Leaving cert paper 2 project maths (mainstream schools) q3

whiteandlight

Anyone here a higher level teacher manage to get this out? It's wrecking my head!

derb12

I get k=5.828 (just using phone calculator so may be rounding error).

The intersection with x+y=2 must occur at 1,1
The distance between centre of circle h,h and 1,1 must be = h so solve equation (distance formula) and discard answer that will turn out to be <2 to give h = 3.414
Then translate 1,1 -> 3.414,3.414 -> 5.828,5.828 to find intersection with parallel tangent, giving you x+y=2k and hence k=5.828.

Nice question. I'll be curious to see the marking scheme.

whiteandlight

derb12 wrote: »

I get k=5.828 (just using phone calculator so may be rounding error).

The intersection with x+y=2 must occur at 1,1
The distance between centre of circle h,h and 1,1 must be = h so solve equation (distance formula) and discard answer that will turn out to be <2 to give h = 3.414
Then translate 1,1 -> 3.414,3.414 -> 5.828,5.828 to find intersection with parallel tangent, giving you x+y=2k and hence k=5.828.

Nice question. I'll be curious to see the marking scheme.

Dang missed the (1,1)! I used the distance from a point to a line saying that centre (-g,-g) to the line x+y=2 must be the same as the distance to the x-axis and got the same 3.414 or 2+root2 for the centre. Used gsqd + fsqd +c =rsqd to get c but when I then tried to solve the circle with the line to get another point I got an unsolvable Quadratic so couldn't find the point to do the translation!
Not too far off which isn't bad given I havent been near a HL class since I did my own leaving cert 10 years ago!

sullanefc

Several methods to attack this question.
My first instinct was to use f, g, c to find the equation of the circle, or at least the centre. We know that g^2 = c and f^2 = c and thus f = g, so the centre would be (-g, -g). You can then say that r = perpendicular distance from centre to bottom line, where r = mod(g). So I found the centre and radius to both be 2+sqrt(2). You can then say that the radius, 2 + sqrt(2) = perpendicular distance from centre to x + y -2k = 0 and solve for k.

Another method is this:
As both lines have a slope of 1, you can draw a line from the origin to the upper point of contact (which will also have a slope of 1 and make a 45 degree angle with the horizontal) and all the points on your line will have the same x and y values. Then you pythagoras the hell out of it so to speak. You then have the co-ords of the upper point of contact and can sub in to line equation to find k.

And finally, you could use a combination of pythagoras and translations as the poster above has used.

What I don't like about this question, is that the answer is decimal, but the question does not say to give your answer to a number of decimal places. Higher level students of the old course are always taught to give their answers in an exact form like an integer or surd where it would not be specified. You cannot do this here as you have to get the root of a root.

In short, the question should have said to find k to 3 decimal places to make it a bit easier for those students who would have persisted with surds until the bitter end, making it more difficult.

sullanefc

Musicmental85 wrote: »

Dang missed the (1,1)! I used the distance from a point to a line saying that centre (-g,-g) to the line x+y=2 must be the same as the distance to the x-axis and got the same 3.414 or 2+root2 for the centre. Used gsqd + fsqd +c =rsqd to get c but when I then tried to solve the circle with the line to get another point I got an unsolvable Quadratic so couldn't find the point to do the translation!
Not too far off which isn't bad given I havent been near a HL class since I did my own leaving cert 10 years ago!

Hey bud, didn't see your post when I replied. Your method would have been my first instinct as well, and judging by what work you did, I'd say you would get a good chunk of the 25 marks for your work.

whiteandlight

sullanefc wrote: »

Musicmental85 wrote: »

Dang missed the (1,1)! I used the distance from a point to a line saying that centre (-g,-g) to the line x+y=2 must be the same as the distance to the x-axis and got the same 3.414 or 2+root2 for the centre. Used gsqd + fsqd +c =rsqd to get c but when I then tried to solve the circle with the line to get another point I got an unsolvable Quadratic so couldn't find the point to do the translation!
Not too far off which isn't bad given I havent been near a HL class since I did my own leaving cert 10 years ago!

Hey bud, didn't see your post when I replied. Your method would have been my first instinct as well, and judging by what work you did, I'd say you would get a good chunk of the 25 marks for your work.

Well I'm a teacher but not teaching HL so it was more of an interest thing. Id agree with the above poster tho, the root thing threw me when they hadnt mentioned decimals. I can only assume it probably confused candidates too.

Dougery1

Does anyone have the solution for q 8 b, c and d. I got a= 44 and b= 100. Everything so abstract after that. My head is melting. I can't rest until I know how to do it. Please help. 

sullanefc

double post

sullanefc

Dougery1 wrote: »

Does anyone have the solution for q 8 b, c and d. I got a= 44 and b= 100. Everything so abstract after that. My head is melting. I can't rest until I know how to do it. Please help. 

For b and c, you have to imagine a sector with an angle of 1 degree.



(b) The longer the radius, the bigger the arc length and thus the greater the error in the position of R. As the distance |PR| (25cm) is greater than |QR| (12cm), then an error in the angle alpha will have a greater effect on the position of R. You can use arc length = 1/360(2*pie*r) to show this.

(c) For this you have to think about the situations where the |PR| < |QR|. One such situation would be when the arm is doubled back on itself. i.e. |PR| = 8cm and |QR| = 12cm. If you draw a triangle of the various positions where |QR| = |PR| = 12 cm, and solve the angle beta, you can find the range of values where beta would have a bigger influence on the error.

(d) This was posted in the Leaving Cert section and it looks right to me:
http://www.boards.ie/vbulletin/showpost.php?p=79147523&postcount=341

I hope this helps.

geenamae

see that's what i cannot stand about the new course! In the old course questions were very clear and well defined. In the new paper 2 it is hard to be certain of anything. They didn't tell you that the tangents are parallel and i for one was always thought never to assume anything in Maths.
Of course in the end after spending mabye 40 minutes on it i had no choice but to assume they had to be parallel. Now that sort of thinking in Maths just doesn't seem right to me.

sullanefc

geenamae wrote: »

see that's what i cannot stand about the new course! In the old course questions were very clear and well defined. In the new paper 2 it is hard to be certain of anything. They didn't tell you that the tangents are parallel and i for one was always thought never to assume anything in Maths.
Of course in the end after spending mabye 40 minutes on it i had no choice but to assume they had to be parallel. Now that sort of thinking in Maths just doesn't seem right to me.

I'm no fan of PM, but to be fair, the lines ARE parallel because if you write both in the form y = mx + c then you can see that both have a slope of 1. This bit of knowledge was required to do the question and so was being examined as well.

My problem with it would be, that if you missed this bit, then you could be snookered for the rest of the question.

Dougery1

If you double the arm back onto itself then is QR still not 12 and now PR is 8 or am I totally missing the point here!!!!!

sullanefc

Dougery1 wrote: »

If you double the arm back onto itself then is QR still not 12 and now PR is 8 or am I totally missing the point here!!!!!

Apologies, yes you are right, I wasn't minding my P's and Q's I suppose. Got them mixed up. Post above edited to correct.