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    the parametric equations of a curve are:
    x=at² , y = 2at

    where a is a real constant and t can take any real value
    a) find the cartesian equation of the curve in the form y² = f(x)
    b) sketch the curve showing the points where the curve intercepts the axes
    c) find the area of the finite region bounded by the curve and the line x=4a
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    LOL! hey wendy! im doin the homework too...

    for a) thats easy just substitute in the value when u worked out wat t equals
    b) should be alright....just put it into your graphic calculator....altho nothin would come up for me! and c) well i don't know how to do that yet! hehe tryin to work it out tho...

    anyone kno about c?? im kinda half way though it at the moment but a bit stuck with the a in the equation....do we substitute a number in place of a or wat??
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    use 'a' in part c.
    a) y^2 = 4ax

    b)calculate "sensible" values of y, x and plot them,

    c) integrate with limits, 4a, and 0
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    hmm i dont seem to be able to get a graph on my calculator
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    Its a y = kx2 graph but you've swapped the x and y axis around.
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    so for c do we intergrate the answer we get for a
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    (Original post by Mr. Pink)
    use 'a' in part c.
    a) y^2 = 4ax

    b)calculate "sensible" values of y, x and plot them,

    c) integrate with limits, 4a, and 0
    is there anyway of using a graphic calulator to plot this?
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    for this question when we intergrate would the answer be in terms of a
    as 4a is one of the limits
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    You can plot this by plotting y = \frac{1}{4a}x^{2} and then reflecting it in the y=x line.

    y = \sqrt{4ax}
    I = \int_{0}^{4a}\sqrt{4ax}dx = \sqrt{4a}\int_{0}^{4a}x^{\frac{1  }{2}}dx = \sqrt{4a}\frac{2}{3}\Big[ x^{\frac{3}{2}} \Big]_{0}^{4a} = \sqrt{4a}\frac{2}{3}\Big[ (4a)^{\frac{3}{2}} \Big] = \frac{2}{3}(4a)^{2} = \frac{32a^{2}}{3}
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    (Original post by AlphaNumeric)
    You can plot this by plotting y = \frac{1}{4a}x^{2} and then reflecting it in the y=x line.

    y = \sqrt{4ax}
    I = \int_{0}^{4a}\sqrt{4ax}dx = \sqrt{4a}\int_{0}^{4a}x^{\frac{1  }{2}}dx = \sqrt{4a}\frac{2}{3}\Big[ x^{\frac{3}{2}} \Big]_{0}^{4a} = \sqrt{4a}\frac{2}{3}\Big[ (4a)^{\frac{3}{2}} \Big] = \frac{2}{3}(4a)^{2} = \frac{32a^{2}}{3}

    thanks for the help
 
 
 
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