### Geometry: Parametric curves

### Lissajous figures

A special kind of parametric equations goes by the name of Lissajous figures. These curves describe harmonic motions in physics.

A **Lissajous figure** is a curve #\orange{C}# described by a parametric equation of the form

\[\begin{array}{rcl}\blue{x(t)}&=& \blue{A \sin ( at + b)}\\\green{y(t)}&=& \green{B \sin (ct +d) }\end{array}\]

where #A, B, a, b, c# and #d# are numbers.

The **period **of the curve is the length of the smallest interval of #t# such that the full curve is displayed.

**Example**Determine whether the parametric curve given by equations

\[\begin{array}{rcl}\blue{x(t)}&=& \blue{\sin(t) \cos(2) + \cos(t) \sin(2) }\\ \green{ y(t)}&=& \green{4 \sin( \frac{2t}{\pi} + \pi) } \end{array}\]

is a Lissajous figure.

Yes it is.

We use trigonometric identities to rewrite the equations. For #\blue{x(t)}# we use the identity #\sin( \alpha + \beta ) = \sin(\alpha) \cos (\beta) + \cos(\alpha) \sin(\beta)#. For #\blue{x(t)}# we get

\[\begin{array}{rcl}\blue{ x(t)} & = & \blue{\sin(t) \cos(2) + \cos(t) \sin(2) }\\ & = & \sin(t + 2). \end{array}\]

We see that #\green{y(t)}# is already of the desired form.

**Question**

Let #\orange P# be a point whose orbit is described by parametric equations

\[\begin{array}{rcl}\blue{x(t)}&=&\blue{ 2\sin ( \pi t )}\\ \green{y(t)}&=& \green{2 \cos (t)} \end{array}\]

where #t# lies between #- \pi # and # \pi#. Determine the values of #t# where #\orange P# passes through the #y#-axis.

**Solution**

We solve #\blue{x(t)} = 0# for #t \in \ivcc{- \pi}{\pi}#. We get

\[\begin{array}{rcl}\blue{2 \sin (\pi t )} &=&0 \\\pi t &=& k\cdot \pi \\t &= & k\end{array}\]

where #k# is an integer.

Therefore, for #t = -3, -2, -1, 0, 1, 2, 3# in the interval the point #\orange P# passes through the #y#-axis.

To determine how often the point #P# passes through the origin we simultaneously have to solve #x(t) = 0# and #y(t) = 0#. We have that #x(t) = 3\cdot \sin(2\cdot t+{{\pi}\over{4}}) = 0# if and only if #2\cdot t+{{\pi}\over{4}} # is a multiple of #\pi#. We get

\begin{array}{rcl}

2\cdot t+{{\pi}\over{4}} & = & k \cdot \pi \\

t & = & {{\pi\cdot k}\over{2}}-{{\pi}\over{8}}

\end{array}

for all integers #k#. The values of #t# which lie in the interval are those in the list \[\left[ t={{7\cdot \pi}\over{8}} , t={{3\cdot \pi}\over{8}} , t=-{{\pi}\over{8}} , t=-{{5\cdot \pi}\over{8}} \right] \]

The ones that also have #y(t) = 0# are

\[ \left[ t=-{{\pi}\over{8}} \right] \]

So #P# passes through the origin once.

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