Parametric equations are a set of functions of one or more independent variables called parameters and are used to express the coordinates of the points that make up a geometric object such as a curve or surface. We begin by sketching the graph of a few parametric equations. Tangent lines to parametric curves and motion along a curve is discussed. We end with parametric equations expressed in polar form.

## Sketching Parametric Curves

Let $C$ be a curve defined by $$ P(t)=(f(t),g(t)) $$ where $f$ and $g$ are defined on an interval $I.$ The equations $$ x=f(t) \qquad \text{and}\qquad y=g(t) $$ for $t\in I$ are ** parametric equations** for $C$ with parameter $t.$ The

**of a parameterized curve $C$ is the direction determined by increasing values of the parameter.**

*orientation***Example**. (a) Sketch the graph of the curve $$ C_1: x=t, y=1-t $$ on $[0,1]$ by plotting values for $t;$ and then check your graph by finding an equation in $x$ and $y$ only and then graphing. Show the orientation of the curve. (b) Sketch the graph of the curve $$ C_2: x=1-t^2, y=t^2 $$ on $[0,1]$ by plotting values for $t;$ and then check your graph by finding an equation in $x$ and $y$ only and then graphing.

Show the orientation of the curve.

**Solution**. (a) The graph starts at the point $(0,1)$ and follows the line ${y=1-x}$ until it reaches the other endpoint at $(1,0).$ (b) The graph starts at the point $(1,0)$ and follows the line $x=1-y$ until it reaches the other endpoint at $(0,1).$

Comparing (a) and (b) we see that the graph of a curve may have more than one parametrization. Can you think of another set of parametric equations that gives the same graph?

**Example**. Sketch the graph of the curve $$ C_3: x=\cos 2t, y=\sin 2t $$ on $\left[0,\frac{\pi }{2}\right]$ by plotting values for $t;$ and then check your graph by finding an equation in $x$ and $y$ only and then graphing. Show the orientation of the curve.

**Solution**. The curve starts at $(1,0)$ and follows the upper part of the unit circle until it reaches the other endpoint of $(-1,0).$ Can you think of another set of parametric equations that give the same graph?

**Example**. (a) Find a rectangular equation whose graph contains the curve $C$ with the parametric equations $$ x=\frac{2t}{1+t^2} \qquad \text{and}\qquad y=\frac{1-t^2}{1+t^2} $$ and (b) sketch the curve $C$ and indicate its orientation.

**Solution**. Since \begin{equation} x^2+y^2 =\frac{4t^2+(1-2t^2+t^4)}{(1+t^2)^2} =\frac{1+2t^2+t^4}{(1+t^2)^2}=1 \end{equation} and also $x(0)=0$, $y(0)=0$ and $x(1/2)=4/5$, $y(1/2)=3/5$ we see the graph of the given parametric equations represents the unit circle with orientation counterclockwise.

**Example**. The position of a particle at time $t$ is $(x,y)$ where $x=\sin t$ and $y=\sin^2 t.$ Describe the motion of the particle as $t$ varies over the time interval $[a,b].$

**Solution**. We can eliminate $t$ to see that the motion of the object takes place on the parabola, $y=x^2.$ The orientation of the curve is from $(\sin a, \sin^2 a)$ to $(\sin b, \sin^2 b).$

## Tangent Lines to Parametric Curves

**Theorem**. If $f'(t)$ is continuous and $f'(t)\neq 0$ for $a\leq t \leq b$, then the parametric curve defined by $x=f(t)$ and $y=g(t)$ for $a\leq t \leq b$, can be put into the form $y=F(x).$ Moreover, \begin{equation} F'(x)=\frac{g'(t)}{f'(t)} \quad \text{ and in Leibniz notation } \quad \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \end{equation} whenever, $\frac{dx}{dt}\neq 0.$ Also, if the parametric equations $x(t)$ and $y(t)$ define $y$ as a twice differentiable function over some suitable interval, then \begin{equation} \frac{d^2 y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{ \frac{d}{dt}\left( \frac{dy}{dx}\right)}{ \frac{dx}{dt}}. \end{equation}

