7.6 Parametric Equations

Recall that Section 6.5 presented two different ways to “identify” or represent points on the unit circle—by angle and by slope, as in Figure 7.6.1:

tikzpicture[>=latex,every node/.style=font=] [->,dashed] (0:0.6) arc (0:60:0.6); (60:1.5) circle (2.5pt); [black!60,line width=1pt,->] (-2.2,0) -- (2.5,0) node[right] x; [black!60,line width=1pt,->] (0,-2) -- (0,2.2) node[above] y; [linecolor,line width=1.5pt] (0,0) circle (1.5); [dashed] (0,0) -- (60:1.5) node[black,midway,above] 1; at (35:0.4) ; [below right] at (0:1.5) 1; [below left] at (0:0) 0; [above right] at (60:1.5) (x,y)=( \, , \, ); tikzpicture — Figure 7.6.1 Points on the unit circle $x^2 + y^2 = 1$

tikzpicture[>=latex,every node/.style=font=] [->,dashed] (0:0.6) arc (0:53.13:0.6); [black!60,line width=1pt,->] (-2.5,0) -- (2.8,0) node[right] x; [black!60,line width=1pt,->] (0,-2) -- (0,2.2) node[above] y; [name path=cir,linecolor,line width=1.5pt] (0,0) circle (1.5); [name path=s1,dashed] (1.5,1.5) node[above] slope = t -- (-1.5,0); [below right] at (1.5,-0.1) 1; [below left] at (-1.5,-0.1) -1; [name intersections=of=cir and s1,by=a,b] (a) circle (2.5pt) (b) circle (2.5pt); [dashed,black!60] (0,0) -- (a); (a) circle (2.5pt); at (26:0.4) ; [right] at (47:1.55) (x,y); [fill=white] (-1.5,0) circle (1.5pt); [below left] at (0,0) 0; tikzpicture — Figure 7.6.1 Points on the unit circle $x^2 + y^2 = 1$

When identifying by the angle $\theta$, all points $(x,y)$ on the unit circle can be written as

\[ x ~=~ \cos\,\theta \quad\text{and}\quad y ~=~ \sin\,\theta \]

for any angle $\theta$. When identifying by the slope $t$ of lines through the point $(-1,0)$, recall from the derivation of the half-angle substitution that $\sin\,\theta = \frac{2t}{1+t^2}$ and $\cos\,\theta = \frac{1-t^2}{1+t^2}$. So all points $(x,y)$ on the unit circle except $(-1,0)$ can be written as

\[ x ~=~ \frac{1-t^2}{1+t^2} \quad\text{and}\quad y ~=~ \frac{2t}{1+t^2} \]

for any slope $t$. These two distinct “identifications” are called parametrizations of the unit circle, with parameter $\theta$ in the first case and parameter $t$ in the second.

In general, one way to describe a plane curve $C$ (i.e. a curve in the $xy$-plane) is to write its $x$ and $y$-coordinates as functions of a variable $t$:

\[ x ~=~ x(t) \quad\text{and}\quad y ~=~ y(t) \]

These are parametric equations of $C$, which consists of all points $(x,y)$ such that $x=x(t)$ and $y=y(t)$ for the parameter $t$ in some interval $I$. The shorthand for this is:

\[ C: x=x(t),~y=y(t),~t~\text{in}~I \]

Notice the flexibility that parametric equations provide, since plane curves can take any shape, not limited to the graph of a single function $y=f(x)$. In fact, a curve $y=f(x)$ is the special case where the parametric equations are $x=t$ and $y=f(t)$. In physical settings the parameter $t$ often denotes time, but it can represent anything and any symbol can be used in its place. A curve can have many parametrizations.

Example 7.13

Show that for any constants $\omega \ne 0$ and $r > 0$, and for $t$ measured in radians,

\[ x ~=~ h ~+~ r\,\cos\,\omega t \quad\text{and}\quad y ~=~ k ~+~ r\,\sin\,\omega t \quad\text{for $-\infty < t < \infty$}\quad \]

is a parametrization of the circle $(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$.

