so the term is
we want to find for which term, the next is smaller
but r integer
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.
Hello jiboom
Welcome to Math Help Forum!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.
So the gradient of the normal
So its equation isCan you take it from here?
Grandad
Hello jiboomFor number 2, I assume you mean( not )in which case I should think you'd start like this:
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
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 !
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.
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.
Hello jiboomAh, 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 is:Therefore the normal at is:
The normal at is normal to the curve again at , if these are one and the same line; i.e. if (making the subject and equating coefficients):
(1)and
(2)Equation (1) gives:
Substitute into (2):
(Note: gives ; i.e. and represent the same point; and gives repeated values.)
There's a bit of work to do now with the three different values of , but I think when you simplify everything, you get:
And it's then a question of choosing the values of that give points in the first quadrant. I think these are:
I attach an Excel spreadsheet that shows the curve (with )and a normal (in fact the one where ), which you can (if you have Excel) play around with to confirm that these values seem to be correct.
Grandad