The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

FP2 June 10 last question

This:


Can't do last part. Really bugging me =/. Also spent AGES on part ii), and got angry enough to tear a few pages out of my notebook!

Kind of need some help, thanks!
Original post by jamie092
This:


Can't do last part. Really bugging me =/. Also spent AGES on part ii), and got angry enough to tear a few pages out of my notebook!

Kind of need some help, thanks!


Since the line OB is inclined at an angle of tan⁡−1(34)\tan ^{-1}(\frac{3}{4}) to the horizontal, it is given by the line θ=tan⁡−1(34)\theta = \tan ^{-1}(\frac{3}{4}) in polar coordinates. The line OA represents the line θ=π\theta = \pi in polar coordinates. Therefore the area you found in part (i) between OB, OA and the curve is equivalent to the area under the curve in polar coordinates between the two bounding lines.

It follows that 12∫tan⁡−1(34)π1(1−cos⁡θ)2dθ=(∫−1242x+1dx)−6\dfrac{1}{2} \displaystyle\int _{\tan ^{-1}(\frac{3}{4})} ^{\pi} \dfrac{1}{(1-\cos \theta)^2} d\theta = \left(\displaystyle\int ^{4}_{-\frac{1}{2}} \sqrt{2x+1} dx\right) - 6

Note that cos⁡2θ=1−2sin⁡2θ  ⟹  cos⁡θ=1−2sin⁡2(12θ)\cos 2\theta = 1-2\sin ^2\theta \implies \cos \theta = 1-2\sin ^2 (\frac{1}{2}\theta)

  ⟹  csc⁡2(12θ)=2(1−cos⁡θ) \implies \csc ^2(\frac{1}{2}\theta) = \dfrac{2}{(1-\cos \theta)}

Can you put that all together?
(edited 12 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you