Solution: Since $\omega t$ is similar to the angle $\theta$ in Figure 7.6.1(a), it suffices to show that $(x-h)^2+(y-k)^2=r^2$:

\[ (x-h)^2 ~+~ (y-k)^2 ~=~ r^2 \cos^2 \omega t ~+~ r^2 \sin^2 \omega t ~=~ r^2 \quad\checkmark \]

The constant $\omega$ determines how fast and in which direction the circle is traced as the parameter $t$ varies. For example, for $\omega=2$ the circle $C$ is traced counterclockwise at twice the speed of the parametrization $C: x=h+r \cos\,t$, $y=k+r \sin\,t$. In all cases the circle is re-traced every $2\pi/\omega$ radians. For that reason the interval for $t$ is often restricted to the interval $\ival{0}{2\pi/\omega}$, so that the circle is traced only once.

Example 7.14

[r] $tikzpicture[>=latex,every node/.style=font=] [->,black!60,line width=1pt] (-2,0) -- (2,0) node[right] x; [->,black!60,line width=1pt] (0,-1.2) -- (0,1.2) node[above] y; [line width=1.5pt,linecolor] (0,0) ellipse (1.5 and 0.6); [above right] at (75:1.5 and 0.6) (x,y)=(a\, \,t,b\, \,t); (75:1.5 and 0.6) circle (2.5pt); [below left] at (-1.5,-0.1) -a; [below right] at (1.5,-0.1) a; [above left] at (0,0.6) b; [below left] at (0,-0.6) -b; tikzpicture$

Show that for $a>0$ and $b>0$ the parametric equations

\[ x ~=~ a\,\cos\,t \quad\text{and}\quad y ~=~ b\,\sin\,t \]

for $0 \le t \le 2\pi$ describe an ellipse.

Solution: Since

\[ \frac{x^2}{a^2} ~+~ \frac{y^2}{b^2} ~=~ \frac{a^2\,\cos^2 t}{a^2} ~+~ \frac{b^2\,\sin^2 t}{b^2} ~=~ \cos^2 t ~+~ \sin^2 t ~=~ 1 \qquad\qquad\qquad\qquad \]

for all $t$, then the points $(x,y) = (a\,\cos\,t,b\,\sin\,t)$ lie on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. It should be obvious that the entire ellipse is traced, but it is left as an exercise to show what the parameter $t$ represents.

Example 7.15

[r] $tikzpicture[>=latex,every node/.style=font=] [clip] (-0.5,-1.5) rectangle (5.2,2.3); [name path=hyper] (2.25,2) to[out=210,in=90] (1,0) to[out=-90,in=150] (2.25,-2); [name path=d2] (0.5,1.5) -- (2,1.5); [name intersections=of=d2 and hyper,by=P] (0,0) -- (P); [fill=fillcolor] (0,0) -- (P) -- ++(-90:1.5); (0,0) -- (P); [black!60,line width=1pt,->] (0,-2) -- (0,2) node[above] y; [fill=white,line width=1.5pt,draw=linecolor] (2.25,2) to[out=210,in=90] (1,0) to[out=-90,in=150] (2.25,-2); [black!60,line width=1pt,->] (-0.5,0) -- (3,0) node[right] x; [line width=1.5pt,linecolor] (2.25,2) to[out=210,in=90] (1,0) to[out=-90,in=150] (2.25,-2); (P) circle (2.5pt); [below left] at (0,0) O; [right] at (P) (x,y)=( \,t, \,t); [below left] at (1,0) 1; [right] at (1.4,-1) x^2-y^2=1; [right] at (1.8,0.6) t2 = shaded area; tikzpicture$ Show that the parametric equations

\[ x ~=~ \cosh\,t \quad\text{and}\quad y ~=~ \sinh\,t \]

for $-\infty < t < \infty$ describe one branch of a hyperbola.

Solution: Since

\[ x^2 ~-~ y^2 ~=~ \cosh^2 t ~-~ \sinh^2 t ~=~ 1 \]

for all $t$, then the points $(x,y) = (\cosh\,t,\sinh\,t)$ lie on the unit hyperbola $x^2-y^2=1$. This was in fact shown in Section 7.5, where $t$ is half the area of the shaded region in the above figure for $t>0$. For $t<0$ the shaded region is reflected below the $x$-axis. Since $\cosh\,t \ge 1$ and $\sinh\,t$ can take any value, then the entire right branch of the hyperbola is traced as $t$ varies. Likewise the left branch has parametric equations $x=-\cosh\,t$ and $y=\sinh\,t$. The hyperbola is thus parametrized by area (or negative area for $t <0$). In general, for $a>0$ and $b>0$ the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has parametric equations $x=\pm a\,\cosh\,t$ and $y=b\,\sinh\,t$ for all $t$.

