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Year 13 Maths Help Thread

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Original post by kiiten
Screenshot 2017-01-18 14.02.56.png

why does the version on the RHS have 1/3 outside the bracket. Shouldnt it be 3?

EDIT: Its because 3 has to me multiplied by the power right? So 3^-1 = 1/3


13−2x=13⋅11−23x\frac{1}{3-2x}=\frac{1}{3}\cdot \frac{1}{1-\frac{2}{3}x}
Can anyone help me with FP2 Complex numbers- Loci and transformations? It's the only part of FP2 that I just can't get my head around. Thanks in advance


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Original post by Lucymariexoxo
Can anyone help me with FP2 Complex numbers- Loci and transformations? It's the only part of FP2 that I just can't get my head around. Thanks in advance


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You'll have to post specific questions if you want help on things in particular.
Original post by Lucymariexoxo
Can anyone help me with FP2 Complex numbers- Loci and transformations? It's the only part of FP2 that I just can't get my head around. Thanks in advance
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http://www.examsolutions.net/a-level-maths/edexcel/fp2-tutorials/

have you tried these vids yet?
Hi, I need some M2 help.

For question 5, I got the centre of mass of the large triangle to be (12,12) and the small triangle to be (15,15). I did the alternative method in the mark scheme but I don't see how they got their CoMs.

http://qualifications.pearson.com/co...e_20110128.pdf
http://qualifications.pearson.com/co...s_20110309.pdf
Original post by amelienine
Hi, I need some M2 help.

For question 5, I got the centre of mass of the large triangle to be (12,12) and the small triangle to be (15,15). I did the alternative method in the mark scheme but I don't see how they got their CoMs.

http://qualifications.pearson.com/co...e_20110128.pdf
http://qualifications.pearson.com/co...s_20110309.pdf

Can you please be specific and explain what you don't understand?
Original post by notnek
Can you please be specific and explain what you don't understand?


My large triangle's COM is (12,12). My small triangle's COM is (15,15). The mark scheme's big triangle's COM is (15.15) and it's small triangle's COM is (12,12). I don't see how our COM's switched.

It's number 5a. I did the alternative method but our COM's switched.
(edited 7 years ago)
Reply 1427
Avatar for solC
solC
14
M3.JPG

So I've just started chapter 2 of M3, elastic springs and strings, and I'm finding it quite hard to draw the diagrams, particularly for questions like the above.

I've had a look at the diagram drawn in the solution, but I'm still not too sure what they did and why. I don't understand why the spring has been spliit into 2 smaller ones each of length L :s-smilie:

Thanks

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Where have i gone wrong solving this?

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Original post by kiiten
Where have i gone wrong solving this?

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You've forgotten how to solve quadratics. You can only set each bracket equal to the right-hand-side if the RHS is equal to 0.

e.g. (x+1)(x-2) = 0 --> x + 1 = 0 or x - 2 = 0

This is fine.

(x+1)(x-2) = 1 --> x + 1 = 1 or x - 2 = 1

This is not fine.
Original post by kiiten
Where have i gone wrong solving this?

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you cannot write cos⁡θ(1+2cos⁡θ)=1 \displaystyle \cos \theta (1 + 2 \cos \theta ) = 1 and then say cos⁡θ=1 \displaystyle \cos \theta = 1 or 1+2cos⁡θ=1 \displaystyle 1 + 2 \cos \theta = 1
Original post by DylanJ42
you cannot write cos⁡θ(1+2cos⁡θ)=1 \displaystyle \cos \theta (1 + 2 \cos \theta ) = 1 and then say cos⁡θ=1 \displaystyle \cos \theta = 1 or 1+2cos⁡θ=1 \displaystyle 1 + 2 \cos \theta = 1


Original post by notnek
You've forgotten how to solve quadratics. You can only set each bracket equal to the right-hand-side if the RHS is equal to 0.

e.g. (x+1)(x-2) = 0 --> x + 1 = 0 or x - 2 = 0

This is fine.

(x+1)(x-2) = 1 --> x + 1 = 1 or x - 2 = 1

This is not fine.


Why?
Original post by kiiten
Why?

You need to understand why you can set each bracket to zero in the first place.


(x+1)(x−2)=0(x+1)(x-2)=0

If (x+1)=0(x+1)=0 then this will make the whole of the left equal to 0 since 0 times anything is 0. You can see that when you plug in the solution x=−1x=-1 into the LHS, the left bracket becomes 0 so the whole thing becomes zero.


(x+1)(x−2)=1(x+1)(x-2)=1

In this case if e.g. (x+1)=1(x+1)=1 then that doesn't guarantee that the whole thing will be equal to 11.
Original post by notnek
You need to understand why you can set each bracket to zero in the first place.


(x+1)(x−2)=0(x+1)(x-2)=0

If (x+1)=0(x+1)=0 then this will make the whole of the left equal to 0 since 0 times anything is 0. You can see that when you plug in the solution x=−1x=-1 into the LHS, the left bracket becomes 0 so the whole thing becomes zero.


(x+1)(x−2)=1(x+1)(x-2)=1

In this case if e.g. (x+1)=1(x+1)=1 then that doesn't guarantee that the whole thing will be equal to 11.


Ah i see, ok :smile:

Yeah i see it now - you factorise the quadratic into (2cosx-1)(cosx+1)=0 (note: x = theta)
Original post by kiiten
Ah i see, ok :smile:

Yeah i see it now - you factorise the quadratic into (2cosx-1)(cosx+1)=0 (note: x = theta)

That's correct.
Original post by notnek
That's correct.


i got 2 solutions as 1/3pi and pi which i put into a CAST circle to get:

1.05, 3.14, 5.24 and 6.28

All are correct apart from 6.28. Why? (i know it doesnt fit the equation but why do i get it on the CAST circle :s-smilie:)
Original post by kiiten
i got 2 solutions as 1/3pi and pi which i put into a CAST circle to get:

1.05, 3.14, 5.24 and 6.28

All are correct apart from 6.28. Why? (i know it doesnt fit the equation but why do i get it on the CAST circle :s-smilie:)

cos⁡θ=−1\cos \theta = -1

θ=π\theta = \pi is a solution but you shouldn't have got θ=2π\theta = 2\pi using CAST.

It should be easier to consider the cos graph and you'll see that there's only one solution to cos⁡θ=−1\cos \theta = -1 in the relevant range.
Original post by notnek
cos⁡θ=−1\cos \theta = -1

θ=π\theta = \pi is a solution but you shouldn't have got θ=2π\theta = 2\pi using CAST.

It should be easier to consider the cos graph and you'll see that there's only one solution to cos⁡θ=−1\cos \theta = -1 in the relevant range.


Yeah according to the graph it shouldnt. I think i might have made a mistake with my CAST circle. Can you see any obvious errors? The green line from 0 to pi (and reflected) is the one for theta= pi

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Original post by kiiten
Yeah according to the graph it shouldnt. I think i might have made a mistake with my CAST circle. Can you see any obvious errors? The green line from 0 to pi (and reflected) is the one for theta= pi

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With cos⁡θ=−1\cos \theta = -1, you need cos to be negative and it isn't in the 'A' and 'C' quadrants.
Original post by notnek
With cos⁡θ=−1\cos \theta = -1, you need cos to be negative and it isn't in the 'A' and 'C' quadrants.


Oh yeah oops :3 - how would i draw it in the A and T segments??

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