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Edexcel C3,C4 June 2013 Thread

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Reply 1820
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Story
Original post by usycool1
:hat2:

Remember: ln(3) is also just a constant (it is just a number). So if you're asked to differentiate it, the answer is just 0.

If, however, you had a letter/variable in the bracket e.g:

y = ln(3t)

Then y is not just a constant. :nah:

In this case, the answer would be 1t\dfrac{1}{t} (use the chain rule to see why: let y = ln(u) and u = 3t). :smile:


Thank you, I appreciate the help!

I think what I was getting confused was using the rule of y=ln(x) thus dy/dx= 1/x. So I assumed y=ln(3)t would follow the same pattern...and forgot to use the chain rule.

Also is ln(3)t the same as the ln(3t)... as if you differentiate them you get the same answer?
Original post by Story
Thank you, I appreciate the help!

I think what I was getting confused was using the rule of y=ln(x) thus dy/dx= 1/x. So I assumed y=ln(3)t would follow the same pattern...and forgot to use the chain rule.

Also is ln(3)t the same as the ln(3t)... as if you differentiate them you get the same answer?


If you differentiate ln(3t)ln(3t) you get 1t\frac{1}{t}.

If you differentiate ln(3)tln(3)t you get ln(3)ln(3).

As someone else said, ln(3t)=ln3+lntln(3t)=ln3+lnt so they aren't the same.

You should also remember that ddx(lnf(x))=f′(x)f(x)\dfrac{d}{dx}(lnf(x))=\dfrac{f'(x)}{f(x)}
(edited 10 years ago)
Original post by m4ths/maths247
Often each topic is not really exploited in an exam to be 'the hardest example they could find'
The textbook will often grade and build up questions so that some appear harder than the exams to prep you for something really quite nasty that could appear on the paper.
This is why I'm not a fan of past papers all the time. For me it's about a pupil/student recognising the topics they need to work on rather than constantly doing papers as you don't have the hardest questions on every topic in an exam.


Hello, do u think the jan 2013 c3 paper was harder than the average c3 paper? :smile:
Reply 1823
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Original post by brittanna
If you differentiate ln(3t)ln(3t) you get 1t\frac{1}{t}.

If you differentiate ln(3)tln(3)t you get ln(3)ln(3).

As someone else said, ln(3t)=ln3+lntln(3t)=ln3+lnt so they aren't the same.

You should also remember that ddx(lnf(x))=f′(x)f(x)\dfrac{d}{dx}(lnf(x))=\dfrac{f'(x)}{f(x)}


So if the letter is in brackets you use the chain rule but if the letter is outside of the brackets like y = ln(3)t you use the normal rules of differentiation by bringing the power of t (which is 1) infront of the function then reduces the power by one so you are left with y=ln(3).

I just did two similar questions and for y= 2ln(4t^2) I got dy/dt= 4/t .

and

For y=2ln(4)t^2 I got dy/dt= 4ln(4)t.

Am I along the correct lines?
Original post by Story
So if the letter is in brackets you use the chain rule but if the letter is outside of the brackets like y = ln(3)t you use the normal rules of differentiation by bringing the power of t (which is 1) infront of the function then reduces the power by one so you are left with y=ln(3).

I just did two similar questions and for y= 2ln(4t^2) I got dy/dt= 4/t .

and

For y=2ln(4)t^2 I got dy/dt= 4ln(4)t.

Am I along the correct lines?


Yes! That's correct :smile:.
Reply 1825
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Story
Original post by brittanna
Yes! That's correct :smile:.


Thank you, when something can actually be explained and you know why it happens it just makes it that much easier.

Thanks again for the help!​
Original post by laurawoods
Hello, do u think the jan 2013 c3 paper was harder than the average c3 paper? :smile:


Not as such.
I think the grade boundaries were a touch high for the paper though.

Often students focus on 1/2 tough questions on a paper after the exam and think the paper is the hardest thing ever when in reality the tough questions (i) accounted for very few marks and (ii) were in the middle of many straight forward questions.

Not a crazy paper but I feel 2 marks over in terms of 80 UMS. :smile:
Reply 1827
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kronca
does anyone know how to solve 3^(x)+(x)^3=0 for x?

have no idea :/
Original post by kronca
does anyone know how to solve 3^(x)+(x)^3=0 for x?

have no idea :/


Factorise it. :smile:
Reply 1829
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kronca
Original post by Wick3d
Factorise it. :smile:



cant figure out the main factor to take out://
Original post by kronca
cant figure out the main factor to take out://


Are you sure it isn't 3x instead of 3^x?
Original post by nm786
has anyone got C3 and C4 papers that were leaked about two weeks ago?


What's this? Please elaborate...
Original post by kronca
does anyone know how to solve 3^(x)+(x)^3=0 for x?

have no idea :/


Newton-Raphson would probably be your best bet.

I'm pretty sure this isn't a C3/C4 question, I've tried playing around with logs and stuff but can't find a different way to solve it.
Reply 1833
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nm786
2
Original post by Wick3d
What's this? Please elaborate...

A box of Papers were lost during dispatch process, so we will be sitting replacement papers for both C3 and C4, see this: http://www.edexcel.com/Aboutus/press-room/Pages/june13-gcemaths-replacepapers.aspx This is why i asked because someone on this site may have access to those papers.
(edited 10 years ago)
Original post by justinawe
Newton-Raphson would probably be your best bet.

I'm pretty sure this isn't a C3/C4 question, I've tried playing around with logs and stuff but can't find a different way to solve it.


Me too. Then I had a eureka moment and took the exponential and still didn't get it! Ah well....
Original post by justinawe
Newton-Raphson would probably be your best bet.

I'm pretty sure this isn't a C3/C4 question, I've tried playing around with logs and stuff but can't find a different way to solve it.


I thought the same so I assumed that he might of made a mistake when typing it up, seeing as it could of been 3x instead of 3^x which is just odd.
Original post by nm786
A box of Papers were lost during dispatch process, so we will be sitting replacement papers for both C3 and C4, see this: http://www.edexcel.com/Aboutus/press-room/Pages/june13-gcemaths-replacepapers.aspx This is why i asked because someone on this site may have access to those papers.


What happens if the replacement papers are breached as well? Would be a pretty awkward situation for Edexcel. :tongue:
Original post by kronca
does anyone know how to solve 3^(x)+(x)^3=0 for x?

have no idea :/


I think you'll have to use an iteration.
Reply 1838
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nm786
2
Original post by Wick3d
What happens if the replacement papers are breached as well? Would be a pretty awkward situation for Edexcel. :tongue:

Yeah that would be really awkward, i don't think that's happened before though but I'm sure they are prepared for that too.
(edited 10 years ago)
Original post by brittanna
I think you'll have to use an iteration.


That's probably it. Hadn't done C3 in a while, forgot about iterations :facepalm2:

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