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Edexcel A2 C4 Mathematics June 2016 - Official Thread watch

1. (Original post by zarzaidi)
Hey, thanks for the reply. Yeah I'm pretty sure my method is right, my steps conform to the mark scheme up until they keep it in ln form and I apply e and arrange in y=f(x) form but we still get different values for C which I don't get.
When you apply e
You also apply e to C which then becomes e^C
U think that might be why

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2. (Original post by XxKingSniprxX)
How do you do part (a) to that question?

V = 1/3 . pi . r^2 . h

do you divide the r/h = 16/24 = 2/3

Can't remember.

Where did you get that question from?
part a is similar shapes (proportionality)
3. (Original post by candol)
part a is similar shapes (proportionality)
can you show me a worked solutions of how you would have achieved that answer.

Still a bit until I see how you reached the answer in part (a).
4. (Original post by lordoftheties)
Thanks for helping! Heres an example (part b):

Attachment 551816
You are asked to find dh/dt
You have been given dv/dt (8cm^3/s) - use the units to help you realise this is dv/dt
so dh/dt = dv/dt x something - That something must be dh/dv (so lhs=rhs)
You need to work out dh/dv. This will involve differentiating a formula involving volume and height. This is given in part a). so differentiate that, and then flip it over so it becomes dh/dv.
Now you can multiply dv/dt x dh/dv, sub in the values and job done
5. (Original post by lordoftheties)
Thanks for helping! Heres an example (part b):

Attachment 551816
i recognise it from a madmaths paper?
6. (Original post by XxKingSniprxX)
can you show me a worked solutions of how you would have achieved that answer.

Still a bit until I see how you reached the answer in part (a).
r/h =16/24 = 2/3 so r = 2/3 x h
Now sub into v=1/3r^2 h and we get their answer
7. (Original post by candol)
r/h =16/24 = 2/3 so r = 2/3 x h
Now sub into v=1/3r^2 h and we get their answer
OHHH. That's how they got h^2.

I did r/h = 16/24 = 2/3, then went blank.

Ty
8. Hey guys, is the cross sectional area of a cylinder just the top bit of it? So the cross sectional area of a cylinder would just be pir^2 ?
9. (Original post by XxKingSniprxX)
OHHH. That's how they got h^2.

I did r/h = 16/24 = 2/3, then went blank.

Ty
Still confused about part a. Why would you do r/h? Thanks for the help
10. (Original post by ImNervous)
Still confused about part a. Why would you do r/h? Thanks for the help
You have to compare the ratio of the 2 cones. The big cone and the small cone inside.

r/h = comparing the ratios essentially.

If you notice in the formula they gave you to prove, there is no 'r' so you have to make r the subject from v = 1/3. pi. r^2 .h .
11. (Original post by XxKingSniprxX)
You have to compare the ratio of the 2 cones. The big cone and the small cone inside.

r/h = comparing the ratios essentially.

If you notice in the formula they gave you to prove, there is no 'r' so you have to make r the subject from v = 1/3. pi. r^2 .h .
Aaah I see. Thank you

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12. How woukd you go avout integrating 4/((4+3x)^2)
13. (Original post by metrize)
How woukd you go avout integrating 4/((4+3x)^2)
Spoiler:
Show

14. (Original post by metrize)
How woukd you go avout integrating 4/((4+3x)^2)
This is very standard. Consider the derivative of 1/(4 + 3x) and multiply by appropriate constants
15. (Original post by XxKingSniprxX)
You have to compare the ratio of the 2 cones. The big cone and the small cone inside.

r/h = comparing the ratios essentially.

If you notice in the formula they gave you to prove, there is no 'r' so you have to make r the subject from v = 1/3. pi. r^2 .h .
Actually if you sub in r= 2\3h into that equation would you not get h^3 instead oh h^2
Or am I missing something really obvious.
Again sorry for the questions but it's annoying me. Thanks
Spoiler:
Show

Funny how I didn't spot that and just used the substitution u = 4+3x
but obviously got the same answer.
17. (Original post by XxKingSniprxX)
Funny how I didn't spot that and just used the substitution u = 4+3x
but obviously got the same answer.
Damn that's probably way lengthy.

I could have easily done via recognition aswell, skipping pretty much all the working.
Damn that's probably way lengthy.

I could have easily done via recognition aswell, skipping pretty much all the working.
Yeah true. Its a bit more lengthy in terms of work but more methodical imo.
Doesn't take long when you've done countless examples.

What exams you got left btw?
19. (Original post by ImNervous)
When you apply e
You also apply e to C which then becomes e^C
U think that might be why

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Ahhhh genius! Thank you x
20. (Original post by zarzaidi)
Ahhhh genius! Thank you x
No prob

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