# Core maths help!!

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#1
This is a solomon C4 paper. Im struggling with question 7, from the mark scheme it appears that they have used the show that equation to find the rate when x =0.25? Are you allowed to do this in the exam because your trying to prove it so why are you using it to find values as part of your proof or have i miss interpreted the MS?Thank you!
Paper: http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
MS: http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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3 years ago
#2
(Original post by examstudy)
This is a solomon C4 paper. Im struggling with question 7, from the mark scheme it appears that they have used the show that equation to find the rate when x =0.25? Are you allowed to do this in the exam because your trying to prove it so why are you using it to find values as part of your proof or have i miss interpreted the MS?Thank you!
Paper: http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
MS: http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
No, they haven't used the show that equation.

When 1/4 had been destroyed, 3/4 is left, it is given that this 3/4 gets destroyed at a constant rate, until it is completely gone in 6 hours. So it gets destroyed at a rate of (3/4)/6 per hour. None of these things use the equation.
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#3
(Original post by Kevin De Bruyne)
No, they haven't used the show that equation.

When 1/4 had been destroyed, 3/4 is left, it is given that this 3/4 gets destroyed at a constant rate, until it is completely gone in 6 hours. So it gets destroyed at a rate of (3/4)/6 per hour. None of these things use the equation.
Ahhh that makes much more sense so like (change i amount destroyed) divided by (change in time) thank you!
Please could you help with another q which is on this paper q8,:
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
Ms:
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
This is Q8, in c we worked out the time to get to the maximum height of the ferris wheel (which I'm presuming is half way round), the answer was t=4.65
So then in q d they ask for the time needed to complete two revolutions, which I thought would just be 4xt so 18.6 mins but the answer is 20. Could you please help with where I have gone wrong or what their working is? Kevin De Bruyne
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3 years ago
#4
(Original post by examstudy)
This is Q8, in c we worked out the time to get to the maximum height of the ferris wheel (which I'm presuming is half way round), the answer was t=4.65
So then in q d they ask for the time needed to complete two revolutions, which I thought would just be 4xt so 18.6 mins but the answer is 20.
You made a critical assumption which may not hold (the initial height is when t = 0). If it were true then you would do 4*... as you did.

Knowing that it is not true, try explaining the working in the MS.
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#5
(Original post by Kevin De Bruyne)
You made a critical assumption which may not hold (the initial height is when t = 0). If it were true then you would do 4*... as you did.

Knowing that it is not true, try explaining the working in the MS.
okay.. so i understand a rotation of a circle is 2pi so 4pi is twice but i dont understand why necessarily making that equal to theta in the equation helps
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3 years ago
#6
(Original post by examstudy)
okay.. so i understand a rotation of a circle is 2pi so 4pi is twice but i dont understand why necessarily making that equal to theta in the equation helps
When is cosx's first positive value for which it is a maximum? When is the next one? What's the difference in x? (Radians, or degrees if you prefer)
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#7
(Original post by Kevin De Bruyne)
When is cosx's first positive value for which it is a maximum? When is the next one? What's the difference in x? (Radians, or degrees if you prefer)
cosx's first positive max value is when it equals 1 so when x=0 or when x=360
so the difference is 360
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3 years ago
#8
(Original post by examstudy)
cosx's first positive max value is when it equals 1 so when x=0 or when x=360
so the difference is 360
Yup, so what is this example's relevance to the question?
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#9
(Original post by Kevin De Bruyne)
Yup, so what is this example's relevance to the question?
two revolutions is 720 (4pi)? I just don't get why we use the equation?
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3 years ago
#10
(Original post by examstudy)
two revolutions is 720 (4pi)? I just don't get why we use the equation?
If I had f(t) = cos(t/4), how long does it take to reach from one maximum to other? Given that you know for cosx, the maximum is at x = 0 and then x = 2pi.
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#11
(Original post by Kevin De Bruyne)
If I had f(t) = cos(t/4), how long does it take to reach from one maximum to other? Given that you know for cosx, the maximum is at x = 0 and then x = 2pi.
I guess first maximum is when t/4=0 and second when it equals 2pi so you can solve for t. I understand I just dont understand why we use the H equation in this way??
Thank you

I guess what Im trying to say is how to we know that the equation H=10-9cos(pit/5)+ 2sin(pit/5) is going to be 2 revolutions when (pi t/5) = 4pi??
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3 years ago
#12
(Original post by examstudy)
I guess first maximum is when t/4=0 and second when it equals 2pi so you can solve for t. I understand I just dont understand why we use the H equation in this way??
Thank you

I guess what Im trying to say is how to we know that the equation H=10-9cos(pit/5)+ 2sin(pit/5) is going to be 2 revolutions when (pi t/5) = 4pi??
You've missed a classic trick - this is an extremely difficult paper, but R formulae and combining two trig functions into one and using it is a common feature of C3. And also spotting links between multipart questions

Of course it's difficult to interpret -acosx+bsinx but youve already combined it into Cos(...) at the start of the question and cos on its own, as you've shown, is interpretable easily.

