Problem 136 - Calculus 3

Question

Find the curvature of \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t\rangle\)

Step-by-Step Solution

Answer

The curvature is \(\frac{\sqrt{116}}{29\sqrt{29}}\).

Step 1: Differentiate the Position Vector

To find the curvature, we first need the first derivative of the position vector \(\mathbf{r}(t)\). Differentiate \(\mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle\) with respect to \(t\). This gives us \(\mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle\).

Step 2: Find the Second Derivative

Now, differentiate \(\mathbf{r}'(t)\) to find the second derivative \(\mathbf{r}''(t)\). Differentiate each component: \(\langle 2 \cos t, 5, -2 \sin t \rangle\) gives \(\mathbf{r}''(t) = \langle -2 \sin t, 0, -2 \cos t \rangle\).

Step 3: Compute the Magnitude of the Cross Product

Compute the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\). Calculate this as:\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \2\cos t & 5 & -2\sin t \-2\sin t & 0 & -2\cos t \\end{array} \right| \]Calculate the determinant to find the resulting vector:\[= \mathbf{i}(5(-2\cos t) - 0) - \mathbf{j}(2\cos t(-2\cos t) + 2\sin t(-2\sin t)) + \mathbf{k}(2\cos t(0) + 5(2\sin t))\]\[= \langle -10\cos t, -4\cos^2 t - 4\sin^2 t, 10\sin t \rangle\]Simplify to \(\langle -10\cos t, -4, 10\sin t \rangle\).Now we find the magnitude:\[|\mathbf{r}'(t) \times \mathbf{r}''(t)| = \sqrt{(-10\cos t)^2 + (-4)^2 + (10\sin t)^2} \]\[= \sqrt{100\cos^2 t + 16 + 100\sin^2 t} = \sqrt{100 + 16} = \sqrt{116} \].

Step 4: Compute the Magnitude of the First Derivative

Find the magnitude of the first derivative \(\mathbf{r}'(t)\):\[ |\mathbf{r}'(t)| = \sqrt{(2\cos t)^2 + 5^2 + (-2\sin t)^2} \]\[= \sqrt{4\cos^2 t + 25 + 4\sin^2 t} = \sqrt{4(\cos^2 t + \sin^2 t) + 25} = \sqrt{4 + 25} = \sqrt{29} \].

Step 5: Calculate the Curvature

The formula for curvature \(\kappa\) is given by:\[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]Plug in the magnitudes:\[ \kappa(t) = \frac{\sqrt{116}}{(\sqrt{29})^3} = \frac{\sqrt{116}}{29\sqrt{29}} \]

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with the differentiation and integration of vector fields. It's a powerful tool that helps us understand how vectors behave in space, especially when they're functions of several variables, like in physics or engineering. In our exercise, we're dealing with a vector function \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \), which is a parametric representation of a curve in three-dimensional space.
This curve can represent anything from the path of a moving object to the shape of an electric field.

• The vector components \( \langle 2 \sin t, 5t, 2 \cos t \rangle \) suggest a trajectory or a path traced out in a 3D coordinate system.
• Understanding these paths using vector calculus requires calculating derivatives and integrals of vector functions.

In this exercise, by using vector calculus, we're finding how sharply our path bends at any given moment—the curvature. This involves steps like differentiating the vector to find its tangent and understanding how the path twists or changes direction.

Differentiation

Differentiation is the process of finding the derivative of a function, which tells us the rate at which a function is changing at any given point. In this context, we're applying differentiation to our vector function \( \mathbf{r}(t) \).
To find the curvature, we begin by finding the first derivative, \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \). This represents the velocity or tangent vector of the curve, showing how the position changes over time.

• The first derivative gives us critical insight into the direction and speed of the moving point along the curve.

Moving further, the second derivative \( \mathbf{r}''(t) = \langle -2 \sin t, 0, -2 \cos t \rangle \) helps us find the acceleration, describing how the velocity itself changes.
In simpler terms, just as the first derivative tells us how position changes, the second derivative tells us how the velocity vector changes, which is crucial for understanding curvature.

Curvature Formula

Curvature is a measure of how sharply a curve bends, and in calculus, it's a crucial concept for understanding the behavior of paths or curves in space.
The curvature \( \kappa(t) \) formula involves both the magnitude of the cross product of the first and second derivatives and the cube of the magnitude of the first derivative:
\[\kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\]
Here’s a breakdown of what's happening:

• The cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) gives a vector that is perpendicular to the plane formed by the tangent and acceleration vectors, indicating how the curve twists in space.
• Taking the magnitude of this vector, \( |\mathbf{r}'(t) \times \mathbf{r}''(t)| \), gives a sense of the intensity of this twisting or bending.
• The magnitude of the first derivative \( |\mathbf{r}'(t)| \) is raised to the third power to scale our measure of bending by how fast the curve is being traced out.

In the solution provided, the specific computations yield \( \sqrt{116} \) for the cross product's magnitude and \( \sqrt{29} \) for the first derivative's magnitude, leading to the final curvature result \( \kappa(t) = \frac{\sqrt{116}}{29\sqrt{29}} \).
This quantifies how tightly the curve bends at any given point on the path.