Chapter 17: Problem 8
Sketch the vector field and the flow. Then find the system of differential equations associated with the vector field and check that the flow satisfies the system. $$\vec{v}=y \vec{i}-x \vec{j} ; \quad x(t)=a \sin t, y(t)=a \cos t$$
Short Answer
Expert verified
The flow is a counterclockwise circle; the system is \(\frac{dx}{dt} = y, \frac{dy}{dt} = -x\).
Step by step solution
01
Understanding the Vector Field
The vector field is given by \( \vec{v} = y \vec{i} - x \vec{j} \). This means the vector at any point \((x, y)\) has a horizontal component \(y\) and a vertical component \(-x\). To sketch the vector field, draw vectors at various points \((x, y)\), where the vector points right if \(y > 0\) and left if \(y < 0\), while it points downward if \(x > 0\) and upward if \(x < 0\). The lengths of the vectors are proportional to the values of \(y\) and \(-x\).
02
Sketching the Flow
The flow is given by the parametric equations \(x(t) = a \sin t\) and \(y(t) = a \cos t\). Notice that these represent a circle of radius \(a\), because \(x^2 + y^2 = a^2(\sin^2 t + \cos^2 t) = a^2\). The flow is a circular path around the origin, counterclockwise, given the parametric origin of cosine from the y-component and sine from the x-component.
03
Establishing the System of Differential Equations
To find the system of differential equations, relate the given vector field to the derivatives of \(x(t)\) and \(y(t)\). The field \(\vec{v} = y \vec{i} - x \vec{j}\) gives \(\frac{dx}{dt} = y\) and \(\frac{dy}{dt} = -x\). These equations describe how \(x\) and \(y\) change over time, hence form the differential system:\[\begin{align*}\frac{dx}{dt} &= y,\\frac{dy}{dt} &= -x.\end{align*}\]
04
Verifying the Flow with Differential Equations
Verify that \(x(t) = a \sin t\) and \(y(t) = a \cos t\) satisfy the differential equations. Differentiating \(x(t) = a \sin t\) gives \(\frac{dx}{dt} = a \cos t\). Differentiating \(y(t) = a \cos t\) gives \(\frac{dy}{dt} = -a \sin t\). These are equal to \(\frac{dx}{dt} = y = a \cos t\) and \(\frac{dy}{dt} = -x = -a \sin t\). Thus, the calculated derivatives match the system of differential equations, confirming that the flow is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical expressions that relate a function with its derivatives. These equations are a key part of understanding the behavior of physical systems over time. In our vector field exercise, we find a system of differential equations from the vector field defined by \(\vec{v} = y \vec{i} - x \vec{j}\). This tells us how the components \(x\) and \(y\) change over time.
By differentiating the parametric equations of the flow \(x(t) = a \sin t\) and \(y(t) = a \cos t\), we arrive at:
By differentiating the parametric equations of the flow \(x(t) = a \sin t\) and \(y(t) = a \cos t\), we arrive at:
- \(\frac{dx}{dt} = y\), which shows that the rate of change of \(x\) depends on \(y\).
- \(\frac{dy}{dt} = -x\), indicating the rate at which \(y\) changes is influenced by \(-x\).
Parametric Equations
Parametric equations allow us to represent a path or a curve using a parameter, typically \(t\), which can be thought of as time or a progression variable. In this exercise, the parametric equations \(x(t) = a \sin t\) and \(y(t) = a \cos t\) describe a circle with radius \(a\).
This form of representation provides a clear view of how both \(x\) and \(y\) vary together as \(t\) changes.
By expressing both components in terms of \(t\), it's easy to visualize the path of a moving point, such as the flow represented in the vector field. The equations combine to maintain the relationship \(x^2 + y^2 = a^2\), confirming the circular path.
This form of representation provides a clear view of how both \(x\) and \(y\) vary together as \(t\) changes.
By expressing both components in terms of \(t\), it's easy to visualize the path of a moving point, such as the flow represented in the vector field. The equations combine to maintain the relationship \(x^2 + y^2 = a^2\), confirming the circular path.
- \(x(t)\) models the horizontal displacement as \(t\) progresses.
- \(y(t)\) reflects the vertical displacement over time.
Vector Components
Vector components are the horizontal and vertical parts of a vector. They allow us to break down a vector into understandable segments that can be analyzed. For our vector field \(\vec{v} = y \vec{i} - x \vec{j}\), the components \(y\) and \(-x\) are crucial in mapping out directions and magnitudes of vectors at different points.
The horizontal component, or \(y\), dictates how much a vector moves along the \(x\)-axis, while the vertical component, \(-x\), indicates its movement along the \(y\)-axis.
This decomposition is useful in multiple contexts:
The horizontal component, or \(y\), dictates how much a vector moves along the \(x\)-axis, while the vertical component, \(-x\), indicates its movement along the \(y\)-axis.
This decomposition is useful in multiple contexts:
- Helps in sketching vector fields by visualizing what directions vectors point at any location.
- Assists in solving problems involving forces, velocities, or other vector quantities in physics and engineering.