Problem 220 - Calculus Volume 3

Question

Find the curvature for the following vector functions. \(\mathbf{r}(t)=(2 \sin t) \mathbf{i}-4 t \mathbf{j}+(2 \cos t) \mathbf{k}\)

Step-by-Step Solution

Answer

The curvature \( \kappa(t) \) is \( \frac{1}{10} \), constant and independent of \( t \).

Step 1: Find the First Derivative

Differentiating the vector function \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} - 4t \mathbf{j} + (2 \cos t) \mathbf{k} \) with respect to \( t \), we get:\( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \).

Step 2: Compute the Second Derivative

Now find the second derivative by differentiating \( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \):\( \mathbf{r}''(t) = (-2 \sin t) \mathbf{i} + 0 \mathbf{j} - (2 \cos t) \mathbf{k} \).

Step 3: Find the Magnitudes

Calculate the magnitude of \( \mathbf{r}'(t) \):\[ \| \mathbf{r}'(t) \| = \sqrt{(2 \cos t)^2 + (-4)^2 + (-2 \sin t)^2} = \sqrt{4 \cos^2 t + 16 + 4 \sin^2 t} = \sqrt{4(\cos^2 t + \sin^2 t) + 16} = \sqrt{20} = 2\sqrt{5}. \]Calculate the magnitude of \( \mathbf{r}''(t) \):\[ \| \mathbf{r}''(t) \| = \sqrt{(-2 \sin t)^2 + (0)^2 + (-2 \cos t)^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t} = 2. \]

Step 4: Compute the Curvature

Curvature \( \kappa(t) \) is given by the formula:\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}. \]First, find \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2\cos t & -4 & -2\sin t \ -2\sin t & 0 & -2\cos t \end{vmatrix} \].Calculating the determinant:\[ = \mathbf{i}((-4)(-2\cos t) - (0)(-2\sin t)) - \mathbf{j}((2\cos t)(-2\cos t) - (-2\sin t)(-2\sin t)) + \mathbf{k}((2\cos t)(0) - (-4)(-2\sin t)) \]\[ = 8\cos t \mathbf{i} - (4\cos^2 t + 4\sin^2 t) \mathbf{j} - 8\sin t \mathbf{k}. \]This simplifies to:\[ 8\cos t \mathbf{i} - 4 \mathbf{j} - 8\sin t \mathbf{k}. \]Calculate its magnitude:\[ \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| = \sqrt{(8\cos t)^2 + (-4)^2 + (-8\sin t)^2} = \sqrt{64\cos^2 t + 16 + 64\sin^2 t} = \sqrt{80} = 4\sqrt{5}. \]Finally, compute the curvature:\[ \kappa(t) = \frac{4\sqrt{5}}{(2\sqrt{5})^3} = \frac{4\sqrt{5}}{40\sqrt{5}} = \frac{1}{10}. \]

Step 5: Conclusion: Interpret the Result

The curvature, \( \kappa(t) = \frac{1}{10} \), is constant and independent of \( t \), indicating that the curve has a uniform shape along its path.

Key Concepts

First Derivative

In vector calculus, the first derivative of a vector function gives us the velocity vector. This indicates the rate of change of the vector with respect to a parameter, typically time \( t \). For a vector function \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} - 4t \mathbf{j} + (2 \cos t) \mathbf{k} \), the first derivative \( \mathbf{r}'(t) \) is calculated by differentiating each component with respect to \( t \). Here, we differentiate:

• \( 2 \sin t \) with respect to \( t \), resulting in \( 2 \cos t \).
• \( -4t \) with respect to \( t \), yielding \( -4 \).
• \( 2 \cos t \) with respect to \( t \), resulting in \( -2 \sin t \).

Thus, the first derivative \( \mathbf{r}'(t) \) is \( (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \). This derivative represents the direction and speed the point on the curve is moving at time \( t \).

Second Derivative

The second derivative of a vector function, denoted as \( \mathbf{r}''(t) \), offers insight into the acceleration of the vector. It reflects how the velocity is changing over time. Differentiating the velocity vector \( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \), we find:

• The derivative of \( 2 \cos t \) is \( -2 \sin t \).
• The derivative of \( -4 \) is 0 since it is constant.
• The derivative of \( -2 \sin t \) is \( -2 \cos t \).

As such, the second derivative becomes \( \mathbf{r}''(t) = (-2 \sin t) \mathbf{i} + 0 \mathbf{j} - (2 \cos t) \mathbf{k} \). This vector tells us how the curve is curving or changing direction as time progresses.

Cross Product

The cross product is a crucial tool in vector calculus, as it helps find a vector perpendicular to two given vectors. In the context of curvature, the cross product of the first and second derivatives \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) gives a vector related to how the original curve twists in space.To compute \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):

• We set up a determinant with the unit vectors and the components of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
• Compute the determinant using cofactor expansion, which involves the interactions between corresponding terms of the vectors to find each coordinate of the result.

This results in the cross product \( 8\cos t \mathbf{i} - 4 \mathbf{j} - 8\sin t \mathbf{k} \), a vector that helps determine the curvature by orienting along the axis of rotation of the curve.

Magnitude Calculation

Calculating the magnitude of a vector is vital as it provides the length of the vector, showing the size irrespective of direction. For any vector \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} \), its magnitude is given by \( \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).Consider the magnitudes for both derivatives:

• The magnitude of the first derivative \( \mathbf{r}'(t) \) ensures it is normalized before use in curvature formulae.
• The magnitude of the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) relates directly to the twisting potential of the curve.

In our example, the separate magnitudes lead to streamlined curvature formula calculations. The curvature is ultimately concerned with the ratio of these computed magnitudes, highlighting a dynamic property of the curve itself.