Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - help needed to finish exercises in a book

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    83

    help needed to finish exercises in a book

    As a refresher i set myself a task of completing all the misc exercises in the bostock and chandler A-level book (red ). I have ploughed through what must be in region of 500 past A-level questions (from 1970s-1980s) and out of them i am unable to give a solution to the follwing.

    If anyone could be so kind to help me finish off this book i would be grateful.

    It is not necessarily the case they are hard problems.
    They may well have been met when i'd already done 30-40 questions from the chapter and i ran out of mental power or im just not able to do them


    1) a curve is
    x=a(5cosp+cos5p)
    y=a(5sinp-sin5p)

    Find equation of normal at point with parameter p and find the point in
    1st quadrant of which normal is also normal to curve at another point.

    2) If Sn=a^r(1+a+a^2+.....a^r) by considering (1-a)Sn show that

    (1-a)Sn=(1-a^(2n+2))/(1-a^2) - a^(n+1)[1+a^(n+1)]/(1-a)

    3) In the binomial expansion of (p+q)^n write down the term containing p^r. If p=1/6, q=5/6 and n=30, find the value of r for which this term is greatest in value.

    book answer: 5. is there quick way apart from working the values out upto 5 then seeing they go down at 6?

    4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

    book answer: rt(3)e/30.

    5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jul 2009
    Posts
    192
    Thanks
    4
    so the term is

    (nCr)(p^r)(q^{n-r})=(30Cr)((\frac{1}{6})^r(\frac{5}{6})^{30-r}

    =(30Cr)((\frac{1}{5})^r(\frac{5}{6}^{30})

    we want to find for which term, the next is smaller

    (30Cr)((\frac{1}{5})^r(\frac{5}{6})^{30}>(30Cr+1)(  (\frac{1}{5})^{r+1}(\frac{5}{6})^{30}

    (30Cr)>(30Cr+1)(\frac{1}{5})

    \frac{30!}{r!(30-r)!}>\frac{30!}{(r+1)!(30-r-1)!}(\frac{1}{5})

    \frac{1}{30-r}>\frac{1}{r+1}(\frac{1}{5})

    r>4.1

    but r integer

    5\leq r
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom

    Welcome to Math Help Forum!
    Quote Originally Posted by jiboom View Post
    As a refresher i set myself a task of completing all the misc exercises in the bostock and chandler A-level book (red ). I have ploughed through what must be in region of 500 past A-level questions (from 1970s-1980s) and out of them i am unable to give a solution to the follwing.

    If anyone could be so kind to help me finish off this book i would be grateful.

    It is not necessarily the case they are hard problems.
    They may well have been met when i'd already done 30-40 questions from the chapter and i ran out of mental power or im just not able to do them


    1) a curve is
    x=a(5cosp+cos5p)
    y=a(5sinp-sin5p)

    Find equation of normal at point with parameter p and find the point in
    1st quadrant of which normal is also normal to curve at another point.

    2) If Sn=a^r(1+a+a^2+.....a^r) by considering (1-a)Sn show that

    (1-a)Sn=(1-a^(2n+2))/(1-a^2) - a^(n+1)[1+a^(n+1)]/(1-a)

    3) In the binomial expansion of (p+q)^n write down the term containing p^r. If p=1/6, q=5/6 and n=30, find the value of r for which this term is greatest in value.

    book answer: 5. is there quick way apart from working the values out upto 5 then seeing they go down at 6?

    4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

    book answer: rt(3)e/30.

    5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.
    You will find it easier to get help if you post questions separately, each in the appropriate sub-forum - see Rule 14.

    But here's a start for the first one.

    \frac{dx}{dp}=-5a(\sin p +\sin 5p)

    \frac{dy}{dp}=5a(\cos p -\cos 5p)

    So the gradient of the normal = -\frac{dx}{dy}=\frac{\sin p +\sin 5p}{\cos p -\cos 5p}

    So its equation is
    y-a(5\sin p -\sin 5p) = \frac{\sin p +\sin 5p}{\cos p -\cos 5p}\Big(x -a(5\cos p + \cos 5p)\Big)
    Can you take it from here?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    83
    posted questions in different sections.
    Thanks for replies so far,

    regarding 3)
    are we certain that once the term goes down it does not increase again?

    Any ideas for number 2?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom
    Quote Originally Posted by jiboom View Post
    posted questions in different sections.
    Thanks for replies so far,

    regarding 3)
    are we certain that once the term goes down it does not increase again?

