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The Edexcel C4 (21/06/12 - PM) Revision Thread

Reply 380

wam-bam

1

Original post by James A

Dude that question is integration.

1. I'm female :colondollar:

2. part (a) is differentiation and part (b) is integration, i'm stusk on part a

Reply 381

Pritesh.Mistry

Does anyone know what the grade boundaries were in January 2012?

Also I just did the Jan 11 paper. Got 66/75 only to realise you need 69/75 for an A! Crazy... I'm hoping it's only high because January is usually when further mathematicians take the exam.

Hoping for a moderately hard paper.. boundaries in June have been around 61 for an A in recent years.

Reply 382

Glacier

Original post by wam-bam

OKay so now i'm stuck on another differentiation Q, sorry, basically i've got the differential Eqn right but i can't seem to equate it to 0 right. Please could you have a go and explain it.

Thanks

here's the answer

3 ans.JPG

First differentiate using the chain rule (remember that

\left( \sin x\right) ^{-\dfrac {1} {2}}=\dfrac {1} {\sqrt {\sin x}}

):

\dfrac {dy} {dx}=\sqrt {\sin x}+\dfrac {\dfrac {1} {2}x\cos x} {\sqrt {\sin x}}

You're looking for a stationary point so:

\dfrac {dx} {dx}=0

Multiply by

2\sqrt {\sin x}

to get

2\sin x+x\cos x=0

Then divide by

\cos x

to get the answer:

2\tan x+x=0

Reply 383

Glacier

Original post by Pritesh.Mistry

Does anyone know what the grade boundaries were in January 2012?

Also I just did the Jan 11 paper. Got 66/75 only to realise you need 69/75 for an A! Crazy... I'm hoping it's only high because January is usually when further mathematicians take the exam.

Hoping for a moderately hard paper.. boundaries in June have been around 61 for an A in recent years.

70 for an A*, 65 for an A and 58 for a B in January 2012.

Reply 384

F1's Finest

21

Original post by Pritesh.Mistry

Does anyone know what the grade boundaries were in January 2012?

Also I just did the Jan 11 paper. Got 66/75 only to realise you need 69/75 for an A! Crazy... I'm hoping it's only high because January is usually when further mathematicians take the exam.

Hoping for a moderately hard paper.. boundaries in June have been around 61 for an A in recent years.

http://www.edexcel.com/iwantto/I%20want%20to%20%20Tasks/1201%20GCE%20Units%20Grade%20Boundaries%20(Home%20Int).pdf

Jan 2012 you needed 65 for an A.

Reply 385

House3

Original post by wam-bam

OKay so now i'm stuck on another differentiation Q, sorry, basically i've got the differential Eqn right but i can't seem to equate it to 0 right. Please could you have a go and explain it.

Thanks

here's the answer

3 ans.JPG

Find attached my working. The top line of the scan was cut out but it just said y = x√sinx as in the question

Hope that helps.

Edit: Remember that A is at a stationary point on the graph, i.e. where dy/dx = 0. That's why I differentiated y = x√sinx then let the answer = 0. (I differentiated using the product rule, explaining why I've got 'u' and 'v' etc)

(edited 11 years ago)

REMEMBER

[br]y = a^x means [br][br]\ln y = \ln a^x[br][br]\ln y = x\ln a[br][br]\dfrac {1}{y}\dfrac{dy}{dx} =\ln a \dfrac{dy}{dx}=y\ln a[br][br][br][br][br]

(edited 11 years ago)

Reply 387

wam-bam

1

Original post by House3

Find attached my working. The top line of the scan was cut out but it just said y = x√sinx as in the question

Hope that helps.

