Introduction to Curves

Defining a curve

Level Curves

A 2D curve can be described by means of their Cartesian equation \(f(x, y) = c\) and the curve is a set of points

\[C = \{(x,y)\in\mathbb R^2 : f(x, y) = c\}\]

A 3D curve can be described by 2 equations \(f_1(x, y, z) = c_1, f_2(x,y,z) = c_2\) s.t.

\[C = \{(x,y, z)\in\mathbb R^3 : f_1(x, y, z) = c_1, f_2(x,y,z) = c_2\}\]

A level curve is a set of points which the quantity \(f(x, y)\) reaches the 'level' \(c\).

Parameterized Curve

A parameterized curve is a map

\[\gamma: (a, b)\rightarrow \mathbb R^n, -\infty \leq a < b\leq \infty\]

A parameterization of level curve \(C\) is a parametrized curve, whose image is contained in \(C\).

The parameterization of a curve is not unique, for example, the image of \((\cos t, \sin t)\) is equal to \((\cos 2t, \sin 2t)\)

Example If \(\gamma(t) = (t^2, t^4)\) is parameterization of \(y = x^2\)?

No, note that \(1 = (-1)^2\), while for any \(t \in \mathbb R\), \(t^2 \geq 0\) so that we cannot parameterize \(y=x^2\).

Example Find the parameterization of \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).
Take \((2\cos t, 3\sin t)\), note that \(\frac{(2\cos t)^2}{4} + \frac{(3\sin t)^2}{9} = \cos t^2 + \sin t^2 = 1\)

Example Find the Cartesian equations of \(\gamma(t) = (e^t, t^2)\).
Let \(x = e^t\), then \(t = \log x,\implies y = t^2 = (\log x)^2\).

Smooth Curve

A parameterized curve \(\gamma\) is smooth if \(\gamma\) is infinitely differentiable for all \(t \in (a, b)\). Note that since \(\gamma: (a, b)\rightarrow\mathbb R^n\), its derivative is defined as a vector-valued function and each derivative on each component. We will often denote \(\dot\gamma:=\frac{d\gamma}{dt}, \ddot\gamma := \frac{d^2\gamma}{dt^2}\).

Tangent Vector

Let \(\gamma(t)\) be a parameterized curve, then \(\dot\gamma(t)\) is the tangent vector of \(\gamma\) at point \(\gamma(t)\).

Theorem If the tangent vector of a parameterized curve is constant, then the image of the curve is (part of) a straight line.

proof. Assume \(\dot\gamma(t) = \mathbf a\) for all \(t\). Then, integrate \(\dot\gamma\) component-wise,

\[\gamma (t) = \int \dot\gamma(t)dt = \int \mathbf a dt = t\mathbf a + \mathbf b\]

Example Consider the parameterized curve \(\gamma(t) = ((1 + 2\cos t)\cos t, (1 + 2 \cos t) \sin t)\). Its tangent vector is

\[\dot\gamma(t) = (-\sin t - 2\sin 2t, \cos t + 2\cos 2t)\]

Note that \(\gamma\) has a self-intersection at \(\gamma(2\pi / 3) = \gamma(4\pi / 3) = (0, 0)\), while

\[\dot\gamma(2\pi/3) = (\sqrt 3/ 2, -3/2)\neq (-\sqrt 3/ 2, -3/2) = \dot\gamma(4\pi/3)\]

Source code

import numpy as np
import matplotlib.pyplot as plt
t = np.arange(-4 * np.pi, 4 * np.pi, 0.001)
x = (1 + 2 * np.cos(t)) * np.cos(t); y = (1 + 2 * np.cos(t)) * np.sin(t)
dx1 = 3 ** 0.5 / 2 * t; dy1 = -3 / 2 * t
dx2 = -3 ** 0.5 / 2 * t; dy2 = -3 / 2 * t
plt.figure(figsize=(4, 4)); plt.xlim(-1, 4); plt.ylim(-3, 3)
plt.xticks([]); plt.yticks([])
plt.plot(x, y); plt.plot(dx1, dy1); plt.plot(dx2, dy2)
plt.savefig("../assets/intro_curves.jpg")

Example: Foci of ellipse

Consider the ellipse \(\frac{x^2}{p^2}+ \frac{y^2}{q^2} = 1, p > q > 0\), using the parameterization \(\gamma(t) = (p\cos t, q\sin t)\).
Define the eccentricity of the ellipse as \(\epsilon = \sqrt{1 - \frac{q^2}{p^2}}\) and the foci of the ellipse as \(\mathbf f_1 = (- \epsilon p, 0), \mathbf f_2 = (\epsilon p, 0)\).

