A smidgen of a favour..

daggy

I realise all of you are prob very busy, but any math lovers out there?
What with school finished i am going crazy trying to figure out some maths problems. I missed a lot of maths this year and the book i have is in irish -- pure pointless!

1999 paper1, Higher level
Question 8 (c)

I got as far as substituting in 2Sin(theta) and factorising the 4.. and now i am baffled. If i take "u" to be sin(theta) ,then where do i go?

Also 1998 P1 , 3 (c) .
6 (b)
8 (c)

Anybody?

AliAD

Hi...
1999 8c goes a follows...x=2Sin(theta) dx=2Cos(theta)d(theta)
4-xsquared= 4-4Sinsquared= 4cos squared
then square root of 4-xsquared= 2sintheta
then change the limits, and sub them in...

As for 1998 3c use log tables pg9 to write cos theta= 1-t2
1+t2
and sin theta as 2t over 1+t2 with t= tan theta.
you should get something like 2 all over 1+cos theta+isintheta and then sub in.

1998 6B y=kx2
dy (dy)
dx= 2kx and (dx)2= 4k2x2...then sub into formula

Hope that helps you...good luck

*belle*

Fill the value given for x into the sum.

then you must get rid of the dx by differentiating x=2sin(theta)

and work on from there! there is a bit more to it .
sorry i tried writing it all out but its hard to do on the computer!:o

ha im too late.what harm.

MathsManiac

1998. 3(c):
(I'm going to write A for theta)
Sub in for z.
Mult above and below by conjugate of bottom. After a couple of steps, you get numerator as:
2[(1+cosA) - isinA]
and denominator as:
2(1+cosA).

The 2s cancel, and the real part divides in to get 1, and the imaginary part is sinA/(1+cosA). It remains to put that into the form tan(kA), where k€Q.

Use the formulas in the tables, there's a couple of ways to do this, but basically, you've to cop that you need "half-angles":
The formula sin2A=2sinAcosA means that sinA=2sin(A/2)cos(A/2).
And the formula: 1+cos2A = 2cos^2(A) means that 1+cosA = cos^2(A/2). (That's a rearrangement of the 5th formula in the left column uder the triangle on p.9, by the way.)
Sub these in and cancel to get tan(A/2). (In other words, k=1/2).

MathsManiac

1998. 6(b)
Sub. kx^2 for y and 2kx for dy/dx, simplify, and then think:

If this has to be true for all x, what is the value of k?

MathsManiac

1998. 8(c)

Draw a sketch and it should be clear.

(The curve and line intersect at (1,1), so you'll be integrating from 1 to 4.)

Peleus

i'm supposed to be studying english but i did 1999 paper II Q 8 90(c) so i scanned it. It's attatched to this. oh and so is the y=1/x graph one.

MathsManiac

Those area integrals are often easier than you think.

If f(x) lies above g(x) all the way between x=a and x=b, then the area between them is the integral of [f(x)-g(x)] from a to b. (And, by the way, this is true no matter where they cross the axes.)

In the case of the 1998 question, you just need to integrate [2x - 1 - 1/x] from 1 to 4.

Also, by the way, if you drew a sketch of the 1999 question. you'd see that it's just a piece of a circle, and you could do it by trig.

Michael Collins

MathsManiac wrote: »

Also, by the way, if you drew a sketch of the 1999 question. you'd see that it's just a piece of a circle, and you could do it by trig.

Interesting, you reckon they'd accept evaluating an intergral the cheeky way? I mean given they give you an integral symbol and say "evaluate" rather than "Find the area..."

MathsManiac

Dunno. But it is interesting anyway.

(And the end result is that you have evaluated the integral.)

daggy

MathsManiac wrote: »

1998. 8(c)

Draw a sketch and it should be clear.

(The curve and line intersect at (1,1), so you'll be integrating from 1 to 4.)

I did that but blanked when i came to the curve. how the hell ?!

and Also : Jesus lads! Thanks a bunch! Wasnt expecting this kind of a reply.

MathsManiac

y = 1/x is one of the curves whose shape you really should know. Anyway, see Peleus's post for a picture of it.

(And you're welcome!)

daggy

[QUOTE
Use the formulas in the tables, there's a couple of ways to do this, but basically, you've to cop that you need "half-angles":
The formula sin2A=2sinAcosA means that sinA=2sin(A/2)cos(A/2).
And the formula: 1+cos2A = 2cos^2(A) means that 1+cosA = cos^2(A/2). (That's a rearrangement of the 5th formula in the left column uder the triangle on p.9, by the way.)
Sub these in and cancel to get tan(A/2). (In other words, k=1/2).[/QUOTE]

Could you elaborate on that? I got as far as 1-iTan(kA) , but how do you know you need "half-angles"?
apologies..

Parsley

Our teacher told us they dumbed down the honours maths integration/differentiation after about 2001 or so, and that doing questions from before then was pointless as we wouldn't be able for em.

MathsManiac

daggy wrote: »

Could you elaborate on that? I got as far as 1-iTan(kA) , but how do you know you need "half-angles"?
apologies..

It's really just from becoming familiar with things that are likely to work, from practice. A couple of things might help as clues:

• The "1+cosA" might be reminiscent of "1+cos2A", which you might recognise as something from a formula.
• The question said k€Q, which means that it might be a fraction (since if it was going to be a whole number, they probably would have said k€N or k€Z).
• There's not much else you can do: if it had said sin2A on top, you'd automatically think of turning it into 2sinAcosA, probably, but since "sinA" is already as basic as it gets, you have to think of something else, and thinking of A as being twice of a half A is one of the tricks that sometimes works.
• You could try other things instead. For example, if you multiplied above and below by 1-cosA, that might have got you somewhere; sometimes you have to just give something a lash and see what happens, and learn to spot which things are going somewhere and which are going nowhere.

All of this, of course, is usually more helpful in the trig questions than in calculus, but, of course, you need to be able to ransfer your knowledge between areas of the course.

Not sure if that helps.