# Thread: Cartesian Equation for a particle path.

1. ## Cartesian Equation for a particle path.

I have this example that I found quite easy

$r(t)= 4 \cos t i+5 \sin t j$

using $\cos^2t+\sin^2t= 1$ I concluded that

$25x^2+16y^2=400$

But anyone shed some light on this one?

$r(t)= \frac{1-t^2}{1+t^2} i+\frac{2t}{1+t^2} j$

2. Originally Posted by Bushy
I have this example that I found quite easy

$r(t)= 4 \cos t i+5 \sin t j$

using $\cos^2t+\sin^2t= 1$ I concluded that

$25x^2+16y^2=400$

But anyone shed some light on this one?

$r(t)= \frac{1-t^2}{1+t^2} i+\frac{2t}{1+t^2} j$
$x = \frac{1-t^2}{1+t^2}$ .... (1)

$y = \frac{2t}{1+t^2}$ .... (2)

Square each equation and add ....

3. Hello, Bushy!

Mr. F is absolutely correct!

$r(t)\:=\:\dfrac{1-t^2}{1+t^2}i +\dfrac{2t}{1+t^2} j$

This problem involves one of my favorite "puzzles".
I can't resist sharing it, so I must solve the problem . . .

We have: . $\begin{Bmatrix}x &=& \dfrac{1-t^2}{1+t^2} & [1] \\ \\[-3mm]y &=& \dfrac{2t}{1+t^2} & [2]\end{Bmatrix}$

Square the equations: . $\begin{Bmatrix}x^2 &=& \dfrac{(1-t^2)^2}{(1+t^2)^2} \\ \\[-3mm] y^2 &=& \dfrac{4t^2}{(1+t^2)^2} \end{Bmatrix}$

Add:. $x^2 + y^2 \;=\;\dfrac{1 - 2t^2 + t^4}{(1+t^2)^2} + \dfrac{4t^2}{(1+t^2)^2} \;=\;\dfrac{1+2t^2+t^4}{(1+t^2)^2} \;=\;\dfrac{(1+t^2)^2}{(1+t^2)^2}$

Hence, we have: . $x^2+y^2\;=\;1$

This is a circle centered at the Origin with radius 1.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's the puzzle . . .

To find the $\,y$-intercepts, let $\,x=0$ and solve.

In [1], let $x = 0\!:$ . $\dfrac{1-t^2}{1+t^2} \:=\:0 \quad\Rightarrow\quad t \:=\:\pm1$

Then: . $y \:=\:\dfrac{2(\pm1)}{1+(\pm1)^2} \:=\:\pm1$

The $\,y$-intercepts are: . $(0,1),\;(0,\text{-}1)$
. .
We already knew that, didn't we?

To find the $\,x$-intercepts, let $\,y = 0$ and solve.

In [2], let $\,y = 0\!:$ . $\dfrac{2t}{1+t^2} \:=\:0 \quad\Rightarrow\quad t = 0$

Then: . $\,x \;=\;\dfrac{1-0^2}{1+0^2} \;=\;1$

The $\, x$-intercept is: . $(1,0)$

But we know that the circle has two $\,x$-intercepts.

Where is the other one? .Where is our error?

4. Originally Posted by Soroban
But we know that the circle has two $\,x$-intercepts.

Where is the other one? .Where is our error?
Thanks for your explanation. That was amazing!

To be honest, I cannot see the error.

My guess at the next x-intercept is (-1,0) as

$x^2+y^2= 1$ is a circle with centre at (0,0) and radius 1.

Not sure where your method falls down.

5. Originally Posted by Bushy
Thanks for your explanation. That was amazing!

To be honest, I cannot see the error.

My guess at the next x-intercept is (-1,0) as

$x^2+y^2= 1$ is a circle with centre at (0,0) and radius 1.

Not sure where your method falls down.
Hint: Draw the graph of $\displaystyle x = \frac{1 - t^2}{1 + t^2}$ for all real values of t. What do you notice ....?

6. Mr. F has a great approach!

Here's an algebraic explanation.

We have: . $\begin{Bmatrix}x &=& \dfrac{1-t^2}{1+t^2} \\ \\[-3mm] y &=& \dfrac{2t}{1+t^2} \end{Bmatrix}$

To find the $\,x$-intercepts, find the value(s) of $\,t$ which produces . $y = 0.$

We found one value, $\,t = 0$, which gave us: $(1,0)$

What is the other value?

Consider letting $\,t$ grow extremely large.

We have: . $\displaystyle \lim_{t\to\infty} y \;=\;\lim_{t\to\infty}\frac{2t}{1+t^2}$

$\displaystyle \text{Divide top and bottom by }t^2\!:\;\;\lim_{t\to\infty}\frac{\frac{2}{t}}{\fr ac{1}{t^2} + 1} \;=\;\frac{0}{0+1} \;=\;0$

Hence, $\,y = 0$ when $\,t$ "equals" Infinity.

$\displaystyle \text{Then the }x\text{ -value is: }\;\lim_{t\to\infty} x \;=\;\lim_{t\to\infty}\frac{1-t^2}{1+t^2}$

$\displaystyle \text{Divide top and bottom by }t^2\!:\;\;\lim_{t\to\infty}\frac{\frac{1}{t^2} - 1}{\frac{1}{t^2}+1} \;=\;\frac{0-1}{0+1} \;=\;-1$

Hence, $\,x = -1$ when $\,t$ "equals" Infinity.

Therefore, when $\,t$ "equals" Infinity,
. . we have the other $\,x$-intercept: . $(\text{-}1,0)$

Of course, $\,t$ can never equal Infinity, can it?

The graph of the parametric equations is a unit circle
. . centered at the Origin with a "hole" at (-1, 0).

Okay, okay, I'm busted . . . It was a trick question.

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