Claim \(\forall a \in C. d(\mathbf f_1, a) + d(\mathbf f_2, a)= \mathbf c\) where \(\mathbf c\) does not depend on \(a\).

proof. Let \(a\in C\), take some \(t_0\) s.t. \(a = (p\cos t_0, q\sin t_0)\). Consider \(d(a, \mathbf f_1)\)

\[\begin{align*} d(a, \mathbf f_1)^2 &= (p\cos t_0 - \epsilon p)^2 + (q\sin t_0)^2 \\ &= p^2\cos^2 t_0 - 2\epsilon p^2 \cos t_0 + \epsilon^2 p^2 + q^2\sin^2 t_0\\ &= (p^2 -q^2)\cos^2 t_0 - 2\epsilon p^2 \cos t_0 + (p^2 - q^2) + q^2\cos^2 t_0 + q^2\sin^2 t_0\\ &= (p^2 -q^2)\cos^2 t_0 - 2\epsilon p^2 \cos t_0 + p^2\\ &= p^2(\epsilon^2 \cos^2 t_0 - 2\epsilon\cos t_0 + 1)\\ &= p^2(\epsilon\cos t_0 - 1)^2 \end{align*}\]

Then, taking the square root, notice that \(\epsilon < 1. -1 < \cos t_0 < 1\) so that \(1 - \epsilon\cos t_0\) > 0. \(d(a, \mathbf f_1) = p(1 - \epsilon \cos t_0)\). Similarly, \(d(a, \mathbf f_2) = p(1 + \epsilon \cos t_0)\). Therefore,

\[d(\mathbf f_1, a) + d(\mathbf f_2, a) = 2p\]

Claim \(\forall a \in C\). The product of the distances from \(\mathbf f_1\) and \(\mathbf f_2\) to the tangent line at \(a\) does not depend on \(a\).

proof. For some \(t\), the tangent line is

\[\gamma'(t) = (-p\sin t, q\cos t)\]

Hence the normal is \(\mathbf n(t) = (q\cos t, p\sin t)\) so that \(\gamma'(t)\cdot \mathbf n(t) =\mathbf 0\). Note that the distance between some point \(\mathbf f_1\) and the tangent line is \((\mathbf f_1 - a)\cdot \frac{\mathbf n}{\|\mathbf n\|}\). Therefore, the product is

\[\begin{align*} d(\mathbf f_1, l(t))&=\quad (\mathbf f_1 - a)\cdot \frac{\mathbf n}{\|\mathbf n\|}\\ &= \frac{(\epsilon p - p\cos t)q\cos t + q\sin t p \sin t}{(q^2\cos^t + p^2 \sin^2 t)^{1/2}}\\ &= \frac{\epsilon pq\cos t - pq\cos^2 t + pq\sin^2 t}{(q^2\cos^2 t + p^2 \sin^2 t)^{1/2}}\\ &= \frac{pq(\epsilon \cos t - 1)}{(q^2\cos^2 t + p^2 \sin^2 t)^{1/2}}\\ d(\mathbf f_2, l(t))&=\frac{pq(-\epsilon \cos t - 1)}{(q^2\cos^2 t + p^2 \sin^2 t)^{1/2}}\\ d(\mathbf f_1, l(t))d(\mathbf f_2, l(t))&= \frac{p^2q^2(1 + \epsilon \cos t)(1 - \epsilon \cos t)}{q^2\cos^2 t + p^2 \sin^2 t}\\ &= q^2\frac{p^2-(p^2 - q^2)\cos^2 t)}{p^2\sin^2 t + p^2\cos^2 t + (q^2 - p^2)\cos^2 t}\\ &= q^2 \end{align*}\]

Example (Viviani's Curve)

Show that

\[\gamma(t) = (\cos^2 t - 1/2, \sin t\cos t, \sin t)\]

is a parameterization of the curve of intersection of the cylinder of radius \(1/2\) and axis the z-axis with the sphere of radius \(1\) and center \((-1/2, 0, 0)\).

The surface of the circle can be represented as

\[C_1 = \{(x, y, z) : (x + 1/2)^2 + y^2 + z^2 = 1\}\]

The surface of the cylinder is

\[C_2 = \{(x, y, z): x^2 + y^2 = 1/4\}\]

Let \(z = \sin t\), we have the equations

\[\begin{cases}x^2 + x + 1/4 + y^2 + \sin^2 t = 1\\x^2 + y^2 = 1/4\end{cases}\]

solves to have \(x + 1/2 = 1 - \sin^2 t\Rightarrow x = \cos^2 t - 1/2\),
then \(y = \sqrt{1/4 - (\cos^2 t - 1/2)^2} = \sqrt{\cos^2 t(1-\cos^2 t)} = \sin t\cos t\).