Question

Curvature for plane curves Use the result of Exercise 55 to find the curvature function of the following curves. $$\mathbf{r}(t)=\left\langle a \cos ^{3} t, a \sin ^{3} t\right\rangle \text { (astroid) }$$

Short answer

Question: Find the curvature function of an astroid defined by the vector function $$\mathbf{r}(t)=\left\langle a \cos ^{3} t, a \sin ^{3} t\right\rangle$$ using the formula $$\kappa(t) = \frac{\lVert \mathbf{r'}(t) \times \mathbf{r''}(t) \rVert}{\lVert \mathbf{r'}(t) \rVert ^3}$$ Answer: The curvature function \(\kappa(t)\) of the astroid is given by $$\kappa(t) = \frac{-3 (\sin^2{t}-2\cos^2{t})(\cos^2{t}-2\sin^2{t})}{a(\sin{t}\cos{t})^2}.$$

Step-by-step solution

Find the first derivative of r(t)

Given the function, $$\mathbf{r}(t)=\left\langle a \cos ^{3} t, a \sin ^{3} t\right\rangle$$ Differentiate it with respect to time t to obtain the first derivative: $$\mathbf{r'}(t) =\left\langle -3a \cos^{2}{t}\cdot \sin{t}, 3a \sin^{2}{t}\cdot \cos{t}\right\rangle$$

Find the second derivative of r(t)

Again differentiating with respect to time, we obtain: $$\mathbf{r''}(t) =\left\langle -3a(\cos^2{t}-2\sin^2{t})\cos{t}, 3a(\sin^2{t}-2\cos^2{t})\sin{t}\right\rangle$$

Calculate the cross product of r'(t) and r''(t)

Since these are planar curves, their cross product will be in the z-direction only. So we are interested in the z-component of the cross product: $$\mathbf{r'}(t)\times\mathbf{r''}(t) = \left(-3a\cos^{2}{t}\sin{t}\right)(3a(\sin^2{t}-2\cos^2{t})\sin{t}) - \left(3a\sin^{2}{t}\cos{t}\right)(-3a(\cos^2{t}-2\sin^2{t})\cos{t}) $$

Calculate the magnitude of r'(t) and r'(t)xr''(t)

First, we find the magnitude of r'(t): $$\lVert \mathbf{r'}(t) \rVert = \sqrt{(-3a\cos^{2}{t}\sin{t})^2+(3a\sin^{2}{t}\cos{t})^2} = 3a\sin{t}\cos{t}$$ Next, the magnitude of the cross product of r'(t) and r''(t): $$\lVert \mathbf{r'}(t) \times\mathbf{r''}(t) \rVert = \lvert -9a^2\sin^2{t}\cos^2{t}(\sin^2{t}-2\cos^2{t})(\cos^2{t}-2\sin^2{t})\rvert $$

Find the curvature function

Now that we have the magnitudes, we can compute the curvature function using the formula: $$\kappa(t) = \frac{\lVert \mathbf{r'}(t) \times \mathbf{r''}(t) \rVert}{\lVert \mathbf{r'}(t) \rVert ^3} = \frac{-9a^2\sin^2{t}\cos^2{t}(\sin^2{t}-2\cos^2{t})(\cos^2{t}-2\sin^2{t})}{(3a\sin{t}\cos{t})^3}$$ We now have the curvature function \(\kappa(t)\): $$\kappa(t) = \frac{-3 (\sin^2{t}-2\cos^2{t})(\cos^2{t}-2\sin^2{t})}{a(\sin{t}\cos{t})^2}$$

Key concepts

Plane Curves

A plane curve is a curve that lies entirely on a single plane in a two-dimensional space. This means every point on the curve can be defined with two coordinates, typically denoted as \( x \) and \( y \). Plane curves can take various forms like lines, circles, ellipses, and more complex shapes. They are often described using parametric equations, involving a parameter like \( t \). For instance, the equation \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \) represents a plane curve in parametric form, where \( x(t) \) and \( y(t) \) are functions of \( t \).

Understanding plane curves is essential in geometry because they help in depicting the motion of points in two-dimensional space. These curves can be simple or very complex, like the astroid discussed in this exercise. Plane curves are foundational to many fields of study, including physics, engineering, and computer graphics.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any point. When applied to plane curves, differentiation allows us to find the tangent vectors, which are crucial for other calculations such as curvature.

When we differentiate a parametric curve \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), we do it separately for each component to obtain the first derivative or the velocity vector: \( \mathbf{r'}(t) = \langle x'(t), y'(t) \rangle \). Taking a further differentiation gives us the second derivative or the acceleration vector: \( \mathbf{r''}(t) = \langle x''(t), y''(t) \rangle \). These derivatives are used to explore properties like velocity and acceleration of motion represented by the curve.

Differentiation also plays a key role in finding the curvature of plane curves, which is the measure of how sharply the curve bends at a point.

Astroid

An astroid is a particular type of hypocycloid, which is a type of plane curve. It is generated by a point on a circle of radius \( R \) rolling inside a fixed circle of radius \( 4R \). The astroid has a distinctive star-like shape with four cusps and is symmetric about both the x-axis and y-axis.

Astroids have parametric equations given by \( \mathbf{r}(t) = \langle a \cos^3{t}, a \sin^3{t} \rangle \), where \( a \) is a constant defining the size of the astroid. The parameter \( t \) typically ranges from 0 to \( 2\pi \).

Studying astroids and their properties is useful in various fields such as engineering design and physics, as they can model certain motion paths and shapes accurately. The astroid is also an example discussed in the context of computing curvature, which measures how much a curve deviates from a straight line.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in a third vector, which is perpendicular to the plane of the given vectors. For plane curves, which are two-dimensional, we often "extend" vectors to three dimensions by adding a zero \( z \)-component. However, we're really only interested in the magnitude of the cross product rather than the entire vector for curvature calculations.

In the context of plane curves like our astroid, the cross product helps us find the curvature. We take the cross product of the first derivative \( \mathbf{r'}(t) \) with the second derivative \( \mathbf{r''}(t) \). The result's magnitude is used to determine how sharply the curve bends at a particular point.

Computing the cross product involves a specific formula that uses determinants. For example, the scalar z-component of the cross product of \( \mathbf{r'}(t) = \langle x', y', 0 \rangle \) and \( \mathbf{r''}(t) = \langle x'', y'', 0 \rangle \) gives \( x'y'' - y'x'' \), which is crucial for evaluating curvature.