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Math Help - simple parameter trig equations HELP !

  1. #1
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    simple parameter trig equations HELP !

    just realli confused with parameterss especially with trig equations

    a) x = 2costheta y=2sintheta 0<theta<pi

    b) x=cos t, y=cos t 0< t < 2pi

    c) x=2cos theta, y = root 3 sin theta 0<theta<2pi

    d) x = 2 cos t, y = sin t , 0< theta < pi

    e) 2t/[(1+t)squared], y = [(1-t)squared]/[(1+t)squared]

    i have the answers to these questions i jst dont haf the solution and i dont no how to get there
    im also realli confused about the restrictions

    any help is GREATLY appreciated
    thankyouu !
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  2. #2
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    Quote Originally Posted by THSKluv View Post
    just realli confused with parameterss especially with trig equations

    a) x = 2costheta y=2sintheta 0<theta<pi Mr F says: Square each of them and then add together.

    b) x=cos t, y=cos t 0< t < 2pi Mr F says: Square each of them and then add together.

    c) x=2cos theta, y = root 3 sin theta 0<theta<2pi Mr F says: {\color{red} x = \cos t} and {\color{red} \frac{y}{\sqrt{3}} = \sin t}. Square each of these and then add together.

    d) x = 2 cos t, y = sin t , 0< theta < pi Mr F says: {\color{red} \frac{x}{2} = \cos t} and {\color{red} y = \sin t}. Square each of these and then add together.

    e) 2t/[(1+t)squared], y = [(1-t)squared]/[(1+t)squared]

    i have the answers to these questions i jst dont haf the solution and i dont no how to get there
    im also realli confused about the restrictions

    any help is GREATLY appreciated
    thankyouu !
    e) x = \frac{2t}{(1 + t)^2} \Rightarrow 2x = \frac{4t}{(1 + t)^2} .... (1)

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

    (1) + (2): 2x + y = \frac{4t}{(1 + t)^2} + \frac{1 - 2t + t^2}{(1 + t)^2} = \frac{1 + 2t + t^2}{(1 + t)^2} = 1.
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  3. #3
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    thankyouu !!!
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  4. #4
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    uhm just a question
    for a) , the answer in my textbook says root (4 - x squared) except when i square and add together i get x squared plus y squared = 4
    is there a difference??
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  5. #5
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    Quote Originally Posted by THSKluv View Post
    uhm just a question
    for a) , the answer in my textbook says root (4 - x squared) except when i square and add together i get x squared plus y squared = 4
    is there a difference??
    Yes.

    The path lies on the circle x^2 + y^2 = 4. But there is a restriction on the values of x and y since 0 < \theta < \pi. This restiction means that -2 < x < 2 and 0 < y < 2.

    So you only want the top half of this circle, that is, y = \sqrt{4 - x^2}.
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  6. #6
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    Hello, THSKluv!

    If we are to graph these functions, it's more work . . .


    a)\;\;x \:=\: 2\cos\theta\qquad y\:=\:2\sin\theta\qquad  0\le\theta\le\pi
    We have: . \begin{array}{ccc}\dfrac{x}{2} &=& \cos\theta \\ \\[-3mm] \dfrac{y}{2} &=& \sin\theta \end{array}


    Square: . \begin{array}{ccc}\left(\dfrac{x}{2}\right)^2 &=& \cos^2\!\theta \\ \\[-3mm] \left(\dfrac{y}{2}\right)^2 &=& \sin^2\!\theta \end{array}


    \text{Add: }\;\frac{x^2}{4} + \frac{y^2}{4} \;=\;\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}} \qquad \Rightarrow\qquad \frac{x^2}{4} + \frac{y^2}{4} \;=\;1


    Hence, we have: . x^2+y^2 \:=\:4

    This is a circle: center (0,0), radius 2.

    Because of the domain, [0, \pi], the graph is the upper semicircle.




    b)\;\;x\:=\:\cos t \qquad y\:=\:\cos t\qquad 0\le  t \le 2\pi
    We see that: . y \:=\:x

    This is a straight line, through the origin, slope 1.


    The domain, [0, 2\pi], makes for an interesting graph.

    When t = 0\!:\;\;x \:=\:y\:=\:\cos 0 \:=\:1
    . . The graph starts at (1,1).

    When t = 2\pi\!:\;\;x \:=\:y\:=\:\cos2\pi \:=\:1
    . . The graph ends at (1,1) . . . same point!


    What happens in-between?

    When t = \tfrac{\pi}{2}\!:\;\;x\:=\:y\:=\:\cos\tfrac{\pi}{2  } \;=\;0
    . . The graph is at the Origin.
    When t = \pi\!:\;\;x\:=\:y\:=\:\cos\pi \:=\:-1
    . . The graph is at (-1,-1).
    When t = \tfrac{3\pi}{2}\!:\;\;x\:=\:y\:=\:\cos\tfrac{3\pi}  {2} \:=\:0
    . . The graph is at the Origin.


