# Parametric Equations and Calculus

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

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

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 $$x^2+y^2 =\frac{4t^2+(1-2t^2+t^4)}{(1+t^2)^2} =\frac{1+2t^2+t^4}{(1+t^2)^2}=1$$ 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, $$F'(x)=\frac{g'(t)}{f'(t)} \quad \text{ and in Leibniz notation } \quad \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ 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 $$\frac{d^2 y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{ \frac{d}{dt}\left( \frac{dy}{dx}\right)}{ \frac{dx}{dt}}.$$

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 $\frac{dy}{dt}=-e^{-t}$, $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-e^{-t}}{e^t}=-\frac{1}{e^{2t}}.$$ 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 $\frac{dy}{dt}=1+\ln t$ $$\frac{dy}{dx}=\frac{ dy/dt}{ dx/dt}=\frac{1+\ln t}{t/\sqrt{t^2+1}}=\frac{\sqrt{t^2+1}(1+\ln t)}{t}.$$ 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
\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}.

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 $$\label{bereq} \left(x^2+y^2\right)^2=4\left(x^2-y^2\right)$$ 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 $$\label{paracurderpol} \frac{dy}{dx} =\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin \theta+ r \cos \theta}{\frac{dr}{d\theta} \cos \theta- r\sin \theta} \qquad \text{ whenever }\ \frac{dx}{d\theta} \neq 0$$ 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 $f'(g(-5))?$