Example 7.16

Bézier curves^[1] are used in Computer Aided Design (CAD) to join the ends of an open polygonal path of noncollinear control points with a smooth curve that models the “shape” of the path. The curve is created via repeated linear interpolation, illustrated in Figure 7.6.2 and described below for $n=3$ points:

tikzpicture[>=latex,every node/.style=font=,rotate=-90] (B0) at (3,0); (B1) at (0,0); (B2) at (3,4); [line width=1pt] (B0) -- (B1) -- (B2); (B0) circle (2.5pt); (B1) circle (2.5pt); (B2) circle (2.5pt); [left] at (B0) B_0; [left] at (B1) B_1; [right] at (B2) B_2; (A0) at ((B0)!0.4!(B1)); (A1) at ((B1)!0.4!(B2)); [red!40,opacity=0.5,line width=1pt] (A0) -- (A1); [red!40,opacity=0.5] (A0) circle (2.5pt); [red!40,opacity=0.5] (A1) circle (2.5pt); (P) at ((A0)!0.4!(A1)); [linecolor] (P) circle (2.5pt); [left] at (A0) A_0; [right] at ([yshift=1mm] A1) A_1; [above] at ([xshift=-1mm] P) P; [dashed,->] ([xshift=-1mm,yshift=-2mm] B0) -- ([xshift=2mm,yshift=-2mm] A0) node[midway,left] 40%; (aux1) at ((A0)!7pt!-90:(P)); (aux2) at ((P)!7pt!90:(A0)); [dashed,->] (aux1) -- (aux2) node[midway,below] 40%; (aux3) at ((B1)!7pt!90:(A1)); (aux4) at ((A1)!7pt!-90:(B1)); [dashed,->] (aux3) -- (aux4) node[midway,above right] 40%; tikzpicture — Figure 7.6.2 Bézier curve $B$ with 3 control points $B_0$, $B_1$, $B_2$

tikzpicture[>=latex,every node/.style=font=] (B0) at (0,0); (B1) at (0,3); (B2) at (4,0); [line width=1pt] (B0) -- (B1) -- (B2); (B1) circle (2.5pt); [left] at (B0) B_0; [left] at (B1) B_1; [right] at (B2) B_2; in 0.1,0.2,...,0.9 (A0) at ((B0)! !(B1)); (A1) at ((B1)!0 !(B2)); [red!40,opacity=0.5] (A0) -- (A1); [red!40,opacity=0.5] (A0) circle (2.5pt); [red!40,opacity=0.5] (A1) circle (2.5pt); (P) at ((A0)! !(A1)); [linecolor] (P) circle (2.5pt); [linecolor,line width=1.5pt,domain=0.0:1.0,samples=500,variable= ] plot (4*pow( ,2), -6*pow( ,2) + 6* ); (B0) circle (2.5pt); (B2) circle (2.5pt); [below] at (1,1.4) B; tikzpicture — Figure 7.6.2 Bézier curve $B$ with 3 control points $B_0$, $B_1$, $B_2$

For three control points $B_0$, $B_1$, $B_2$, pick $0 \le t \le 1$. Say $t=0.4$, which you can think of as a percentage: $0.4 = 40\%$. Let $A_0$ and $A_1$ be the points $40\%$ of the way from $B_0$ to $B_1$ and from $B_1$ to $B_2$, respectively, as in Figure 7.6.2(a). Then the point $P$ that is $40\%$ of the way from $A_0$ to $A_1$ is on the Bézier curve $B$ joining $B_0$ and $B_2$. Do this for every $t$ in $\ival{0}{1}$ to fill out the curve $B$, as in Figure 7.6.2(b).

It can be shown via de Casteljau’s algorithm that the Bézier curve $B$ for any three control points $B_0=(x_0,y_0)$, $B_1=(x_1,y_1)$ and $B_2=(x_2,y_2)$ in the $xy$-plane has parametric equations

\begin{equation} x ~=~ (1-t)^2 x_0 ~+~ 2t\,(1-t)\,x_1 ~+~ t^2 x_2 \quad\text{and}\quad y ~=~ (1-t)^2 y_0 ~+~ 2t\,(1-t)\,y_1 ~+~ t^2 y_2 \tag{7.13}\end{equation}

for $0 \le t \le 1$. Write out and simplify the explicit parametric equations for the Bézier curve $B$ with control points $B_0=(1,2)$, $B_1=(2,4)$ and $B_2=(4,1)$.

