## Parametric Equations |

Sometimes it is impossible to write the equation of a curve in the form \(y=f(x)\) or \(x=f(y)\). But we may
be able to write \(x\) and \(y\) in terms of a third variable or parameter, say \(t\), in a set of equations
called * parametric equations*:
\[ x=f(t),\; y=g(t),\; a\leq t\leq b. \]
A curve given by parametric equations is called a * parametric curve* which is sometimes written as
\[ c(t)=(f(t),g(t)),\; a\leq t\leq b. \]
Note that \(c(t)\) is drawn with an orientation using the values of \(t\), starting from the initial point
\((f(a),g(a))\) and ending at the terminal point \((f(b),g(b))\).

**Example.**

- The parametric equations of the circle \(x^2+y^2=9\) are
\[ x=3\cos t,\; y=3\sin t,\; 0\leq t\leq 2\pi. \]
It can be verified as follows:
\[ x^2+y^2=(3\cos t)^2 +(3\sin t)^2=9(\cos^2 t+\sin^2 t)=9. \]
The top and bottom halves of the circle are given by \(0\leq t\leq \pi\) and \(\pi\leq t\leq 2\pi\) respectively.
The left semicircle of the preceding circle can be written as
\[ x=3\cos t,\; y=3\sin t,\; \frac{\pi}{2}\leq t\leq \frac{3\pi}{2}. \]
The initial point of the semicircle is \(\left(3\cos \left( \frac{\pi}{2} \right),3\sin \left( \frac{\pi}{2} \right) \right)=(0,3)\)
and the terminal point is \(\left(3\cos \left( \frac{3\pi}{2} \right),3\sin\left( \frac{3\pi}{2} \right)\right)=(0,-3)\).

- The parametric equations of the circle \((x-a)^2+(y-b)^2=r^2\) are \[ x=a+r\cos t,\; y=b+r\sin t,\; 0\leq t\leq 2\pi. \]
- The parametric equations of the ellipse \(\displaystyle\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\) are \[ x=\alpha+a\cos t,\; y=\beta+b\sin t,\; 0\leq t\leq 2\pi. \]
- The parametric equations of the line segment from \((a_1,b_1)\) to \((a_2,b_2)\) are \[ x=(1-t)a_1+ta_2,\; y=(1-t)b_1+tb_2,\; 0\leq t \leq 1. \]
- The parametric equations of the line \(y=mx+b\) are
\[ x=t,\; y=mt+b,\; -\infty < t <\infty. \]
The above is a natural parametrization of the line where \(x\) is the parameter \(t\) and \(y\) is a function
of \(t\).

** Implicitization of Parametric Curves:**

Sometimes we can eliminate the parameter \(t\) form parametric equations \(x=f(t),\; y=g(t)\) of a curve and
write an implicit equation in \(x\) and \(y\).

**Example.**
Find an equation for each of the following parametric curves:
(a) \(x=t^2+2,\,y=t+1\), \(-1\leq t < \infty\), (b) \(c(t)=(\tan t,\sec^2 t)\), \(-\displaystyle\frac{\pi}{2} < t < \frac{\pi}{2}\).

* Solution.* (a) By solving for \(t\) from \(y=t+1\), we get \(t=y-1\). Substituting \(t=y-1\) in
\(x=t^2+2\), we get \[ x=(y-1)^2+2. \]
Note that \(t=-1\) corresponds to \((3,0)\) and \(y=t+1\) increases with \(t\). Thus the above parabola starts
from \((3,0)\) and goes to the increasing direction of \(y\).

(b) Here \(x=\tan t\) and \(y=\sec^2 t\). Since \(\sec^2 t-\tan^2 t=1\),
\[y-x^2=1 \implies y=x^2+1. \]
Since \(-\infty < \tan t < \infty\) for \(-\displaystyle\frac{\pi}{2}< t < \frac{\pi}{2}\),
\(-\infty < x < \infty\). Thus \(c(t)=(\tan t,\sec^2 t)\), \(-\displaystyle\frac{\pi}{2}< t < \frac{\pi}{2}\)
represents the parabola \(y=x^2+1\).

