For Time \[$ T \geq 0 \$\], A Particle Moves In The \[$ Xy \$\]-plane With A Velocity Vector Given By $\[ V(t) = \left\ \textless \ \ln(t^3 - T + 3), 1 - \ln(t + 5) \right\ \textgreater \ . \\]At Time \[$ T = 0 \$\],

For Time $t \geq 0$, a Particle Moves in the $xy$-Plane with a Velocity Vector Given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$

In this article, we will explore the motion of a particle in the $xy$-plane with a velocity vector given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$ for time $t \geq 0$. We will analyze the velocity vector and determine the position of the particle at any given time $t$.

Velocity Vector

The velocity vector of the particle is given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$. This vector has two components: the $x$-component, which is given by $\ln(t^3 - t + 3)$, and the $y$-component, which is given by $1 - \ln(t + 5)$.

X-Component

The $x$-component of the velocity vector is given by $\ln(t^3 - t + 3)$. This component represents the rate of change of the particle's position in the $x$-direction. To understand the behavior of this component, we need to analyze the function $t^3 - t + 3$.

The function $t^3 - t + 3$ is a cubic function, which means it has a single inflection point. The inflection point occurs when the second derivative of the function is equal to zero. To find the inflection point, we need to calculate the second derivative of the function.

Let's calculate the first derivative of the function:

$$\frac{d}{dt}(t^3 - t + 3) = 3t^2 - 1$$

Now, let's calculate the second derivative of the function:

$$\frac{d^2}{dt^2}(t^3 - t + 3) = 6t$$

To find the inflection point, we need to set the second derivative equal to zero and solve for $t$:

$$6t = 0 \Rightarrow t = 0$$

Since the inflection point occurs at $t = 0$, we need to analyze the behavior of the function $t^3 - t + 3$ in the neighborhood of $t = 0$.

Y-Component

The $y$-component of the velocity vector is given by $1 - \ln(t + 5)$. This component represents the rate of change of the particle's position in the $y$-direction. To understand the behavior of this component, we need to analyze the function $1 - \ln(t + 5)$.

The function $1 - \ln(t + 5)$ is a logarithmic function, which means it has a vertical asymptote at $t = -5$. The vertical asymptote occurs when the argument of the logarithm is equal to zero.

To understand the behavior of the function $1 - \ln(t + 5)$, we need to analyze the behavior of the logarithm function in the neighborhood of $t = -5$.

Velocity Vector at Time $t = 0$

At time $t = 0$, the velocity vector is given by:

$$v(0) = \left\langle \ln(0^3 - 0 + 3), 1 - \ln(0 + 5) \right\rangle = \left\langle \ln(3), 1 - \ln(5) \right\rangle$$

To simplify the expression, we can use the fact that $\ln(3) \approx 1.0986$ and $\ln(5) \approx 1.6094$.

$$v(0) \approx \left\langle 1.0986, 1 - 1.6094 \right\rangle = \left\langle 1.0986, -0.5094 \right\rangle$$

Position of the Particle

To determine the position of the particle at any given time $t$, we need to integrate the velocity vector with respect to time.

Let's integrate the $x$-component of the velocity vector:

$$x(t) = \int \ln(t^3 - t + 3) dt$$

To integrate the logarithmic function, we can use the fact that $\int \ln(x) dx = x\ln(x) - x + C$.

$$x(t) = (t^3 - t + 3)\ln(t^3 - t + 3) - (t^3 - t + 3) + C$$

Now, let's integrate the $y$-component of the velocity vector:

$$y(t) = \int (1 - \ln(t + 5)) dt$$

To integrate the logarithmic function, we can use the fact that $\int \ln(x) dx = x\ln(x) - x + C$.

$$y(t) = (t + 5) - (t + 5)\ln(t + 5) + C$$

In this article, we analyzed the motion of a particle in the $xy$-plane with a velocity vector given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$ for time $t \geq 0$. We determined the position of the particle at any given time $t$ by integrating the velocity vector with respect to time.

The $x$-component of the velocity vector is given by $\ln(t^3 - t + 3)$, which represents the rate of change of the particle's position in the $x$-direction. The $y$-component of the velocity vector is given by $1 - \ln(t + 5)$, which represents the rate of change of the particle's position in the $y$-direction.