Edit: Remember that A is at a stationary point on the graph, i.e. where dy/dx = 0. That's why I differentiated y = x√sinx then let the answer = 0. (I differentiated using the product rule, explaining why I've got 'u' and 'v' etc)

thanks a lot, could you help me with part b of this Q as well please :colondollar:

thanks

Reply 388

ViralRiver

16

Original post by Glacier

First differentiate using the chain rule (remember that

\left( \sin x\right) ^{-\dfrac {1} {2}}=\dfrac {1} {\sqrt {\sin x}}

):

\dfrac {dy} {dx}=\sqrt {\sin x}+\dfrac {\dfrac {1} {2}x\cos x} {\sqrt {\sin x}}

You're looking for a stationary point so:

\dfrac {dx} {dx}=0

Multiply by

2\sqrt {\sin x}

to get

2\sin x+x\cos x=0

Then divide by

\cos x

to get the answer:

2\tan x+x=0

When solving trig identities, don't ever divide through by common terms, factor them out otherwise you could lose a solution!

Reply 389

geditor

3

If you write down a particular solution (that works and valid) but isn't listed in the mark scheme, can you still get full marks for it if you get the correct answer? (Also, what happens if you use this solution but do not get the correct answer?) Thanks :smile:

Reply 390

King_Arthur

6

Original post by House3

Find attached my working. The top line of the scan was cut out but it just said y = x√sinx as in the question

Hope that helps.

Edit: Remember that A is at a stationary point on the graph, i.e. where dy/dx = 0. That's why I differentiated y = x√sinx then let the answer = 0. (I differentiated using the product rule, explaining why I've got 'u' and 'v' etc)

this method is much easier

Reply 391

Glacier

Original post by ViralRiver

When solving trig identities, don't ever divide through by common terms, factor them out otherwise you could lose a solution!

So how do I get from

2\sin x+x\cos x=0

to

2\tan x+x=0

by factoring?

Reply 392

ViralRiver

16

Original post by Glacier

So how do I get from

2\sin x+x\cos x=0

to

2\tan x+x=0

by factoring?

cosx(2tanx +x)=0

Then you have cosx = 0, or 2tanx+x = 0.

Reply 393

Glacier

Original post by ViralRiver

cosx(2tanx +x)=0

Then you have cosx = 0, or 2tanx+x = 0.

Thanks for explaining. :smile:

Reply 394

House3

Original post by wam-bam

thanks a lot, could you help me with part b of this Q as well please :colondollar:

thanks

2 docs attached with my working.
I've put every step so you can follow my working haha, you could prob do the final stages in less steps.

Could you tell me if i give the same answer as in your textbook? Cos obviously i dont have the answers with me. It should give the same as me, which is roughly equal to 18.44. If it's wrong, sorry in advance, tell me and i'll see where i went wrong.

Also, forgot to add: say "units^3" after your answer if it's volume, doesn't hurt (and units^2 for area)

come on cant we get some really challenging questions so board :boring:

Reply 396

Brainiac

how do you get from line 1 to line 2?

(edited 11 years ago)

c4.png8.8KB

Reply 397

House3

Original post by Brainiac

how do you get from line 1 to line 2?

I've attached the working below (you could do it in less steps but I included every step in case there was something you didn't follow).

It's based on the double angle formulae (the ones for cos2A, sin2A, tan2A).

Memorise them well as they almost always come up in C3 / C4.

Does anyone know how to answer this question. I thought i knew and then saw the mark scheme. It from jan 08 if that helps:

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm3s–1 and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm2.

(a) Show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential equation where k is a positive constant. (3)

(dh/dt) = 0.4 - K(h)^0.5

When h = 25, water is leaking out of the hole at 400 cm3s–1. (1)

(b) Show that k = 0.02.

THANKS :biggrin:

(edited 11 years ago)

Reply 399

Brainiac

Original post by House3

I've attached the working below (you could do it in less steps but I included every step in case there was something you didn't follow).

It's based on the double angle formulae (the ones for cos2A, sin2A, tan2A).

Memorise them well as they almost always come up in C3 / C4.

cheers, yeah that makes perfect sense, i knew the formula, just was silly not to realise A can be 2x and then it works, the 4x in the cos threw me a little :smile:

wont happen again

The Edexcel C4 (21/06/12 - PM) Revision Thread

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