Question

Curvature for plane curves Use the result of Exercise 55 to find the curvature function of the following curves. $$\mathbf{r}(t)=\langle a \sin t, a \cos t\rangle(\text { circle })$$

Short answer

The curvature function of the given curve is \(\kappa(t) = \frac{1}{a}\).

Step-by-step solution

Compute the first derivative

To find the curvature, we first need to compute the first derivative of the position vector \(\mathbf{r}(t)\). The first derivative, \(\mathbf{r}'(t)\), is found by taking the derivative of each component with respect to \(t\): $$\mathbf{r}'(t) = \langle \frac{d}{dt} (a \sin t), \frac{d}{dt} (a \cos t) \rangle$$ Taking the derivatives of the components, we get: $$\mathbf{r}'(t) = \langle a \cos t, -a \sin t \rangle$$

Compute the second derivative

Now we need to compute the second derivative of the position vector \(\mathbf{r}(t)\). The second derivative, \(\mathbf{r}''(t)\), is found by taking the derivative of each component of \(\mathbf{r}'(t)\) with respect to \(t\): $$\mathbf{r}''(t) = \langle \frac{d}{dt} (a \cos t), \frac{d}{dt} (-a \sin t) \rangle$$ Taking the derivatives of the components, we get: $$\mathbf{r}''(t) = \langle -a \sin t, -a \cos t \rangle$$

Calculate the cross product of the first and second derivatives

To apply the curvature formula, we need to find the cross product of the first and second derivatives, \(\mathbf{r}'(t) \times \mathbf{r}''(t)\). In two dimensions, the cross product of two vectors is equivalent to the determinant of a 2x2 matrix formed by the components of the vectors: $$\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = \left|\begin{matrix} a \cos t & -a \sin t \\ -a \sin t & -a \cos t \end{matrix}\right|$$ Computing the determinant, we get: $$\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = a^2$$

Calculate the magnitude of the first derivative cubed

To apply the curvature formula, we also need the magnitude of the first derivative cubed, \(\|\mathbf{r}'(t)\|^3\). First, find the magnitude of \(\mathbf{r}'(t)\): $$\|\mathbf{r}'(t)\| = \sqrt{(a \cos t)^2 + (-a \sin t)^2} = \sqrt{a^2(\cos^2 t + \sin^2 t)}$$ Since \(\cos^2 t + \sin^2 t = 1\), $$\|\mathbf{r}'(t)\| = a$$ Now cube the magnitude: $$\|\mathbf{r}'(t)\|^3 = a^3$$

Find the curvature function

Now we can apply the curvature formula: $$\kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} = \frac{a^2}{a^3} = \frac{1}{a}$$ So the curvature function of the given curve is \(\kappa(t) = \frac{1}{a}\).

Key concepts

Plane curves

A plane curve is a path traced out by a moving point in two-dimensional space. You can think of it as a gentle line drawn on a flat piece of paper. Plane curves can take various forms, from simple to complex.

Common examples of plane curves include:

• Circles: Perfectly round shapes where each point is equidistant from the center.
• Ellipses: Oval shapes that resemble stretched circles.
• Parabolas: U-shaped curves commonly seen in quadratic graphs.
• Spirals: Curves that wind around a point, moving further away with each loop.

In the context of this exercise, the plane curve we are studying is a circle, represented by the position vector \( \mathbf{r}(t) = \langle a \sin t, a \cos t \rangle \). This formula describes how the position changes with the parameter \(t\), tracing the circle's path as \(t\) varies.

Derivatives

Derivatives are fundamental concepts in calculus that describe how a function changes as its input changes. For plane curves, derivatives help us understand the curve's behavior at different points.

Here's how derivatives relate to plane curves in our example:

• First Derivative \( \mathbf{r}'(t) \): This tells us the velocity or rate of change of the curve at any point \(t\). We've calculated the first derivative as \( \langle a \cos t, -a \sin t \rangle \).
• Second Derivative \( \mathbf{r}''(t) \): This represents the curve's acceleration or how its direction of change is changing. For our circle, it turns out to be \( \langle -a \sin t, -a \cos t \rangle \).

Understanding the derivatives allows us to analyze the curve's dynamics, specifically how it bends and twists as it evolves. In the setting of curvature, these derivatives help us quantify the curve's sharpness.

Cross product

The cross product is a mathematical operation used mostly in three dimensions. However, when dealing with plane curves in two dimensions, it simplifies to calculating determinants.

To find curvature, we need the determinant (or cross product in 2D terms) of the first and second derivative vectors:

• The determinant gives us a scalar but crucial number that helps determine the interaction between two vectors \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
• For our circle, the determinant is calculated as: \( a^2 \).

This operation serves an important role in the curvature computation by providing a measure of how much the curve "twists" in the plane. This scalar value is critical for applying the curvature formula effectively.

Curvature formula

The curvature formula allows us to measure how sharply a curve bends at any given point. For plane curves, curvature is commonly defined as follows:

\[ \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \]

• \(\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|\) represents the magnitude of the cross product (or determinant) of the first and second derivatives.
• \(\|\mathbf{r}'(t)\|^3\) stands for the magnitude of the velocity vector cubed.

For our circle example, we have determined:

• \(\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = a^2\)
• \(\|\mathbf{r}'(t)\|^3 = a^3\)

By applying the curvature formula, we find the curvature of the circle as:\[ \kappa(t) = \frac{1}{a} \]This indicates that for a circle, the curvature at any point is constant and inversely proportional to the radius \(a\). The smaller the circle, the sharper the curve, and thus, the larger the curvature.