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2- given the velocity field ; u = y2-x2 , v = 2xy sketch the field, find the velocity and acceleration components at point (2,2) and (2,-2).

User Tepu
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1 Answer

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Answer:

To sketch the velocity field, we can plot a set of velocity vectors at various points in the domain. Here, we will plot the vectors at a grid of points in the xy-plane.

First, let's plot the vector field using Python:

import numpy as np

import matplotlib.pyplot as plt

# Define the velocity field functions

def u_func(x, y):

return y**2 - x**2

def v_func(x, y):

return 2*x*y

# Define the grid of points

x = np.linspace(-3, 3, 20)

y = np.linspace(-3, 3, 20)

X, Y = np.meshgrid(x, y)

# Compute the velocity components at each point in the grid

U = u_func(X, Y)

V = v_func(X, Y)

# Plot the vector field

fig, ax = plt.subplots(figsize=(6, 6))

ax.quiver(X, Y, U, V)

ax.set_xlabel('x')

ax.set_ylabel('y')

ax.set_xlim(-3, 3)

ax.set_ylim(-3, 3)

plt.show()

----------------------------

To find the velocity and acceleration components at points (2,2) and (2,-2), we first need to evaluate the velocity field functions at these points:

At (2,2):u = y^2 - x^2 = 2^2 - 2^2 = 0

v = 2xy = 2*2*2 = 8

So the velocity vector at (2,2) is (0, 8).

To find the acceleration components, we need to compute the partial derivatives of the velocity field functions with respect to x and y:

a_x = ∂u/∂x = -2x

a_y = ∂u/∂y = 2y

So at (2,2), the acceleration vector is (-4, 4).

At (2,-2):u = y^2 - x^2 = (-2)^2 - 2^2 = -4

v = 2xy = 2*2*(-2) = -8

So the velocity vector at (2,-2) is (-4, -8).

To find the acceleration components, we again need to compute the partial derivatives of the velocity field functions:a_x = ∂u/∂x = -2x

a_y = ∂u/∂y = 2y

So at (2,-2), the acceleration vector is (-4, -4).

Step-by-step explanation:

User YaronK
by
8.3k points

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