Question

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=\langle a(\omega t-\sin \omega t), a(1-\cos \omega t)\rangle $$

Short answer

The curvature \(K(t)\) of the curve is given by \(\frac{\sqrt{(\omega t - \sin \omega t)^2 + (\cos \omega t)^2}}{a^2\omega}\)

Step-by-step solution

Find the derivative of the vector function

First, we must compute the derivative of the vector function \(\mathbf{r}\), which is given by: \(\mathbf{r}'(t) = \langle a\omega(1-\cos \omega t), a\omega \sin \omega t\rangle\)

Calculate the second derivative of the vector function

Next, we calculate the second derivative of the vector function \(\mathbf{r}\), which should yield: \(\mathbf{r}''(t) = \langle -a\omega^2(\omega t - \sin \omega t), -a\omega^2(\cos \omega t)\rangle\)

Find the norm of both derivatives

Compute the magnitudes of the first and second derivative vectors. We know that the magnitude (or norm) of a vector \(\mathbf{a} = \langle x,y \rangle\) is \(\|\mathbf{a}\| = \sqrt{x^2 + y^2}\), Thus, \(|\mathbf{r}'(t)| = a\omega\) and \(|\mathbf{r}''(t)| = a\omega^2\sqrt{(\omega t - \sin \omega t)^2 + (\cos \omega t)^2}\).

Find the curvature of the curve

Finally, substituting the magnitudes of the first and second derivative into the curvature formula will yield \(K(t) = \frac{|\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} = \frac{a\omega^2\sqrt{(\omega t - \sin \omega t)^2 + (\cos \omega t)^2}}{a^3\omega^3} = \frac{\sqrt{(\omega t - \sin \omega t)^2 + (\cos \omega t)^2}}{a^2\omega}\).

Key concepts

Vector Function Derivatives

When dealing with the motion of an object along a path, a vector function can represent the object's position as a function of time. In calculus, finding the derivative of a vector function is similar to finding the derivative of a standard function, but with the difference that each component of the vector is differentiated independently.

The derivative of the vector function \( \mathbf{r}(t) \) tells us about the object's velocity. For our curve \( \mathbf{r}(t) = \langle a(\omega t - \sin \omega t), a(1 - \cos \omega t) \rangle \), the first derivative \( \mathbf{r}'(t) \) signifies the instantaneous rate of change of the position, which provides information about how fast and in what direction the object is moving at any given time. In applications, understanding these derivatives is essential for predicting future positions and velocities of moving objects.

Norm of a Vector

The norm or magnitude of a vector is a measure of its length and is found by taking the square root of the sum of the squares of its components. It is denoted by \( \|\mathbf{a}\| \) for a vector \(\mathbf{a} \) with components \( x \) and \( y \) such that \( \mathbf{a} = \langle x, y \rangle \). The norm is calculated as \( \|\mathbf{a}\| = \sqrt{x^2 + y^2} \).

The concept of norm is crucial in many areas of mathematics and physics, as it quantifies the size of vectors – a fundamental concept when analyzing forces, velocities, and other vector quantities. When calculating curvature, the norm of the derivative vectors represents the speed of an object moving along the curve, which is necessary to determine how sharply the curve bends at each point.

Calculus in Motion

Calculus plays a pivotal role in describing and understanding motion. When studying the curvature of a curve, calculus offers tools to measure how an object's path bends and twists in space. To quantify this, we cannot solely rely on the object's position \( \mathbf{r}(t) \) – we also need its derivatives.

The curvature \( K \) of a curve at a given point, which is a good indicator of how rapidly the curve is changing direction at that point, is found using the derivatives. It tells us the rate of change of the tangent vector to the curve, providing insights into the curve's geometry and an object's trajectory. In essence, calculus equips us with the methods to analyze dynamic systems and the motion of an object along a curve, which is a cornerstone concept in disciplines such as physics, engineering, and robotics.