    The graph looks like this:
    Code:
                    |     P
                    |     *(1,1)
                    |   *
                    | *
        - - - - - - + - - - - -
                  * |
                *   |
       (-1,-1)*     |
              Q     |

    As t goes from 0 to \pi, line segment \overrightarrow{PQ} is plotted.

    As t goes from \pi to 2\pi, line segment \overrightarrow{QP} is plotted.




    c)\;\;x\:=\:2\cos\theta \qquad y\:=\:\sqrt{3}\sin\theta \qquad  0< \theta < 2\pi
    Same procedure as (a) . . .


    We have: . \begin{array}{ccc}\dfrac{x}{2} &=&\cos\theta \\ \\[-3mm]\dfrac{y}{\sqrt{3}} &=& \sin\theta \end{array}


    Square: . \begin{array}{ccc}\left(\dfrac{x}{2}\right)^2 &=& \cos^2\!\theta \\ \\[-3mm] \left(\dfrac{y}{\sqrt{3}}\right)^2 &=& \sin^2\!\theta \end{array}


    \text{Add: }\;\frac{x^2}{4} + \frac{y^2}{3} \:=\:\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}} \qquad\Rightarrow \qquad \frac{x^2}{4} + \frac{y^2}{3} \:=\:1

    This is an ellipse: center (0,0), semimajor axis 2, semiminor axis √3.

    Because of the domain, [0,2\pi], we have the entire ellipse.




    d)\;\;x \:=\:2\cos t\quad  y\:=\:\sin t\quad 0< \theta < \pi
    Same procedure as c).



    e)\;\;x\:=\:\frac{2t}{(1+t)^2} \quad y \:=\:\frac{(1-t)^2}{(1+t)^2}

    mr fantastic is correct . . . The equation is: . 2x + y \:=\:1
    . . This is a line with intecepts: \left(\tfrac{1}{2},0\right),\;(0, 1)

    But I am having difficulty determining the extent of this graph.


    Since no domain is mentioned, I assume: t \in (-\infty,\infty)

    Then I did some "eyeball" limits . . .

    If t = \text{-}\infty\!:\;\;x \:=\:\frac{\text{-}\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)

    If t = \text{-}1\!:\;\text{ unde{f}ined.}

    If t = 0\!:\;\; x \:=\:0,\;y \:=\:1 \quad\Rightarrow\quad (0,1)

    If t = 1\!:\;\;x \:=\:\tfrac{1}{2},\;y \:=\:0 \quad\Rightarrow\quad \left(\tfrac{1}{2},0\right)
    If t = +\infty\!:\;\;x \:=\:\frac{\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)


    I have most of it figured out, but there are some fuzzy areas.

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  7. #7
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    Quote Originally Posted by Soroban View Post
    [snip]
    mr fantastic is correct . . . The equation is: . 2x + y \:=\:1
    . . This is a line with intecepts: \left(\tfrac{1}{2},0\right),\;(0, 1)

    But I am having difficulty determining the extent of this graph.


    Since no domain is mentioned, I assume: t \in (-\infty,\infty)

    Then I did some "eyeball" limits . . .

    If t = \text{-}\infty\!:\;\;x \:=\:\frac{\text{-}\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)

    If t = \text{-}1\!:\;\text{ unde{f}ined.}

    If t = 0\!:\;\; x \:=\:0,\;y \:=\:1 \quad\Rightarrow\quad (0,1)

    If t = 1\!:\;\;x \:=\:\tfrac{1}{2},\;y \:=\:0 \quad\Rightarrow\quad \left(\tfrac{1}{2},0\right)
    If t = +\infty\!:\;\;x \:=\:\frac{\infty}{\infty^2} \:=\:0, \quad y \:=\:\frac{\infty^2}{\infty^2} \:=\:1 \quad\Rightarrow\quad (0,1)


    I have most of it figured out, but there are some fuzzy areas.
    Adding some more details:

    The graph of x = \frac{2t}{(t + 1)^2} has a vertical asymptote at t = -1, a horizontal asymptote x = 0, passes through (0, 0) and has a turning point at t = 1 \Rightarrow x = \frac{1}{2}. So the range of this graph is \left(-\infty , \, \frac{1}{2}\right]. Hence the domain of the parametric curve is -\infty < x \leq \frac{1}{2}.


    Similar analysis can be applied to the graph of y = \frac{(1 - t)^2}{(1 + t)^2}. It has a horizontal asymptote y = 1, vertical asymptote at t = -1, turning point at t = 1 \Rightarrow y = 0 and y-intercept at y = 1. The range of this graph is [0, +\infty). Hence the range of the parametric curve is 0 \leq y < +\infty.


    Hence the path is the part of line 2x + y = 1 over the domain x \leq \frac{1}{2}.
    Last edited by mr fantastic; June 6th 2009 at 05:59 PM. Reason: Fixed an error
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  8. #8
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    thank you guyz so much !
    realli helped me =D
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