Chapter 12: Problem 25
Sketch the curve traced out by the vectorvalued function. Indicate the direction in which the curve is traced out. $$ \mathbf{F}(t)=3 \sin t \mathbf{i}+3 \sin t \mathbf{j}-3 \sqrt{2} \cos t \mathbf{k} $$
Short Answer
Expert verified
Sketch an elliptical helix along \( x = y \) with direction as \( t \) increases.
Step by step solution
01
Identify the Components
Examine the components of the vector function \( \mathbf{F}(t) = 3 \sin t \mathbf{i} + 3 \sin t \mathbf{j} - 3 \sqrt{2} \cos t \mathbf{k} \). The \( \mathbf{i} \) and \( \mathbf{j} \) components are both \( 3 \sin t \), and the \( \mathbf{k} \) component is \( -3 \sqrt{2} \cos t \).
02
Identify the Plane of the Curve
Since both \( x(t) = 3 \sin t \) and \( y(t) = 3 \sin t \) are equal, the curve lies in a plane intersecting equally along the \( x \, \) and \( y \, \) axes. This suggests that the projection onto the \( xy \)-plane is a line \( x = y \).
03
Parametrize the Curve in 3D
Rewrite the function using the equality \( x = y \): \( x(t) = y(t) = 3 \sin t \). We have the third component as \( z(t) = -3 \sqrt{2} \cos t \). This leads to the parametric equations \[ x(t) = 3 \sin t, \quad y(t) = 3 \sin t, \quad z(t) = -3 \sqrt{2} \cos t. \]
04
Analyze the Curve's Shape
Recognizing that both the \( x \) and \( y \) components are \( 3 \sin t \), these components describe a diagonal line on the \( x = y \) plane. For \( z(t) = -3 \sqrt{2} \cos t \), the curve's projection onto the \( xz \) or \( yz \) planes will be an ellipse, with \( z = 0 \) when \( \cos t = 0 \).
05
Sketch the Curve
The curve traces out an elliptical helical path along the plane \( x = y \), given by these ellipses in the \( xz \) and \( yz \) planes. The ellipse oscillates between maxima and minima \( z = \pm 3 \sqrt{2} \) as \( cos(t) \) takes these values.
06
Determine the Direction of Traversal
Since \( t \) represents time, the direction of the curve is determined by \( t \) increasing; here, this means \( (x,y) \) increases and decreases periodically, and \( z \) oscillates from \(-3\sqrt{2}\) to \(3\sqrt{2}\) and back, indicating a wave moving along z when \( \sin t \) is increasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way to represent a curve in mathematics using parameters, typically designated by a variable like "t". These equations help us define points in a coordinate system as functions of this parameter. Instead of describing a curve explicitly with an equation like \(y = f(x)\), parametric equations express both x and y in terms of a third variable, usually time.For instance, in our exercise, both the x and y coordinates are represented by \(3 \sin t\). Here:
- \(x(t) = 3 \sin t\)
- \(y(t) = 3 \sin t\)
3D Vector Functions
3D vector functions are crucial in representing and analyzing curves in three-dimensional space. A vector function maps a scalar parameter, like 't' in our exercise, to a vector in 3D. Here, the function \(\mathbf{F}(t) = 3 \sin t \mathbf{i} + 3 \sin t \mathbf{j} - 3 \sqrt{2} \cos t \mathbf{k}\) defines the curve.Let's break down the components:
- \(3 \sin t \mathbf{i}\) and \(3 \sin t \mathbf{j}\) together describe movement in the x-y plane along a line where x equals y.
- \(-3 \sqrt{2} \cos t \mathbf{k}\) describes motion up and down in the z direction.
Vector-Valued Functions
Vector-valued functions extend the concept of parametric functions by using vectors to encapsulate more complex movement in space. In a vector-valued function, each component of the vector is a function of a separate parameter, frequently time 't', which allows the representation of curves and surfaces in a concise, function-based form.In the given function \(\mathbf{F}(t) = 3 \sin t \mathbf{i} + 3 \sin t \mathbf{j} - 3 \sqrt{2} \cos t \mathbf{k}\):
- The x and y components formed by \(3 \sin t\) signify synchronized oscillations along the x-y diagonal plane.
- The z component, \(-3 \sqrt{2} \cos t\), introduces oscillatory motion vertically.