Question

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) 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. $$\begin{aligned}\mathbf{r}(t)=&(2 \cos t+2 \sin t) \mathbf{i}+(-\cos t+2 \sin t) \mathbf{j} \\\&+(\cos t-2 \sin t) \mathbf{k}\end{aligned}$$

Short answer

Answer: The given curve is a straight line and lies in the plane \(z = 1 - y\).

Step-by-step solution

(Step 1: Identify the functions for x, y, and z components)

The given parametric curve is $$\mathbf{r}(t) =(2 \cos t + 2 \sin t) \mathbf{i} + (-\cos t + 2 \sin t) \mathbf{j} + (\cos t - 2 \sin t) \mathbf{k} .$$ We can rewrite it as $$ x(t) = 2 \cos t + 2 \sin t, \\ y(t) = -\cos t + 2 \sin t, \\ z(t) = \cos t - 2 \sin t. $$

(Step 2: Eliminate the parameter t)

To eliminate the parameter t, we will first find the expression of \(\sin t\) and \(\cos t\) in terms of x, then substitute these expressions into expressions for y and z. From the expression of x(t), $$ x(t) = 2 \cos t + 2 \sin t \\ \Rightarrow \cos t + \sin t = \frac{x}{2} \\ \Rightarrow \cos^2 t + 2 \sin t \cos t + \sin^2 t = \frac{x^2}{4} \\ \Rightarrow 1 + 2 \sin t \cos t = \frac{x^2}{4} \\ \Rightarrow \sin t \cos t = \frac{x^2}{8} - \frac{1}{2}.$$

(Step 3: Finding sin(t) and cos(t) and substitute)

Now, let's rewrite the expressions for \(y(t)\) and \(z(t)\) using the expression for \(\sin t \cos t \) We know, $$y(t) = -\cos t + 2 \sin t = \cos t (2 \sin t - 1) \\ z(t) = \cos t - 2 \sin t = \cos t (1 - 2 \sin t)$$ We have, $$\sin t \cos t = \frac{x^2}{8} - \frac{1}{2}$$ Now, substituting the values, $$y= (2 \sin t - 1)(\cos t) \\ z= (1 - 2 \sin t)(\cos t)$$ We can substitute the expression of \(\sin t \cos t\), $$y = 2\left(\frac{x^2}{8} - \frac{1}{2}\right) - \cos t \\ z = 1 - 2\left(\frac{x^2}{8} - \frac{1}{2}\right) - \cos t$$

(Step 4: Express z in terms of x and y)

Now we express z in terms of x and y. From the above expressions, we have $$y + \cos t = 2\left(\frac{x^2}{8} - \frac{1}{2}\right) \\ z - \cos t = 1 - 2\left(\frac{x^2}{8} - \frac{1}{2}\right)$$ Adding the two equations: $$y + z = 1 \\ \Rightarrow z = 1 - y$$

(Step 5: Describe the curve)

The curve lies in the plane with the equation \(z = 1 - y\). This is a plane parallel to the x-axis, and every value of x corresponds to a unique combination of y and z values. As t varies, the curve oscillates within this plane, and since it's a linear equation, it forms a straight line. The curve is a straight line within the plane \(z = 1-y\). In conclusion, the given curve is a straight line and lies in the plane \(z=1-y\).

Key concepts

Vector Functions

Vector functions are a fascinating concept in mathematics, extending our understanding from scalar functions to multi-dimensional space. They assign a vector to each value in their domain. In essence, a vector function takes a parameter, typically denoted as \( t \), and returns a vector. Consider a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), where \( x(t) \), \( y(t) \), and \( z(t) \) are the component functions determining the behavior in the 3D space spanned by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).

• **Domain:** The set of all possible values of \( t \)
• **Image:** The curve described by the function in space

In the original exercise, our vector function is \( \mathbf{r}(t) = (2 \cos t + 2 \sin t) \mathbf{i} - (\cos t - 2 \sin t) \mathbf{j} + (\cos t - 2 \sin t) \mathbf{k} \). Each component describes the trajectory in the respective direction, and together they form a path, or curve, in the three-dimensional space.

Plane Curves

A plane curve is a curve that resides entirely within a single plane. In mathematics, this concept is crucial for understanding how curves behave in two-dimensional planes even when described by parameters in three-dimensional space. When dealing with vector functions like \( \mathbf{r}(t) \), it becomes essential to determine whether the path described lies on a specific plane. In our original exercise, we're looking at a specific case where the curve is shown to lie in a plane described by the relation \( z = 1 - y \). This indicates the curve takes on a specific trajectory that doesn't stray into a third dimension independently.

• The plane’s equation, \( z = 1 - y \), illustrates that every point on the curve satisfies this linear relation.
• The curve does not twist out of this plane, even though it is described in three-dimensional space.

Understanding this helps simplify the problem by reducing dimensions, which makes graphing and analysis more manageable.

Eliminating Parameters

Eliminating parameters from equations simplifies complex forms, making it easier to interpret the geometry of a curve. When given a parametric equation, the aim of eliminating the parameter \( t \) is often to find a direct relationship between variables without the auxiliary parameter.In the given exercise, parameter elimination involves expressing \( y \) and \( z \) in terms of \( x \) alone, effectively removing \( t \). This conversion explores algebraic manipulation to achieve such simplification:

• Starting with \( x(t) = 2 \cos t + 2 \sin t \), find expressions for \( \sin t \) and \( \cos t \).
• Substitute these into expressions for \( y(t) \) and \( z(t) \).
• This reduction process results in a direct equation, like \( z = 1 - y \), which describes how \( y \) and \( z \) relate without the need for \( t \).

The ability to eliminate the parameter is powerful. It transforms a parameter-based approach into a more straightforward form, aiding in better visualization and comprehension of the plane in which the curve resides.