Dr. Lalitha Subramanian
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Aksharam
A parametric relation is a relation where both $x$ and $y$ are defined in terms of a third variable. This third variable is called a Parameter. A plane curve is defined by the set of points $(x,y)$ such that $x=f(t)$ and $y=g(t)$ where $f$ and $g$ are both defined on an interval $I$. The equations $x=f(t)$ ane $y=g(t)$ are the parametric equations of the curve, with the parameter $t$.
Consider the set of all ordered pairs $x=t+1$ and $y=t^2$, where $t$ is any real number. Find the points determined by $t=-2, \quad -1, \quad 0, \quad 1, \quad 2$. Also, find an algebraic relationship between the variables $x$ and $y$, and graph this relation.
Solution:
\begin{array}{cccc} t & x=t+1 & y=t^2 & (x,y)\\ \hline -2 & -1 & 4 & (-1,4)\\ -1 & 0 & 1 & (-1,1)\\ 0 & 1 & 9 & (0,1) \\ 1 & 2 & 1 & (2,1) \\ 2 & 3 & 4 & (2,3)\\ \hline \end{array}To find the relationship between $x$ and $y$, we eliminate $t$ between the two equations: \begin{eqnarray} x &=& t+1\\ t &=& x-1\\ y &=& t^2\\ y &=& (x-1)^2\\ \end{eqnarray} Graph of this relationship is shown below:
Find coordinates of the point represented by $t= -1$ in the relationship $x=3t-5$ and $y=6-2t$.
Solution:Substituting the given value of $t$ in the two equations, we get \begin{eqnarray} x &=& 3(-1)-5\\ &=& -8\\ y &=& 6 - 2(-1)\\ &=& 8\\ \end{eqnarray} Hence, the coordinates of the point is $(-8,8)$.
Parametric equations for a given relationship is not unique. We can assume either $x$ of $y$ as a parameter either alone as an equation involving the parameter, and then substitute in the given relationship to get the corresponding parametric equation for $y$. A parameter is represented by any symbol of though commonly used parameters are $t$ and $\theta$.
Find a pair of parametric equations for the relationship $x^2 + y^2 = 4, \quad x,y \geq 0$
Solution: First, we solve the above equation for $y$: \begin{eqnarray} y^2 &=& 4 - x^2\\ y &=& \sqrt{4-x^2}\\ \end{eqnarray}
As per the given conditions, $y$ is positive. Also, from the solution for $y$, we see that the restriction of $x$ is $-4 \leq x \leq 4$ Now, let the parameter be $t$, and let $x = t$. So, we get the second parametric equation as $y = \sqrt{4-t^2}$, with $-4 \leq t \leq 4$. If we take the first equation as $x = t^2$, then, the second equation would be $y = \sqrt{4-t}$, with $t \leq 4$.
$(6,0)$
$(2,1)$
$(y+5)^2=4x$
$(-4,-9), (-2,-4), (0,1), (2,6), (4,1)$
$(0,-3), (-1,-2), (0,-1), (3,0), (8,1)$
If we take $x=t$, then, $y=2t+7$
$x=2 \cos \theta$ and $y=2 \sin \theta$
For problems #8-12, for each plane curve, find the rectangular form and then graph the curve:
$y=(x-2)^2+1$
$\frac{x^2}{4}+\frac{y^2}{9}=1$.
$y=(x-1)^3-1$
$y= \sqrt{x^2+2}$
$y=\frac{1}{x}$