With amazon I usually scroll down to the product reviews first.. and they look alright.(Original post by DarkEnergy)
No problem, cheers for looking anyways. Reckon this will be alright? link
Year 13 Maths Help Thread
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 31082016 22:38

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 31082016 22:41
I tried that aforementioned STEP question (it's STEP I 2014 Q4) and my 'solution' to the first part currently contains an unjustified statement.
An accurate clock has an hour hand of length and a minute hand of length (where ), both measured from the pivot at the center of the clock face. Let be the distance between the ends of the hands when the angle between the hands is , where .
Show that the rate of increase of is greatest when .
The locus of points touched by the hour hand is the circle with radius and the locus of points touched by the minute hand is the circle with radius . The center of the two circles is
Let us hold the minute hand fixed at a certain point and vary from to .
Let be the point the hour hand is currently at. A diagram can show that the maximum rate of increase of occurs when the line joining A to B is a tangent to so that AOB is a right angled triangle, and by Pythagoras' theorem, .
I think I'm just restating the question in a different form right now, but don't leave me any hints at the moment as I'm trying to work out a breakthrough in the coming days. I will leave this up here in case I'm still stuck after a while.Last edited by Palette; 31082016 at 22:53. 
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 01092016 01:46
(Original post by Palette)
I tried that aforementioned STEP question (it's STEP I 2014 Q4) and my 'solution' to the first part currently contains an unjustified statement.
An accurate clock has an hour hand of length and a minute hand of length (where ), both measured from the pivot at the center of the clock face. Let be the distance between the ends of the hands when the angle between the hands is , where .
Show that the rate of increase of is greatest when .
The locus of points touched by the hour hand is the circle with radius and the locus of points touched by the minute hand is the circle with radius . The center of the two circles is
Let us hold the minute hand fixed at a certain point and vary from to .
Let be the point the hour hand is currently at. A diagram can show that the maximum rate of increase of occurs when the line joining A to B is a tangent to so that AOB is a right angled triangle, and by Pythagoras' theorem, .
I think I'm just restating the question in a different form right now, but don't leave me any hints at the moment as I'm trying to work out a breakthrough in the coming days. I will leave this up here in case I'm still stuck after a while.
Spoiler:Showcosine rule 
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 01092016 18:37
(Original post by ValerieKR)
Hint below if you come back later and need oneSpoiler:Showcosine rule
.
I'll try another approach after the many failed implicit differentiation attempts: just substituting
and see what happens.
If then of course
so so which doesn't particularly help either aside from the fact that the triangle formed is a right angled triangle and the line connecting the minute and hour hands is a tangent to the circle.
But I have yet to prove why the rate of increase of is greatest when this happens which is something I have to figure out...
I can gladly type up the second (much easier) half of the question for marking.
P.S. I know this should go in the STEP prep thread but I am a bit of a wimp. I hope that I'll improve at STEP over the coming months after practice.Last edited by Palette; 01092016 at 18:59. 
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 01092016 19:21
(Original post by Palette)
I tried the cosine rule and then used implicit differentiation countless times before and it just leads to dead ends (or I get something which resembles a second order differential equation but 1) second order ODEs aren't in the C1C4 syllabus 2) it's not actually a second order ODE)...
.
I'll try another approach after the many failed implicit differentiation attempts: just substituting
and see what happens.
If then of course
so so which doesn't particularly help either aside from the fact that the triangle formed is a right angled triangle and the line connecting the minute and hour hands is a tangent to the circle.
But I have yet to prove why the rate of increase of is greatest when this happens which is something I have to figure out...
I can gladly type up the second (much easier) half of the question for marking.
P.S. I know this should go in the STEP prep thread but I am a bit of a wimp. I hope that I'll improve at STEP over the coming months after practice.Spoiler:Showlet theta = y because I don't know latex
x^2 = a^2 + b^2 2ab cos(y)
2x(dx/dy)= 2absin(y)
dx/dy = absin(y)/x
d^2x/dy^2 = (abcos(y)x  (dx/dy)absin(y))/x^2= 0
cos(y)x(dx/dy)sin(y)= 0
dx/dy=cot(y)x and from earlier = absin(y)/x
therefore cot(y)x^2= absin(y)
and x^2 = ab(sec(y)cos(y))
abcos^2(y) +x^2cos(y)ab=0
cos(y) = x^2+sqrt(x^4 +4a^b^2)/2ab (quadratic formula)
(b^2 + a^2  x^2)/2ab = x^2+sqrt(x^4 +4a^2b^2)/2ab
b^2 +a^2 = sqrt(x^4 +4a^2b^2)
a^4 +b^4 + 2a^2b^2=x^4 +4a^2b^2
(a^2b^2)^2 = x^4
b^2a^2 = x^2
x = sqrt(b^2a^2)
= sqrt(b^2a^2) (b>a)Last edited by ValerieKR; 01092016 at 22:59. 
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 01092016 19:24
(Original post by Palette)
I tried the cosine rule.... 
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 01092016 20:55
Im stuck on a coordinate geometry Q! !
"The diagram shows a rectangle (all sides are 90 degree) ABCD. The point A is (2,14), B is (2,8) and C lies on the xaxis. Find (i) The equation of BC (ii) the coordinates of C and D"
I have done (i) and its correct the answer is y=2/3+20/3 and ive found the coordinates of C for (ii) which are C(10,0) but ive got no clue for coordinates of D. 
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 01092016 21:20
(Original post by Coolsul98)
Im stuck on a coordinate geometry Q! !
"The diagram shows a rectangle (all sides are 90 degree) ABCD. The point A is (2,14), B is (2,8) and C lies on the xaxis. Find (i) The equation of BC (ii) the coordinates of C and D"
I have done (i) and its correct the answer is y=2/3+20/3 and ive found the coordinates of C for (ii) which are C(10,0) but ive got no clue for coordinates of D.
also mod(BD) = mod(AC)
(mod being the length of the line segment)
You can use those three things to work it out.
OR
you could just use vectors and say coordinates of D = coordinates of C + (BA)Last edited by ValerieKR; 01092016 at 21:23. 
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 01092016 22:50
(Original post by ValerieKR)
x
Starting with the cosine rule:
Implicitly differentiating both sides:
Implicitly differentiating again (the greatest rate of increase happens when :
Let us cancel the 2:
.
Let :
.
Substituting
:
.
Cancelling :
.
.
Am I on the right track so far? 
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 01092016 22:53
(Original post by Palette)
Are you Zacken's sibling ?
Starting with the cosine rule:
Implicitly differentiating both sides:
Implicitly differentiating again (the greatest rate of increase happens when :
Let us cancel the 2:
.
Let :
.
Substituting
:
.
Cancelling :
.
.
Am I on the right track so far?
Jeez sorry I didn't realise the spoiler thing shrunk my maths
You are  next step is in the spoiler if you need it and I will amend the format of that last spoiler tagSpoiler:Showquadratic in cos(theta)Last edited by ValerieKR; 01092016 at 22:59. 
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 01092016 23:00
How do you leave a line inside the spoiler?

