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Math Help - Cartesian equations

  1. #1
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    Cartesian equations

    Given x=2 \sin (nt+\frac{\pi}{3}) and y=4 \sin (nt+ \frac{\pi}{6}), express x and y in terms of \sin nt and \cos nt. Find the Cartesian equation of the locus of the point (x,y) as t varies.

    I have expressed x and y in terms of \sin nt and \cos nt already.
    x=2 \sin (nt+\frac{\pi}{3})
    x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)
    x= \sin nt+ \sqrt{3} \cos nt
    and
    y=4 \sin (nt+ \frac{\pi}{6})
    y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)
    y= 2\sqrt{3} \cos nt +2  \sin nt
    Now my problem is how do i form the Cartesian equation in x and y as t varies?
    I take n as constant and t varies? Can anyone explain?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    Given x=2 \sin (nt+\frac{\pi}{3}) and y=4 \sin (nt+ \frac{\pi}{6}), express x and y in terms of \sin nt and \cos nt. Find the Cartesian equation of the locus of the point (x,y) as t varies.

    I have expressed x and y in terms of \sin nt and \cos nt already.
    x=2 \sin (nt+\frac{\pi}{3})
    x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)
    x= \sin nt+ \sqrt{3} \cos nt
    and
    y=4 \sin (nt+ \frac{\pi}{6})
    y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)
    y= 2\sqrt{3} \cos nt +2 \sin nt
    Now my problem is how do i form the Cartesian equation in x and y as t varies?
    I take n as constant and t varies? Can anyone explain?
    Thanks
    x=2 \sin (nt+\frac{\pi}{3}) and y=4 \sin (nt+ \frac{\pi}{6})

    x=2 \sin (nt+\frac{\pi}{6}+\frac{\pi}{6})

    let \alpha=nt+\frac{\pi}{6}

    x=2 \sin (\alpha+\frac{\pi}{6}) and y=4 \sin (\alpha)

    \frac{x}{2}=\sin (\alpha)\cos(\frac{\pi}{6}) + \cos (\alpha)\sin(\frac{\pi}{6})

    and

    \frac{y}{4}=\sin (\alpha)

    use \sin^2 (\alpha) + \cos^2(\alpha) = 1 to eliminate \alpha and get your equation
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  3. #3
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    Hello, arze!

    Given: . \begin{array}{ccc}x &=&2 \sin \left(nt+\frac{\pi}{3}\right) \\ y &=& 4 \sin \left(nt+ \frac{\pi}{6}\right)\end{array}

    Express x and y in terms of \sin nt and \cos nt.

    Find the Cartesian equation of the locus of the point (x,y) as t varies.


    I have expressed x and y in terms of \sin nt and \cos nt already.

    . . \begin{array}{ccccc}x\:=\: 2\sin\left(nt+\tfrac{\pi}{3}\right) \:=\:2\left(\tfrac{1}{2}\sin nt+\tfrac{\sqrt{3}}{2}\cos nt\right) &\Rightarrow&  x\:=\: \sin nt+ \sqrt{3} \cos nt \\ \\[-4mm] y\:=\:4\sin\left(nt+ \tfrac{\pi}{6}\right) \:=\:4\left(\tfrac{\sqrt{3}}{2} \cos nt+\tfrac{1}{2} \sin nt\right) &\Rightarrow& y \:=\: 2\sqrt{3} \cos nt +2  \sin nt \end{array}

    . . . . . Good work!

    Now my problem is how do i form the Cartesian equation in x and y as t varies?

    We have: . \begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}


    \begin{array}{ccccc}\text{Square }{\color{blue}(1)}: & x^2 &=& \sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + 3\cos^2\!nt \\<br />
\text{Square }{\color{blue}(2)}: & \dfrac{y^2}{4} &=& 3\sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + \cos^2\!nt \end{array}

    Add: . x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt

    . . . . x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}

    . . . . x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, arze!


    We have: . \begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}



    Add: . x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt

    . . . . x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}

    . . . . x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)
    if you want to do it that way, solve (1) and (2) for sin and cos:

    2\sin nt = \frac{\sqrt3y}{2}-x

    2\cos nt = \sqrt3x-\frac{y}{2}

    then use

    \sin^2nt + \cos^2nt = 1

    to get an equation in x and y
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  5. #5
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    Hello, Tesla!

    Absolutely right!


    It's an old habit.

    Whenever I see something like: . \begin{array}{ccc}x &=& a\sin\theta \\ y &=& b\cos\theta\end{array}

    . . I tend to use the square-and-add approach to simplify.

    I must learn to get out of my "box".

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