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Original post by moheat13
I got:
(1/root5)arctan((3tan(x/2) + 1)/root5)
Or something like that (can't fully remember).
You had to complete the square for the denominator and then do u = (t+2/3)^2 to have it ready for the arctan integration.

I got 1/3root(5) arctan(t+(1/3)/root(5))
Reply 81
Original post by TommyKing
I got root(7)/4 for e.

Yeah mb just put that into calculator and realised. I got different coefficients of cos and sin theta but had the same eccentricity of root7/4.
Reply 82
Original post by Maximus_2005
Guys what was your final answer to the integration question? Does anyone remember? T formula one

1/Sqrt (5) ( arctan ((3t+1)/Sqrt 5) +c
Original post by parth_132005
I got 1/3root(5) arctan(t+(1/3)/root(5))

Same but ugh I forgot my +c
Reply 84
Can anyone tell me how to do the vectors question for Part A? Really annoyed me, I knew how to do everything in the question except the first bit, but everything else relied on having the coordinates for B and C...
Original post by moheat13
Can anyone tell me how to do the vectors question for Part A? Really annoyed me, I knew how to do everything in the question except the first bit, but everything else relied on having the coordinates for B and C...

You needed to find the general equation of the line l to A and then say that 15 is equal to the square root of the three conponents squared, solve for lambda and then sub that back in....i got lambda is -2 and -1
Literally same, no clue how to find the co-ordinates so i made my own up lol. I tried 5 different ways to find B and C. I had 30 mins for that questions ffs. Praying for method marks

Original post by moheat13
Can anyone tell me how to do the vectors question for Part A? Really annoyed me, I knew how to do everything in the question except the first bit, but everything else relied on having the coordinates for B and C...
Reply 87
Original post by Labradoodle1
Same but ugh I forgot my +c

"+c" yknow, that thing don't exist in my books
can anyone at least remember the equation of the line that you lot got in part a of the vectors question?
(r - (7i +13j - 24k)) x (3i + ?j + 9k)

i think but probs wrong

Original post by Maximus_2005
can anyone at least remember the equation of the line that you lot got in part a of the vectors question?
Did anyone else get 4.4 for the last part of the vectors question (shortest distance)
Reply 91
YARE YARE. If the equation of a line is (r-a)×b=0, then r-a is parralel to b, r-a=cb, c is a scalar. r=a+cb, a,b vectors and c scalar.
(edited 10 months ago)
Reply 92
Original post by Anonymous :p
Did anyone else get 4.4 for the last part of the vectors question (shortest distance)


ye
The vectors question being 15 marks that you couldn't get if you didn't understand how to do part a was extremely harsh imo
Original post by LucrativeHelper
(r - (7i +13j - 24k)) x (3i + ?j + 9k)

i think but probs wrong


Were you supposed to put it in (r-a)xb form????? I just left it as r = a + lambda(b)
Reply 95
Original post by parth_132005
Were you supposed to put it in (r-a)xb form????? I just left it as r = a + lambda(b)


that was given in the question (cross prodcut form)... i thought FP1 STUDENTS would be smarter than this...
(edited 10 months ago)
oh nah lol thats just the question ahaha thought u asking what the line was

Original post by parth_132005
Were you supposed to put it in (r-a)xb form????? I just left it as r = a + lambda(b)
Do you guys think marks would be possible if you made up B and C lol asking for a friend....
Original post by qftroon
that was given in the question (cross prodcut form)... i thought FP1 STUDENTS would be smarter than this...


No need to be so condescending.
Reply 99
Original post by LucrativeHelper
Do you guys think marks would be possible if you made up B and C lol asking for a friend....


if its correct, 1 mark. if incorrect, zero marks. (Assuming no working shown, just a random guess)

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