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Thread: Cartesian Equation for a particle path.

  1. #1
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    Cartesian Equation for a particle path.

    I have this example that I found quite easy

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

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

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

    But anyone shed some light on this one?

    $\displaystyle r(t)= \frac{1-t^2}{1+t^2} i+\frac{2t}{1+t^2} j$
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  2. #2
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    Quote Originally Posted by Bushy View Post
    I have this example that I found quite easy

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

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

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

    But anyone shed some light on this one?

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

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

    Square each equation and add ....
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  3. #3
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    Hello, Bushy!

    Mr. F is absolutely correct!


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

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


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

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


    Add:. $\displaystyle 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: .$\displaystyle x^2+y^2\;=\;1$

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


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


    Here's the puzzle . . .


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

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

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

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



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

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

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

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


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

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

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  4. #4
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    Quote Originally Posted by Soroban View Post
    But we know that the circle has two $\displaystyle \,x$-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

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

    Not sure where your method falls down.
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  5. #5
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    Quote Originally Posted by Bushy View Post
    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

    $\displaystyle 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 $\displaystyle \displaystyle x = \frac{1 - t^2}{1 + t^2}$ for all real values of t. What do you notice ....?
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  6. #6
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    Mr. F has a great approach!


    Here's an algebraic explanation.


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


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

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

    What is the other value?



    Consider letting $\displaystyle \,t$ grow extremely large.

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

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

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


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

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

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


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



    Of course, $\displaystyle \,t$ 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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