10.2 | Plane Curves and Parametric Equations


Let \(x = f(t) \) and \(y = g(t)\) then the curve obtained by plotting the ordered pairs \((x,y)\) for all \(t\) in the domain of \(f\) and \(g\) is called a parametric curve, or plane curve, while the equations which define it are called parametric equations. The following are equivalent:

  • \(x = f(t) \), \(y = g(t)\)
  • \((f(t), g(t))\)
  • \( \langle f(t), g(t) \rangle \)
  • \( f(t)\mathbf{i} + g(t)\mathbf{j} \)
Definition of a Smooth Curve
A curve \(C\) represented by \(x = f(t)\) and \(y = g(t)\) on an interval \(I\) is called smooth when \(f’\) and \(g’\) are continuous on \(I\) and not simultaneously \(0\), except possibly at the endpoints of \(I\). The curve \(C\) is called piecewise smooth when it is smooth on each subinterval of some partition of \(I\).

Match the parametric equations with the graphs on the left labled I-VI.

  1. \( x = \frac{ \sin 2t}{4+t^2}\), \(y = \frac{ \cos 2t}{4+t^2} \)
  2. \( x= \cos 5t\), \( y = \sin 2t\)
  3. \(x = \sin 2t \), \( y = \sin(t + \sin 2t) \)
  4. \( x= t^2-2t\), \( y = \sqrt{t}\)
  5. \( x= t + \sin 4t \), \( y = t^2 + \cos 3t\)
  6. \(x = t^4 -t+1\), \(y =t^2\)

Graph the curve with parametric equations
$$
x= t \text{ and } y = t^2
$$


$$
\begin{array}{c|c}
t & (x,y) \\ \hline
-2 & (-2,4) \\
-1 & (-1,1) \\
0 & (0,0) \\
1 & (1,1)
\end{array}
$$
The ordered pairs in the second column of the table all seem to lie on the parabola with equation \(y = x^2\). To see this, note that
$$
y = t^2 = x^2
$$
since \(x = t\).

Graph the curve with parametric equations
$$
x= 2\cos \theta \text{ and } y = 2 \sin \theta; \; \; 0 \leq \theta \leq 2 \pi
$$


$$
\begin{array}{c|c}
\theta & (x,y) \\ \hline
0 & (2,0) \\
\pi/4 & (\sqrt{2},\sqrt{2}) \\
\pi/2 & (0,2) \\
3\pi/4 & (-\sqrt{2},\sqrt{2})\\
\pi & (-2,0)\\
\end{array}
$$
The ordered pairs in the second column of the table all seem to lie on the circle with equation \(x^2+y^2=4\). To see this, note that
the parametric equations satisfy the rectangular equation:
$$
x^2+y^2=4.
$$
That is,
$$
\begin{aligned}
x^2 + y^2 & \overset{?}{=} 4 \\
(2 \cos \theta)^2 + (2 \sin \theta)^2 & \overset{?}{=} 4 \\
4(\cos^2 \theta + \sin^2 \theta) & \overset{?}{=} 4 \\
4 &= 4
\end{aligned}
$$
Since \(\theta\) ranges from \(0\) to \(2 \pi\), the graph will be the graph of a complete circle of radius \(2\) centered at the origin. If, instead, \(\theta\) ranged from \(0\) to \(\pi/2\), then the graph would be the portion of the full circle in the first quadrant.

Graph the curve with parametric equations
$$
x= \cos(2\theta) \text{ and } y = \sin \theta; \; \; 0 \leq \theta \leq 2 \pi.
$$


$$
\begin{array}{c|c}
\theta & (x,y) \\ \hline
0 & (1,0) \\
\pi/4 & (0,\sqrt{2}/2) \\
\pi/2 & (-1,1) \\
3\pi/4 & (0,\sqrt{2}/2)\\
\pi & (1,0)\\
5\pi/4 & (0,-\sqrt{2}/2)\\
3\pi/2 & (-1,1)\\
\end{array}
$$
The ordered pairs in the second column of the table all seem to lie on a parabola, opening sideways, with equation \(x=1-2y^2\). To see this, note that the parametric equations satisfy the rectangular equation:
$$
x=1-2y^2.
$$
That is,
$$
\begin{aligned}
x & \overset{?}{=} 1-2y^2 \\
\cos(2 \theta) & \overset{?}{=} 1-2(\sin \theta)^2 \\
2- 2 \sin^2 \theta & \overset{?}{=} 1-2(\sin \theta)^2 \\
2- 2 \sin^2 \theta & = 1-2\sin^2 \theta \\
\end{aligned}
$$
Since \(\theta\) ranges from \(0\) to \(2 \pi\), the graph will only be a portion of the parabola. In fact, as \(\theta \) ranges from \(0\) to \(2 \pi\) portions of the graph are retraced.

E 10.2 Exercises