**Example**. Find the slope of the tangent line to the curve defined by the parametric equations $x=2(\theta-\sin \theta)$ and $y=2(1-\cos \theta)$ at the point corresponding to the value of the parameter $\theta=\pi/6.$

**Solution**. Since $\frac{dx}{d\theta}=2(1-\cos \theta)$ and $\frac{dy}{d\theta}=2\sin \theta.$ The slope of the tangent line at $\theta=\pi/6$ is

\begin{align} \left.\frac{dy}{dx}\right|{\theta=\pi/6} & =\left.\frac{dy/d\theta}{dx/d\theta}\right|{\theta=\pi/6} \\ & =\left.\frac{2\sin \theta}{2(1-\cos \theta)}\right|_{\theta=\pi/6} =\frac{1}{2\left.(1-\frac{\sqrt{3}}{2}\right)}=2+\sqrt{3}. \end{align} as desired.

**Example**. Find an equation of the tangent line to the curve defined by the parametric equations $x=e^t$ and $y=e^{-t}$ at the point $(1,1).$ Then sketch the curve and the tangent line(s).

**Solution**. Since $\frac{dx}{dt}=e^t$ and $\frac{dy}{dt}=-e^{-t}$, \begin{equation} \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-e^{-t}}{e^t}=-\frac{1}{e^{2t}}. \end{equation} The point $(1,1)$ corresponds to $t=0.$ The slope of the tangent line is $$ \left.\frac{dy}{dx}\right|_{t=0}=-1, $$ so an equation is $y-1=-1(x-1)$ or just $-x+2.$

**Example**. Find the points on the curve defined by parametric equations $x=t^3-3t$ and $y=t^2$ at which the tangent line is either horizontal or vertical. Then sketch the curve.

**Solution**. First we find the derivatives of $x$ and $y$ with respect to $t$: $\frac{dx}{dt}=3t^2-3$ and $\frac{dy}{dt}=2t.$ To find the point(s) where the tangent line is horizontal, set $\frac{dy}{dt}=0$ obtaining $t=0.$ Since $\frac{dx}{dt} \neq 0$ at this $t$ value, the required point is $(0,0).$ To find the point(s) where the tangent line is vertical, set $\frac{dx}{dt}=0$ obtaining $t=\pm 1.$ Since $\frac{dy}{dt}\neq 0$ at either of these $t$-values, the required points are $(2,1)$ and $(-2,1).$

**Example**. Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ given the parametric equations $x=\sqrt{t^2+1}$ and $y=t \ln t.$

**Solution**. Since $\frac{dx}{dt}=\frac{t}{\sqrt{t^2+1}}$ and $\frac{dy}{dt}=1+\ln t$ \begin{equation} \frac{dy}{dx}=\frac{ dy/dt}{ dx/dt}=\frac{1+\ln t}{t/\sqrt{t^2+1}}=\frac{\sqrt{t^2+1}(1+\ln t)}{t}. \end{equation} Next,

\begin{align} & \frac{d}{dt}\left[ \frac{(t^2+1)^{1/2}(1+\ln t)}{t}\right] \\ & \qquad = \frac{t\left[ \frac{1}{2} (t^2+1)^{-1/2} (2t) (1+\ln t)+\frac{ \left( t^2+1\right)^{1/2}}{t}\right]-(t^2+1)^{1/2}(1+\ln t)}{t^2} \\ & \qquad =\frac{ \frac{t^2(1+\ln t)}{(t^2+1)^{1/2}}+(t^2+1)^{1/2}-(t^2+1)^{1/2}(1+\ln t)}{t^2} \\ & \qquad = \frac{ \frac{t^2(1+\ln t)}{(t^2+1)^{1/2}}+\frac{t^2+1}{(t^2+1)^{1/2}}-\frac{(t^2+1)(1+\ln t)}{(t^2+1)^{1/2}}}{t^2} \\ & \qquad = \frac{t^2+t^2\ln t+t^2+1-t^2-t^2\ln t-1-\ln t}{t^2(t^2+1)^{1/2}} \\ & \qquad = \frac{t^2-\ln t}{t^2\sqrt{t^2+1}} \end{align} and so \begin{equation}