Note that the parabola \(y=x^2+1\) can be parametrized in multiple ways. For example, its natural parametrization
is
\[c(t)=(t,t^2+1),\; -\infty < t < \infty. \]

** Derivatives from Parametric Equations:**

Consider the following parametric equations:
\[ x=f(t),\; y=g(t),\; a\leq t\leq b, \]
where \(x\) and \(y\) are differentiable functions of \(t\) and \(y\) is a differentiable function of \(x\).
Then \(y\) is a differentiable function of \(t\). By the chain rule,
\[ \frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt} \]
which implies
\[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)},\; \text{ provided } x'(t)\neq 0. \]

\(c(t)=(t^3-3t,4-t^2)\)

**Example.**
Find the equations of the tangent lines to the parametric curve \(c(t)=(t^3-3t,4-t^2)\) at \((0,1)\).

* Solution.*
First note that \(t^3-3t=t(t^2-3)=0 \implies t=0,\pm \sqrt{3}\). Since \(t=0\implies y=4-t^2=4\) and
\(t=\pm \sqrt{3}\implies y=4-t^2=1\), both \(t=\sqrt{3}\) and \(t=-\sqrt{3}\) correspond to \((0,1)\)
where \(c(t)=(t^3-3t,4-t^2)\) self-intersects.
\[ \frac{dy}{dx}=\frac{y'(t)}{x'(t)}
=\frac{-2t}{3t^2-3}. \]
The slope of the tangent line corresponding to \(t=\sqrt{3}\) is
\[ \frac{dy}{dx}=\frac{y'(t)}{x'(t)}
=\left. \frac{-2t}{3(t^2-1)} \right\vert_{t=\sqrt{3}}=\frac{-\sqrt{3}}{3}. \]
Thus the equation of the tangent line corresponding to \(t=\sqrt{3}\) is
\[ y-1=\frac{-\sqrt{3}}{3}x. \]
Similarly the slope of the tangent line corresponding to \(t=-\sqrt{3}\) is
\[ \frac{dy}{dx}=\frac{y'(t)}{x'(t)}
=\left. \frac{-2t}{3(t^2-1)} \right\vert_{t=-\sqrt{3}}=\frac{\sqrt{3}}{3}\]
and the equation of the tangent line corresponding to \(t=-\sqrt{3}\) is
\[ y-1=\frac{\sqrt{3}}{3}x. \]

** Integration from Parametric Equations:**

Suppose that \(y\) is an integrable function of \(x\) on \([a,b]\) given by the following parametric equations:
\[ x=f(t),\; y=g(t),\; \alpha \leq t\leq \beta, \]
where \(a=f(\alpha)\), \(b=f(\beta)\), and \(x\) is a differentiable function of \(t\). Since \(dx=f'(t)\;dt\),
\[ \int_a^b y \;dx=\int_{\alpha}^{\beta} g(t)f'(t)\;dt=\int_{\alpha}^{\beta} y(t)x'(t)\;dt. \]

**Example.**
Find the area of the region enclosed by the ellipse \(\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

* Solution.* Suppose that the equation of the top half of the ellipse is \(y=f(x)\), \(-a\leq x \leq a\).
Then \(f(x)\geq 0\) for \(-a\leq x \leq a\). By the symmetry of the ellipse, the required area is
\[2\int_{-a}^a f(x) \;dx.\]
Now \(y=f(x)\), \(-a\leq x \leq a\) is parametrized as
\[ x=a\cos t,\; y=b\sin t, \]
where \(t\) goes from \(t=\pi\) to \(t=0\) so that \(-a=a\cos (\pi)\) and \(a=a\cos (0)\).
Therefore, the required area is
\[\begin{align*}
2\int_{-a}^a f(x) \;dx &=2\int_{\pi}^0 y(t)x'(t)\;dt\\
&=2\int_{\pi}^0 b\sin t (-a\sin t)\;dt\\
&=-ab\int_{\pi}^0 2\sin^2 t \;dt\\
&=ab\int_0^{\pi} (1-\cos(2t)) \;dt\\
&=\left.ab \left( t-\frac{\sin(2t)}{2} \right) \right\vert_0^{\pi}\\
&=\pi ab.
\end{align*}\]
In case you do not like \(t\) going backward from \(t=\pi\) to \(t=0\) in the above parametrization of the
top half of the ellipse, you may use the following alternative parametrization:
\[ x=a\cos(\pi-t),\; y=b\sin (\pi-t),\; 0\leq t\leq \pi.\]
Then
\[ 2\int_{-a}^a f(x) \;dx =2\int_0^{\pi}y(t)x'(t)\;dt=2\int_0^{\pi} b\sin (\pi-t) a\sin(\pi-t)\;dt=\pi ab.\]

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