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    Curve C has parametric equations x= sint y=sin2t 0<t<pi/2

    Find the area bounded by C and the x axis

    Thanks in advance!
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    dx/dt = costt dy/dt = -2cost2t

    dy/dx = dy/dt x 1/(dx/dt)

    = -2cos2t
    costt

    Anyone know if this is right so far? Are the limits pi/2 and 0?
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    Shouldn't dx/dt be -cost?
    Also, as it's parametric integration, don't you use

    dy/dx = ydx/dt ?

    I could be wrong -I'm not getting the right answer after all.

    Limits are right, I think
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    to find the area under a parametric curve, you integrate y.dx

    so i get (hm don't know how to do an integral sign thingy.. so will just say I= )

    I = sin2t cost with the limits being pi/2 and 0

    then using trig identity, this equals

    I = 1/2 (sin 3t + sin t)

    integrated..
    y = [1/2 (-1/3 cos3t - cost) ] with limits.. pi/2, 0

    but uhh this is obviously very wrong, as when you work it out i get a minus third. haha....eek okay so that wasn't very helpful sorry
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    Ahhh, you legend. You've done it, you must just have put the numbers in wrnog

    That gives you

    0 - 1/2(-1/3 - 1)

    =2/3

    Which is the answer. I was missing the 1/3 on the cos3t, that's all. Grr.
    I'll rep you tomorrow!
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    (Original post by _luce_)
    then using trig identity, this equals

    I = 1/2 (sin 3t + sin t)
    What identity is that?
    It's been a while since I done C4 revision...bad considering exam is in few days
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    ah don't worry I got it

    Sin A + Cos B = Sin([A+B]/2)Cos([A-B]/2)
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    ooh yay

    and its from the identity in c3 (the one that i never seemed to use or see the use for...):

    2sinAcosB = sin(A + B) + sin (A - B)

    we have sin2t cost so you take out 1/2 as a factor, and make A = 2t and B = t
    = 1/2 sin (2x + x) + sin(2x - x)
    if you actually expand the whole of the RHS out using sin (A + B) and sin (A- B) identity then you get...

    sin2xcosx + cos2x sinx + sin2xcosx - cos2xsinx
    = 2sin2xcosx
    (you don't actually need to do that, it just proves it so you know where it comes from.. sorry if i've stated the obvious there!)
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    haha damn slow typing
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    (Original post by Apagg)
    Curve C has parametric equations x= sint y=sin2t 0<t<pi/2

    Find the area bounded by C and the x axis

    Thanks in advance!
    As mentioned by luce -> Area = ∫y dx

    dx/dt = cos t so dx = cos t dt
    y = sin 2t

    Area = ∫ (sin 2t)(cos t dt)
    = 2 ∫ (sin t cos^2 t) dt
    = 2 ∫ (sin t (1-sin^2 t)) dt {cos^2 t = (1 - sin^2 t)}
    = 2 ∫ (sin t - sin^3 t) dt
    = 2 (cos t - 3 sin^2 t cos t) {within limits of 0 and π/3}
    = 2 [(cos π/3 - 3 ((sin π/3)^2 cos π/3)) - (cos 0 - 3 ((sin 0)^2 cos 0))]
    = 2 [(0.5 - 3(0.75*0.5)) - (1 - 3(0))]
    = - 3.25 (square units)

    I feel I may have done it the slightly long winded way ...but hopefully it should be correct.
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    (Original post by Seth)
    As mentioned by luce -> Area = ∫y dx

    dx/dt = cos t so dx = cos t dt
    y = sin 2t

    Area = ∫ (sin 2t)(cos t dt)
    = 2 ∫ (sin t cos^2 t) dt
    = 2 ∫ (sin t (1-sin^2 t)) dt {cos^2 t = (1 - sin^2 t)}
    = 2 ∫ (sin t - sin^3 t) dt
    = 2 (cos t - 3 sin^2 t cos t) {within limits of 0 and π/3}
    = 2 [(cos π/3 - 3 ((sin π/3)^2 cos π/3)) - (cos 0 - 3 ((sin 0)^2 cos 0))]
    = 2 [(0.5 - 3(0.75*0.5)) - (1 - 3(0))]
    = - 3.25 (square units)

    I feel I may have done it the slightly long winded way ...but hopefully it should be correct.
    You set it up perfectly right here and could have integrated to 2*-1/3cos³t= -2/3cos³t.
    The limits are π/2 and 0 and cosπ/2= 0 anyway so you get 0-(-2/3)= 2/3.
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    (Original post by silent ninja)
    You set it up perfectly right here and could have integrated to 2*-1/3cos³t= -2/3cos³t.
    The limits are π/2 and 0 and cosπ/2= 0 anyway so you get 0-(-2/3)= 2/3.
    Ah thanks a lot for that, that definitely is the slightly more time-saving method:p:. I'm a bit rusty these days it seems..
 
 
 
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