\frac{d^2y}{dx^2}=\frac{ \frac{d}{dt}\left( \frac{dy}{dx}\right)}{ \frac{dx}{dt}} = \frac{t^2-\ln t}{t^2\sqrt{t^2+1}}\cdot \frac{\sqrt{t^2+1}}{t}=\frac{t^2-\ln t}{t^3}. \end{equation}

**Example**. Show that the curve defined by parametric equations $x=t^2$ and $y=t^3-3t$ crosses itself. Find equations of the tangent lines to the curve at that point.

**Solution**. Observe that the curve is at the point $(3,0)$ when $t_1=-\sqrt{3}$ and $t_2=\sqrt{3}$, so the curve crosses itself at the point $(3,0).$

Since $\frac{dx}{dt}=3t^2-3$ and $$ \frac{dy}{dt}=\frac{3(t^2-1)}{2t}, $$ at the first point of intersection, $$ m_1=\left.\frac{dy}{dx}\right|{t=-\sqrt{3}} =\frac{3(3-1)}{2(-\sqrt{3})} =-\sqrt{3} $$ and an equation of the tangent line is $y-0=-\sqrt{3}(x-3)$ or $y=-\sqrt{3}(x-3).$ At the second point, $$ m_2 = \left.\frac{dy}{dx}\right|{t=\sqrt{3}}=\frac{3(3-1)}{2(\sqrt{3})}=\sqrt{3} $$ and an equation of the tangent line is $y-0=\sqrt{3}(x-3)$ or $y=\sqrt{3}(x-3).$

## Motion Along a Curve

We can think of $t$ on the closed interval $[a, b]$ as representing time, and in doing so we can interpret the parametric equations in terms of the motion of a particle as follows: at time $t=a$ the particle is at the initial point $(f(a), g(a))$ of the curve or trajectory $C$. As $t$ increases from $t=a$ to $t =b$, the particle traverses the curve in a specific direction called the ** orientation of a curve**, eventually ending up at the terminal point $(f(b), g(b))$ of the curve.

**Example**. Use implicit differentiation to find all points on the lemniscate of Bernoulli \begin{equation} \label{bereq} \left(x^2+y^2\right)^2=4\left(x^2-y^2\right) \end{equation} where the tangent line is horizontal.

**Solution**. Using implicit differentiation we have, $$ 2 \left(x^2+y^2\right) \left(2x+2y\frac{dy}{dx}\right)=8x-8y\frac{dy}{dx}$$ and so $$ \frac{dy}{dx}=-\frac{x \left(-2+x^2+y^2\right)}{y \left(2+x^2+y^2\right)}\ $$ we need to find all $(x,y)$ where $dy/dx=0.$ Clearly, the point $(0,0)$ is ruled out and so $-2+x^2+y^2=0$; that is $x^2+y^2=2.$ Using $x^2+y^2=2$ with the original we see $x^2-y^2=1$ also. Therefore, $2x^2=3$ and so $x=\pm \sqrt{3/2}$ and $y=\pm \sqrt{1/2}.$

## Parametric Equations in Polar Form

To find the slope of the tangent line to the graph of $r=f(\theta)$ at the point $P(r, \theta)$, let $P(x, y)$ be the rectangular representation of $P$. Then \begin{align} & x=r\cos \theta = f(\theta)\cos \theta \\ & y=r\sin \theta = f(\theta)\sin \theta \end{align} We can view these equations as parametric equations for the graph of ${r=f(\theta)}$ with parameter $\theta$. We have \begin{equation} \label{paracurderpol} \frac{dy}{dx} =\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin \theta+ r \cos \theta}{\frac{dr}{d\theta} \cos \theta- r\sin \theta} \qquad \text{ whenever }\ \frac{dx}{d\theta} \neq 0 \end{equation} and this gives the slope of the tangent line to the graph of $r=f(\theta)$ at any point $P(r, \theta)$.