Solution: The parametric equations for $B_0=(x_0,y_0)=(1,2)$, $B_1=(x_1,y_1)=(2,4)$, $B_2=(x_2,y_2)=(4,1)$ are:

\begin{align*} x ~&=~ (1-t)^2 (1) ~+~ 2t\,(1-t)\,(2) ~+~ t^2 (4) ~=~ 1 - 2t + t^2 + 4t - 4t^2 + 4t^2 ~=~ t^2 + 2t + 1\\ y ~&=~ (1-t)^2 (2) ~+~ 2t\,(1-t)\,(4) ~+~ t^2 (1) ~=~ 2 - 4t + 2t^2 + 8t - 8t^2 + t^2 ~=~ -5t^2 + 4t + 2 \end{align*}

[r] The Bézier curve $B: x=t^2 + 2t + 1,$ $y=-5t^2 + 4t + 2$, $0 \le t \le 1$ for $B_0$, $B_1$, $B_2$ is shown in the figure on the right. It is left as an exercise to show that this curve is part of a parabola. In general Bézier curves can be created for $n \ge 3$ control points in the plane, with the parametric equations being polynomials of degree $n-1$ in the parameter $t$. In the exercises you will be guided in how to derive the parametric equations in the cases $n=3$ and $n=4$. Bézier curves can also be constructed for control points in three-dimensional space. A similar construct—a Bézier surface—is used in three dimensions to model the boundary of a polyhedron (i.e. a solid whose faces are polygons).

A curve with parametric equations $x=x(t)$ and $y=y(t)$ might not be the graph of a single function $y=f(x)$, but the derivative $\dydx$ can still be found by using the differentials of $x$ and $y$ as functions of $t$: $\dy=y'(t)\,\dt$ and $\dx=x'(t)\,\dt$, so that

\begin{equation} \dydx ~=~ \frac{y'(t)\,\dt}{x'(t)\,\dt} ~=~ \frac{y'(t)}{x'(t)} ~=~ \frac{\Dydt}{\Dxdt} \tag{7.14}\end{equation}

when $\dxdt \ne 0$. The second derivative $\frac{d^2y}{\dx^2}$ can then be found via the Chain Rule:

\begin{equation} \ddt\,\left(\dydx\right) ~=~ \left(\ddx\,\left(\dydx\right)\right)\,\cdot\,\dxdt ~=~ \frac{d^2y}{\dx^2} \,\cdot\,\dxdt \quad\Rightarrow\quad \frac{d^2y}{\dx^2} ~=~ \frac{\Ddt\,\left(\Dydx\right)}{\Dxdt} \tag{7.15}\end{equation}

For $t$ in $\ival{a}{b}$ with $x_1=x(a)$ and $x_2=x(b)$, the integral $\int_{x_1}^{x_2} y\,\dx$ is given by:

\begin{equation} \int_{x_1}^{x_2} y~\dx ~=~ \int_a^b y(t)\,x'(t)~\dt \tag{7.16}\end{equation}

Example 7.17

A cycloid is the path of a point $P$ on a circle rolling along a straight line. Figure 7.6.3 shows the cycloid $C$ traced by a circle of radius $a$ rolling along the $x$-axis so that $P$ touches the origin during the roll:

tikzpicture[>=latex,every node/.style=font=] (O) at (2.2,1); (x0) at (0,0); (y1) at (0,2); (a0) at (2.2,0); [name path=circ,line width=1pt] (O) circle (1); [name path=cycloid,linecolor,domain=-pi:3*pi,samples=200,line width=1.5pt] plot ( - sin( r),1 - cos( r)); [name intersections=of=cycloid and circ,by=P,R] (P) circle (2.5pt); [above left] at (P) P; (O) circle (1pt); [dashed] (2.2,0) -- (O) -- (P) -- ((x0)!(P)!(a0)); [dashed] (P) -- ((x0)!(P)!(y1)) node[left] y; [dashed] (O) -- ++(1,0) node[pos=0.7,below] a; [black!60,->,line width=1pt] (0,0) -- (0,2.5) node[above] y; [black!60,->,line width=1pt,anchor=base] (-pi,0) -- (3*pi,0) node[right] x node[black,shift=(0,-0.3)] at (x0) 0 node[black,shift=(0,-0.3)] at ((x0)!(P)!(a0)) x node[black,shift=(0,-0.3)] at (a0) a node[black,shift=(0,-0.3)] at (2*pi,0) 2 a; [->,shift=(O)] (125:1.3) arc (125:85:1.3); [dashed,->,shift=(O)] (0:0.3) arc (0:132:0.3); [shift=(2.2,1),right] at (60:0.4) t; [dashed,->,shift=(O)] (270:0.3) arc (270:137:0.3); [shift=(2.2,1),left] at (240:0.4) ; [shift=(O)] (0.2,0) -- (0.2,-0.2) -- (0,-0.2); [above right] at (3.4,2) C; tikzpicture — Figure 7.6.3 Cycloid $C$ for a circle of radius $a$

Find parametric equations for $C$ and find $\dydx$.

