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FP2 Revision Thread

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Reply 80

zrancis there is a mistake in your last line.

jcaotz as I said you can use chain rule to find what your looking for very easily, if you can't see it directly from zrancis' formula. Alternatively consider d/dx arccos(ba) which you can see above, then let b =1/a.

Reply 81

username80328

clever, thanks

Reply 82

blue_n_white

Hi got a question here from a past paper which i dont have answers to, so i have no idea if ive got any of it right and was just wondering if any1 has any answers?

(i) Given that y=cos^-1(2x), for -1/2=<x=<1/2, find dy/dx (2 marks)

(ii) Use the substitution x=1/2sinθ to find ∫(1/(√(1-4x²))) dx (3 marks)

(iii) Hence show that cos^-1(2x) + sin^-1(2x) = a, where a is a constant to be found (3 marks)

Thanks

(Plus, has ne1 got any additional past OCR P5 or P6 papers that they could send me? I need all the help i can get)

Reply 83

datr

Part 1 can be solved by looking at the above posts $\frac{dy}{dx} = \frac{-2}{\sqrt{1-4x^2}}$ .

ii) $\int \frac{1}{\sqrt{1 - 4x^2}} dx$

$x = \frac{1}{2}\sin\theta$
$\frac{dx}{d\theta} = \frac{1}{2}\cos\theta$

$\frac{1}{2} \int \frac{\cos\theta}{\sqrt{1-\sin^2\theta}} d\theta = \frac{1}{2} \int 1 d\theta = \frac{\theta}{2} + c = \frac{\arcsin2x}{2} + c$

iii) $\arccos 2x = -2\int \frac{1}{\sqrt{1-4x^2}} dx = -arcsin2x - 2c$

$\therefore \arccos 2x + \arcsin 2x = -2c$

$\arccos 0 = \frac{\pi}{2}$
$\arcsin 0 = 0$

$\therefore a = \frac{\pi}{2}$

It's actually quite a meal this question, the same result can be accomplished by considering transformations of graphs much more quickly.

Reply 84

blue_n_white

datr

cheers, now to make sense of that.....

Reply 85

blue_n_white

im stuck again!!!

Use the substitution t=tan(x/2) to show that

(between pi/2 and 0) ∫1/(2-cosx) dx = (2(pi).sqrt(3))/9.

Reply 86

insparato

$\displaystyle \int_0^{\frac{\pi}{2}} \frac{1}{2-cosx} \hspace5 dx$

$t = tan \frac{x}{2}$

$\frac{dt}{dx} = \frac{1}{2}sec^2\frac{x}{2}$

$\frac{2.dt}{1+t^2} = dx$

$cosx = \frac{1-t^2}{1+t^2}$

$\displaystyle \int_0^1 \frac{1}{2 - \frac{1-t^2}{1+t^2}} . \frac{2}{1+t^2} \hspace5 dt$

$\displaystyle \int_0^1 \frac{2}{2(1+t^2)-(1-t^2)} \hspace5 dt$

$\displaystyle \int_0^1 \frac{2}{1+3t^2}\hspace5 dt$

$\displaystyle \frac{2}{3} \int_0^1 \frac{1}{\frac{1}{3}+t^2}$

$= [\frac{2}{3}.\sqrt3 arctan \sqrt3t]_0^1$

$= \frac{2\sqrt3}{3}\frac{\pi}{3} = \frac{2\sqrt3 \pi}{9}$

Reply 87

blue_n_white

cheers 4 that

Reply 88

babydoll_89

PLEASE help me with this problem (coordinate systems):-
Show that s(3^0.5 , 0) is a focus of the ellipse with the equation 3x^2 + 4y^2 = 36 ( ih ve done that part)
The origin is O and P is any point on the ellipse. A line is drawn from O perpendicular to the tangent to the ellipse at P and this line meets the line SP, produced if necessary, in the point Q. Show that the locus of Q is a circle

How would you rate this question (easy, medium ,difficult, or very difficult?)

Reply 89

blue_n_white

hey im stuck again! this is really frustrating.

Find the solution of the differential equation

√(4-x^2 ) dy/dx = √(4-y^2 )

for which y= -1 when x=1. Express your answer in a form not involving trigonometric functions.

(P.S sorry babydoll_89 i would help but i havent covered that in my course)

Reply 90

Rabite

Well, separate and integrate for a start :tongue:

Reply 91

blue_n_white

Rabite

Well, separate and integrate for a start :tongue:

yeh i did that and got 2 an aswer bt its wrong....

Reply 92

Rabite

Okay.
$\int_{-1}^{y} \frac{dy}{4-y^2} = \int _1 ^{x} \frac{dx}{4-x^2}$

$[\arcsin y/2]^{y}_{-1} = [\arcsin x/2]^{x}_{1}$

$\arcsin y/2 + \pi /6 = \arcsin x/2 - \pi /6$
I think. xD

$\arcsin y/2 + \pi/3 = \arcsin x/2$
Then sine both sides and draw some triangles!

Reply 93

blue_n_white

yeh i got that, but the answer is......

y = x/2 - 1/2√(3(4-x^2))

{i got y = x - √3 } :s-smilie:

Reply 94

Breadster™

Why are you doing a definite integral the y and x values are just there so that you can determine the constant right?

I got y=x but it's probably wrong :frown:

bleh

you got the answer?

Reply 95

Breadster™

actually i got y=x - pi/2

but thats probably wrong too

Reply 96

Rabite

Work from
$\arcsin (y/2) =\arcsin (x/2)-\pi/3$
And use the sin(A+B) result on the RHS. You'll need cos{arcsin{x/2}}, which you can find by drawing triangles.

And yes, I did the definite integrals because if I didn't, I'd just forget about the +c. >>;;