I'm not 100% sure if this is the specific step you're stuck with - if not, please say so Once you get that, it's just an extension of the cos example. We know that with cosx, we pick 0 as a starting point and know that it's a max (which I think may have confused things), we know that because of the properties of cos, 2pi later it's also a max, hence it's gone through one revolution (a complete cycle)

Equally if we look for where it's 4pi, then that'll be two complete cycles of cos, hence two revolutions
It doesn't have to be a max though, we're just using that cos(k+2pi) = cos(k) where k is some number, and an easy case is where k = 0
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#13
(Original post by Kevin De Bruyne)
You've missed a classic trick - this is an extremely difficult paper, but R formulae and combining two trig functions into one and using it is a common feature of C3. And also spotting links between multipart questions

Of course it's difficult to interpret -acosx+bsinx but youve already combined it into Cos(...) at the start of the question and cos on its own, as you've shown, is interpretable easily.

I'm not 100% sure if this is the specific step you're stuck with - if not, please say so Once you get that, it's just an extension of the cos example. We know that with cosx, we pick 0 as a starting point and know that it's a max (which I think may have confused things), we know that because of the properties of cos, 2pi later it's also a max, hence it's gone through one revolution (a complete cycle)

Equally if we look for where it's 4pi, then that'll be two complete cycles of cos, hence two revolutions
It doesn't have to be a max though, we're just using that cos(k+2pi) = cos(k) where k is some number, and an easy case is where k = 0

Right okay, again thank you for taking the time to reply Can you pls just explain why they are making 4pi= pi(t)/5? I feel maybe like I'm missing something?
I understand the 4pi bit I just dont understand why the pi(t)/5?.. I can see it comes from the equation( with H and not the R formula?) but why does that mean they are equal and that thats the time after two revolutions?
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3 years ago
#14
(Original post by examstudy)
Right okay, again thank you for taking the time to reply Can you pls just explain why they are making 4pi= pi(t)/5? I feel maybe like I'm missing something?
I understand the 4pi bit I just dont understand why the pi(t)/5?.. I can see it comes from the equation( with H and not the R formula?) but why does that mean they are equal and that thats the time after two revolutions?
That's what I'm trying to explain with a more simple example but maybe the link isn't clear.

You know that at times a and b, if b - a = 2pi, that is, the difference between the two values is 2pi, then cos(b) = cos(a). For example, for cosx, two values are a = 0 and b = 2pi, in which case b-a satisfies the condition. This means that between a and b, there has been exactly one period of cos, which is exactly the same as one revolution of the wheel.

The function you're dealing with is Rcos(pit/5 + c) where the R has no bearing on the length it takes for a revolution because, well, its constant. And you already know that if b - a = 2pi then there's one revolution between times a and b. What's a and b in this case? The variable will be time, and it will look like where t1 and t2 correspond to values a and b such that b - a = 2pi. So we need to find the values of t.

First, the constants cancel, so you get b - a = 0.2pi(T2 - T1) = 2pi. If this is true then T2 - T1 is the time taken for one revolution. It doesn't matter what they are as long as its true, so for simplicity set T1 = 0 and solve instead of trying to find two T1s and T2s by trial and error
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#15
isn't the function we are dealing with root85cos( theta + 0.2187)?
I'm really not following I'm sorry.. its a two mark question surely it doesn't require so much working,
I get the fact that 2 pi is the period and one revolution! And i get therefore why 4pi is used... can you just explain why its equal to pi(t)/5 without any examples cause i feel like they are confusing me further??
I dont understand just simply why they made it equal to pit/5... thats not what the R formula we found earlier its from the H equation?
I'm sorry that I'm not understanding im a very visual person and find reading lots of words hard to take in

(Original post by Kevin De Bruyne)
That's what I'm trying to explain with a more simple example but maybe the link isn't clear.

You know that at times a and b, if b - a = 2pi, that is, the difference between the two values is 2pi, then cos(b) = cos(a). For example, for cosx, two values are a = 0 and b = 2pi, in which case b-a satisfies the condition. This means that between a and b, there has been exactly one period of cos, which is exactly the same as one revolution of the wheel.

The function you're dealing with is Rcos(pit/5 + c) where the R has no bearing on the length it takes for a revolution because, well, its constant. And you already know that if b - a = 2pi then there's one revolution between times a and b. What's a and b in this case? The variable will be time, and it will look like where t1 and t2 correspond to values a and b such that b - a = 2pi. So we need to find the values of t.

First, the constants cancel, so you get b - a = 0.2pi(T2 - T1) = 2pi. If this is true then T2 - T1 is the time taken for one revolution. It doesn't matter what they are as long as its true, so for simplicity set T1 = 0 and solve instead of trying to find two T1s and T2s by trial and error
0
3 years ago
#16
(Original post by examstudy)
isn't the function we are dealing with root85cos( theta + 0.2187)?
I'm really not following I'm sorry.. its a two mark question surely it doesn't require so much working,
I get the fact that 2 pi is the period and one revolution! And i get therefore why 4pi is used... can you just explain why its equal to pi(t)/5 without any examples cause i feel like they are confusing me further??
I dont understand just simply why they made it equal to pit/5... thats not what the R formula we found earlier its from the H equation?
I'm sorry that I'm not understanding im a very visual person and find reading lots of words hard to take in
Afraid not, because that's what I tried to do in my last post. I explained exactly why the formula disappears into only looking at theta (which is now tpi/5) + 0.21... (which cancels out)

It is a lot of words and difficult to take in, yes. I would suggest writing out the equations / ideas that I listed step by step on paper and hopefully it will make more sense.

