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Math Help - 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

    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
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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

    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 ....
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  3. #3
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    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?

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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 \,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.
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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

     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 ....?
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  6. #6
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    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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