Parametric Equations

Some plane curves are not the graphs of functions $y = f(x).$ In particular, graphs of functions cannot fail the vertical line test: for each $a$ there can be at most one point on the curve with $x$-coordinate $a.$ In order to describe more curves, it is convenient to consider $x$ and $y$ as functions of a separate variable $t$ (called a parameter), i.e.

$$x=f(t), \quad y=g(t).$$

This is known as a parametric equation for the curve that is traced out by varying the values of the parameter $t.$

Show that the parametric equation $x=\cos t$ and $y=\sin t$ $(0 \leqslant t\leqslant 2\pi)$ traces out a circle.

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Eliminating $t$ gives

$$x^2+y^2= \cos^2 t+\sin^2 t=1,$$

which in fact is the equation of a circle with radius $1$. $_\square$

A circle centered at $(h,k)$ with radius $r$ can be described by the parametric equation

$$x=h+r\cos t, \quad y=k+r\sin t.$$

Eliminating $t$ as above leads to the familiar formula

$$(x-h)^2+(y-k)^2=r^2.$$

What are the radius $r$ and center $(h,k)$ of

$$\begin{array}{c}&x=3+8\cos 4t, &y=-2+8\sin 4t, &0 \leq t\leq 2\pi? \end{array}$$

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From the general equation above, we have

$$r=8, \quad c(h,k) = (3,-2). \ _\square$$

In the parametric equation

$$\begin{array}{c}&x=8\cos At, &y=8\sin At, &0 \leqslant t\leqslant 2\pi, \end{array}$$

how does $A$ affect the circle as $A$ changes?

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Eliminating $t$ gives

$${ x }^{ 2 }+{ y }^{ 2 }={ \cos }^{ 2 } At+{ \sin }^{ 2 } At=1,$$

which is still a circle with radius $1$ and center at the origin. If we have $A=\frac{1}{2}$, $(x,y)=\left(\cos\frac{1}{2}t, \sin \frac{1}{2}t \right)$, i.e. as $t$ ranges from $0$ to $2\pi,$ the equation starts at $(1,0)$ and stops at $(-1,0)$. This means that it goes halfway through the circle. So $A$ governs the rate at which the equation traces out a circle.

Similarly, if $A=2,$ the equation moves twice around the circle. $_\square$

A line that passes through point $(h,k)$ with slope $m$ can be described by the parametric equation

$$x = h + t, \quad y = k + mt.$$

More generally, let $m = \tan \alpha,$ where $\alpha$ is the tilt angle. Changing $t$ to $t\cos\alpha,$ the parametric equation will become

$$x = h+t\cos \alpha, \quad y = k+t\sin \alpha.$$

Let $P, Q$ be two points of interception between line $L:x-y+2=0$ and parabola $y=x^2$. What is the length $|PQ|?$

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$L : x-y+2=0$ passes through point $(0,2)$ and has a tilt angle of $\alpha=\frac{\pi}{4}$ and hence the parametric equations

$$x=t\frac{\sqrt{2}}{2}, \quad y = 2 + t\frac{\sqrt{2}}{2}.$$

Substituting into the parabola gives

$$\begin{aligned} y &= x^2 \\ 2+t\frac{\sqrt{2}}{2}&=t^2\frac{1}{2} \\ t^2-\sqrt{2}t-4 &= 0\\\\ |PQ| &= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \\ &= \sqrt{(t_1 - t_2)^2}\\ &= \sqrt{(t_1+t_2)^2-4t_1t_2} \\ &= \sqrt{2-4(-4)} \\ &= 3\sqrt{2}.\ _\square \end{aligned}$$

Converting from a parametric equation to an equation in terms of Cartesian coordinates involves eliminating $t$:

What does the parametric equation

$$\begin{array}{c}&x=t^{2}+t, &y =2t-1\end{array}$$

describe?

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Plugging the value of $t$ in $y,$ which is $t=\frac{1}{2} (y+1),$ into $x$ gives

$$x=\frac{1}{4}y^{2}+y+\frac{3}{4}.$$

This is the equation of a parabola opening to the right. $_\square$

The slope of the tangent line to a curve $y=f(x)$ at a point $(x_0,y_0)$ is $f'(x_0).$ When $x$ and $y$ are defined parametrically, the derivative $\frac{dy}{dx}$ can be computed as $\small \frac{\hspace{3mm} \frac{dy}{dt}\hspace{3mm} }{\frac{dx}{dt}},$ or more symmetrically as follows:

If a curve is given by parametric equations $x=f(t)$ and $y=g(t),$ the tangent line to the curve at the point $\big(f(t_0),g(t_0)\big)$ is given by

$$f'(t_0)\big(y-g(t_0)\big) = g'(t_0)\big(x-f(t_0)\big),$$

as long as either $f'(t_0)\ne 0$ or $g'(t_0) \ne 0.$

Here are some special cases:

• Parabola $y^2=4ax:$ $(x,y) \mapsto (at^2, 2at)$
  Equation of tangent at the point with parameter $t:$ $y=\frac{x}{t}+at$
  Equation of normal at the point with parameter $t:$ $y=-xt+2at+at^3$
• Ellipse (possibly a circle) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1:$ $(x,y) \mapsto (a \cos \theta , b \sin \theta)$
  Equation of tangent at the point with parameter $\theta:$ $\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
  Equation of normal at the point with parameter $\theta:$ $\frac{ax}{\cos \theta}-\frac{by}{\sin \theta}=a^2-b^2$
• Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1:$ $(x,y) \mapsto (a \sec \theta , b \tan \theta)$
  Equation of tangent at the point with parameter $\theta:$ $\frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1$
  Equation of normal at the point with parameter $\theta:$ $\frac{ax}{\sec \theta}+\frac{by}{\tan \theta}=a^2+b^2$

Let $C$ be the curve given by the parametric equations

$$\begin{aligned} x &= \cos(t)\big(1-\cos(t)\big) \\ y &= \sin(t)\big(1-\cos(t)\big) \end{aligned}$$

for $0 \le t \le 2\pi.$ What is the equation of the tangent line at the points $\left(\frac14,\frac{\sqrt{3}}{4}\right)$ and $\left(-\frac34,\frac{3\sqrt{3}}4\right)$ on the curve?

