FP3 June 14th Discussion Thread

I think I've done very badly, I've realised that I messed up the first question. Can't believe that, it's so damn easy. I think to make things worse that I actually crossed out the correct one. I've lost quite a few marks on this, really need an A* but I'll probably be lucky to even get an A because FP3 seems to always have high boundaries.

Original post by nebelbon

Same guess, i got 1.08 for the obtuse angle; anyone get this.

Also what was the limit for the improper integral i got 2? i think i integrated wrong.

Yeah I got that value for the angle, I then subtracted it from 2pi because 1.08 is acute.
Unfortunately, that angle, 5.2 radians is reflex. So, both 1.08 and 5.2 are wrong. Though I'm hoping I'll pick up some method marks, I didn't have time to fix it as this point so left it at 5.2. I hated that question!
I also got 2 for the integral.

Reply 101

scottb3scott

1

Original post by jellybeaaans

can anyone remember how many marks the maclaurin's Q was from the part after proving -e^-y?

3marks

Reply 102

scottb3scott

1

Original post by RPA101

For the very last question... I got something like 144 degrees as well. I think into radians it was about 2.53. Sorry to say this, but if you got 1.08 radians that's less than 90 degrees so it can't be right :frown:

Original post by nebelbon

I don't agree with 8d but then again I wasn't confident in my answer.

Original post by actuarialmaestro:p

Yup 144.7 degrees... :biggrin:

Ive made a thread for FP4, Im assuming most people doing FP3 are doing FP4.

Lets get the discussion going

http://www.thestudentroom.co.uk/showthread.php?t=2031919&p=38113852#post38113852

Original post by kirstyy93

How did you actually work out N? I couldn't do it :frown:

obviously meaning I couldn't do the last part, I got P and Q though.

Reply 103

Z1G

Hi I've done question 6 a different way to everyone, do you think this would be ok instead? I never saw the quicker way of implicit differentiation!

Basically:

y=ln(1+sinx)

dy/dx = cosx/(1+sinx)

d2y/dx2 = -1/(1+sinx) = -e^-y

d3y/dx3 = cosx/(1+sinx)^2

d4y/dx4 = -sinx/(1+sinx)^2 -2(1/1+sinx)(cosx/(1+sinx))^2

Basically keeping it all in terms of x

Then I said e^y = 1+sinx, therefore sinx = e^y-1

then swapping in for x,

-(e^y-1)/(e^2y)-2e^-y(dy/dx)^2

And tidying up to get e^-2y - e^-y - 2e^-y(dy/dx)^2

Im not sure, do you think this will get any marks and is it even correct? I feel so stupid missing that trick but oh well!

Reply 104

Z1G

Original post by member910132

Here

Which makes the expansion in q6

x-\frac{x^2}{2} +\frac{x^3}{6} - \frac{x^4}{12}

This was posted from The Student Room's iPhone/iPad App

I know you said you were leaving but...

in your working it says d3y/dx3 = e^-y dy/dx which I agree with

Then d4y/dx4 = d/dy (e^-y dy/dx) dy/dx which I also agree with

But when you open up the brackets you use the product rule, but how do you differentiate dy/dx with respect to x, if you know what I mean? Sorry if its just me being daft!

Reply 105

f1mad

13

Further pure on AQA is all about tricks :colone:

.

Reply 106

member910132

OP

Original post by Z1G

I know you said you were leaving but...

in your working it says d3y/dx3 = e^-y dy/dx which I agree with

Then d4y/dx4 = d/dy (e^-y dy/dx) dy/dx which I also agree with

But when you open up the brackets you use the product rule, but how do you differentiate dy/dx with respect to x, if you know what I mean? Sorry if its just me being daft!

NP:

\dfrac{d}{dx} (\frac{dy}{dx}) = \dfrac{d^2y}{dx^2}

Reply 107

Z1G

Original post by member910132

NP:

\dfrac{d}{dx} (\frac{dy}{dx}) = \dfrac{d^2y}{dx^2}

sorry I actually meant how do you differentiate dy/dx with respect to y haha, where it says d/dy (e^-y dy/dx)

Reply 108

member910132

OP

Original post by Z1G

sorry I actually meant how do you differentiate dy/dx with respect to y haha, where it says d/dy (e^-y dy/dx)

Now that you mention it... the best person to ask is probably f1

Original post by f1mad

Further pure on AQA is all about tricks :colone:

.

So after we get

Unparseable latex formula:

y^{(3)} = e^{-y} \dfrac{dy}{dx}[br]\[br]y^{(4)} = \dfrac{d}{dx} (e^{-y} \dfrac{dy}{dx})

Are we correct so far ?

Then we go

Unparseable latex formula:

y^{(4)} = \dfrac{d}{dy} (e^{-y} \dfrac{dy}{dx}) \times \dfrac{dy}{dx}[br]\[br]=[\dfrac{d}{dy} (e^{-y}) \times \dfrac{dy}{dx} + \dfrac{d}{dy} (\dfrac{dy}{dx}) \times e^{-y} ] \times \dfrac{dy}{dx}

Right ? So from here can we go :

y^{(4)} = [-e^{-y} \dfrac{dy}{dx}+e^{-y}\dfrac{d^2y}{dx^2} ]\times \dfrac{dy}{dx}

I think we all get the same expansion but to get the 3 marks for finding y^(4) we need to have the above correct. My only concern is where I have

\dfrac{d^2y}{dx^2}

I thinking it should be

\dfrac{d}{dy} (\dfrac{dy}{dx}) = \dfrac{d}{dx} (\dfrac{dy}{dx}) \times \dfrac{dx}{dy}

Which would give:

\dfrac{d^2y}{dx^2} \dfrac{dx}{dy}

what was your expression for y^(4) ?

(edited 11 years ago)

Reply 109

jellybeaaans

Original post by scottb3scott

3marks

ohhh was it 3 marks for finding y^3, y^4 and expanding ln(1+sinx)? :s-smilie:

(edited 11 years ago)

Reply 110

H2SO4

Original post by member910132

How did people get alpha ? Can someone post a solution ? Or just tell me how we get PN or QN ?

This was posted from The Student Room's iPhone/iPad App

OP=OQ

ANGLE =pie/3

Cosine rule

FP3 June 14th Discussion Thread

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you