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# P3 today watch

1. how did everyone find the P3 exam today??

i HATED the vector question. *urgh*
2. can any 1 remember what they got?

and working for qu 6
3. (Original post by [email protected])
can any 1 remember what they got?

and working for qu 6
q 6 was??
4. differentiation,
5. hm... think i was pretty unsuccessful with that! only managed to make it thru a couple of the questions, most of them i had to leave the last parts of it joyous!
6. Question 2 was also exteremly difficult esp. the first part
7. left out most of q 3
8. does anyone remember the answer for the last part of the very last ques of the paper.....it was to find the value of y
9. the only thing i know is that the x cord of q was 11.75. 6/75 for me!!!
10. I got y=1/[e^(pi/6)-1]
Area of parallelogram = root(675)
On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
11. (Original post by SsEe)
I got y=1/[e^(pi/6)-1]
Area of parallelogram = root(675)
On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
i got the same area for the parallelogram as well i think

yup i got the same x-cor-ord too

but the paper was messed up.
12. (Original post by Sinan)
does anyone remember the answer for the last part of the very last ques of the paper.....it was to find the value of y
thought the exam was well hard.... the vector question part c and d were well dodgy.

So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????

i think i screwed up the other integration question as well even tho it was simple.... the last part of that one i got something like 9/8 for the area????

ther was a nine mark question as well wasn't there? n e one remember what that was about?
13. (Original post by DaKe)
thought the exam was well hard.... the vector question part c and d were well dodgy.

So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????

i think i screwed up the other integration question as well even tho it was simple.... the last part of that one i got something like 9/8 for the area????

ther was a nine mark question as well wasn't there? n e one remember what that was about?
*sigh*
14. It was harder than any past paper I'd done. For the parallelogram I cheated my ass off. As I'm a big swot I read ahead and I've read the vectors chapter in P6. The area of the parallelogram was 2*(0.5|OB||OD|sin(t)) or something. I recognised that as the modulus of the vector product (like dot product only you get a vector as your answer) and there's a way of calculating that from the determinant of a matrix. I did the question in about 3 lines (3 lines of a 3x3 matrix that is)
15. (Original post by SsEe)
I got y=1/[e^(pi/6)-1]
Area of parallelogram = root(675)
On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
for the integration one i integrated up to P(5,9) or something, then added the area of the triangle from 5 to 11.75 with height 9.... any one know if that sounds rite???
16. (Original post by DaKe)
thought the exam was well hard.... the vector question part c and d were well dodgy.

So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????
y/(y+1) = e

y = ey + e

y - ey = e

y(1-e) = e

y = e/(1-e)

I don't know what the question was, or if it would matter, though.

Ben
17. (Original post by Ben.S.)
y/(y+1) = e

y = ey + e

y - ey = e

y(1-e) = e

y = e/(1-e)

I don't know what the question was, or if it would matter, though.

Ben
at least we're not alone with messing up. hope the grade boundaries would go down.
18. I left all Q2 except half the first part blank. I needed more time!

In the last question I got y=sqrt[2]/(e^pi/6 - sqrt[2]), but I probably made an arithmetic slip somewhere cause I did it in like 5mins.
19. (Original post by Ben.S.)
y/(y+1) = e

y = ey + e

y - ey = e

y(1-e) = e

y = e/(1-e)

I don't know what the question was, or if it would matter, though.

Ben
shite...... how did i not see that?????
20. The area can't have been 9/8. Far too small. let's see if I remember:
x = t² + 1
y = 3t + 3
Area = 9[(5 + 47/4)/2] - Int (from y = 0 to 9) x dy
= 603/8 - Int(from t = -1 to 2) x dy/dt dt
= 603/8 - Int(from t = -1 to 2) (3t²+3) dt
= 603/8 - [ ((2)³+3(2)) - ((-1)³+3(-1))
= 603/8 - [ 14 - (-4)]
= 459/8

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