Year 13 Maths Help Thread

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    (Original post by DarkEnergy)
    No problem, cheers for looking anyways. Reckon this will be alright? link
    With amazon I usually scroll down to the product reviews first.. and they look alright.
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    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 a and a minute hand of length b (where b>a), both measured from the pivot at the center of the clock face. Let x be the distance between the ends of the hands when the angle between the hands is \theta, where 0\leq\theta<\pi.

    Show that the rate of increase of x is greatest when x=(b-a)^{\frac{1}{2}}.

    The locus of points touched by the hour hand is the circle C_1 with radius a and the locus of points touched by the minute hand is the circle C_2 with radius b. The center of the two circles is O

    Let us hold the minute hand fixed at a certain point B and vary \theta from {0} to \pi.

    Let A be the point the hour hand is currently at. A diagram can show that the maximum rate of increase of x occurs when the line joining A to B is a tangent to C_1 so that AOB is a right angled triangle, and by Pythagoras' theorem, x=(b^2-a^2)^{\frac{1}{2}}.

    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.
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    (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 a and a minute hand of length b (where b>a), both measured from the pivot at the center of the clock face. Let x be the distance between the ends of the hands when the angle between the hands is \theta, where 0\leq\theta<\pi.

    Show that the rate of increase of x is greatest when x=(b-a)^{\frac{1}{2}}.

    The locus of points touched by the hour hand is the circle C_1 with radius a and the locus of points touched by the minute hand is the circle C_2 with radius b. The center of the two circles is O

    Let us hold the minute hand fixed at a certain point B and vary \theta from {0} to \pi.

    Let A be the point the hour hand is currently at. A diagram can show that the maximum rate of increase of x occurs when the line joining A to B is a tangent to C_1 so that AOB is a right angled triangle, and by Pythagoras' theorem, x=(b^2-a^2)^{\frac{1}{2}}.

    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.
    Hint below if you come back later and need one
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    cosine rule
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    (Original post by ValerieKR)
    Hint below if you come back later and need one
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    cosine rule
    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 C1-C4 syllabus 2) it's not actually a second order ODE)...

    x^2=a^2+b^2-2ab\cos\theta.

    I'll try another approach after the many failed implicit differentiation attempts: just substituting
    x=(b-a)^{\frac{1}{2}} and see what happens.
    If x=(b-a)^{\frac{1}{2}} then of course b^2-a^2= a^2+b^2-2ab\cos\theta
    so -2a^2=-2ab\cos\theta so a=b\cos\theta 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 x 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.
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    (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 C1-C4 syllabus 2) it's not actually a second order ODE)...

    x^2=a^2+b^2-2ab\cos\theta.

    I'll try another approach after the many failed implicit differentiation attempts: just substituting
    x=(b-a)^{\frac{1}{2}} and see what happens.
    If x=(b-a)^{\frac{1}{2}} then of course b^2-a^2= a^2+b^2-2ab\cos\theta
    so -2a^2=-2ab\cos\theta so a=b\cos\theta 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 x 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.
    You can avoid differential equations by using simultaneous equations - I've got a line by line solution in the spoiler tab if you're really really stuck at a specific point
    Spoiler:
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    let 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^2-b^2)^2 = x^4
    b^2-a^2 = x^2
    x = sqrt(b^2-a^2)
    = sqrt(b^2-a^2) (b>a)
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    (Original post by Palette)
    I tried the cosine rule....
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    You're on the right track. You are also correct to use implicit differentiation to find  dx/d\theta . You find to show that the value of  x is  (b^2-a^2)^{1/2} when  \mathbf{\frac{dx}{d\theta }} is a maximum.
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    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 x-axis. 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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    (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 x-axis. 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.
    The line BD is perpendicular to AC and passes through both B and 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 = co-ordinates of C + (B-A)
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    (Original post by ValerieKR)
    x
    Are you Zacken's sibling ?

    Starting with the cosine rule:
    x^2=a^2+b^2-2ab\cos\theta
    Implicitly differentiating both sides:

    2x\frac{dx}{d\theta}=2ab\sin \theta

    Implicitly differentiating again (the greatest rate of increase happens when \frac{d^2x}{d{\theta}^2}=0:

    2x\frac{d^2x}{d{\theta}^2}+2(\frac{dx}{d\theta})^{2}

    =2ab\cos \theta

    Let us cancel the 2:
    x\frac{d^2x}{d{\theta}^2}+(\frac{dx}{d\theta})^{2}=ab\cos \theta.

    Let \frac{d^2x}{d{\theta}^2}=0:

    (\frac{dx}{d\theta})^{2}=ab\cos \theta.

    Substituting
    \frac{dx}{d\theta}=\frac{ab\sin \theta}{x}:

    (\frac{a^2b^2{\sin}^2\theta}{x^2  })=ab\cos \theta.

    Cancelling ab:

    (\frac{ab{\sin}^2\theta}{x^2})=\cos \theta.

    (\frac{ab(1-{\cos}^2\theta)}{x^2})=\cos \theta.

    Am I on the right track so far?
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    (Original post by Palette)
    Are you Zacken's sibling ?

    Starting with the cosine rule:
    x^2=a^2+b^2-2ab\cos\theta
    Implicitly differentiating both sides:

    2x\frac{dx}{d\theta}=2ab\sin \theta

    Implicitly differentiating again (the greatest rate of increase happens when \frac{d^2x}{d{\theta}^2}=0:

    2x\frac{d^2x}{d{\theta}^2}+2(\frac{dx}{d\theta})^{2}

    =2ab\cos \theta

    Let us cancel the 2:
    x\frac{d^2x}{d{\theta}^2}+(\frac{dx}{d\theta})^{2}=ab\cos \theta.

