Problem 1
Question
Graph the curve traced by the given vector function. $$ \mathbf{r}(t)=2 \sin t \mathbf{i}+4 \cos t \mathbf{j}+t \mathbf{k} ; t \geq 0 $$
Step-by-Step Solution
Verified Answer
The vector function traces a helix in 3D space.
1Step 1: Understand the Vector Function
The vector function is given as \( \mathbf{r}(t) = 2 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + t \mathbf{k} \). It represents a position in 3D space for any given value of \( t \geq 0 \). The components are parameterized by \( t \) for the \( x \), \( y \), and \( z \) axes respectively: \( x(t) = 2 \sin t \), \( y(t) = 4 \cos t \), \( z(t) = t \).
2Step 2: Identify Projection in the XY-plane
Looking at the \( x(t) = 2 \sin t \) and \( y(t) = 4 \cos t \) components, recognize these form an ellipse. This is an ellipse centered at the origin, with a semi-major axis of 4 (along the \( y \)-axis) and a semi-minor axis of 2 (along the \( x \)-axis).
3Step 3: Identify Behavior in the Z-direction
The \( z \)-component is \( t \), indicating a linear increase in the \( z \) direction as \( t \) increases, essentially adding a twist that takes the form of a helix.
4Step 4: Sketch the Helical Curve
Understanding the projections, the function forms a helical structure. As \( t \) increases, the curve traces an ellipse in the \( xy \)-plane while moving linearly upward, giving the shape of a three-dimensional helix. The curve moves counter-clockwise when viewed from above and elevates linearly along the \( z \)-axis.
Key Concepts
Parametric EquationsHelical Curves3D GraphingEllipse in xy-plane
Parametric Equations
Parametric equations are a way of describing a curve in space by expressing the coordinates of points on the curve as functions of a variable, often denoted as \( t \). Instead of using a single equation involving \( x \), \( y \), and \( z \), we use separate equations for each coordinate. This allows us to express complex shapes that cannot be easily described by standard equations. For a vector function like \( \mathbf{r}(t) = 2 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + t \mathbf{k} \), each component corresponds to one of the coordinates:
- \( x(t) = 2 \sin t \)
- \( y(t) = 4 \cos t \)
- \( z(t) = t \)
Helical Curves
A helical curve is a three-dimensional spiral, similar to the shape of a spring or a twirling ribbon. The vector function \( \mathbf{r}(t) \) in the exercise exhibits this helical nature. In the given example, the helical shape arises because the curve maintains a circular motion in one plane while advancing steadily in a perpendicular direction. For our vector function:
- The motion in the \( xy \)-plane forms an ellipse, rotating around the origin.
- The \( z \)-component \( z(t) = t \) adds a linear increase in height.
3D Graphing
3D graphing allows us to visualize curves and surfaces in three dimensions. When dealing with parametric vector functions, such as \( \mathbf{r}(t) = 2 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + t \mathbf{k} \), 3D graphing helps us comprehend how the curve behaves over different dimensions.To graph a helical curve:
- Focus on the projections in each plane: the ellipse in the \( xy \)-plane, and the line in the \( z \)-direction.
- Combine these perspectives to illustrate the complete spiral path.
- Use software tools or sketching methods to draw the 3D path, seeing how it rises as it revolves.
Ellipse in xy-plane
An ellipse in the \( xy \)-plane is a closed curve that resembles an elongated circle. In the given vector function, the terms \( x(t) = 2\sin t \) and \( y(t) = 4\cos t \) define this elliptical shape. The properties of this ellipse are:
- Semi-major axis along the y-axis of length 4
- Semi-minor axis along the x-axis of length 2
Other exercises in this chapter
Problem 1
Compute the gradient for the given function. $$ f(x, y)=x^{2}-x^{3} y^{2}+y^{4} $$
View solution Problem 1
For the given position function, find the unit tangent. $$ \mathbf{r}(t)=(t \cos t-\sin t) \mathbf{i}+(t \sin t+\cos t) \mathbf{j}+t^{2} \mathbf{k}, t>0 $$
View solution Problem 2
$$ \text { Verify the divergence theorem. } $$ $$ \begin{aligned} &\mathbf{F}=6 x y \mathbf{i}+4 y z \mathbf{j}+x e^{-y} \mathbf{k} ; D \text { the region bound
View solution Problem 2
Consider a transformation \(T\) defined by \(x=\sqrt{v-u}\) \(y=v+u\). Find the images of the points \((1,1),(1,3)\), and \((\sqrt{2}, 2)\) in the \(x y\) -plan
View solution