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

I agree, Integration is doing in my head!!!

Supposedly we finished the course today, now it's revision - or rather in my case learning it all from scratch.

C4 :angry:

Maybe you should take some of your own advice? Have you never read "The Power of Positive Thinking"? I haven't, but everything you need is in the title!

Reply 81

mld

7

Original post by grazie

Maybe you should take some of your own advice? Have you never read "The Power of Positive Thinking"? I haven't, but everything you need is in the title!

Thanks grazie. Good idea. I've still got time to do all the work. :wink:

Reply 82

Dan4444

Does anyone have a link to the solomon papers?

Reply 83

This Honest

3

4 weeks or so to the exam!
My penultimate exam.

Does anyone have any tips to do rate of change and vectors questions?

Reply 84

jlnbello

Hey can someone pls show me what the grade boundaries are for 70 Ums and 80 Ums. thanks

Reply 85

chickpe

The diagram shows the curve C with parametric equations
x=asin2t, y=acost, 0 ≤ t ≤ 1/2pi

where a is a positive constant. The point P lies on C and has coordinates (3/4a, 1/2a)

(a) Find dy/dx, giving your answer in terms of t.

(b) Find an equation of the tangent to C at P.

(c) Show that a cartesian equation of C is y2=a2−ax.

The shaded region is bounded by C, the tangent at P and the x-axis. This shaded region is rotated through 2π radians about the x-axis to form a solid of revolution.

(d) Use calculus to calculate the volume of the solid revolution formed, giving your answer in the form kπa3, where k is an exact fraction

i got a, b and c but d is death
can anyone help me out?

Reply 86

grazie

2

Original post by chickpe

The diagram shows the curve C with parametric equations
x=asin2t, y=acost, 0 ≤ t ≤ 1/2pi

where a is a positive constant. The point P lies on C and has coordinates (3/4a, 1/2a)

(a) Find dy/dx, giving your answer in terms of t.

(b) Find an equation of the tangent to C at P.

(c) Show that a cartesian equation of C is y2=a2−ax.

The shaded region is bounded by C, the tangent at P and the x-axis. This shaded region is rotated through 2π radians about the x-axis to form a solid of revolution.

(d) Use calculus to calculate the volume of the solid revolution formed, giving your answer in the form kπa3, where k is an exact fraction

i got a, b and c but d is death
can anyone help me out?

Not sure why you're struggling as you've worked out most of what you need earlier in the question. Do you remember how to calculate volumes of revolution?

V = \pi \displaystyle\int y^2 dx

Work out the limits and you're sorted. BTW, what's the source of the question? It's helpeful for those following the thread.

Reply 87

grazie

2

Original post by Dan4444

Does anyone have a link to the solomon papers?

http://www.sci2.co.uk/maths.html. Enjoy

BTW, some of the Edexcel question papers on this site are not the final versions and there are a few errors (which would have been there when set!!!). Get the latest versions from the Edexcel website.

(edited 11 years ago)

Reply 88

sukhys

2

guys i dont understand why, for the answer to question 3c on exercise 6c (integration). is -1/8cos(8x) + 1/2cos(2x) + c

why is it positive 2x and not -2x.

Reply 89

NutterFrutter

Original post by sukhys

guys i dont understand why, for the answer to question 3c on exercise 6c (integration). is -1/8cos(8x) + 1/2cos(2x) + c

why is it positive 2x and not -2x.

Look at example c on the page before. Remember sin integrates to -cos.

Reply 90

sukhys

2

Original post by NutterFrutter

Look at example c on the page before. Remember sin integrates to -cos.

i understand that sin integrates to -cos. i dont understand the formulae for difference of sines or cosines which is

2sinAcosB = sin(A+B) + sin(A-B)

A=3x
B=5x

sub them in

2sin3xcos5x = sin(3x+5x) + sin(3x-5x)
2sin3xcos5x = sin8x + sin(-2x)
they got 2x not -2x WHY

Reply 91

Venomilys

13

C3 is ****ing easy. Did a past paper after 4 months of no work and got 74/75, I lost one mark on a silly mistake when transforming graphs. :teehee:

C4 is alright also, I scored 73/75 in my class mock like a don, barely revised too... probably going to do a C4 past paper today, aiming for 100 percent. Wish me luck people!

:gangster:

Reply 92

Venomilys

13

Original post by sukhys

2sin3xcos5x = sin(3x+5x) + sin(3x-5x)
2sin3xcos5x = sin8x + sin(-2x)
they got 2x not -2x WHY

Brother, think carefuly! If I hinted as to why I would just give it away. If you can't spot it now, I recommend coming back to the problem tomorrow. That's the most effective way you'll learn, 'self-induced' learning so to speak. :yes:

Reply 93

NutterFrutter

Original post by sukhys

i understand that sin integrates to -cos. i dont understand the formulae for difference of sines or cosines which is

2sinAcosB = sin(A+B) + sin(A-B)

A=3x
B=5x

sub them in

2sin3xcos5x = sin(3x+5x) + sin(3x-5x)
2sin3xcos5x = sin8x + sin(-2x)
they got 2x not -2x WHY

sin(-x) = -sinx

possibly?

Reply 94

sukhys

2

Original post by Ilyas

Brother, think carefuly! If I hinted as to why I would just give it away. If you can't spot it now, I recommend coming back to the problem tomorrow. That's the most effective way you'll learn, 'self-induced' learning so to speak. :yes:

bro, i have no idea why its positive please give me a hint. the only thing i can think of is u cant have a negative angle?

Reply 95

KamilS

1

How does it work to get an A*? I heard that you only need an A overall for EVERY module for maths apart from C3 and C4 where you need 180 UMS across the two for an A*. Is this true?

Reply 96

F1Addict

12

Original post by sukhys

i understand that sin integrates to -cos. i dont understand the formulae for difference of sines or cosines which is

2sinAcosB = sin(A+B) + sin(A-B)

A=3x
B=5x

sub them in

2sin3xcos5x = sin(3x+5x) + sin(3x-5x)
2sin3xcos5x = sin8x + sin(-2x)
they got 2x not -2x WHY

Don't know the full question, but I suspect there's a double negative somewhere in the question. sin(-2x)=-sin(2x) so with the double negative you'd have -(-sin(2x)) = sin(2x).

Just a shot in the dark.

__________________

Original post by sukhys

the only thing i can think of is u cant have a negative angle?

You can.

Angles are sort of like vectors (difference is that angles are restricted to one axial direction, vectors are not). They indicate how far in a certain direction to go, so if it was negative, you'd just go the other way.

(edited 11 years ago)

Reply 97

sukhys

2

Original post by NutterFrutter

sin(-x) = -sinx

possibly?

yh your right, thanks :biggrin:

Reply 98

sukhys

2

Original post by KamilS

How does it work to get an A*? I heard that you only need an A overall for EVERY module for maths apart from C3 and C4 where you need 180 UMS across the two for an A*. Is this true?

no, you need 480 ums and of those 480 ums, 180 MUST come from c3 and c4.

Reply 99

sukhys

2

can anyone confirm that the answer to question 2a on exercise 6D is wrong in the back of the book and the CD solution bank.
and also question 2d answer is wrong in the book and CD solution bank. if these are NOT wrong can some1 please explain how they got that answer.

(edited 11 years ago)

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

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