this pathline with the streamline through the same point.

Short Answer

Expert verified
Also, the parametric equations given were verified. The equation obtained for the pathline is \(x_p = 2 e^{2t}\) and \(y_p = 2 e^{4t}\) and for the streamline is \(x y^2 = 8\), which is different than the one for pathline, indicating different directions for the path of the particle and the flow at that point.

Step by step solution

01

Verify the Parametric Equations

The given velocity vector is \(\vec{V}=2x\hat{i}+4y\hat{j}\). Starting with \(x\), the motion equation is \(\frac{dx}{dt} = 2x\). Solving this differential equation, we'll find \(x(t)=c_1 e^{2t}\), which matches with \(x_p\) given. Now proceeding with \(y\), the motion equation is \(\frac{dy}{dt} = 4y\). Solving this, we'll find \(y(t) = c_2 e^{4t}\), which matches \(y_p\). This verifies the given parametric equations.
02

Obtain Equation for Pathline

To find the equations of the pathline, we substitute \(c_1\) and \(c_2\) using the initial conditions at \(t=0\), where \((x, y) = (2,2)\). Hence, \(c_1 = x(0) = 2\) and \(c_2 = y(0) = 2\). Then we can obtain the equations for the pathline as \(x_p = 2 e^{2t}\) and \(y_p = 2 e^{4t}\).
03

Obtain Equation for Streamline

The streamline equation can be obtained by integrating the formula \(dy/dx = V_y/V_x = 4y/2x = 2y/x\). Simplifying gives us the streamline equation \(x y^2 = k\), where \(k\) is an arbitrary constant. Applying the initial condition \((x, y) = (2,2)\) gives \(k = 8\). So the equation for the streamline is \(x y^2 = 8\).

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