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Since $\frac{dx}{dt} = \sin t(2\cos t-1)$ and $\frac{dy}{dt} = \cos t-\cos^2 t+\sin^2 t,$ plugging in $t = \frac{\pi}3$ gives

$$0\left(y-\frac{\sqrt{3}}4\right) = 1\left(x-\frac14\right) \implies x =\frac14.$$

So the tangent line at $\left(\frac14,\frac{\sqrt{3}}{4}\right)$ is vertical.

On the other hand, at $\left(-\frac34,\frac{3\sqrt{3}}4\right),$ $t=\frac{2\pi}3,$ so the equation is

$$-\sqrt{3}\left(y-\frac{3\sqrt{3}}4\right) = 0\left(x+\frac34\right) \implies y= \frac{3\sqrt{3}}4.$$

So the tangent line is horizontal. $_\square$

An ellipse with center at the origin and axes coinciding with the coordinate axes is usually described by the following parametrization:

$$(x,y) \mapsto (a \cos \theta , b \sin \theta ),$$

where $a$ and $b$ are the lengths of the semi-major axis and the semi-minor axis, respectively.

Similarly, the parametric equation of an ellipse is

$$\begin{array}{c}&x=h+a\cos t, &y=k+b\sin t.\end{array}$$

Eliminating $t$ gives

$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1.$$

Note that when $a=b,$ the equation becomes that of a circle.

Given the parametric equation of an ellipse

$$\begin{array}{c}&x=4\cos 3t, &y =7\sin 3t,\end{array}$$

what will be the length of the semi-major and semi-minor axes of the ellipse?

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Eliminating the parameter, we have

$$\begin{aligned} \frac{x^{2}}{16}+\frac{y^2}{49} &=\frac{4^{2}\cos^{2}3t}{16}+\frac{7^{2}\sin^{2}3t}{49}\\\\ &=\cos^{2}3t+\sin^{2}3t\\\\ &=1. \end{aligned}$$

Thus, the length of the semi-major axis is $7$ and the length of the semi-minor axis is $4$. $_\square$

A useful formula is the following equation of the line joining the points with parameters $\alpha$ and $\beta$:

$$\frac{x}{a} \cos \frac{\alpha+\beta}{2} + \frac{y}{b} \sin \frac{\alpha+\beta}{2} = \cos \frac{\alpha-\beta}{2}.$$

An east-west opening hyperbola centered at $(h,k)$ can be described by the parametric equation

$$x = h + a\sec t, \quad y = k + b\tan t,$$

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

Eliminating $t$ gives

$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.$$

Similarly, for a north-south opening hyperbola, the parametric equation is

$$x = h+b\tan t, \quad y = k+a\sec t.$$

Let a curve $C = \left \{ \begin{array}{lr} x= f(t) \\ y = g(t) \end{array} \right.$, where $t\in [a,b]$. If $f$ is differentiable and $g$ is continuous, the area bounded by $C, x = f(a), x = f(b),$ and the $x$-axis is

$$A = \int\limits_a^b y\, dx = \int \limits_a ^ b g(t) f'(t) \text{ }\mathrm{d}t.$$

Note that this is signed area; the area below the $x$-axis is counted as negative area.

Show that the area of an ellipse with axis lengths $a$ and $b$ is

$$A = \pi ab.$$

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The parametric equation of an ellipse centered at $(0,0)$ is

$$f(t) = a\cos t, \quad g(t) = b\sin t.$$

Our approach is to only consider the upper half, then multiply it by two to get the area of the entire ellipse.

First, we need to find the left and right bounds in terms of $t$, such that

$$\begin{aligned} f(t_1)&=-a &&\implies t_1 = \pi \\ f(t_2)&= a &&\implies t_2 = 0. \end{aligned}$$

Hence,

$$\begin{aligned} A &= \int \limits_0 ^\pi g(t)f'(t) \text{ }\mathrm{d}t \\ &= ab \int \limits_0 ^\pi \sin ^2 t \text{ }\mathrm{d}t \\ &= ab \int\limits_0 ^\pi \frac{1}{2} (1-\cos 2t) \text{ }\mathrm{d}t \\ &= \frac{1}{2} ab \bigg [ t-\frac{1}{2} \sin 2t \bigg ] _0 ^\pi \text{ }\mathrm{d}t \\ &= \frac{1}{2} ab \pi.\ _\square \end{aligned}$$

Show that the parametric equation of a projectile traces out a parabola.

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We have

$$\begin{aligned} x&=v\cos \theta t &\qquad (1)\\ y&=v\sin \theta t-\frac { 1 }{ 2 } gt^2. &\qquad (2) \end{aligned}$$

Substituting the value of $t$ from parameter $1$ into parameter $2$, we have

$$y=x\tan \theta -\frac { 1 }{ 2 } \frac { g }{ v^{2}\cos^2 \theta} x^2.$$

Let $a=\tan \theta$ and $b=\frac { 1 }{ 2 } \frac { g }{ v^{2}\cos^2 \theta }$, then

$$y=-bx^2+ax,$$

which indeed is the equation of a parabola opening downward. $_\square$