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 01092016 23:44
I now know why I kept messing up my implicit differentiationit was a stupid mistake which I somehow repeated each and every time at the same stage. Usually is the letter used for the independent variable but here it is the letter used for the dependent variable so when I differentiated , I ended up with a wrong expression.
The universities I'm applying to don't require STEP but apparently STEP is a good preparation for a maths degree._{As I see I'm guessing the next stage involves the use of our original expression involving from the cosine rule or we can exploit the derived equation from the cosine rule , though the latter seems too messy. Will therefore try the former:}_{Starting off from before:}
_{Rearranging:}
Exploiting the cosine rule at the start:
Let's see if we can get a quadratic in terms of :
We're getting close!
Multiplying by :
We're getting closer...
Rearranging gives us:
And by means of the quadratic formula:. Our values in terms of are and and when these values are inserted into the cosine rule, we get and . As and , only the former equation makes sense. So .J'ai fini.I extend a MASSIVE thank you for your help!Last edited by Palette; 01092016 at 23:51.Post rating:3 
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 01092016 23:44
What resources can i use to revise and selfteach s3

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 01092016 23:50
(Original post by youreanutter)
What resources can i use to revise and selfteach s3 
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 01092016 23:51
(Original post by Palette)
I now know why I kept messing up my implicit differentiationit was a stupid mistake which I somehow repeated each and every time at the same stage. Usually is the letter used for the independent variable but here it is the letter used for the dependent variable so when I differentiated , I ended up with a wrong expression.
The universities I'm applying to don't require STEP but apparently STEP is a good preparation for a maths degree._{As I see I'm guessing the next stage involves the use of our original expression involving from the cosine rule or we can exploit the derived equation from the cosine rule , though the latter seems too messy. Will therefore try the former:}_{Starting off from before:}
_{Rearranging:}
Exploiting the cosine rule at the start:
Let's see if we can get a quadratic in terms of :
We're getting close!
Multiplying by :
We're getting closer...
Rearranging gives us:
And by means of the quadratic formula:. Our values in terms of are and and when these values are inserted into the cosine rule, we get and . As and , only the former equation makes sense. So .J'ai fini.
That's an interesting way to do that final bit ^.^
You've spent the first few lines turning a cos quadratic... into another (sort of identical) cos quadratic and then solving that second one :s
Slightly quicker to solve the first quadratic (abcos^2(y) + x^2cos(y)  ab = 0), but your way does work (I think)!Last edited by ValerieKR; 02092016 at 05:31. 
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 02092016 00:00
(Original post by ValerieKR)
Great!
That's an interesting way to do that final bit ^.^
Slightly quicker to solve the quadratic for cos(y) (abcos^2(y) + x^2cos(y)  ab = 0) then set it equal to what the cosine rules gives, but your way does work!
I find this question rather long by the standards of the STEP I questions I've done. It's still going to be a while until I am brave enough to join the STEP Prep thread though. I might try STEP I Q5 or Q6 2014 and then move on to some 1990s STEP I questions. 
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 02092016 00:02
(Original post by Palette)
I find this question rather long by the standards of the STEP I questions I've done. It's still going to be a while until I am brave enough to join the STEP Prep thread though. I might try STEP I Q5 or Q6 2014 and then move on to some 1990s STEP I questions.
Just make sure you keep track of which ones you do! In the last few weeks I was scouring up and down all the papers for what must have summed to hours looking for ones I hadn't donePost rating:1 
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 02092016 00:27
(Original post by dugdugdug)
I've not done school level maths for many years but do have query with regard to the notation.
Suppose f is a function of x, then the first differential is written f'. The second diff is written f'' or f(ii).
So shouldn't the sixth diff be written f(vi)?
* 
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 02092016 00:29
(Original post by ValerieKR)
I've always seen f^6(x)  never seen your roman numeral version
** 
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 02092016 00:58
(Original post by Palette)
I find this question rather long by the standards of the STEP I questions I've done. It's still going to be a while until I am brave enough to join the STEP Prep thread though. I might try STEP I Q5 or Q6 2014 and then move on to some 1990s STEP I questions.
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Updated: October 24, 2016
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