# Help with finding Cartesian Equations from Parametric Equations using Chain Rule

• Sep 1st 2007, 05:54 AM
macho_maggot2001
Help with finding Cartesian Equations from Parametric Equations using Chain Rule
Hi!

This is my first post here :)

I have encountered some trouble in my math homework.

A curve is defined parametrically by the equations:
x= t^3 - 6t + 4 and y= t - 3 + 2/t

Find:
i. the equations of the normals to the curve at the points where the curve meets the x-axis.
ii. the coordinates of their point of intersection.

I am simply stumped...
• Sep 1st 2007, 06:35 AM
Soroban
Hello, macho_maggot2001!

Welcome aboard!

Quote:

A curve is defined parametrically by the equations: . $\begin{array}{ccc}x & = & t^3 - 6t + 4\\
y & = & t - 3 + \frac{2}{t}\end{array}$

Find:

1) the equations of the normals to the curve at the points
. . .where the curve meets the x-axis.

2) the coordinates of their point of intersection.

1) The curve meets the x-axis when $y = 0$
. . $t - 3 + \frac{2}{t}\:=\:0\quad\Rightarrow\quad t^2-3t + 2 \:=\:0\quad\Rightarrow\quad (t-1)(t-2) \:=\:0\quad\Rightarrow\quad t\,=\,1,\,2$

Then: . $x\,=\,-1,\,0$
. . The x-intercepts are: . $(-1,\,0),\;(0,\,0)$

Then we need the derivative, $\frac{dy}{dx}$, to find the slope of the tangents.

We have: . $\begin{Bmatrix}\frac{dy}{dt} & = & 1 - \frac{2}{t^2} \\ \\
\frac{dy}{dx}\;=\;\frac{1-\frac{2}{t^2}}{3t^2-6} \;=\;\frac{t^2-2}{t^2(3t^2-6)}$

When $t = 1\!:\;\frac{dy}{dx} \:=\:\frac{1}{3}$
When $t = 2\!:\;\frac{dy}{dx} \:=\:\frac{1}{12}$

One normal has point $(-1,0)$ and $m = -3$
. . Its equation is: . $y - 0 \:=\:-3(x + 1)\quad\Rightarrow\quad\bf{{\color{blue} y \:=\:-3x-3}}$

The other has point $(0,0)$ and $, = -12$
. . Its equation is: . $y - 0 \:=\:-12(x-0)\quad\Rightarrow\quad\bf{{\color{blue} y \:=\:-12x}}$

2) They intersect when: . $-3x - 3 \:=\:-12x\quad\Rightarrow\quad9x\:=\:3$

. . . $x \,=\,\frac{1}{3}\quad\Rightarrow\quad y \,=\,-4$

Their intersection is: . $\bf{{\color{blue}\left(\frac{1}{3},\,-4\right)}}$