    Any ideas for number 2?
    For number 2, I assume you mean
    S_n=a^n(1+a + a^2+...+a^n) ( n not r)
    in which case I should think you'd start like this:
    aS_n = a^n(a+a^2+...+a^n+a^{n+1})

    \Rightarrow (1-a)S_n=a^n(1 - a^{n+1})
    As far as number 3 is concerned, it is well know that the binomial distribution has a single maximum peak, so, yes, I am certain that once the term goes down it does not increase again. But I realise that's not a proof.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2009
    Posts
    192
    Thanks
    4
    for 3 we showed that the next term is smaller than the previous for all values larger than 5 so if this was not correct for all r>=5 then proof by induction would have a serious problem.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    83
    Quote Originally Posted by Grandad View Post
    Hello jiboomFor number 2, I assume you mean
    S_n=a^n(1+a + a^2+...+a^n) ( n not r)
    in which case I should think you'd start like this:
    aS_n = a^n(a+a^2+...+a^n+a^{n+1})

    \Rightarrow (1-a)S_n=a^n(1 - a^{n+1})
    As far as number 3 is concerned, it is well know that the binomial distribution has a single maximum peak, so, yes, I am certain that once the term goes down it does not increase again. But I realise that's not a proof.

    Grandad
    sorry, ive mistyped the question. it should read

    Sn=sum_{r=0 to N} a^r(1+a+a^2+...a^r)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom
    Quote Originally Posted by jiboom View Post
    sorry, ive mistyped the question. it should read

    Sn=sum_{r=0 to N} a^r(1+a+a^2+...a^r)
    Please confirm the question, which you have now written as:
    Given S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r), prove that (1-a)S_n= \frac{1-a^{2n+2}}{1-a^2}-\frac{a^{n+1}(1+a^{n+1})}{1-a}
    The problem is that I don't think this is correct. Checking it when n=0 does not give the correct answer.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom

    This is what question 3 should be, I think:

    S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r)

    \Rightarrow aS_n = \sum_{r=0}^na^r(a+a^2+a^3+...+a^{r+1})


    \Rightarrow (1-a)S_n = \sum_{r=0}^na^r(1-a^{r+1})

    =\sum_{r=0}^na^r-\sum_{r=0}^na^{2r+1}
    Each of these is a GP. I'll leave you to work out what the common ratios, etc, are. The result, then, is:

    (1-a)S_n = \frac{1-a^{n+1}}{1-a}-\frac{a(1-a^{2n+2})}{1-a^2}


    Grandad
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2009
    Posts
    83
    Quote Originally Posted by Grandad View Post
    Hello jiboomPlease confirm the question, which you have now written as:
    Given S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r), prove that (1-a)S_n= \frac{1-a^{2n+2}}{1-a^2}-\frac{a^{n+1}(1+a^{n+1})}{1-a}
    The problem is that I don't think this is correct. Checking it when n=0 does not give the correct answer.

    Grandad
    what you have posted is the question from the book. i agree with you, the question looks wrong. It seemed strange to me that we get 2 gps which we can sum but one term matches and the other is a country mile away. that is why i posted it to see if i was misunderstanding Sn. the chapter on series has more mistakes in 41 questions than the other 16 chapters !
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2009
    Posts
    83
    posted

    1) a curve is
    x=a(5cosp+cos5p)
    y=a(5sinp-sin5p)

    Find equation of normal at point with parameter p and find the point in
    1st quadrant of which normal is also normal to curve at another point.


    4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

    book answer: rt(3)e/30.

    5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.

    in other areas but getting no rplies. Anyone in this section wanna suggest something?

    @grandad: i can get that far with 1). i simplify a bit further using cos A+cos B etc, but the final part alludes me.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Dec 2009
    Posts
    83
    Quote Originally Posted by jiboom View Post
    posted

    1) a curve is
    x=a(5cosp+cos5p)
    y=a(5sinp-sin5p)

    Find equation of normal at point with parameter p and find the point in
    1st quadrant of which normal is also normal to curve at another point.


    4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

    book answer: rt(3)e/30.
    i am getting closer to finishing,5 is done. Iust these 2 to go. Anyone able to help? I especially need a help with 4. i dont get 30 but 48 (i think or something like that). If anyone else agrees i can put that down as another error in book.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom

    I don't have a satisfactory answer to #1, but I agree with your answer to #4:

    y = \left(\frac{1+x}{2+x}\right)^{\frac{1}{2}}

    \Rightarrow y^2 = \frac{1+x}{2+x}

    \Rightarrow 2y\frac{dy}{dx}=\frac{(2+x)1-(1+x)1}{(2+x)^2}
    =\frac{1}{(2+x)^2}
    When x = 2, y = \left(\frac34\right)^{\frac12} and so

    2\left(\frac34\right)^{\frac12}\frac{dy}{dx}= \frac{1}{4^2}=\frac{1}{16}

    \Rightarrow \frac{dy}{dx}=\frac{\sqrt3}{48}\approx\frac{\delta y}{\delta x}

    So when \delta x = e, \delta y \approx \frac{\sqrt3e}{48}

    Grandad
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Dec 2009
    Posts
    83
    thanks for the confirmation. I maybe wondered if they wanted something else but surely the increase in y is the delta y you have.


