Chapter 12: Problem 22
Sketch the curve traced out by the vectorvalued function. Indicate the direction in which the curve is traced out. $$ \mathbf{F}(t)=2 \cos \mathbf{t} \mathbf{i}-\sin t \mathbf{j}-3 \mathbf{k} \text { for }-\pi \leq t \leq 0 $$
Short Answer
Expert verified
The curve is an ellipse lying in the plane \( z = -3 \), traced clockwise from \( x=2, y=0 \).
Step by step solution
01
Identify the Components of the Vector Function
The vector function \( \mathbf{F}(t) = 2 \cos t \mathbf{i} - \sin t \mathbf{j} - 3 \mathbf{k} \) consists of three components:- The \( \mathbf{i} \) component: \( 2 \cos t \)- The \( \mathbf{j} \) component: \( -\sin t \)- The \( \mathbf{k} \) component: \( -3 \).The function's output is a point in space with these components, giving a curve as \( t \) varies from \( -\pi \) to \( 0 \).
02
Understand the Individual Components
Each component represents a direction in 3D space:- \( 2\cos(t) \): This oscillates between \(-2\) and \(2\) as \( t \) changes.- \( -\sin(t) \): This oscillates between \(-1\) and \(1\), with a negative sign flipping the standard sine wave.- \( -3 \): This is constant, indicating that the curve lies in a plane parallel to the \( xy \)-plane at \( z = -3 \).
03
Determine the Curve's Path
Considering the \( \mathbf{i} \) and \( \mathbf{j} \) components:- \( (x(t), y(t)) = (2\cos(t), -\sin(t)) \).These parametric equations suggest an ellipse in the \( xy \)-plane, as the values of \( x(t) \) and \( y(t) \) trace out the path of an ellipse.
04
Understand the Direction and Curve
The parameter \( t \) decreases from \( -\pi \) to \( 0 \). Calculate key points:- At \( t = -\pi \), \( x = 2, y = 0 \). - At \( t = 0 \), \( x = 2, y = 0 \).These calculations confirm the ellipse begins and ends at the same point, \( (2, 0) \), tracing clockwise from \( t = -\pi \) to \( t = 0 \).
05
Sketch and Indicate Direction
Plot the points and trace the elliptical path on the \( xy \)-plane, with a constant \( z \) value of \( -3 \). Mark the directions using arrows starting from \( x = 2, y = 0 \) moving through key points, e.g., \( t = -\frac{3\pi}{2} \) (\(-2, 1\)), highlighting the clockwise direction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parameterized Curves
A parameterized curve in mathematics is a representation of a curve using a parameter, typically denoted as \( t \). Instead of describing a curve directly by a geometric shape or a single equation, we use a set of equations that define each coordinate of the curve separately, typically x, y, and z in 3D space. This allows for greater flexibility, as the curve's path is defined by changing the parameter's value over a specified range.
For example, with our function \( \mathbf{F}(t) = 2 \cos t \mathbf{i} - \sin t \mathbf{j} - 3 \mathbf{k} \), \( t \) is our parameter. Here, \( t \) varies between \( -\pi \) and \( 0 \), creating a parametrization of the path traced by our curve. This approach enables us to easily sketch complex shapes by calculating key points, without solving a single equation involving all coordinates.
For example, with our function \( \mathbf{F}(t) = 2 \cos t \mathbf{i} - \sin t \mathbf{j} - 3 \mathbf{k} \), \( t \) is our parameter. Here, \( t \) varies between \( -\pi \) and \( 0 \), creating a parametrization of the path traced by our curve. This approach enables us to easily sketch complex shapes by calculating key points, without solving a single equation involving all coordinates.
3D Space
Understanding 3D space is essential for examining vector-valued functions and their associated curves. Unlike the 2D plane that involves only the x and y axes, 3D space adds a z-axis, giving a complete picture of motion and positioning. Each vector in 3D has three components, which correspond to movements along these three independent axes.
In the given exercise, the vector \( \mathbf{F}(t) \) is composed of three parts: \( 2 \cos t \mathbf{i}\), \( -\sin t \mathbf{j}\), and \( -3 \mathbf{k}\). The components linking to \( \mathbf{i} \) and \( \mathbf{j} \) determine the path on the xy-plane, while the constant \(-3\) for \( \mathbf{k} \) indicates a fixed position along the z-axis. Essentially, this places our curve on a plane at height \( z = -3 \) in the given space, reducing its need for consideration of vertical movement.
In the given exercise, the vector \( \mathbf{F}(t) \) is composed of three parts: \( 2 \cos t \mathbf{i}\), \( -\sin t \mathbf{j}\), and \( -3 \mathbf{k}\). The components linking to \( \mathbf{i} \) and \( \mathbf{j} \) determine the path on the xy-plane, while the constant \(-3\) for \( \mathbf{k} \) indicates a fixed position along the z-axis. Essentially, this places our curve on a plane at height \( z = -3 \) in the given space, reducing its need for consideration of vertical movement.
Components of Vectors
Vectors can be broken down into components that describe their influence in different directions. In a 3D vector function such as \( \mathbf{F}(t) = 2 \cos t \mathbf{i} - \sin t \mathbf{j} - 3 \mathbf{k} \), each component serves a purpose:
- The \( \mathbf{i} \) component, \( 2 \cos t \), affects the x-axis, altering the horizontal position.
- The \( \mathbf{j} \) component, \( -\sin t \), controls the y-axis, influencing the vertical movement.
- The \( \mathbf{k} \) component, being a constant \(-3\), denotes the z-axis position, fixing the curve in a plane parallel to the xy-plane.
Sketching Curves
Sketching curves from vector-valued functions requires interpreting the parametric equations for each axis. In this instance, the curve described by \( \mathbf{F}(t) = 2 \cos t \mathbf{i} - \sin t \mathbf{j} - 3 \mathbf{k} \) suggests movements in 3D space.
To sketch this, focus on the path traced out on the xy-plane by the \( 2 \cos t \mathbf{i} \) and \( -\sin t \mathbf{j} \) components. These parametric forms imply that the curve traces an ellipse, as the x-component \( 2 \cos t \) ranges from -2 to 2, and y-component \( -\sin t \) spans from -1 to 1. Constant \(-3\) along the z-axis ensures the ellipse is drawn on the plane at z = -3.
Visualizing the direction involves observing how values change as \( t \) adjusts. Starting at \( t = -\pi \), the curve starts at the point (2, 0, -3) and progresses clockwise around the ellipse, ending back at the same coordinate when \( t \) reaches 0. This path can be marked and connected with arrows to illustrate the route the curve takes, providing a clear and comprehensible sketch of the vector's journey.
To sketch this, focus on the path traced out on the xy-plane by the \( 2 \cos t \mathbf{i} \) and \( -\sin t \mathbf{j} \) components. These parametric forms imply that the curve traces an ellipse, as the x-component \( 2 \cos t \) ranges from -2 to 2, and y-component \( -\sin t \) spans from -1 to 1. Constant \(-3\) along the z-axis ensures the ellipse is drawn on the plane at z = -3.
Visualizing the direction involves observing how values change as \( t \) adjusts. Starting at \( t = -\pi \), the curve starts at the point (2, 0, -3) and progresses clockwise around the ellipse, ending back at the same coordinate when \( t \) reaches 0. This path can be marked and connected with arrows to illustrate the route the curve takes, providing a clear and comprehensible sketch of the vector's journey.