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Chapter 12: Problem 12

Describe the graph of the equation. $$ \mathbf{r}=3 \mathbf{i}+2 \cos t \mathbf{j}+2 \sin t \mathbf{k} $$

Short Answer

Expert verified
The graph is a circle of radius 2 centered at (3,0,0) parallel to the yz-plane.

Step by step solution

01

Understand the Equation

The vector equation given is \( \mathbf{r}(t) = 3 \mathbf{i} + 2\cos(t) \mathbf{j} + 2\sin(t) \mathbf{k} \). This represents a parametric equation in 3D space where \( t \) is the parameter, and the coefficients of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) correspond to the x, y, and z coordinates respectively.
02

Analyze the Components

The x-component is constant at 3, which means the graph is fixed at x = 3. The y-component is \( 2\cos(t) \) and the z-component is \( 2\sin(t) \), both having a range from -2 to 2 as \( t \) varies.
03

Recognize the Circle in yz-plane

The y-component and z-component form parametric equations \( y = 2\cos(t) \) and \( z = 2\sin(t) \), which describe a circle centered at origin in the yz-plane with radius 2.
04

Transform to 3D Path

Since x is constantly 3, the path described by the vector \( \mathbf{r} \) is a circle lying in a plane perpendicular to the x-axis (yz-plane) but offset to the position where x is constantly 3.
05

Describe the Graph

The graph is a circle of radius 2 in the yz-plane that is parallel to the yz-plane but shifted 3 units along the x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Equations in 3D
A vector equation is a mathematical statement that provides a vector at each point in space, defined through parameters. In this case, the equation \( \mathbf{r}(t) = 3 \mathbf{i} + 2 \cos(t) \mathbf{j} + 2 \sin(t) \mathbf{k} \) is such a vector equation. This uses the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) to describe movement in three-dimensional space. Here's how each component translates:
  • \( x \)-Component: The equation includes a constant term of 3 for \( \mathbf{i} \), indicating it doesn鈥檛 change with \( t \). Hence, this shows a constant alignment along the x-axis, effectively pinning the path at \( x = 3 \).

  • \( y \)-Component: The \( 2 \cos(t) \) term linked with \( \mathbf{j} \) indicates oscillation along the y-axis, a cyclical pattern that mirrors the cosine wave.

  • \( z \)-Component: Similarly, \( 2 \sin(t) \) attached to \( \mathbf{k} \) shows mobility up and down the z-axis in a sinusoidal manner.
Together, these components describe a set path in 3D space, illustrating both linear and cyclical motions.
Understanding Circular Motion
Circular motion is central to interpreting the graph of the provided vector equation. When movements on a plane exhibit periodic motion (like a circle), this is known as circular motion. This is characterized by:
  • Radius: The distance from the circle's center to any point on its circumference. Here, the values 2 in both \( 2 \cos(t) \) and \( 2 \sin(t) \) denote a radius of 2.

  • Periodicity: Both cosine and sine functions are periodic with cycles of \( 2 \pi \), completing full rotations, which is evident in the graph's circular form.
These trigonometrical functions lead to a repeating cycle in the yz-plane, akin to unending circular motion confined to this plane's surface. Since the x-position remains static, the circle doesn't alter its position along the x-axis, translating the circular motion perpendicularly in its entirety at \( x=3 \).
Coordinate Geometry: YZ-Plane
In coordinate geometry, mapping paths or curves involves looking at how points align in space. Here, the focus is the yz-plane鈥攁 flat surface described by y and z coordinates. For this equation:
  • Each point on the path remains on this plane because of having specific y and z values dictated by \( t \).

  • The circular path described by \( y = 2 \cos(t) \) and \( z = 2 \sin(t) \) can be seen as a circle centered at the origin \((0,0)\) within the yz-plane. This plane's orientation means any linear shift along another axis (in this case, x) doesn鈥檛 alter the plane's essence.
The circle in the yz-plane thus appears as a profile (a projection) from any viewpoint along the x-axis, helping simplify visualizations in coordinate geometry. This plane's circular equilibrium, paired with the constant x value, forms a cross-sectional representation of the rotational path described by these vector elements.

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Most popular questions from this chapter

The nuclear accelerator at the Enrico Fermi Laboratory is circular with a radius of \(1 \mathrm{~km}\). Find the scalar normal component of acceleration of a proton moving around the accelerator with a constant speed of \(2.9 \times 10^{5} \mathrm{~km} / \mathrm{s}\).

Find \(\mathbf{T}(t), \mathbf{N}(t)\), and \(\mathbf{B}(t)\) for the given value of \(t\). Then find equations for the osculating, normal, and rectifying planes at the point that corresponds to that value of \(t\). $$ \mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\mathbf{k} ; t=\pi / 4 $$

Find an arc length parametrization of the curve that has the same orientation as the given curve and for which the reference point corresponds to \(t=0 .\) $$ \mathbf{r}(t)=\cos ^{3} t \mathbf{i}+\sin ^{3} t \mathbf{j} ; 0 \leq t \leq \pi / 2 $$

The formula $$ \mathbf{r}(t)=\left(v_{0} \cos \alpha\right) t \mathbf{i}+\left(s_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}\right) \mathbf{j} $$ models a position function for projectile motion [Formula (26)]. Identify the various quantities \(\left(v_{0}, \alpha, s_{0}\right.\), and \(g\) ) in this formula and discuss how the formula is derived, including any assumptions that are made.

As illustrated in the accompanying figure on the next page, suppose that the equations of motion of a particle moving along an elliptic path are \(x=a \cos \omega t\), \(y=b \sin \omega t\) (a) Show that the acceleration is directed toward the origin. (come.). (b) Show that the magnitude of the acceleration is proportional to the distance from the particle to the origin.
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