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Higher Maths :: Post-Exam Discussion :: 2010 :: Did you make it out alive?

theblackswan

ukdragon fancy putting our minds to rest?

We found the coordinates of A, yeah?

So draw a line x=A
Find the area of top curve(1/3x) - bottom "curve"(0) which would be 1/3x - 0

So you would intergrate 1/3x dx with 0 and 4.5 (X coordinate of A) or you could fine the area using A- 1/2bh

Then you would find the area between thhe two "proper curves" with limits 4.5 and 9.

Then add the two areas to get the total area

Reply 401

garlicbreadman

12

theblackswan

it was a calculator paper though, but would it have worked?

you said you got:

Reply 402

ukdragon37

14

theblackswan

it was a calculator paper though, but would it have worked?

No by iffy I mean "maths not covered in Higher maths". To see what I mean, if I do the integration from 0 to 9:

Unparseable latex formula:

\begin{array}{l}[br] \displaystyle\int\limits_0^9 {\dfrac{1}{3}x - {{\left( {2x - 9} \right)}^{\frac{1}{2}}}dx} = \left[ {\dfrac{1}{6}{x^2} - \frac{1}{3}{{\left( {2x - 9} \right)}^{\frac{3}{2}}}} \right]_0^9 \\ \\[br] = \dfrac{{81}}{6} - \dfrac{{27}}{3} - 0 + \dfrac{1}{3}{\left( { - 9} \right)^{3/2}} \\ \\[br] \end{array}

Now how did you get rid of the last part? Try it in your calculator, it will give you a maths error most probably.

Reply 403

theblackswan

garlicbreadman

you said you got:

well all of my answers were the same, just wondering if the working i have done to reach that answer is viable

Reply 404

theblackswan

ukdragon37

No by iffy I mean "maths not covered in Higher maths". To see what I mean, if I do the integration from 0 to 9:

Unparseable latex formula:

\begin{array}{l}[br] \displaystyle\int\limits_0^9 {\dfrac{1}{3}x - {{\left( {2x - 9} \right)}^{\frac{1}{2}}}dx} = \left[ {\dfrac{1}{6}{x^2} - \frac{1}{3}{{\left( {2x - 9} \right)}^{\frac{3}{2}}}} \right]_0^9 \\ \\[br] = \dfrac{{81}}{6} - \dfrac{{27}}{3} - 0 + \dfrac{1}{3}{\left( { - 9} \right)^{3/2}} \\ \\[br] \end{array}

Now how did you get rid of the last part? Try it in your calculator, it will give you a maths error most probably.

Dunno, but i got 4.5 somehow

Reply 405

garlicbreadman

12

ukdragon37

No by iffy I mean "maths not covered in Higher maths". To see what I mean, if I do the integration from 0 to 9:

Unparseable latex formula:

\begin{array}{l}[br] \displaystyle\int\limits_0^9 {\dfrac{1}{3}x - {{\left( {2x - 9} \right)}^{\frac{1}{2}}}dx} = \left[ {\dfrac{1}{6}{x^2} - \frac{1}{3}{{\left( {2x - 9} \right)}^{\frac{3}{2}}}} \right]_0^9 \\ \\[br] = \dfrac{{81}}{6} - \dfrac{{27}}{3} - 0 + \dfrac{1}{3}{\left( { - 9} \right)^{3/2}} \\ \\[br] \end{array}

Now how did you get rid of the last part? Try it in your calculator, it will give you a maths error most probably.

well you can't find the square root of a -ve number so it can't be done

Here's a problem, how many faces do you see:

Reply 406

theblackswan

Dunno, but i got 4.5 somehow

woops

Reply 407

ukdragon37

14

Blu3j4yw4y

Yeah, I just wrote that there were no solutions > no area, which would have made sense considering the curve didn't cut from 0 > A so you would have expected a potion with no apparant solutions I would have thought...I think

That's because you are essentially trying to take the square root of a negative number, which gives you an imaginary part (you do this in "complex numbers" in AH). This "imaginary part" does not affect the real numbers part but it is certainly wrong maths to say it just equals 0.

It all depends on how the SQA will mark it. Either they could just let it go or they could judge it as "eased working" and deduct marks. Personally I wouldn't do it that way but that's because I know it's wrong maths. The SQA however may give you the benefit of the doubt.

Reply 408

alex282

20

how annoying is it when you get the wright answer but put the wrong letter in multiple choice?

