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    a) sketch the curve c whose parametric equations are
    x= 2 cos t, y = 5 sin t, 0<t< pi

    b) calculate the area of the finite region bounded by the curve c and the x axis.

    c) find the equation of c in the form y=f(x)
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    x/2 = cos t y/5 = sin t

    x²/4 = cos²t y²/25 = sin²t
    x²/4 + y²/25 = 1
    25x² + 4y² = 100

    --------------

    y = (sqrt(100-25x^2)) /2
    y = 2.5sqrt(4-x²)
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    (Original post by Undry1)
    x/2 = cos t y/5 = sin t

    x²/4 = cos²t y²/25 = sin²t
    x²/4 + y²/25 = 1
    25x² + 4y² = 1

    --------------

    y = (sqrt(1-25x^2)) /2
    is this for part B?
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    no that is the cartesian equation, he got it wrong on the second last line when he multiplied by 100.

    the function looks like an ellipse, it crosses the x axis at 2, -2, crosses the y axis at 5 and -5.

    integrate it to find the area.
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    the area of an ellipse is π*a*b where 2a and 2b are the lengths of the axes.

    you are asked for half of this.

    a = 2 and b = 5 so the area you require is π*2*5/2 = 5π
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    (Original post by the bear)
    the area of an ellipse is π*a*b where 2a and 2b are the lengths of the axes.

    you are asked for half of this.

    a = 2 and b = 5 so the area you require is π*2*5/2 = 5π
    my answer doesnt seem to be anything near this
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    i cant seem to be able to plot the graph on my graphic calculator, nothing seems to come up. anyone know what ive done wrong
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    are there any limits when intergrating?
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    Area = ∫5sint.-2sint.dt Limits: 0,π/2
    = -∫10sin²t dt
    =-∫10(½-½cos2t)dt
    =-∫5-5cos2t dt
    = -5t-2.5sint

    Area = -(-5(π/2)-2.5sinπ/2)
    Area = 2.5π+2.5
 
 
 
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Updated: November 21, 2005

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