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 [fig:paramcircle]:

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

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

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

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

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 \(t\):

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

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:

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

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 \(\PageIndex{1}\): Circles

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

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

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 [fig:paramcircle](a), it suffices to show that \((x-h)^2+(y-k)^2=r^2\):

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

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 \(\PageIndex{1}\): Ellipses

Show that for \(a>0\) and \(b>0\) the parametric equations

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

for \(0 \le t \le 2\pi\) describe an ellipse.

Solution

Since

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

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 \(\PageIndex{1}\): Hyperbolas

Show that the parametric equations

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

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

Solution

Since

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

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 \(\PageIndex{1}\): Bezier Curves

Bézier curves¹³ 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 [fig:bezier] and described below for \(n=3\) points:

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 [fig:bezier](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 [fig:bezier](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

\[\label{eqn:bezier3} x ~=~ (1-t)^2 x_0 ~+~ 2t\,(1-t)\,x_1 ~+~ t^2 x_2 \quad\text{and}\quad y ~=~ (1-t)^2 y_0 ~+~ 2t\,(1-t)\,y_1 ~+~ t^2 y_2 \]

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:

\[\begin{aligned} x ~&=~ (1-t)^2 (1) ~+~ 2t\,(1-t)\,(2) ~+~ t^2 (4) ~=~ 1 - 2t + t^2 + 4t - 4t^2 + 4t^2 ~=~ t^2 + 2t + 1\\ y ~&=~ (1-t)^2 (2) ~+~ 2t\,(1-t)\,(4) ~+~ t^2 (1) ~=~ 2 - 4t + 2t^2 + 8t - 8t^2 + t^2 ~=~ -5t^2 + 4t + 2\end{aligned} \nonumber \]

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

\[\label{eqn:paramderiv1} \dydx ~=~ \frac{y'(t)\,\dt}{x'(t)\,\dt} ~=~ \frac{y'(t)}{x'(t)} ~=~ \frac{\Dydt}{\Dxdt} \]

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

\[\label{eqn:paramderiv2} \ddt\,\left(\dydx\right) ~=~ \left(\ddx\,\left(\dydx\right)\right)\,\cdot\,\dxdt ~=~ \frac{d^2y}{\dx^2} \,\cdot\,\dxdt \quad\Rightarrow\quad \frac{d^2y}{\dx^2} ~=~ \frac{\Ddt\,\left(\Dydx\right)}{\Dxdt} \]

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:

\[\label{eqn:paramint} \int_{x_1}^{x_2} y~\dx ~=~ \int_a^b y(t)\,x'(t)~\dt \]

Example \(\PageIndex{1}\): cycloid

A cycloid is the path of a point \(P\) on a circle rolling along a straight line. Figure [fig:cycloid] 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:

Find parametric equations for \(C\) and find \(\dydx\).

Solution

For the angles \(t\) and \(\theta\)—measured in radians—shown in Figure [fig:cycloid], \(t+\theta + \pi/2 = 2\pi\), 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 \(\PageIndex{1}\), but with center \((h,k)=(a \theta,a)\), radius \(r=a\), and \(\omega=1\),

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

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 ([eqn:paramderiv1]) the derivative \(\dydx\) is given by:

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

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 [fig:cycloid] that the cycloid has cusps at those values of \(x\).

A cycloid appears in the solution of the famous brachistochrone problem:¹⁴ 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.¹⁵

[sec7dot6]

1. [exer:ellipminor] 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 [fig:ellipminor]. 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 \(\PageIndex{1}\).
2. [exer:ellipmajor] Similar to Exercise [exer:ellipminor], 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 [fig:ellipmajor]. 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 \(P=(a \cos\,t,b \sin\,t)\).
3. Show that the cycloid in Example Example \(\PageIndex{1}\): has global maxima at \(x=(2k+1)\pi a\) for all integers \(k\), and that the cycloid is always concave down.
4. Show that the area under the cycloid in Example \(\PageIndex{1}\) over the interval \(\ival{0}{2\pi a}\) is \(3\pi a^2\).
5. [exer:ratcurve] 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.
6. [exer:paramsegment] 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

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

   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 \(P_0P_1\): \(\frac{P_0P(t)}{P_0P_1} = t\).
7. [[1.]]
8. [exer:involute] 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 [fig:involute]. Show that the path \(C\) that the end traces—called the involute of the circle—has parametric equations \(x=a(\cos \theta + \theta \sin \theta)\), \(y=a(\sin \theta - \theta \cos \theta)\).
9. 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

   Example \(\PageIndex{1}\): bezier

   Add text here.

   Solution

   .
   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 ([eqn:degreetwo]).)
   3. Find the point where the curve \(B\) has a global maximum.
10. Use Exercise [exer:paramsegment] to derive the formulas ([eqn:bezier3]) for the parametric equations of the general Bézier curve for \(n=3\) control points.
11. 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:

    \[x ~=~ (1-t)^3x_0 + 3t(1-t)^2x_1 + 3t^2(1-t)x_2 + t^3x_3 \quad\text{and}\quad y ~=~ (1-t)^3y_0 + 3t(1-t)^2y_1 + 3t^2(1-t)y_2 + t^3y_3 \nonumber \]

12. 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 [exer:ratcurve]). (Hint: For all real \(t\), find the intersections of the lines \(y=tx\) with the curve. Consider also the special case of the line \(x=0\).)
13. Sketch the graph of the curve \(C: x=2t-4t^3\), \(y=t^3-3t^4\), \(-\infty < t < \infty\).