    For the curve question i simplify dy/dx as

    dy/dx= 5a(5cos t-cos 5t)/[-5a(sin t+ sin 5t)]
    =cos 5t-cos t/[sin 5t+sin t]
    = -2sin3tsin2t/[2sin3tcos2t]
    =-sin2t/cos2t

    so grad of normal is cos(2t)/sin(2t) hence normal is given by

    y-a(5 sin t -sin 5t)= cos(2t)/sin(2t)[x-a(5cos t+cos 5t)]

    sin(2t)y-cos(2t)x=a[5sin(t)sin(2t)-sin(5t)sin(2t)-cos(5t)cos(2t)-5cos(t)cos(2t)
    .......................=a[-5(cos(2t)cos(t)-sin(2t)sin(t))-(cos(5t)cos(2t)+sin(5t)sin(2t)
    sin(2t)y-cos(2t)x=a[-5cos(3t)-cos(3t)]=-6acos(3t)

    so need slick way of finding the other points in first quadrant which normal is normal at another point.
    Last edited by jiboom; December 23rd 2009 at 03:41 AM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jiboom
    Quote Originally Posted by jiboom View Post
    thanks for the confirmation. I maybe wondered if they wanted something else but surely the increase in y is the delta y you have.


    For the curve question i simplify dy/dx as

    dy/dx= 5a(5cos t-cos 5t)/[-5a(sin t+ sin 5t)]
    =cos 5t-cos t/[sin 5t+sin t]
    = -2sin3tsin2t/[2sin3tcos2t]
    =-sin2t/cos2t

    so grad of normal is cos(2t)/sin(2t) hence normal is given by

    y-a(5 sin t -sin 5t)= cos(2t)/sin(2t)[x-a(5cos t+cos 5t)]

    sin(2t)y-cos(2t)x=a[5sin(t)sin(2t)-sin(5t)sin(2t)-cos(5t)cos(2t)-5cos(t)cos(2t)
    .......................=a[-5(cos(2t)cos(t)-sin(2t)sin(t))-(cos(5t)cos(2t)+sin(5t)sin(2t)
    sin(2t)y-cos(2t)x=a[-5cos(3t)-cos(3t)]=-6acos(3t)

    so need slick way of finding the other points in first quadrant which normal is normal at another point.
    Ah, I see you spotted the mistake you'd made in the sign, and you've corrected it. This is a very tricky question (unless I'm overlooking something obvious - which is entirely possible!).

    Yes, I agree that the equation of the normal at p is:
    y\sin2p=x\cos2p-6a\cos3p
    Therefore the normal at q is:
    y\sin2q=x\cos2q-6a\cos3q
    The normal at p is normal to the curve again at q, if these are one and the same line; i.e. if (making y the subject and equating coefficients):
    \frac{\cos2p}{\sin2p}=\frac{\cos2q}{\sin2q} (1)
    and
    \frac{\cos3p}{\sin2p}=\frac{\cos3q}{\sin2q} (2)
    Equation (1) gives:
    \tan2q=\tan2p

    \Rightarrow 2q = 2p + n\pi, n = 1, 2, 3 (Note: n =4 gives p = q +2\pi; i.e. p and q represent the same point; and n > 4 gives repeated values.)

    \Rightarrow q = p + n\pi/2
    Substitute into (2):
    \frac{\cos3p}{\sin2p}= \frac{\cos (3p+3n\pi/2)}{\sin(2p +n\pi)}
    There's a bit of work to do now with the three different values of n, but I think when you simplify everything, you get:

    n=1:
    \tan3p = -1

    \Rightarrow p = \pi/4, 7\pi/12, 11\pi/12, ...
    n=2:
    \cos 3p = 0

    \Rightarrow p = \pi/6, \pi/2, 5\pi/6, ...
    n=3:
    \tan3p = 1

    \Rightarrow p = \pi/12, 5\pi/12, 3\pi/4, ...
    And it's then a question of choosing the values of p that give points in the first quadrant. I think these are:
    p = \pi/12, \pi/6, \pi/4, 5\pi/12
    I attach an Excel spreadsheet that shows the curve (with a = 1)and a normal (in fact the one where p = \pi/6), which you can (if you have Excel) play around with to confirm that these values seem to be correct.

    Grandad
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Some exercises
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: March 30th 2010, 07:32 AM
  2. A little help needed with Artin's book
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 16th 2010, 05:49 PM
  3. Please help me with these exercises!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 7th 2009, 08:24 AM
  4. Need help with few exercises. Please help!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 26th 2009, 01:10 AM
  5. Can anyone help me finish this
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 12th 2008, 03:47 PM

Search Tags


/mathhelpforum @mathhelpforum