Question

Special formula: Curvature for plane curves Show that the curve \(\mathbf{r}(t)=\langle f(t), g(t)\rangle,\) where \(f\) and \(g\) are twice differentiable, has curvature $$ \kappa(t)=\frac{\left|f^{\prime} g^{\prime \prime}-f^{\prime \prime} g^{\prime}\right|}{\left(\left(f^{\prime}\right)^{2}+\left(g^{\prime}\right)^{2}\right)^{3 / 2}} $$ where all derivatives are taken with respect to \(t.\)

Short answer

In this exercise, we showed that the curvature formula for plane curves is given by: $$ \kappa(t) = \frac{\left|f^{\prime}(t) g^{\prime\prime}(t)-f^{\prime\prime}(t) g^{\prime}(t)\right|}{\left(\left(f^{\prime}(t)\right)^{2}+\left(g^{\prime}(t)\right)^{2}\right)^{3 / 2}} $$ We did this by first finding the first and second derivatives of the vector function \(\mathbf{r}(t) = \langle f(t), g(t)\rangle\). Then, we computed the curvature formula of plane curves and showed that it matches the given expression.

Step-by-step solution

Compute \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime\prime}(t)\)

We start by finding the first and second derivatives of \(\mathbf{r}(t) = \langle f(t), g(t)\rangle\). To take the derivative of a vector-valued function, we differentiate each component separately with respect to the parameter \(t\): $$ \mathbf{r}^{\prime}(t) = \langle f^{\prime}(t), g^{\prime}(t) \rangle \\ \mathbf{r}^{\prime\prime}(t) = \langle f^{\prime\prime}(t), g^{\prime\prime}(t) \rangle $$

Compute the curvature formula

The curvature formula for plane curves can be found using the following formula: $$ \kappa(t) = \frac{|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t)|}{\|\mathbf{r}^{\prime}(t)\|^3} $$

Find \(\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t)\) and \(\|\mathbf{r}^{\prime}(t)\|^3\)

Since \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime\prime}(t)\) are 2D vectors, their cross product will be a scalar: $$ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) = f'(t)g''(t) - f''(t)g'(t) $$ Now, we compute \(\|\mathbf{r}^{\prime}(t)\|^3\): $$ \|\mathbf{r}^{\prime}(t)\| = \sqrt{(f'(t))^2 + (g'(t))^2} \\ \|\mathbf{r}^{\prime}(t)\|^3 = \left(\sqrt{(f'(t))^2 + (g'(t))^2}\right)^3 = \left((f'(t))^2 + (g'(t))^2\right)^{3/2} $$

Compute \(\kappa(t)\)

Using the curvature formula from Step 2, we can now compute \(\kappa(t)\): $$ \kappa(t) = \frac{|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t)|}{\|\mathbf{r}^{\prime}(t)\|^3} = \frac{|f'(t)g''(t) - f''(t)g'(t)|}{\left((f'(t))^2 + (g'(t))^2\right)^{3/2}} $$ The result matches the given expression, so we have shown that the curvature formula for plane curves is: $$ \kappa(t) = \frac{\left|f^{\prime}(t) g^{\prime\prime}(t)-f^{\prime\prime}(t) g^{\prime}(t)\right|}{\left(\left(f^{\prime}(t)\right)^{2}+\left(g^{\prime}(t)\right)^{2}\right)^{3 / 2}} $$

Key concepts

Plane Curves

A plane curve can be thought of as a smooth, continuous line that lies on a flat surface, often described in terms of a parameter. This parameter, usually denoted by \(t\), helps us pinpoint locations along the curve as it changes. For instance, if you imagine drawing a loop on a piece of paper with a pencil, the movement of the pencil can be considered as tracing out a plane curve. The power of plane curves lies in their mathematical descriptions, which allow us to explore their properties, like curvature.Curvature basically tells us how sharply a curve bends. It's a way of quantifying the twisty nature of a path. For plane curves, understanding curvature helps in analyzing how the path behaves at each point, whether the curve is tight or gentle, using the formula provided in the exercise.

Vector-Valued Functions

Vector-valued functions are a core tool in describing curves in a mathematical setting. They assign a vector to each point on a curve, capturing both direction and magnitude. This is essential for representing movements in space.For example, given a vector-valued function \(\mathbf{r}(t) = \langle f(t), g(t) \rangle\), this function describes the position of a particle moving in a plane over time:

• \(f(t)\) indicates the horizontal position.
• \(g(t)\) indicates the vertical position.

By viewing these components together, we can map out paths like a rollercoaster track or a winding road on a flat landscape. The differentiation of these functions, as seen in the exercise, unlocks insights into how curves shift and evolve, notably for calculating curvature.

Differentiation

Differentiation is a process that lets us discover the rate of change of variables. In simpler terms, it tells us how fast something is changing at any given instant. For vector-valued functions, we differentiate each component separately to find the derivatives of the function.In the example of \(\mathbf{r}(t) = \langle f(t), g(t) \rangle\), by differentiating, we get:

• \(\mathbf{r}^{\prime}(t) = \langle f^{\prime}(t), g^{\prime}(t) \rangle\)
• \(\mathbf{r}^{\prime\prime}(t) = \langle f^{\prime\prime}(t), g^{\prime\prime}(t) \rangle\)

These derivatives have significant roles:

• \(\mathbf{r}^{\prime}(t)\) provides us with directions and speeds as the curve moves.
• \(\mathbf{r}^{\prime\prime}(t)\) reveals information about how the direction itself changes, useful for finding the curvature like in the exercise.

Differentiation, therefore, is pivotal in helping us understand the dynamics of curves, and is critical to the computation of curvature because it shows how rapidly these curves bend and twist.

Cross Product

The cross product is a mathematical operation used to determine a vector perpendicular to two given vectors. However, when they're two-dimensional, like in our exercise, the cross product results in a scalar instead of a vector.For example, consider two vectors \( \mathbf{r}^{\prime}(t) = \langle f^{\prime}(t), g^{\prime}(t) \rangle \) and \( \mathbf{r}^{\prime\prime}(t) = \langle f^{\prime\prime}(t), g^{\prime\prime}(t) \rangle \). Their cross product is computed as:\[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) = f^{\prime}(t)g^{\prime\prime}(t) - f^{\prime\prime}(t)g^{\prime}(t)\]This result provides a scalar (a single number), which plays a key role in determining the curvature of the curve. In particular, the magnitude of this scalar helps to tell us about the twisting characteristics of the plane curve, as it appears in the numerator of the curvature formula. In this way, the cross product facilitates understanding complex planar motions in simple, scalar terms.