Reply 409

garlicbreadman

12

ukdragon37

That's because you are essentially trying to take the square root of a negative number, which gives you an imaginary part (you do this in "complex numbers" in AH). This "imaginary part" does not affect the real numbers part but it is certainly wrong maths to say it just equals 0.

It all depends on how the SQA will mark it. Either they could just let it go or they could judge it as "eased working" and deduct marks. Personally I wouldn't do it that way but that's because I know it's wrong maths. The SQA however may give you the benefit of the doubt.

I must admit though, that Q 6)c) was a tricky one because it looks ok to find the area just using limits 0 and 9 and using each of the y=...... equations given. But since there are 3 "walls" or "borders" involved, you have to split it.

Most of the time when we find the area between two curves using intergration, it is only between the x axis and a curve or between two curves.
Here, the area is bounded by 3 surfaces: the x-axis, and the straight line and the curve
I didn't have enough time to do it like that so just did what I could asap.

Reply 410

Enzo-259

15

I can't believe it. After being really leniant, I barely scrape a B. I've just ****** up everything. It's not even that I couldn't do it! I lost 14 marks
on the MC, 10of which were becuase I didn't bother reading the question properly.

So there goes medicine, and the next five years of my life. :frown:

Reply 411

LuhLah

OP

3

garlicbreadman

well you can't find the square root of a -ve number so it can't be done

Here's a problem, how many faces do you see:

Every time you post pictures, my eyes bleed.

Reply 412

Livesindreams

Okay, so I'm going to need some help here to find out if I've passed or not.

Paper 1

In Q21a) I got the wrong midpoint, but I kept working through and got the answer to be 6x+11y=152. Would I get any marks at all? (3)

21c) I guessed it to be BT:TQ = 2:1. But because I got the wrong midpoint and so wrong value of Q I would have lost the marks, but I did show how I would have got it if I had the right point. So again any marks? (2)

23b) I couldn't get what x and y were, yeah I know silly mistakes. So I made up some values, x=4 and y=1. And then got sinb=1/root17 and cosb=4/root17. So any marks there? (4)

c)i) Because of my mistake in b, I messed this one up as well. And got sin(a-b)= 10/root15root17 (2)
ii) Got sin(b-a)= -10/root15root17, so again, any marks in these questions? (2)

Paper 2

2b)ii) I put down that the min/max for cos= 0. Again stupid mistake, but carried it through like you normally would and got max=67.4 and min= 247.4. (2)

3b) I wrote down the radius of the larger circle as root72 and the centre of it to be (-7,-2) (6)

5b) For some reason, I gave the equation and extra x and so made it;
A(x)=x(12x-2x^3) = 12x^2-2x^4. But I worked this out normally and got the max value to be root3. (6)

6c) I screwed this up badly, but I knew to integrate but used A and 0 as the limits, to get 47/3 units^2. (I made my A=5)(7)

Please help and tell me if there is anywhere I could get marks in these questions. Please :puppyeyes:

Please to be helping? Even if it's just a rough estimate.

Reply 413

garlicbreadman

12

Enzo-259

I can't believe it. After being really leniant, I barely scrape a B. I've just ****** up everything. It's not even that I couldn't do it! I lost 14 marks
on the MC, 10of which were becuase I didn't bother reading the question properly.

So there goes medicine, and the next five years of my life. :frown:

Tried to PM you but u have disabled it.
I just wanted to say, don't give up. I want to do the same as you and know how you must feel. You have probabaly got a minimum B, just keep working for your other exams. Don't give up no matter what. You never know, you could even have got an A, how sure are you you have lost marks?

Spoiler

What did you get in Section A, Paper 1?

Reply 414

Corpsetaker

11

Crap!

I literally just realised (after looking at ukdragon's latex) how to integrate that to get the area! I had no idea how you'd integrate something to the power of a half, but now I see it's just chain integration!

CURSE YOU SQA!

CURSE YOU!!!!

did anyone use vector technique to find the ratio for section B ?
and for the shaded area question I found the area between 9 and 0, then found the area between 9 and 4.5 and toke it away from the area between 9 and 0 - do you guys think I got 0 marks for that question ?

Reply 417

username165626

13

Corpsetaker

Crap!

I literally just realised (after looking at ukdragon's latex) how to integrate that to get the area! I had no idea how you'd integrate something to the power of a half, but now I see it's just chain integration!

CURSE YOU SQA!

CURSE YOU!!!!

Sorry...

Reply 418

Corpsetaker

11

mysqa

Sorry...

You know, I seen who quoted me and, for a fraction of a second, I nearly crapped myself thinking you were a representative of the SQA or something :p: