∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
Backed by popular demand I bring you:
Vectors
*drum roll*
What can I say, apart from vectors are gay. They are actually quite easy once you understand them, and I am gonna explain why.
We'll start off easy:
Two points A and B have position vectors 2ij+2k and 2j+3k respectively:
i)Find the vector AB, and hence find the length of AB
ii)Find a vector equation for the line through A and B.
What I normally do is write out A and B in columm form, i.e:
i=x
j=y
z=k
would correspond to each other, cause that is all they mean:
therefore vector A would be:
(2 ) 2i
(1) j
(2 ) 2k
B would be:
(0 ) 0i
(2 ) 2j
(3 ) 3k
Now we have the vectors in nice easy forms we can begin the questions:
i)Find the vector AB, and hence find the length of AB
You have to visualise where A and B are, the coordinates are how far away they are from the origin also called 'O' (0i+0j+0k). Therefore to get from the origin to A, its (2ij+2k), this means '+2' across the X axis, '1' along the 'Y' axis and '+2' along the 'Z' axis. Still with me? Good. And to get back from 'A' to 'O' its the opposite signs, i.e. 1*(2ij+2k) = (2i+j2k), this means '2' across the X axis, '+1' along the 'Y' axis and '2' along the 'Z' axis.
Well that's vectors at a glance.
From what I have explained above. which any simpeton should understand vectors A and B are positioned relative to the origin. So we can call them OA and OB, with OA meaning how to get from O to A.
The question is asking us how to get from A to B basically, and you have to work out how to do so using ONLY the information provided.
To get from A to B, you have to think about how to get from the point 'A' to the point 'B'. From the information, we only know how to get from the Origin to A, and how to get from the Origin to B, right? OA and OB.
To get from A to O, its simply OA, and to get to be its OB right, therefore:
OA+OB=AB OR:
OBOA=AB
Therefore if OA is:
(2 )
(1)
(2 )
and OB:
(0 )
(2 )
(3 )
and OBOA=AB then:
(0 )  (2 ) = (2)
(2 )  (1) = (3 )
(3 )  (2) = (1 )
The position vector AB can be expressed as:
(2)
(3 )
(1 )
or (2j+3k+l)
See wasn't so hard now, was it?
i)Find the vector AB, and hence find the length of AB
For the length, its the square root of the sum of the vector squared, aslo called the modulus (the modulus has a different meaning than what you think) , i.e:
√[(i^2+j^2+k^2)]
in this case it would be:
√[(2)^2+(3)^2+(1)^2] = √14
√14 is the length of AB.
ii)Find a vector equation for the line through A and B.
Vector equations are always in the form:
r=a+tp
a=(any point that lies on the line)
t=a letter
p=(a line that is parallel to the vector)
Thus in our example, there are two known points on our line, 'A' and 'B'. so we could use any of those for 'a'.
For P, we just use the vector of AB as its parallel to the line AB (actually they are both one and the same). Therefore if we all these values in, we get:
____(2 ) +t(2)
r = (1) +t(3 )
____(2 ) +t(1 )
or
r = (2ij+2i)+t(2i+3j+i)
x
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Bhaal85
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 08062004 12:38

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 08062004 13:04
(Original post by Bhaal85)
[B]Backed by popular demand I bring you:
Vectors
*drum roll*
What can I say, apart from vectors are gay. They are actually quite easy once you understand them, and I am gonna explain why.
... 
Bhaal85
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 08062004 13:26
Very basic vectors covered.

