Question

Consider the curve. $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$ \begin{aligned} \mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i} \\ &+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j}+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k}\end{aligned}$$

Short answer

Answer: The curve is an ellipse-like shape that lies on a plane in 3D space, without any intersection points with the x, y, and z axes. It can be viewed as a projection of a circle into the plane.

Step-by-step solution

Write the parametric equations for x(t), y(t), and z(t)

The given curve has the parametric equations: $$ x(t) = \frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t \\ y(t) = -\frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t \\ z(t) = \frac{1}{\sqrt{3}} \sin t $$

Find the intersection points with the coordinate axes

To find the intersection points with the x, y, and z axes, we set the other two coordinates to zero and solve for t. Intersection with x-axis (y(t) = 0, z(t) = 0): $$ -\frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t = 0 \\ \frac{1}{\sqrt{3}} \sin t = 0 $$ Intersection with y-axis (x(t) = 0, z(t) = 0): $$ \frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t = 0 \\ \frac{1}{\sqrt{3}} \sin t = 0 $$ Intersection with z-axis (x(t) = 0, y(t) = 0): $$ \frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t = 0 \\ -\frac{1}{\sqrt{2}} \cos t + \frac{1}{\sqrt{3}} \sin t = 0 $$ Solving these systems of equations, we find that there are no intersection points with x, y, and z axes.

Calculate some points on the curve

To see the behavior of the curve, we will calculate the coordinates for some values of t: $$ t=0: \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \\ t=\frac{\pi}{4}: \left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{3}}\right) \\ t=\frac{\pi}{2}: \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \\ t=\frac{3\pi}{4}: \left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{3}}\right) $$

Visualize and describe the curve

Based on the calculated points and the fact that the curve lies in a plane, it can be visualized as an ellipse-like shape (without intersection points with the axes) that lies on a plane in the 3D space. The curve is simply a projection of a circle into the plane.

Key concepts

3D Parametric Curves

In the world of mathematics, a parametric equation describes a curve by expressing the coordinates of the points of the curve as functions of a parameter. In three-dimensional space, 3D parametric curves involve three functions, each representing one coordinate axis.
For the given problem, the parametric curve is defined by \[ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \] where the parameter \( t \) varies within a specified interval. Each function (\( x(t) \), \( y(t) \), and \( z(t) \)) depends on \( t \).
These curves allow tracing more complex shapes than can be achieved with simple equations. In this example, the complexity arises from the coefficients of cosine and sine for each component.
It's intriguing how these combinations affect the shape and orientation of the curve in 3D space.

Intersection Points

The concept of intersection points provides insight into where parametric curves meet or cross certain boundaries, like axes in coordinate systems.
When examining whether a parametric curve intersects any axis, we set the remaining coordinate functions to zero in the parametric equations.

• For the x-axis: Set \( y(t) = 0 \) and \( z(t) = 0 \).
• For the y-axis: Set \( x(t) = 0 \) and \( z(t) = 0 \).
• For the z-axis: Set \( x(t) = 0 \) and \( y(t) = 0 \).

In this exercise, solving the system of equations after setting these conditions shows no intersection with any axes. This result helps us understand better how the curve is positioned in relation to these axes.

Graphical Visualization

Visualizing 3D parametric curves can be challenging, but it's a necessary tool to understand how such a curve behaves in space.

• A basic visualization involves plotting calculated points for various values of \( t \).
• Connecting these points can help trace the path, revealing how the curve looks.

This step helps students form a mental picture of abstract mathematical concepts.
In this scenario, the points calculated suggest an ellipse-like shape lying on a plane. Graphical tools or software can be useful here, allowing students to explore and rotate the curve to examine it from different perspectives. This visualization confirms understanding and interpretation of the curve's characteristics.

Plane Curves

A plane curve is any curve that lies entirely within a single plane in three-dimensional space.
Although the parametric equation is initially in 3D, by proving it lies in a plane, it simplifies the analysis significantly.
The challenge is demonstrating that all points \( \mathbf{r}(t) \) of the curve satisfy an equation of the form \[ Ax + By + Cz = D \] for constants \( A \), \( B \), \( C \), and \( D \).
In the exercise, after reviewing the parametric form, it's clear that the complex coefficients of cosine and sine indicate a reduction to a plane, forming an ellipse-like curve without any axis intersections. The insight gained here is crucial for simpler mathematical and computational analyses.