Problem 221 - Calculus Volume 3

Question

Find the curvature for the following vector functions. \(\mathbf{r}(t)=\sqrt{2} e^{t} \mathbf{i}+\sqrt{2} e^{-t} \mathbf{j}+2 t \mathbf{k}\)

Step-by-Step Solution

Answer

The curvature of the vector function is \( \kappa(t) = \frac{\sqrt{32 + 8 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3} \).

Step 1: Find the First Derivative

The first step is to find the derivative of the vector function \( \mathbf{r}(t) = \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j} + 2t \mathbf{k} \) with respect to \( t \). Differentiate each component function separately:\[ \mathbf{r}'(t) = \left( \sqrt{2} e^{t} \right) \mathbf{i} + \left( -\sqrt{2} e^{-t} \right) \mathbf{j} + 2 \mathbf{k} \]

Step 2: Find the Second Derivative

Now, find the derivative of \( \mathbf{r}'(t) \) to get the second derivative \( \mathbf{r}''(t) \).Differentiate each component of \( \mathbf{r}'(t) = \sqrt{2} e^{t} \mathbf{i} - \sqrt{2} e^{-t} \mathbf{j} + 2 \mathbf{k} \):\[ \mathbf{r}''(t) = \left( \sqrt{2} e^{t} \right) \mathbf{i} + \left( \sqrt{2} e^{-t} \right) \mathbf{j} \]

Step 3: Calculate Cross Product

Compute the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):Firstly, note the vector components:- \( \mathbf{A} = \mathbf{r}'(t) = (\sqrt{2} e^{t}, -\sqrt{2} e^{-t}, 2) \)- \( \mathbf{B} = \mathbf{r}''(t) = (\sqrt{2} e^{t}, \sqrt{2} e^{-t}, 0) \)The cross product is calculated as:\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \sqrt{2} e^{t} & -\sqrt{2} e^{-t} & 2 \ \sqrt{2} e^{t} & \sqrt{2} e^{-t} & 0 \end{array} \right| \]Solving the determinant, we find:\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = (-4 \sqrt{2} \mathbf{j} - 2\sqrt{2} e^{t} \mathbf{k}) \]

Step 4: Compute Magnitudes

Next, calculate the magnitudes.- For \( \mathbf{r}'(t) \): \[ \| \mathbf{r}'(t) \| = \sqrt{(\sqrt{2} e^{t})^2 + (-\sqrt{2} e^{-t})^2 + (2)^2} = \sqrt{2 e^{2t} + 2 e^{-2t} + 4} \]- For \( \mathbf{r}'(t) \times \mathbf{r}''(t) \): \[ \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| = \sqrt{(-4 \sqrt{2})^2 + (-2 \sqrt{2} e^{t})^2} = \sqrt{32 + 8 e^{2t}} \]

Step 5: Calculate Curvature

The curvature \( \kappa(t) \) is calculated using the following formula:\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]Plug in the magnitudes calculated in step 4:\[ \kappa(t) = \frac{\sqrt{32 + 8 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3} \]

Key Concepts

Understanding Vector Functions

Vector functions describe how a vector moves through space over time. They take a parameter, often time denoted as \( t \), and output a vector. For example, the function \( \mathbf{r}(t) = \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j} + 2t \mathbf{k} \) describes a path in 3D space. Each component represents movement along the \( x \), \( y \), and \( z \) axes respectively.
Here's why vector functions are important:

• They allow the description of curves and paths in space, which are crucial in physics and engineering.
• Understanding them is essential for solving problems in mechanics and electromagnetism where movement is involved.

Learning about vector functions will aid in visualizing how objects move and change over time.

Grasping Derivatives in Calculus

Derivatives are all about understanding change. In calculus, when we talk about the derivative of a function, we are looking at how that function changes at any given point. For vector functions, derivatives are computed component-wise, which means differentiating each part of the vector independently.
Steps for finding derivatives of a vector function include:

• Differentiate each component function separately.
• Combine these derivatives back into a single vector.

The first derivative of a vector function gives us the velocity vector, showing the direction and rate of change of the vector over time. For instance, \( \mathbf{r}'(t) = \left( \sqrt{2} e^{t} \right) \mathbf{i} + \left( -\sqrt{2} e^{-t} \right) \mathbf{j} + 2 \mathbf{k} \).
Mastering derivatives is vital because they provide information about motion, force, and energy dynamics.

Getting the Hang of the Cross Product in Vector Calculus

The cross product is a vector operation that takes two vectors and returns a third vector perpendicular to both. It’s unique to vector calculus and has specific rules compared to other types of products.
Properties of the cross product include:

• The magnitude of the resulting vector corresponds to the area of the parallelogram spanned by the original vectors.
• The direction is determined by the right-hand rule.

In curvature calculations, the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) helps ascertain how sharply a curve is changing direction. For instance, this problem gives \( \mathbf{r}'(t) \times \mathbf{r}''(t) = (-4 \sqrt{2} \mathbf{j} - 2\sqrt{2} e^{t} \mathbf{k}) \).
Understanding the cross product assists in analyzing rotational dynamics and torque in physics.

Unpacking Curvature Calculation

Curvature quantifies how fast a curve is changing direction at a given point. In vector calculus, it’s essential for understanding the geometry of paths and surfaces.
To calculate curvature \( \kappa(t) \), use the formula:

• \[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]
• The numerator is the magnitude of the cross product of the first and second derivatives, while the denominator is the magnitude of the first derivative raised to the third power.

In our example, substituting in the magnitudes yields \( \kappa(t) = \frac{\sqrt{32 + 8 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3} \), capturing the bend of the curve in space. Curvature is crucial for designing paths and analyzing trajectories in engineering and physics.

Calculating the Magnitude of Vectors

The magnitude of a vector also known as its length or norm, provides a scalar measure of its size regardless of direction. In three dimensions, it is computed using the Pythagorean theorem.
To find the magnitude \( \| \mathbf{v} \| \) of a vector \( \mathbf{v} = (x, y, z) \), use:

• \[ \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \]

In the curvature problem, these calculations were needed twice: once for the magnitude of the first derivative vector and again for the magnitude of the cross product.
Having a firm grasp on how to calculate magnitudes is useful in determining distances, speeds, and forces represented by vectors.