    Let \frac{d^2x}{d{\theta}^2}=0:

    (\frac{dx}{d\theta})^{2}=ab\cos \theta.

    Substituting
    \frac{dx}{d\theta}=\frac{ab\sin \theta}{x}:

    (\frac{a^2b^2{\sin}^2\theta}{x^2  })=ab\cos \theta.

    Cancelling ab:

    (\frac{ab{\sin}^2\theta}{x^2})=\cos \theta.

    (\frac{ab(1-{\cos}^2\theta)}{x^2})=\cos \theta.

    Am I on the right track so far?
    Zacken is Cinderella and I am his ugly sister

    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 tag
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    quadratic in cos(theta)
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    How do you leave a line inside the spoiler?
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    I now know why I kept messing up my implicit differentiation-it was a stupid mistake which I somehow repeated each and every time at the same stage. Usually x is the letter used for the independent variable but here it is the letter used for the dependent variable so when I differentiated 2x\frac{dx}{d\theta}, 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 x^2 I'm guessing the next stage involves the use of our original expression involving x^2 from the cosine rule or we can exploit the derived equation from the cosine rule \cos \theta=\frac{x^2-a^2-b^2}{2ab}, though the latter seems too messy. Will therefore try the former:Starting off from before:

    (\frac{ab(1-{\cos}^2\theta)}{x^2})=\cos\thet  a
    Rearranging:(\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=x^  2
    Exploiting the cosine rule at the start: x^2=a^2+b^2-2ab\cos\theta

    Let's see if we can get a quadratic in terms of \cos\theta:
    We're getting close!
    (\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=a^  2+b^2-2ab\cos\theta

    Multiplying by \cos\theta:
    ab-ab{\cos}^2\theta=a^2\cos\theta+b  ^2\cos\theta-2ab{\cos}^2\theta

    We're getting closer...

    Rearranging gives us:ab{\cos}^2\theta-(a^2+b^2)\cos\theta+ab=0

    And by means of the quadratic formula:\frac{a^2+b^2 \pm \sqrt{a^4+b^2+2a^2b^2-4a^2b^2}}{2ab}=\frac{a^2+b^2 \pm (a^2-b^2)}{2ab}. Our values in terms of \theta are \frac{b^2}{ab} and \frac{a^2}{ab} and when these values are inserted into the cosine rule, we get x^2=a^2+b^2-2a^2=b^2-a^2 and x^2=a^2+b^2-2b^2. As b>a and x^2>0, only the former equation makes sense. So x=(b^2-a^2)^{\frac{1}{2}}.J'ai fini.I extend a MASSIVE thank you for your help!
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    What resources can i use to revise and selfteach s3
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    (Original post by youreanutter)
    What resources can i use to revise and selfteach s3
    Just do all textbook questions and solve any C3 paper you can find
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    (Original post by Palette)
    I now know why I kept messing up my implicit differentiation-it was a stupid mistake which I somehow repeated each and every time at the same stage. Usually x is the letter used for the independent variable but here it is the letter used for the dependent variable so when I differentiated 2x\frac{dx}{d\theta}, 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 x^2 I'm guessing the next stage involves the use of our original expression involving x^2 from the cosine rule or we can exploit the derived equation from the cosine rule \cos \theta=\frac{x^2-a^2-b^2}{2ab}, though the latter seems too messy. Will therefore try the former:Starting off from before:

    (\frac{ab(1-{\cos}^2\theta)}{x^2})=\cos\thet  a
    Rearranging:(\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=x^  2
    Exploiting the cosine rule at the start: x^2=a^2+b^2-2ab\cos\theta

    Let's see if we can get a quadratic in terms of \cos\theta:
    We're getting close!
    (\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=a^  2+b^2-2ab\cos\theta

    Multiplying by \cos\theta:
    ab-ab{\cos}^2\theta=a^2\cos\theta+b  ^2\cos\theta-2ab{\cos}^2\theta

    We're getting closer...

    Rearranging gives us:ab{\cos}^2\theta-(a^2+b^2)\cos\theta+ab=0

    And by means of the quadratic formula:\frac{a^2+b^2 \pm \sqrt{a^4+b^2a^b^2-4a^2b^2}}{2ab}=\frac{a^2+b^2 \pm (a^2-b^2)}{2ab}. Our values in terms of \theta are \frac{b^2}{ab} and \frac{a^2}{ab} and when these values are inserted into the cosine rule, we get x^2=a^2+b^2-2a^2=b^2-a^2 and x^2=a^2+b^2-2b^2. As b>a and x^2>0, only the former equation makes sense. So x=(b^2-a^2)^{\frac{1}{2}}.J'ai fini.
    Great!
    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)!
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    (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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    (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.
    I think the length is because there's only one 'easy' way to approach it and once you've realised what it is it's mostly algebra
    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 done
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    (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)?

    *
    I've always seen f^6(x) - never seen your roman numeral version
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    (Original post by ValerieKR)
    I've always seen f^6(x) - never seen your roman numeral version
    As I said, it is a long time ago I did maths.

    **
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    (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.
    Yeah I would try easier questions first if I were you. Paper 1 2007 seems to be quite easy (relative).
 
 
 
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