Problem 146 - Calculus 3

Question

Calculate the curvature of the circular helix \(\mathbf{r}(t)=r \sin (t) \mathbf{i}+r \cos (t) \mathbf{j}+t \mathbf{k}\)

Step-by-Step Solution

Answer

The curvature of the helix is \( \kappa = \frac{r}{(r^2 + 1)^{3/2}} \).

Step 1: Compute the first derivative of \( \mathbf{r}(t) \)

First, we need to find the first derivative of the vector function \( \mathbf{r}(t) \). The given helix is \( \mathbf{r}(t)=r \sin(t) \mathbf{i} + r \cos(t) \mathbf{j} + t \mathbf{k} \). The derivative is \( \mathbf{r}'(t) = r \cos(t) \mathbf{i} - r \sin(t) \mathbf{j} + \mathbf{k} \).

Step 2: Compute the magnitude of the first derivative

To find the magnitude of \( \mathbf{r}'(t) \), we use the formula for the magnitude of a vector \( \| \mathbf{v} \| = \sqrt{v_x^2 + v_y^2 + v_z^2} \). Thus, \( \|\mathbf{r}'(t)\| = \sqrt{(r \cos(t))^2 + (-r \sin(t))^2 + 1^2} = \sqrt{r^2 \cos^2(t) + r^2 \sin^2(t) + 1} \). Simplifying gives \( \|\mathbf{r}'(t)\| = \sqrt{r^2 + 1} \).

Step 3: Compute the second derivative of \( \mathbf{r}(t) \)

Next, we find the second derivative of the vector function. The first derivative was \( \mathbf{r}'(t) = r \cos(t) \mathbf{i} - r \sin(t) \mathbf{j} + \mathbf{k} \). The second derivative is \( \mathbf{r}''(t) = -r \sin(t) \mathbf{i} - r \cos(t) \mathbf{j} \).

Step 4: Compute the magnitude of the second derivative

The magnitude of the second derivative is found using the same formula for the vector magnitude. Thus, \( \|\mathbf{r}''(t)\| = \sqrt{(-r \sin(t))^2 + (-r \cos(t))^2} = \sqrt{r^2 \sin^2(t) + r^2 \cos^2(t)} \). Simplifying gives \( \|\mathbf{r}''(t)\| = r \).

Step 5: Calculate the curvature formula

The curvature \( \kappa \) is given by the formula \( \kappa = \frac{\| \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \). We can now substitute the magnitudes computed in steps 2 and 4 into this formula: \( \kappa = \frac{r}{(r^2 + 1)^{3/2}} \).

Key Concepts

Vector Calculus: Understanding the Basics

Vector calculus is a branch of mathematics that deals with vector fields and the differentiation and integration of vector functions. It is fundamental in physics and engineering. In vector calculus, vectors are entities with both magnitude and direction. Vectors are generally represented in a three-dimensional coordinate system using unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). These unit vectors correspond to the x, y, and z axes.

Consider a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), where \( t \) is a parameter. It represents a curve in space. Calculating elements such as curvature or tangents involves differentiating this function with respect to \( t \). Essential operators in vector calculus include the gradient, divergence, and curl.

• **Gradient:** Measures the rate and direction of change in a scalar field.
• **Divergence:** Describes the extent to which a vector field spreads out from a point.
• **Curl:** Measures the rotation of a vector field around a point.

The key is making sense of how vectors change, which leads to valuable insights in science and engineering.

Curvature Calculation: Measuring the Bend

Curvature is a measure of how much a curve deviates from being a straight line. In the context of a vector function, it provides insight into the bending of the path defined by the curve. To compute curvature, we require both the first and second derivatives of the vector function. For a vector curve \(\mathbf{r}(t)\), the first derivative \( \mathbf{r}'(t) \) is the tangent vector, indicating the direction of movement. Its magnitude gives the speed of the moving point.

The second derivative \( \mathbf{r}''(t) \) is the vector itself indicative of how the tangent changes. This reflects the curve's acceleration.

• The formula for curvature \( \kappa \) is given by \( \kappa = \frac{\| \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \).
• This encapsulates how much the curve bends based on its rate of speed change and acceleration.
• In our circular helix exercise, \( \kappa = \frac{r}{(r^2 + 1)^{3/2}} \) gives us the curvature.

The curvature of a space curve gives insight into the geometric nature of paths, valuable in fields such as robotics, computer graphics, and road design.

Differentiation of Vector Functions: Calculating Change

Differentiation of vector functions is a process similar to differentiating scalar functions, focusing on how vector quantities change. When dealing with vectors, differentiation is done component-wise. This concept is crucial when assessing changes in vector fields. For a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is computed as

• \( \frac{d}{dt}(x(t)) \mathbf{i} + \frac{d}{dt}(y(t)) \mathbf{j} + \frac{d}{dt}(z(t)) \mathbf{k} \).

This results in another vector, representing the velocity of the point defined by \( \mathbf{r}(t) \). Differentiation not only yields velocity but helps in calculating acceleration too by further differentiation. In our exercise, the first derivative \( \mathbf{r}'(t) = r \cos(t) \mathbf{i} - r \sin(t) \mathbf{j} + \mathbf{k} \) represents the velocity along the helix, while the second derivative \( \mathbf{r}''(t) = -r \sin(t) \mathbf{i} - r \cos(t) \mathbf{j} \) indicates the acceleration and thus the curvature.
Differentiating vector functions unlocks dynamic insights into movements and interactions within fields across physics and engineering.