At time $t = 0$, the velocity vector is given by $v(0) = \left\langle \ln(3), 1 - \ln(5) \right\rangle \approx \left\langle 1.0986, -0.5094 \right\rangle$.

The position of the particle at any given time $t$ is given by $x(t) = (t^3 - t + 3)\ln(t^3 - t + 3) - (t^3 - t + 3) + C$ and $y(t) = (t + 5) - (t + 5)\ln(t + 5) + C$.

We hope this article has provided a comprehensive analysis of the motion of a particle in the $xy$-plane with a velocity vector given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$ for time $t \geq 0$.
Q&A: For Time $t \geq 0$, a Particle Moves in the $xy$-Plane with a Velocity Vector Given by $v(t) = \left\langle \ln(t^3 - t + 3), 1 - \ln(t + 5) \right\rangle$

Q: What is the velocity vector of the particle at time $t = 0$?

A: The velocity vector of the particle at time $t = 0$ is given by $v(0) = \left\langle \ln(3), 1 - \ln(5) \right\rangle \approx \left\langle 1.0986, -0.5094 \right\rangle$.

Q: What is the position of the particle at time $t = 0$?

A: To determine the position of the particle at time $t = 0$, we need to integrate the velocity vector with respect to time. The position of the particle at time $t = 0$ is given by $x(0) = (0^3 - 0 + 3)\ln(0^3 - 0 + 3) - (0^3 - 0 + 3) + C$ and $y(0) = (0 + 5) - (0 + 5)\ln(0 + 5) + C$.

Q: What is the rate of change of the particle's position in the $x$-direction at time $t = 0$?

A: The rate of change of the particle's position in the $x$-direction at time $t = 0$ is given by the $x$-component of the velocity vector, which is $\ln(3) \approx 1.0986$.

Q: What is the rate of change of the particle's position in the $y$-direction at time $t = 0$?

A: The rate of change of the particle's position in the $y$-direction at time $t = 0$ is given by the $y$-component of the velocity vector, which is $1 - \ln(5) \approx -0.5094$.

Q: How does the velocity vector change as time increases?

A: As time increases, the velocity vector changes due to the changing values of the $x$-component and the $y$-component. The $x$-component changes due to the changing value of the function $t^3 - t + 3$, while the $y$-component changes due to the changing value of the function $1 - \ln(t + 5)$.

Q: Can the velocity vector be expressed in terms of a single function?

A: No, the velocity vector cannot be expressed in terms of a single function. The $x$-component and the $y$-component are given by different functions, and there is no single function that can express both components.

Q: How can the position of the particle be determined at any given time $t$?

A: The position of the particle at any given time $t$ can be determined by integrating the velocity vector with respect to time. The position of the particle is given by $x(t) = (t^3 - t + 3)\ln(t^3 - t + 3) - (t^3 - t + 3) + C$ and $y(t) = (t + 5) - (t + 5)\ln(t + 5) + C$.

Q: What is the significance of the inflection point of the function $t^3 - t + 3$?

A: The inflection point of the function $t^3 - t + 3$ occurs when the second derivative of the function is equal to zero. This point is significant because it represents a change in the behavior of the function.

Q: What is the significance of the vertical asymptote of the function $1 - \ln(t + 5)$?

A: The vertical asymptote of the function $1 - \ln(t + 5)$ occurs when the argument of the logarithm is equal to zero. This point is significant because it represents a change in the behavior of the function.

Q: Can the velocity vector be expressed in terms of a parametric equation?

A: Yes, the velocity vector can be expressed in terms of a parametric equation. The parametric equation is given by $x(t) = (t^3 - t + 3)\ln(t^3 - t + 3) - (t^3 - t + 3) + C$ and $y(t) = (t + 5) - (t + 5)\ln(t + 5) + C$.

Q: How can the velocity vector be used to determine the motion of the particle?

A: The velocity vector can be used to determine the motion of the particle by integrating the velocity vector with respect to time. The position of the particle at any given time $t$ can be determined by integrating the velocity vector with respect to time.

Q: What is the relationship between the velocity vector and the position of the particle?

A: The velocity vector is related to the position of the particle by the equation $v(t) = \frac{d}{dt}x(t)$. This equation represents the relationship between the velocity vector and the position of the particle.