Problem 22 - Essential Calculus Early Transcendentals

Question

Graph the curve with parametric equations \(x=\cos t\) \(y=\sin t, z=\sin 5 t\) and find the curvature at the point \((1,0,0)\)

Step-by-Step Solution

Answer

The curvature at the point (1,0,0) is \( \frac{1}{26} \).

Step 1: Parametric Equations Definition

The parametric equations given are: \( x = \cos t \), \( y = \sin t \), and \( z = \sin 5t \). These equations define a curve in 3D space where \( t \) is the parameter.

Step 2: Identify the Point on the Curve

The point \((1,0,0)\) corresponds to specific parameter values. From \(x = \cos t = 1\), we determine that \(t = 0\) because cosine achieves 1 at \(t = 0\). Re-evaluation for \(y = \sin t\) and \(z = \sin 5t\) gives \(y = 0\) and \(z = 0\), confirming this point.

Step 3: Compute First Derivatives

Find the derivatives with respect to \( t \): \( \frac{dx}{dt} = -\sin t \), \( \frac{dy}{dt} = \cos t \), and \( \frac{dz}{dt} = 5\cos 5t \). Substitute \(t = 0\), resulting in \(\frac{dx}{dt} = 0\), \(\frac{dy}{dt} = 1\), and \(\frac{dz}{dt} = 5\).

Step 4: Compute Second Derivatives

Determine second derivatives: \( \frac{d^2x}{dt^2} = -\cos t \), \( \frac{d^2y}{dt^2} = -\sin t \), and \( \frac{d^2z}{dt^2} = -25 \sin 5t \). At \(t = 0\), these are \(\frac{d^2x}{dt^2} = -1\), \(\frac{d^2y}{dt^2} = 0\), and \(\frac{d^2z}{dt^2} = 0\).

Step 5: Discuss Curvature Formula

The curvature \( \kappa \) at a point on a parametric curve is given by \[ \kappa = \frac{\| \mathbf{T}' \times \mathbf{T}'' \|}{\| \mathbf{T}' \|^3} \] where \( \mathbf{T}' = ( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} ) \) and \( \mathbf{T}'' \) is the vector of second derivatives.

Step 6: Compute Tangent Vector and Derivatives

The tangent vector \( \mathbf{T}' \) at \( t = 0 \) is \((0, 1, 5)\). The second derivative vector \( \mathbf{T}'' \) is \((-1, 0, 0)\). Calculate the cross product \( \mathbf{T}' \times \mathbf{T}'' = (0, -5, -1)\).

Step 7: Calculate Magnitudes and Curvature

The magnitude of the cross product \( \| \mathbf{T}' \times \mathbf{T}'' \| = \sqrt{0^2 + (-5)^2 + (-1)^2} = \sqrt{26}\). The magnitude of \( \mathbf{T}' \) is \(\sqrt{0^2 + 1^2 + 5^2} = \sqrt{26}\). Thus, the curvature \( \kappa = \frac{\sqrt{26}}{(\sqrt{26})^3} = \frac{1}{26} \).

Key Concepts

Curvature

The concept of curvature measures how sharply a curve bends at a given point. When working with parametric equations, curvature provides insight into the nature of the curve formed by these equations. In general, curvature (denoted as \( \kappa \)) is defined through derivatives of the position and tangent vectors.
In the context of 3D curves represented by parametric equations, the curvature formula is \[\kappa = \frac{\| \mathbf{T}' \times \mathbf{T}'' \|}{\| \mathbf{T}' \|^3}\]Where:

• \( \mathbf{T}' \) is the first derivative (or tangent vector)
• \( \mathbf{T}'' \) is the second derivative vector
• The cross product \( \mathbf{T}' \times \mathbf{T}'' \) gives a vector that is orthogonal to both \( \mathbf{T}' \) and \( \mathbf{T}'' \)

The result, when normalized by the cube of the tangent vector's magnitude, gives a scalar measure of curvature. At a point on a curve, small curvature numbers imply a flatter curve, while larger numbers indicate sharper bends.

3D Space

Three-dimensional (3D) space is a geometric model of the physical universe, representing objects with three coordinates: length, width, and height. In parametric systems, each coordinate is defined by a separate function of the variable \( t \), which represents time or a similar parameter.
For example, consider the parametric equations:

• \( x = \cos t \)
• \( y = \sin t \)
• \( z = \sin 5t \)

These equations generate a path or trajectory in 3D space as \( t \) varies. By adjusting \( t \), you can trace out complicated shapes, such as helices or Lissajous figures.
In this framework, understanding how each parameter influences the curve is vital for graphing and analyzing the path's spatial behavior.

Tangent Vector

The tangent vector at a point on a parametric curve helps to describe the direction in which the curve proceeds from that point. It is built from the first derivatives of the parametric equations.
For a position function \( \mathbf{r}(t) = (x(t), y(t), z(t)) \), the tangent vector, often labeled as \( \mathbf{T}' \), can be expressed as:\[\mathbf{T}' = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right)\]This vector points in the direction of increasing \( t \) and has a magnitude equal to the instantaneous speed if considered over time. For the initial point given by the exercise, the tangent vector came out to be \((0, 1, 5)\).
The tangent vector's role is crucial in calculating the curvature and assessing how sharply the curve experiences changes at a specific location.

Cross Product

In vector calculus, the cross product is an operation on two vectors that returns a third vector, orthogonal (perpendicular) to the original two. This operation is crucial when calculating curvature in 3D space, especially for parametric curves.
Given two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), their cross product \( \mathbf{a} \times \mathbf{b} \) is characterized by:

• Magnitude: Equal to the area of the parallelogram spanned by \( \mathbf{a} \) and \( \mathbf{b} \)
• Direction: Perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \)

For our curve, we find \( \mathbf{T}' \times \mathbf{T}'' \) to calculate curvature. This cross product \[(0, 1, 5) \times (-1, 0, 0) = (0, -5, -1)\]represents a vector orthogonal to both \( \mathbf{T}' \) and \( \mathbf{T}'' \) at the computed point. Understanding and calculating cross products is a critical skill for analyzing curves in 3D space, as it helps derive other essential characteristics like curvature.