## Exercises Parametric Equations

**Exercise**. Find the rectangular equations for the curve represented by

$(1) \quad x=4\cos \theta$ and $y=3\sin\theta$, $0\leq \theta \leq 2\pi$.

$(2) \quad x=\sin t$ and $y=\sin2t$, $0\leq t \leq 2\pi$.

$(3) \quad C: x=t^2$, $y=t-1$; $0\leq t \leq 3$

$(4) \quad C: x=t^2+1$, $y=2t^2-1$; $-2\leq t\leq 2$

**Exercise**. Sketch the graph of the parametric equations $x=\cos (\pi -t)$ and $y=\sin (\pi -t)$, then indicate the direction of increasing $t.$

**Exercise**. Sketch the graph of the parametric equations $x=3 t$ and $y=9t^2$, then indicate the direction of increasing $-\infty <t<\infty .$

**Exercise**. Sketch the graph of the parametric equations $x=3-3t$ and $y=2t$, then indicate the direction of increasing $0\leq t\leq 1.$

**Exercise**. Find parametric equations and a parameter interval for the motion of a particle that starts at $(a,0)$ and traces the ellipse $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. $$ (a) once clockwise (b) once counterclockwise (c) twice clockwise (d) twice counterclockwise.

**Exercise**. Find a parametrization for the curve whose graph is the lower half of the parabola $x-1=y^2.$

**Exercise**. Find a parametrization for the curve whose graph is the ray (half-line) with initial point $(-1,2)$ that passes through the point $(0,0).$

**Exercise**. Find an equation for the line tangent to the curve $x=t$ and $y=\sqrt{t}$ at $t=1/4.$ Also, find the value of $\frac{d^2y}{dx^2}$ at

this point.

**Exercise**. Convert $\left(x^2+y^2\right)^2=4\left(x^2-y^2\right)$ to polar form to find all points on the lemniscate of Bernoulli where the tangent line is horizontal.

**Exercise**. Consider the cardioid $r= 1 + \cos \theta$. (a) Find the slope of the tangent line to the cardioid at the point where $\theta=\pi/6$. (b) Find the points on the cardioid where the tangent lines are horizontal and where the tangent lines are vertical.

**Exercise**. Find the slope of the tangent line to the curve $x=2\sin \theta$, $y=3\cos \theta$ at the point corresponding to the value of the parameter $\theta=\pi/4$.

**Exercise**. Find the slope of the tangent line to the curve $x=2(\theta-\sin \theta)$, $y=2(1-\cos \theta)$ at the point corresponding to the value of the parameter $\theta=\pi/6$.

**Exercise**. Find $d^2y/dx^2$ given $x=\sqrt{t}$, $y=1/t$.

**Exercise**. Find $d^2y/dx^2$ given $x=\sin 2t$, $y=\cos 2t$.

**Exercise**. Find an equation of the tangent line to the curve $x=t^2+t$, $y=t^2-t^3$ at the given point $(0,2)$.

**Exercise**. Find an equation of the tangent line to the curve $x=e^t$, $y=e^{-t}$ at the given point $(1,1)$.

**Exercise**. Find an equation for the line tangent to the curve $x=t$ and $y=\sqrt{t}$ at $t=1/4.$ Also, find the value of $\frac{d^2y}{dx^2}$ at this point.

**Exercise**. Find an equation for the line tangent to the curve $x=t-\sin t$ and $y=1-\cos t$ at $t=\pi /3.$ Also, find the value of $\frac{d^2y}{dx^2}$

at this point.

**Exercise**. Suppose that $u=g(x)$ is differentiable at $x=-5,$ $y=f(u)$ is differentiable at ${u=g(-5)}$ and $(f\circ g)'(-5)$ is negative. What, if anything, can be said about the values of $g'(-5)$ and $f'(g(-5))?$