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# C4 integration watch

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

Find the area bounded by C and the x axis

2. 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?
3. 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
4. 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
5. 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!
6. (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
7. ah don't worry I got it

Sin A + Cos B = Sin([A+B]/2)Cos([A-B]/2)
8. 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!)
9. haha damn slow typing
10. (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

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.
11. (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.
12. (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. I'm a bit rusty these days it seems..

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