2005 paper 2, q5 equation of tangent

kimo6

Posted 23 January 2006 - 02:52 PM

[FONT=Courier] plz help me with qu 5 on the 2oo5 past paper PII
i need some solotions

thanx

dondon

Posted 23 January 2006 - 03:20 PM

Q5 in the 2005 paper 2 is a question about intersecting curves, is this what you mean?

Rocky

Posted 23 January 2006 - 03:27 PM

Shame this site doesn't give worked solution to Maths Higher Qs.

George

Posted 23 January 2006 - 04:32 PM

QUOTE(Rocky @ Jan 23 2006, 03:27 PM)
Shame this site doesn't give worked solution to Maths Higher Qs.

We do, through HSN extra for teachers

Nathan

Posted 23 January 2006 - 04:44 PM

QUOTE(Rocky @ Jan 23 2006, 03:27 PM)
Shame this site doesn't give worked solution to Maths Higher Qs.

There was a site for maths solutions (www.mathsrevision.com) but the site appears to have been taken offline...can anyone confirm or deny this?

Edited by nathanm, 23 January 2006 - 04:47 PM.

Steve

Posted 23 January 2006 - 04:45 PM

QUOTE(kimo6 @ Jan 23 2006, 03:52 PM)
i need some solotions

No, you just need a little help!

So the curves intersect where:

Then you can use 'difference of two squares' factorisation and find the x-coordinates of the points of intersection.

Note: when you work out the area, you should notice that the area is symmetrical, so you can simplify the integral.

Posted 23 January 2006 - 05:13 PM

As soon as you hear area and curves, you should think integration.

There 'top' curve is y=x2 and the 'bottom' curve is y=2x2 - 9. You are to find the area enclosed by the curves.

When working out the area, you need to do the 'top' curve - the 'bottom' curve. But if you look at the question you are first going to have to find the points K and L, and work you the enclosed area by the curves between these points.

To work out the points, you should see that the curves meet at these points. So at K and L the functions are equal to each other:
x2 = 2x2 - 9
Solve for x...

So we are restricted on the x-axis between the two values for x you just found, and we are working out the area between the curves between these points.

As I said, the area between the curves is the integral of the top curve minus the bottom curve, so:
(top) - (bottom)
(x2) - (2x2 - 9) dx
Now integrate that (raise the power by one, then divide by the new power). Once you've integrated, you need to substitute in your values for x into two verisons of the integrated equation and subtract the lower (in this case the negative one) away from the higher one. So you'll end up with [integrated function with highest x value] - [integrated function with lowest x value]. Work it out and you should end up with a positive number, and remember that your units here are "units2".

kimo6

Posted 23 January 2006 - 06:16 PM

ok thanx ppl got the answer!!

Ally

Posted 23 January 2006 - 06:51 PM

QUOTE(nathanm @ Jan 23 2006, 04:44 PM)
QUOTE(Rocky @ Jan 23 2006, 03:27 PM)
Shame this site doesn't give worked solution to Maths Higher Qs.

There was a site for maths solutions (www.mathsrevision.com) but the site appears to have been taken offline...can anyone confirm or deny this?

Hmm. I don't get a response when I ping the domain...

Nathan

Posted 23 January 2006 - 06:55 PM

QUOTE(Ally @ Jan 23 2006, 06:51 PM)
Hmm. I don't get a response when I ping the domain...

what does that mean?

Rocky

Posted 24 January 2006 - 08:05 PM

Doesn't work, so that sucks.

Rocky

Posted 24 January 2006 - 08:07 PM

Also good one George, shame pupils such as myself cannot get them. Why not just give a free trial for the 2005 Solutions for pupils too, as that is greedness to charge lol

Mr H

Posted 24 January 2006 - 09:17 PM

QUOTE(kimo6 @ Jan 23 2006, 02:52 PM)
[FONT=Courier] plz help me with qu 5 on the 2oo5 past paper PII
i need some solotions

thanx

Full solutions to the 2005 Higher paper can be found here: http://www.invergordon.highland.sch.uk/dep...higherindex.htm

Rocky

Posted 24 January 2006 - 09:23 PM

Thanks dude.

George

Posted 25 January 2006 - 10:43 PM

QUOTE(Rocky @ Jan 24 2006, 08:07 PM)
Also good one George, shame pupils such as myself cannot get them. Why not just give a free trial for the 2005 Solutions for pupils too, as that is greedness to charge lol

Well, you're entitled to think that I suppose

As it happens, we put a lot of work into this site, particularly for Higher Maths. We've spent hours and hours updating the Higher notes, and our new version (with about 150 pages in total) is nearly ready. Remember, you can have those for free!

So I certainly don't feel greedy charging teachers £100 for the package we offer

PS - if your teacher doesn't know about the site already, you could let them know about the solutions - then they can print them out for you, and everyone else at your school

Ally

Posted 25 January 2006 - 10:50 PM

Tracy

Posted 26 January 2006 - 06:35 PM