Solution: For the angles $t$ and $\theta$—measured in radians—shown in Figure 7.6.3, $t+\theta + \pi/2 = 2\pi$, so $t=3\pi/2 - \theta$. The point $P$ touches the origin as the circle rolls, so the horizontal distance from the circle’s center to the $y$-axis is the length of the circular arc with central angle $\theta$, namely $a \theta$. So by the parametrization of the circle as in Example 7.13, but with center $(h,k)=(a \theta,a)$, radius $r=a$, and $\omega=1$,

\begin{align*} x ~&=~ a\,\theta ~+~ a\,\cos\,t ~=~ a\,\left(\theta ~+~ \cos\,\left(\tfrac{3\pi}{2} - \theta\right)\right) ~=~ a\,(\theta ~-~ \sin\,\theta)\\ y ~&=~ a ~+~ a\,\sin\,t ~=~ a\,\left(1 ~+~ \sin\,\left(\tfrac{3\pi}{2} - \theta\right)\right) ~=~ a\,(1 ~-~ \cos\,\theta) \end{align*}

Thus, $C: x=a\,(\theta \,-\, \sin\,\theta)$, $y=a\,(1 \,-\, \cos\,\theta)$, $-\infty < \theta < \infty$ is a parametrization of the cycloid $C$. As in formula (7.14) the derivative $\dydx$ is given by:

\[ \dydx ~=~ \frac{y'(\theta)}{x'(\theta)} ~=~ \frac{a\,\sin\,\theta}{a\,(1 \,-\, \cos\,\theta)} ~=~ \frac{\sin\,\theta}{1 \,-\, \cos\,\theta} ~=~ \cot\,\tfrac{1}{2}\theta \]

Thus, $\dydx$ is undefined when $\cos\,\theta = 1$, namely, when $\theta = 2\pi k$ for all integers $k$, i.e. when $x=a\,(\theta \,-\, \sin\,\theta) = a\,(2\pi k \,-\, \sin\,2\pi k) = 2\pi ka$. Notice from Figure 7.6.3 that the cycloid has cusps at those values of $x$.

[r] A cycloid appears in the solution of the famous brachistochrone problem:^[2] find the plane curve joining two points $A$ and $B$—where $B$ is at a lower height than $A$ but not directly under it—along which an object slides frictionless under the force of gravity alone from $A$ to $B$ in the shortest time. It turns out that the optimal path is not a straight line, but part of an inverted (upside-down) cycloid with a cusp at $A$, as in the figure on the right.^[3]

Exercises

For $a>b>0$, in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ inscribe a circle of radius $b$ centered at the origin, called the minor auxiliary circle. This circle is parametrized by $x=b \cos\,t$ and $y=b \sin\,t$ for $0 \le t \le 2\pi$, with the angle $t$ shown in Figure 7.6.4. From each point on that circle draw a horizontal line segment to the point $P$ on the ellipse in the same quadrant, as shown. Show that $P=(a \cos\,t,b \sin\,t)$.
Note: The angle $t$ is called the eccentric angle of the ellipse, and it is the parameter for the parametrization of the ellipse in Example 7.14.

Figure 7.6.4 Exercise 1

Figure 7.6.5 Exercise 2
Similar to Exercise 1, circumscribe a circle of radius $a$ (called the major auxiliary circle) around the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. This circle is parametrized by $x=a \cos\,t$ and $y=a \sin\,t$ for $0 \le t \le 2\pi$, with the eccentric angle $t$ shown in Figure 7.6.5. From each point on that circle draw a vertical line segment to the point $P$ on the ellipse in the same quadrant, as shown. Show again that $P=(a \cos\,t,b \sin\,t)$.
Show that the cycloid in Example 7.17 has global maxima at $x=(2k+1)\pi a$ for all integers $k$, and that the cycloid is always concave down.
Show that the area under the cycloid in Example 7.17 over the interval $\ival{0}{2\pi a}$ is $3\pi a^2$.
The parametrization $C: x=\frac{1-t^2}{1+t^2}$, $y=\frac{2t}{1+t^2}$, $-\infty < t < \infty$ of the unit circle $C$ shown earlier makes the unit circle a rational curve, since $x$ and $y$ are rational functions of the parameter $t$. Is the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ a rational curve? Justify your answer.
Let $P_0=(x_0,y_0)$ and $P_1=(x_1,y_1)$ be distinct points in the $xy$-plane. For $t$ in $\ival{0}{1}$ define

