Problem 70 - Calculus III

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 \( \kappa \) is \( \frac{\sqrt{8 e^{-2t} + 16 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3} \).

Step 1: Differentiate the vector function

First, we need to find the derivative of the vector function \( \mathbf{r}(t) = \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j} + 2t \mathbf{k} \). The derivative, \( \mathbf{r}'(t) \), is calculated as follows:\[\mathbf{r}'(t) = \frac{d}{dt}[ \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j} + 2t \mathbf{k} ]\]\[= \sqrt{2} e^{t} \mathbf{i} - \sqrt{2} e^{-t} \mathbf{j} + 2 \mathbf{k}.\]

Step 2: Compute the second derivative

Next, find the second derivative, \( \mathbf{r}''(t) \), which is the derivative of \( \mathbf{r}'(t) \):\[\mathbf{r}''(t) = \frac{d}{dt}[\sqrt{2} e^{t} \mathbf{i} - \sqrt{2} e^{-t} \mathbf{j} + 2 \mathbf{k}]\]\[= \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j}.\]

Step 3: Calculate the cross product

To find the curvature, we need the magnitude of the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):\[\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\sqrt{2} e^{t} & -\sqrt{2} e^{-t} & 2 \\sqrt{2} e^{t} & \sqrt{2} e^{-t} & 0 \end{vmatrix}\]Calculate the determinant:\[= (0 + 2 \sqrt{2} e^{-t})\mathbf{i} - (0 - 2 \sqrt{2} e^{t})\mathbf{j} + (- \sqrt{2} e^{t} - \sqrt{2} e^{t})\mathbf{k}\]\[= 2 \sqrt{2} e^{-t} \mathbf{i} + 2 \sqrt{2} e^{t} \mathbf{j} - 2 \sqrt{2} e^{t} \mathbf{k}.\]

Step 4: Find the magnitude of the cross product

The magnitude of the cross product is:\[|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(2 \sqrt{2} e^{-t})^2 + (2 \sqrt{2} e^{t})^2 + (-2 \sqrt{2} e^{t})^2}\]\[= \sqrt{8 e^{-2t} + 8 e^{2t} + 8 e^{2t}} = \sqrt{8 e^{-2t} + 16 e^{2t}}.\]

Step 5: Find the magnitude of \( \mathbf{r}'(t) \)

Now, calculate the magnitude of the vector \( \mathbf{r}'(t) \):\[|\mathbf{r}'(t)| = \sqrt{(\sqrt{2} e^{t})^2 + (-\sqrt{2} e^{-t})^2 + 2^2}\]\[= \sqrt{2 e^{2t} + 2 e^{-2t} + 4}.\]

Step 6: Calculate the curvature

The curvature \( \kappa \) is given by:\[\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\]Substitute the expressions for the magnitudes obtained in Steps 4 and 5:\[\kappa = \frac{\sqrt{8 e^{-2t} + 16 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3}.\]

Step 7: Final Result

Simplify the expression for curvature if possible. In this case, the curvature expression can be analyzed as is since it is quite complex.

Key Concepts

Vector Functions

Vector functions are a fundamental concept in calculus, particularly when dealing with curves in space. These functions assign a vector to each point in their domain. For example, the given vector function \[ \mathbf{r}(t)=\sqrt{2} e^{t} \mathbf{i}+\sqrt{2} e^{-t} \mathbf{j}+2t \mathbf{k} \] represents a curve in three-dimensional space. Here,

• \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes.
• \( \sqrt{2} e^{t} \), \( \sqrt{2} e^{-t} \), and \( 2t \) are functions of \( t \) that give the x, y, and z components of the vector at any time \( t \).

Understanding vector functions involves recognizing how each component contributes to the position of a point on the curve. This forms the basis for analyzing the curve's properties, such as its curvature.

Derivatives

The derivative of a vector function is crucial in understanding how a curve behaves. It gives us the rate at which the vector's position changes with respect to its parameter. For the given function,\[ \mathbf{r}(t)=\sqrt{2} e^{t} \mathbf{i}+\sqrt{2} e^{-t} \mathbf{j}+2t \mathbf{k} \]the derivative is computed component-wise:

• Differentiate \( \sqrt{2} e^{t} \) to get \( \sqrt{2} e^{t} \).
• Differentiate \( \sqrt{2} e^{-t} \) to get \(-\sqrt{2} e^{-t} \).
• Differentiate \( 2t \) to get \( 2 \).

Thus, the derivative of the vector function is\[ \mathbf{r}'(t) = \sqrt{2} e^{t} \mathbf{i} - \sqrt{2} e^{-t} \mathbf{j} + 2 \mathbf{k}. \]This vector points in the direction of the curve's tangent at each point. It is essential for determining how quickly a point moves along the curve.

Cross Product

The cross product is a vector operation that results in a new vector orthogonal to the two input vectors. It has applications in determining angles and orientations between vectors. When dealing with curvature, we compute the cross product of the first and second derivatives of a vector function.
For this function:

• The first derivative, \( \mathbf{r}'(t) \), is \( \sqrt{2} e^{t} \mathbf{i} - \sqrt{2} e^{-t} \mathbf{j} + 2 \mathbf{k} \).
• The second derivative, \( \mathbf{r}''(t) \), is \( \sqrt{2} e^{t} \mathbf{i} + \sqrt{2} e^{-t} \mathbf{j} \).

The cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) gives a vector that indicates how the curve is bending or twisting at any point in time. Calculating it involves evaluating the determinant:\[\mathbf{r}'(t) \times \mathbf{r}''(t) = 2 \sqrt{2} e^{-t} \mathbf{i} + 2 \sqrt{2} e^{t} \mathbf{j} - 2 \sqrt{2} e^{t} \mathbf{k}. \]

Magnitude

The magnitude of a vector represents its length or size. For a vector in three-dimensional space, its magnitude is calculated using the Pythagorean theorem in three dimensions. The magnitude of the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) tells us how hard the curve bends.
For this curve, we evaluate the magnitude as:\[|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(2 \sqrt{2} e^{-t})^2 + (2 \sqrt{2} e^{t})^2 + (-2 \sqrt{2} e^{t})^2}.\]The calculation simplifies to:\[\sqrt{8 e^{-2t} + 16 e^{2t}}. \]Similarly, finding the magnitude of \( \mathbf{r}'(t) \) is crucial because it provides the length of the tangent vector:\[|\mathbf{r}'(t)| = \sqrt{2 e^{2t} + 2 e^{-2t} + 4}. \]These magnitudes are used directly in the curvature formula.

Curvature Formula

Curvature is a measure of how quickly a curve deviates from being a straight line. The formula for curvature \( \kappa \) is:\[\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}. \]This ratio compares the tendency of the curve to bend versus the tendency to maintain a straight path. To calculate \( \kappa \) for the given vector function, we use:

• The magnitude of the cross product \(|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{8 e^{-2t} + 16 e^{2t}}\).
• The cube of the magnitude of the first derivative \(|\mathbf{r}'(t)|^3 = (\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3\).

By placing these into the formula, we find:\[\kappa = \frac{\sqrt{8 e^{-2t} + 16 e^{2t}}}{(\sqrt{2 e^{2t} + 2 e^{-2t} + 4})^3}. \]This expression describes how the curve bends at each point, providing insight into its geometric properties.