Specific questions to ask are good though.
Again, I can only repeat what was described above, but the formula for H turns into the R formula you found in part a. You then use the period of cos to only be interested in what's inside the brackets of cos(u) (where u is what's inside the brackets) and work out when 2pi time has passed between any two values of ...., and for ease you take the case t = 0 as your initial marker. You then know that if u2 - u1 = 2pi, where u1 is u when t = 0 and where u2 is the time t which is t minutes later from 0, so t-0 is the revolution time... then you can solve u2 - u1 = 2pi, which turns out to be 0.2pit + 0.21... - 0.21... = 0.2pit = 2pi, and t in this equation is the time for one revolution
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#17
(Original post by Kevin De Bruyne)
Afraid not, because that's what I tried to do in my last post. I explained exactly why the formula disappears into only looking at theta (which is now tpi/5) + 0.21... (which cancels out)

It is a lot of words and difficult to take in, yes. I would suggest writing out the equations / ideas that I listed step by step on paper and hopefully it will make more sense.

Specific questions to ask are good though.
Again, I can only repeat what was described above, but the formula for H turns into the R formula you found in part a. You then use the period of cos to only be interested in what's inside the brackets of cos(u) (where u is what's inside the brackets) and work out when 2pi time has passed between any two values of ...., and for ease you take the case t = 0 as your initial marker. You then know that if ....
Thank you.. I really appreciate your help.. Heres what I have taken from it:
H=root85cos(pi(t)/5 +0.2187)
we know 2 revolutions will be 4pi
We have b which is the value of theta isn cos (theta 1) at the end of the second revolution
we have a which is the value of theta in cos(theta 2) at the beginning of the first revolution
therefore b-a = 4pi and cos(theta 1) = cost (theta 2)
so therefore do b-a which = theta 1 ( pi (t1)/5 +0.2187) - theta 2(pi(t2)/5 + 0.2187)
therefore 4pi= pi/5 (t1-t2)
Make t2 equal to 0 because thats when the revolutions start
therefore t1= 20
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3 years ago
#18
(Original post by examstudy)
Thank you.. I really appreciate your help.. Heres what I have taken from it:
H=root85cos(pi(t)/5 +0.2187)
we know 2 revolutions will be 4pi
We have b which is the value of theta isn cos (theta 1) at the end of the second revolution
we have a which is the value of theta in cos(theta 2) at the beginning of the first revolution
therefore b-a = 4pi and cos(theta 1) = cost (theta 2)
so therefore do b-a which = theta 1 ( pi (t1)/5 +0.2187) - theta 2(pi(t2)/5 + 0.2187)
therefore 4pi= pi/5 (t1-t2)
Make t2 equal to 0 because thats when the revolutions start
therefore t1= 20
Excellent work - do you see how it all fits now?

Ultimately it is more difficult a question than for 2 marks and requires more thinking than what the MS suggests, but if it were worth a lot of marks then a lot of people would get 0 anyway. It's important to get from this question the period of cos (and sin, and tan I guess..) and being able to use it.
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#19
(Original post by kevin de bruyne)
excellent work - do you see how it all fits now?

Ultimately it is more difficult a question than for 2 marks and requires more thinking than what the ms suggests, but if it were worth a lot of marks then a lot of people would get 0 anyway. It's important to get from this question the period of cos (and sin, and tan i guess..) and being able to use it.
thank you!! 0
#20
(Original post by Kevin De Bruyne)
Excellent work - do you see how it all fits now?

Ultimately it is more difficult a question than for 2 marks and requires more thinking than what the MS suggests, but if it were worth a lot of marks then a lot of people would get 0 anyway. It's important to get from this question the period of cos (and sin, and tan I guess..) and being able to use it.
Hi
I have another question, would you mind helping me. Its q 3 b :http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
MS
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
the issue isn't so much the question, I understand all the integration and everything but I am at the stage where I have subbed in the limits and I am left with:
[(-2/3+ln3-2ln3)-(-1+ln2)]
My issue is with simplifying the ln's?
So obviously the question wants the answer in the specific form: p-lnq
I did this:
Wrote out the ln terms:
ln3 - ln 9- ln2
Simplifying the first two left me with ln(1/3)-ln(2)
Because the answer wanted the negative infront I went on to do -Ln(2)+ln(1/3) which got me -ln(2/3)
Could you please explain what about this is wrong?.. I follow the correct version completely and that does not require explanation just why mine is incorrect if you could help?
Thank you! Kevin De Bruyne
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