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# Edexcel C4 Revision Thread (28/06 14:00) watch

1. Thanks ljfrugn.
2. The straight lines l1 and l2 have vector equations r = (i + 4j + 2k) + t(8i + 5j + k) and r = (i + 4j + 2k) + s(3i + j) respectively, and P is the point with coordinates (1, 4, 2).

a) Show that the point Q(9, 9, 3) lies on l1.
b) Find the cosine of the acute angle between l1 and l2.

c) Find the possible coordinates of the point R, such that R lies on l2 and PQ = PR.

Part c is the bit I'm stuck on
3. Do you mean |PQ| = |PR|? because of PQ = PR then OQ = OR I think
4. (Original post by TheWolf)
List of papers for practice (Right Click link..then Save As):

You won't have time to do all the papers, so I would do the specimen and mock first, then the Edexcel practice papers, then the Solomon ones. That's already like 20 papers, so if you have spare time, I'd probs have a look at the Zig Zag ones.

PLZ DO U HAVE THE ANSWERS TO THE MOCK PAPER ? VERY ESSENTIAL???
5. (Original post by mik1w)
Do you mean |PQ| = |PR|? because of PQ = PR then OQ = OR I think
It says PQ = PR. I have copied it exactly from the book.
6. basically regarding the VECTORS question:

PQ= 9-1, 9-4, 3-2= (8, 5, 1)= PR
therefore (8,5,1)-(1,4,2) = (7,1,-1)
7. i need 86 in this 4 an A afta 2days m2 massacre
8. guys can you help me on the C4 mock paper question 3b please. Thanks [im sure you have the mock cos its on the first page of this thread]
9. (Original post by mackin boi)
i need 86 in this 4 an A afta 2days m2 massacre
oooh unlucky. i think i have secured my A with C4 in hand (ie with C1-3 M1-2)
10. (Original post by sb1986)
guys can you help me on the C4 mock paper question 3b please. Thanks [im sure you have the mock cos its on the first page of this thread]
You get:

∫ 5/(1-x) - 4/(1+2x) dx, after the first part, right?

= (-5)ln(1-x) - (4)(1/2)ln(1+2x) , from 1/6 to 1/3,

since ∫1/(1-x) dx = -ln(1-x) + c
and ∫1/(1+2x) dx = (1/2)ln(1+2x) + c

Now just simplify and bit, and sub limits in.
11. u should have won the most helpful thing...or runner up atleast
12. Nah. I don't do as much as others. Pretty cool I was nominated though.
13. Find the point of intersection of the line through the points (2,0,1) and (-1,3,4) and the line through the points (-1,3,0) and (4,-2,5). Calculate the angle through the lines.

14. (2,0,1) . (-1,3,4) = (a)x(b)cosθ

-2 + 4 = √(2²+0²+1²) x √(-1²+3²+4²) x cosθ

2 = √5 x √26 x cosθ

cosθ = 2/(√130)

θ= 80°
-
15. can someone please explain to me how to integrate a parametric equation, i cant seem to follow the books example very well.

C4 Page 106

The curve c has parametric equations x=t(1+t) y=1/(1+t) where t is the parameter and t is greater than or equal to 0.
The region are is bounded by the lines x=0 and x=2

a) find the exact area of R
b) Find the exact volume of the solid formed when R is rotated through 2pie radians about the x axis.
16. (Original post by melbourne)
can someone please explain to me how to integrate a parametric equation, i cant seem to follow the books example very well.

C4 Page 106

The curve c has parametric equations x=t(1+t) y=1/(1+t) where t is the parameter and t is greater than or equal to 0.
The region are is bounded by the lines x=0 and x=2

a) find the exact area of R
b) Find the exact volume of the solid formed when R is rotated through 2pie radians about the x axis.
anyone> lol
17. (Original post by melbourne)
can someone please explain to me how to integrate a parametric equation, i cant seem to follow the books example very well.

C4 Page 106

The curve c has parametric equations x=t(1+t) y=1/(1+t) where t is the parameter and t is greater than or equal to 0.
The region are is bounded by the lines x=0 and x=2

a) find the exact area of R
b) Find the exact volume of the solid formed when R is rotated through 2pie radians about the x axis.
A = ∫y dx (limits are 0 and 2)
replace y with y=1/(1+t)

A = ∫1/(1+t) dx

dx/dt = 1 + 2t
dx = 1 + 2t dt
replace dx in the equation with 1 + 2t dt

A = ∫1/(1+t) (1+2t) dt

Now you have to change the limits, as the limits are x atm, and they gotta be t, like the rest of the equation

put x=0 into x=t(1+t), so t=0 or -1, but t>0 , so t=0
oyt x=2 into x=t(1+t), so t=1 or -2, but t>0 , so t=1

now you have the complete equation:

∫1/(1+t) (1+2t) dt (limits 1,0)

simplify this to

∫(2- (1/t) dt (by dividing '2t + 1' by 't + 1')

[2t - ln(1+t)]

= (2-ln2) - (0-ln1)

= 2-ln2
18. mockel - thanks for the help earlier.

I need help on the mock paper question 7B. Thanks!
19. (Original post by sb1986)
mockel - thanks for the help earlier.

I need help on the mock paper question 7B. Thanks!
Do you just want help with the differentiation?

M = 10(0.98)t
lnM = ln10 + tln0.98

differentiating implicitly wrt t:
(1/M)(dM/dt) = ln0.98
dM/dt = M ln0.98
20. What new formulas (other than c3 ones) do we need to learn? I find the half angle ones v.useful.

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