Problem 58 - Calculus Early Transcendentals

Question

Find the curvature and torsion of the curve \(x=\sinh t\) \(y=\cosh t, z=t\) at the point \((0,1,0)\)

Step-by-Step Solution

Answer

Curvature: \(\frac{\sqrt{2}}{2}\), Torsion: \(-1\).

Step 1: Parametric Equation of the Curve

Given the curve defined by \((x = \sinh t, y = \cosh t, z = t)\), we need to evaluate it at the point \((0, 1, 0)\). Determine the parameter \(t\) that places the curve at this point. Set \(x = 0\), \(y = 1\), and \(z = 0\).

Step 2: Determine the Value of Parameter t

The point \((0, 1, 0)\) corresponds to \(x = \sinh t = 0\), which implies \(t = 0\). Verify this with \(y = \cosh t = 1\) and \(z = t = 0\). At \(t = 0\), \(y = 1\) and \(z = 0\), confirming that the parameter value is \(t = 0\).

Step 3: Compute the Tangent Vector

To find the tangent vector, differentiate the position vector \(\mathbf{r}(t) = \langle \sinh t, \cosh t, t \rangle\). The derivative is \(\mathbf{r}'(t) = \langle \cosh t, \sinh t, 1 \rangle\). Evaluate this derivative at \(t = 0\).

Step 4: Evaluate Tangent Vector at t = 0

At \(t = 0\), \(\mathbf{r}'(0) = \langle \cosh(0), \sinh(0), 1 \rangle = \langle 1, 0, 1 \rangle\).

Step 5: Compute the Second Derivative

Find the second derivative, \(\mathbf{r}''(t) = \langle \sinh t, \cosh t, 0 \rangle\). Evaluate this at \(t = 0\).

Step 6: Evaluate Second Derivative at t = 0

At \(t = 0\), \(\mathbf{r}''(0) = \langle 0, 1, 0 \rangle\).

Step 7: Compute Normal Vector and Curvature

The normal vector is the derivative of the unit tangent vector, \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}\). The curvature \(\kappa\) is given by \(\kappa = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}\). Find these quantities at \(t = 0\).

Step 8: Calculate Torsion

Torsion \(\tau\) is given by \(\tau = \frac{(\mathbf{r}'(t) \times \mathbf{r}''(t)) \cdot \mathbf{r}'''(t)}{\|\mathbf{r}'(t)\times \mathbf{r}''(t)\|^2}\). Compute \(\mathbf{r}'''(t) = \langle \cosh t, \sinh t, 0 \rangle\) and evaluate at \(t = 0\).

Step 9: Evaluate and Interpret Results

At \(t = 0\), the cross product \(\mathbf{r}'(0) \times \mathbf{r}''(0) = \langle -1, 0, 1\rangle\). The magnitude is \(\sqrt{2}\), so \(\kappa = \frac{\sqrt{2}}{2}\). For torsion, use \(\mathbf{r}'''(0) = \langle 1, 0, 0 \rangle\), and \((\mathbf{r}'(0) \times \mathbf{r}''(0)) \cdot \mathbf{r}'''(0) = -1\), hence \(\tau = -1\).

Step 10: Final Answer

The curvature \(\kappa\) of the curve at the point \((0, 1, 0)\) is \(\frac{\sqrt{2}}{2}\) and the torsion \(\tau\) is \(-1\).

Key Concepts

Differential Geometry

Differential geometry is a field of mathematics that deals with curves, surfaces, and their properties through differential and integral calculus. It provides a way to understand complex shapes by breaking them down into infinitesimally small pieces, described by mathematical functions. This branch of mathematics is essential when analyzing curves, like the ones studied here, to understand their intrinsic properties such as curvature and torsion. In differential geometry:

• Curvature measures how quickly a curve changes direction. It's like asking, "How sharply is this curve turning?" Curvature is important not only in abstract mathematics but also in physics and engineering for understanding motion and structure stability.
• Torsion tells us how a curve twists out of the plane formed by its tangent and normal vectors. It's comparable to asking, "Is the curve spiraling, and if so, how fast?" Torsion is crucial in applications involving mechanical systems and 3D modeling.

Understanding these elements involves differentiating the curve's position vector to obtain tangent, normal, and binormal vectors, which in turn provide insights into how the curve behaves in space.

Parametric Curves

In mathematics, a parametric curve is one where the x, y, and z coordinates are expressed as functions of a parameter, often denoted as 't'. This approach allows such curves to be easily manipulated and analyzed, offering flexibility in both calculus and geometry.

The given problem involves the parametric equations:

• \(x = \sinh t\): This represents the x-coordinate of the curve, using the hyperbolic sine function, which describes curves that can extend indefinitely.
• \(y = \cosh t\): For the y-coordinate, the hyperbolic cosine function is used, indicating a smooth curve that remains above the x-axis.
• \(z = t\): The z-coordinate is directly equivalent to the parameter itself, which implies linear progression along the z-axis.

Parametric curves are particularly powerful as they allow us to investigate the properties of curves in space without restriction to a Cartesian plane. This makes them invaluable in fields such as computer graphics and robotics, where modeling complex paths is crucial.

Calculus

Calculus is the mathematical study that forms the backbone for understanding change and motion, providing the tools needed to analyze the behavior of parametric curves. In the context of our original problem, calculus is used extensively for:

• Derivatives: Calculating the first, second, and even third derivatives of the position vector is essential. This provides the tangent, normal, and curvature vectors, which are core components of kinematic analysis.
• Cross Products: These are used to find vectors that are perpendicular to two given vectors, such as the tangent and normal vectors. Crucially, the magnitude of this cross product helps determine the curvature and torsion.
• Magnitude and Normalization: Taking the magnitudes of vectors helps in scaling them to unit vectors for further calculations, a common requirement in analyzing properties like torsion.

In summary, calculus allows us to dissect the curve's behavior at any point along its path, revealing deeper insights into its properties and the forces at play that shape it.