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Switching to polar coordinates, $$\begin{array}{l} \frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left(-x y-x^{2} r^{4}+x y-y^{2} r^{4}\right)=-r^{5} \\ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left(y^{2}+x y r^{4}+x^{2}-x y r^{4}\right)=1 \end{array}$$ If we use separation of variables on \(\frac{d r}{d t}=-r^{5}\) we obtain $$r=\left(\frac{1}{4 t+c_{1}}\right)^{1 / 4} \quad \text { and } \quad \theta=t+c_{2}$$ since \(\mathbf{X}(0)=(4,0), r=4\) and \(\theta=0\) when \(t=0 .\) It follows that \(c_{2}=0\) and \(c_{1}=\frac{1}{256} .\) The final solution can be written as $$r=\frac{4}{\sqrt[4]{1024 t+1}}, \quad \theta=t$$ and so the solution spirals toward the origin as \(t\) increases.

Step by step solution

01

Understand the Given Differential Equations

We are given two differential equations expressed in polar coordinates: 1. \( \frac{d r}{d t} = -r^5 \) 2. \( \frac{d \theta}{d t} = 1 \). The first equation tells us how the radius \( r \) changes with time, and the second equation describes how the angle \( \theta \) changes with time.

02

Solve the Radius Equation

We have \( \frac{d r}{d t} = -r^5 \). To solve this, we'll use separation of variables. Separating variables gives us \( \int r^{-5} \, dr = -\int \, dt \). Integrating both sides, we get \( -\frac{1}{4}r^{-4} = t + C_1 \) where \( C_1 \) is an integration constant.

03

Solve for the Radius Function

To express \( r \), we rearrange \( -\frac{1}{4}r^{-4} = t + C_1 \) as \( r^{-4} = -4(t + C_1) \). Taking reciprocal and containing all expressions for \( r \), we have: \( r = \left(\frac{1}{4t + C_1}\right)^{1/4} \).

04

Solve the Angle Equation

The second equation \( \frac{d \theta}{d t} = 1 \) is straightforward. Integrating with respect to \( t \) gives us: \( \theta = t + C_2 \) where \( C_2 \) is another constant of integration.

05

Apply Initial Conditions

We are given that at \( t = 0 \), the initial conditions are \( r = 4 \) and \( \theta = 0 \). Substituting \( t = 0 \) into the equation for \( r (4 = \left(\frac{1}{C_1}\right)^{1/4}) \), we find:\( C_1 = \frac{1}{256} \). For \( \theta \), substituting gives \( 0 = C_2 \), hence \( C_2 = 0 \).

06

Write Down Final Functions for r and θ

Substitute \( C_1 \) and \( C_2 \) into our equations:1. \( r(t) = \frac{4}{\sqrt[4]{1024t + 1}} \)2. \( \theta(t) = t \)Thus, the trajectory spirals towards the origin as \( t \) increases.

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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 a powerful mathematical tool used to describe how quantities change over time. In this exercise, we focus on two specific differential equations that involve the polar coordinate components: the radius function \( r(t) \) and the angle function \( \theta(t) \). The equations provided are:

• \( \frac{d r}{d t} = -r^5 \)
• \( \frac{d \theta}{d t} = 1 \)

\( \frac{d r}{d t} = -r^5 \) shows how the radius decreases over time, implying a shrinking effect towards the central point, or the origin. On the other hand, \( \frac{d \theta}{d t} = 1 \) indicates a constant rate of change of the angle \( \theta \) with respect to time, meaning the rotation is steady and clockwise. Solving these equations helps us understand the motion's nature in polar coordinates.

Separation of Variables

The method of separation of variables is a crucial technique for solving differential equations that can be expressed as a product of functions involving distinct variables. This method is particularly useful when the equation can be split such that each side depends on only one variable.

In the problem, the equation \( \frac{d r}{d t} = -r^5 \) is separated into integrals:

• \( \int r^{-5} \, dr = -\int \, dt \)

By integrating both sides, we isolate \( r \) and \( t \), allowing us to find a solution for the radius \( r(t) \). After integration, the equation becomes \( -\frac{1}{4}r^{-4} = t + C_1 \). Here, \( C_1 \) is a constant of integration that can be determined using initial conditions. Through this separation, we unveil how the function \( r(t) \) evolves with time.

Initial Conditions

Initial conditions are specific values given to a function at a particular point in time, often when \( t = 0 \). They provide crucial information for determining the constants present in the general solution of a differential equation.

In this exercise, the conditions given are \( r = 4 \) and \( \theta = 0 \) when \( t = 0 \). By substituting these values into the expressions obtained from integrating the differential equations, we solve for the integration constants \( C_1 \) and \( C_2 \). These constants are vital for ensuring that the solution accurately represents the situation described:

• For \( r(t) \), finding that \( C_1 = \frac{1}{256} \)
• For \( \theta(t) \), finding that \( C_2 = 0 \)

Initial conditions anchor the solution to the real-world problem we are modeling.

Spiral Trajectories

Spiral trajectories refer to the path an object takes as it rotates around a central point while simultaneously moving closer to or farther from that point. The solution to the given differential equations, \( r(t) = \frac{4}{\sqrt[4]{1024t + 1}} \) and \( \theta(t) = t \), results in such a spiral path.

• The radius \( r(t) \) steadily decreases over time due to the \( -r^5 \) factor, which makes the path spiral inward towards the origin.
• The angle \( \theta(t) = t \) rotates at a constant rate, ensuring a smooth, unchanging twist as time progresses.

As \( t \) increases, the trajectory continues to spiral inward, characterized by the path tightening its loop around the origin. Understanding these trajectories can be useful in various fields such as physics and engineering, where spiraling patterns often emerge in natural phenomena and technological systems.