R(t) is the position of a particle in the x y -plane at time...

${r}(t)$ is the position of a particle in the $x y$ -plane at time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$ \begin{equation} \mathbf{r}(t)=\frac{t}{t+1} \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=-\frac{1}{2} \end{equation}

${r}(t)$ is the position of a particle in the $x y$ -plane at
time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of $t .$
\begin{equation}
\mathbf{r}(t)=\frac{t}{t+1} \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=-\frac{1}{2}
\end{equation}

Key Concepts

Velocity Vector

The velocity vector is defined as the time derivative of the position vector. It provides both the speed and the direction of the particle's motion at any given time. In problems involving motion, computing the velocity vector is crucial for understanding the instantaneous motion of the particle.

Acceleration Vector

The acceleration vector is obtained by differentiating the velocity vector with respect to time, representing the rate of change of velocity. It captures how the motion of the particle is changing, including changes in speed and/or direction, and is essential for analyzing dynamics in a vector-valued setting.

Parametric Representation

This concept involves expressing a geometric object, such as a curve, in terms of one or more parameters. In the context of motion, the position of a particle is given as a function of time via component functions, which describe the particle's trajectory in space without explicitly presenting the relationship between coordinates.

Elimination of the Parameter

Eliminating the parameter is the process of removing the independent variable to obtain an equation that relates the coordinate variables directly. This method is used to derive the explicit equation of a curve from its parametric equations, allowing one to understand the curve's shape in the coordinate plane.

Transcript

00:01 In this question, they tell me that r of t is the position of a particle in the xy plane at time t.

00:06 I'm going to find an equation in x and y whose graph is the path of the particle.

00:11 Then we'll find the particle's velocity and acceleration vectors at the given value of t.

00:18 So here what i have are an x and a y given to me.

00:22 So my x is the component in front of i, and the y is the component in front of the so here my x is equal to t over t plus 1, and my y is equal to 1 over t.

00:40 And so what i want to do is eliminate the parameter first.

00:44 To do that, i'm going to start with the equation here on the right, and i'm going to solve for t.

00:51 I'll say, this implies that t times y is equal to 1, or that t is equal to 1 over y.

01:00 Once i have that, i'm going to substitute it into this equation on the left.

01:08 So then i have x equals 1 over y, and that's being divided by 1 over y plus 1.

01:16 Now, what i'm going to do is to clear the compound fraction away.

01:21 I'm going to multiply by y over y.

01:23 If i do, up top i have 1, in the bottom i have 1 plus y.

01:31 So the eliminating the parameter gave us x equals 1 over 1 plus y.

01:37 That was the first part of this question where we found an equation in x and 1 whose graph is the pathway.

01:46 Now i need my velocity and acceleration vectors.

01:50 So let's see here.

01:52 To do that, my r of t, which was my position vector again, it had components t over t plus 1, followed by 1 over t.

02:04 Now let's get our velocity vector.

02:09 I note my velocity vector is the derivative of my position vector.

02:16 So in that first component, i have to use the quotient rule.

02:19 My denominator, t plus 1, times the derivative of the numerator 1, minus the numerator, t times the derivative of the denominator, 1, all over my denominator being square...

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