\[ x(t) ~=~ (1-t)\,x_0 ~+~ t\,x_1 \quad\text{and}\quad y(t) ~=~ (1-t)\,y_0 ~+~ t\,y_1 \]

and let $P(t)=(x(t),y(t))$.
1. Show that $C: x=x(t)$, $y=y(t)$, $0 \le t \le 1$ is a parametrization of the line segment from $P_0$ to $P_1$.
2. Show that the parameter $t$ is the proportion of the length $P_0P(t)$ to the length $P_0P_1$: $\frac{P_0P(t)}{P_0P_1} = t$.
In an ellipsograph a rod of length $a$ has a peg at one end and another peg a distance $b$ from the other end, so that the rod can slide along two thin perpendicular rails, with the pegs in the rails as in Figure 7.6.6. Show that a marker at the endpoint $P$ traces an ellipse as the rod moves. (Hint: Treat the rails as the $x$ and $y$-axes. Find parametric equations, with a different angle than in Exercise 1.)

Figure 7.6.6 Exercise 7

Figure 7.6.7 Exercise 8

The end $P$ of a thread is held taut as it is unwound from a circle of radius $a$ starting from a point $A$, as in Figure 7.6.7. Show that the path $C$ that the end traces—called the involute of the circle—has parametric equations $x=a(\cos \theta + \theta \sin \theta)$, $y=a(\sin \theta - \theta \cos \theta)$.
Recall that a parabola of the form $y=ax^2+bx+c$ has a constant second derivative $\frac{d^2y}{\dx^2} = 2a$. Consider the Bézier curve $B$ in Example 7.16.
1. Find $\frac{d^2y}{\dx^2}$ for $B$. Is it a constant?
2. Show that $B$ is part of a parabola. Does this contradict part (a)? Explain. (Hint: Show the curve has a second-degree equation of the form (7.10).)
3. Find the point where the curve $B$ has a global maximum.
Use Exercise 6 to derive the formulas (7.13) for the parametric equations of the general Bézier curve for $n=3$ control points.
To form the Bézier curve for $n=4$ control points $B_0=(x_0,y_0)$, $B_1=(x_1,y_1)$, $B_2=(x_2,y_2)$, $B_3=(x_3,y_3)$, for $0 \le t \le 1$ use the three points that are $100t\%$ of the way along the line segments $\overline{B_0B_1}$, $\overline{B_1B_2}$, $\overline{B_2B_3}$ as the control points in the $n=3$ case. Show that the resulting parametrization is:

\[ x ~=~ (1-t)^3x_0 + 3t(1-t)^2x_1 + 3t^2(1-t)x_2 + t^3x_3 \quad\text{and}\quad y ~=~ (1-t)^3y_0 + 3t(1-t)^2y_1 + 3t^2(1-t)y_2 + t^3y_3 \]
Each point on a plane curve lies on some line through the origin. Use that fact to show that the equation $y^2=x^2+x^3$ defines a rational curve (see Exercise 5). (Hint: For all real $t$, find the intersections of the lines $y=tx$ with the curve. Consider also the special case of the line $x=0$.)
Sketch the graph of the curve $C: x=2t-4t^3$, $y=t^3-3t^4$, $-\infty < t < \infty$.

Developed and popularized in the 1960s by two engineers, Pierre Bézier and Paul de Casteljau, for vehicle body modeling at the French automotive manufacturers Renault and Citroën, respectively. ↩
First solved in 1696 by the Swiss physicist and mathematician Johann Bernoulli (1667-1748). ↩
See pp.60-62 in Clegg, J.C., Calculus of Variations, Edinburgh: Oliver & Boyd, Ltd., 1968. For Bernoulli’s proof see pp.644-655 in Smith, D.E., A Source Book in Mathematics, New York: Dover Publications, Inc., 1959. ↩