# 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.