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    Sorry to post another thread. I was trying to do this question around 1AM last night but gave up and went to bed in a bad mood. I'm going away for two weeks later on so i'd really appreciate it if someone could help me out, i don't like questions i can't do -_-.

    The figure shows a sketch of the curve with parametric equations:
    x=(cost)^2. y=(3/2)sint. -pi/2 < t < pi/2
    The sketch shows a curve on one side of the y axis. It passes the x axis once and the y axis twice. It is symmetrical. Some of you will also have the book but the sketch isn't really necessary.

    a) Find cordinates of the point where the curve touches the x axis: Done, (1,0).
    b) Find dx/dt: Done, -sin2t.
    c)i) Here's the part i'm having problems with.
    Show the area A of the finite region bounded by the curve and the y axis is given by A = 6 (integral sign with upper limit pi/2, lower limit 0) (sin^)^2 . cost dt.

    I can integrate that and use the limits to find the area easily but i can't prove that that is the correct formulae.

    I've attempted to rearrange the parameters in terms of sin^2 and cos^2, then put sin^2 t + cos^2 t = 1, but that doesn't get me anywhere.
    I've also attempted to find dx/dt and dy/dt, then find dy/dx and re-integrate to find a cartesian equation but that doesn't work either, and would be missing a constant anyway.

    Thanks for any help.
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    As it's symmetric about the x-axis the area you're looking for is twice that above the x-axis.

    So the answer is

    2 Int[x=0->x=1] y dx

    or if you're going to integrate using the parameter t then its

    2 Int[t=pi/2 -> t = 0 ] y (dx/dt) dt.

    If you put in the expression for y and dx/dt this is the integral you want.
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    this is quite simple...all u need to know is this results (for paremetric equations):

    ∫y dx = ∫ y (dx/dt) dt

    Therefore: we have,

    y = (3/2)sint
    x= (cost)^2
    dx/dt = -2costsint

    As the curve is symmetrical the finite area is given by:

    A = 2∫y dx [pi/2 - 0]

    So, we proceed:

    A = 2∫ (3/2)sint.(-2costsint) dt

    A = 3∫sint.(-2costsint) dt
    A = -6∫(sint)^2.cost

    Ignoring the negative sign as the area is always positive gives the required result.
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    Observe that since it is symmetrical, the area enclosed by the curve and the y-axis is the same as 2 times the area enclosed by the curve and the x-axis, with 0 < x < 1 .

    Thus,
    I = 2 * {(0,1) y dx

    when x = 0, t = pi/2 (taking positive since we can considering the area above the x-axis). when x = 1, t = 0

    Hence, converting limits and everything else into t,
    I = 2 * {(0, pi/2) 2/2*sint * dx/dt * dt
    I = 2 * {(0, pi/2) 3/2*sin t * -2 sin t cos t dt
    I = -6 * {0, pi/2) sin^2 t cos t dt
    I = 6 * {(pi/2, 0) sin^2 t cos t dt

    which is what we wanted to prove.
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    wot a timing RichE... :rolleyes:
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    What a timing Neo
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    (Original post by sephonline)
    What a timing Neo
    lollllll

    for god sake he posted the problem half an hour ago..and we all had to post at the exact minute :rolleyes:
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    Ahh, i never put the 'dx' on the original integral and thus never substituted to find with respect to t.
    I put 'symmetrical' in my own post but didn't even think about it while i was trying to do the question. I see that's why it's beteen pi/2 and 0, as you can just multiply the integral by two >_<.
    Thanks for the help both of you. I understand perfectly now. I guess i need to start to think 'outside the box' more.
    Thanks for your time ^^.
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    anytime mate
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    Just great... Went into town using my mother for a lift and bought the 'revise for you' textbooks for M2/P2/P3 for more question practice.

    We were then coming out of the indoor car park, in the usual que of cars getting ready to use the ticket to pay and open the barrier.

    However, the car in front of us stops for ages and a car park attendant runs across to find out what's up. We didn't know at the time but an 80 year old lady was driving and she had lost her car parking ticket (you take one when you go in and use it to pay when you go out). The car park attendant opens her door, leans in starts rummaging through the glove box to try find it.
    We are in our car, completely stationary at this time.

    All of a sudden, her car immediately starts reversing at top speed (15-25mph), the car parking attendant leaning into the car goes flying. A guy behind the car is knocked onto our bonnet and the car goes flying into the front passenger side door, where i am sat.

    Luckily, it did not hit it at a perpendicular angle, but severely damages our car, her car and knocks down two people, one of whom needed an ambulance.
    What an idiot, no idea why she reversed like that, she must have shifted it into reverse and pressed her foot on the accelerator (hard). I know little about driving but that was NOT poor clutch control.

    Now i'm finally back, everyone was okay apart from one of the attendants, who looked like he had a broken leg.

    That's my excitement for the day.
 
 
 
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Updated: August 20, 2004
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