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.

1. 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 \(\left(3\cos \left( \frac{3\pi}{2} \right),3\sin\left( \frac{3\pi}{2} \right)\right)=(0,-3)\).
   
   
   
2. The parametric equations of the circle \((x-a)^2+(y-b)^2=r^2\) are \[ x=a+r\cos t,\; y=b+r\sin t,\; 0\leq t\leq 2\pi. \]
3. The parametric equations of the ellipse \(\displaystyle\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\) are \[ x=\alpha+a\cos t,\; y=\beta+b\sin t,\; 0\leq t\leq 2\pi. \]
4. The parametric equations of the line segment from \((a_1,b_1)\) to \((a_2,b_2)\) are \[ x=(1-t)a_1+ta_2,\; y=(1-t)b_1+tb_2,\; 0\leq t \leq 1. \]
5. 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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