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 08062004 13:30
Some P3 questions courtesy of Absolution:
∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
1) ∫sec²x.tanx using substitution u=secx
2) ∫(x/x +1)² using subst. of u=x+1
Getting Harder...
3) Finite region bounded by the curve with equation f(x)=x√(4x²) between the lines of x=2 and x=0. Find the VOLUME generated when area is rotated completely about the x axis.
4) Elipse given by: x=3cost, y=2sint
0≤t<2pi
a) find area of finite region bound by the elipse and the positive x and y axis.
b) region rotated around the y axis to form a solid of revolution, find this volume.
OK onto exponential growth and decay...
5) find the general solution of
a) dy/dx = cos² x
b) dy/dx = cos² y
c) (x+1)dy/dx = y + 2 
tammypotato
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 08062004 13:38
(Original post by Bhaal85)
Some P3 questions courtesy of Absolution:
∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
1) ∫sec²x.tanx using substitution u=secx
du/dx=secxtanx du =secxtanxdx
∫udu
u^2/2 so i gt sec ^2 /2 
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 08062004 13:41
(Original post by Bhaal85)
Some P3 questions courtesy of Absolution:
∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
2) ∫(x/x +1)² using subst. of u=x+1
(x/u)^2 du 
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 08062004 13:44
hey peeps, me again.....i need help pls.....
1) INT x/(1+x)
2) INT sec^2x/(1+tanx)^3
thanks 
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 08062004 13:45
Solution for Absolution:
5c)
(x+1)dy/dx = y+2
∫dy/(y+2)=∫dx/(+1)+c
ln(y+2)=ln(x+1)+c
5b)
dy/dx=cos^2y
∫dy/cos^2y=∫dx+c
tany=x+c
5a)
dy/dx=cos^2x
∫dy=∫cos^2x+c dx
cos^2x is the same as (cos2x+1)/2 therefore:
∫dy=∫(cos2x+1)/2 + c dx
y=(sin2x/4)+x/2 
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 08062004 13:50
(Original post by tammypotato)
i cant gt sub can u tell me where i go wrong
du/dx=secxtanx du =secxtanxdx
∫udu
u^2/2 so i gt sec ^2 /2
u=secx, du/dx=tanxsecx dx=du/tanxsecx therefore:
∫sec²x.tanx dx = 1) ∫sec²x.tanx du/tanxsecx, it cancels down to:
∫secx du therefore:
∫u du = (u^2)/2 + c = sec^2x+c or tanx+1 
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 08062004 13:53
(Original post by Bhaal85)
1) ∫sec²x.tanx using substitution u=secx
u=secx, du/dx=tanxsecx dx=du/tanxsecx therefore:
∫sec²x.tanx dx = 1) ∫sec²x.tanx du/tanxsecx, it cancels down to:
∫secx du therefore:
∫u du = (u^2)/2 + c = sec^2x+c or tanx+1 
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 08062004 13:53
2) ∫(x/x +1)² using subst. of u=x+1
u=x+1 therefore du/dx = 1 dx=du.
∫(x/x +1)² dx = ∫(x/x +1)² du
Know that u=x+1 therefore x=u1 producing:
∫(u1)/u² du
lnu+u^1+c
ln(x+1)+1/(x+1)+c 
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 08062004 13:54
(Original post by Bhaal85)
2) ∫(x/x +1)² using subst. of u=x+1
u=x+1 therefore du/dx = 1 dx=du.
∫(x/x +1)² dx = ∫(x/x +1)² du
Know that u=x+1 therefore x=u1 producing:
∫(u1)/u² du
lnu+u^1+c
ln(x+1)+1/(x+1)+c 
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 08062004 13:59
(Original post by Bhaal85)
.........
3) Finite region bounded by the curve with equation f(x)=x√(4x²) between the lines of x=2 and x=0. Find the VOLUME generated when area is rotated completely about the x axis.

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 08062004 14:01
(Original post by Bhaal85)
What is x√(4x²) squared? Is it 4x^2x^4?
4x^2 x^4 
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 08062004 14:08
(x root(4x²))²
x²(4x²) = 4x² x^4 i think 
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 08062004 14:09
∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
3) Finite region bounded by the curve with equation f(x)=x√(4x²) between the lines of x=2 and x=0. Find the VOLUME generated when area is rotated completely about the x axis.
π∫y^2 dx therefore:
y^2=4x^2x^4
π∫4x^2x^4 dx between 2 and 0
=π(4(x^3)/3(x^5)/5]
=π[(4r4r15)(0)]
=π4r4r15 
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 08062004 14:14
(Original post by Bhaal85)
∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π
3) Finite region bounded by the curve with equation f(x)=x√(4x²) between the lines of x=2 and x=0. Find the VOLUME generated when area is rotated completely about the x axis.
π∫y^2 dx therefore:
y^2=4x^2x^4
π∫4x^2x^4 dx between 2 and 0
=π(4(x^3)/3(x^5)/5]
=π[(4r4r15)(0)]
=π4r4r15
Int 4x^2  x^4
4x^3 /3 x^5 /5
pie [8/3  32/5] 
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 08062004 14:56
(Original post by tammypotato)
i gt 56pie/15
Int 4x^2  x^4
4x^3 /3 x^5 /5
pie [8/3  32/5] 
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 08062004 15:16
3) Finite region bounded by the curve with equation f(x)=x√(4x²) between the lines of x=2 and x=0. Find the VOLUME generated when area is rotated completely about the x axis.
π∫y^2 dx therefore:
y^2=4x^2x^4
π∫4x^2x^4 dx between 2 and 0
=π(4(x^3)/3(x^5)/5]
=π[(4r4r15)(0)]
=π4r4r15
∫pi * y² dx = Volume
Pi ∫4x²  x^4
pi [ 4x³/3  x^5/5] between 2 and 0
(4 * 2³/3)  2^5/5  0
32/3 Pi  32/5 Pi = 64/15Pi 
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 08062004 15:28
Hi, hope someone can help me again! This question is on the exam style paper in the back of the P3 revision book, its question 3.
A circle has equation x² + y²  8x + 2y + 8 = 0
Point C on the circle is the nearest point to the origin,O.
Find the distance between O and C.
So far I've worked out the centre and radius of the circle and tried drawing a diagram to help me but I can't quite see what to do.
Emma x x x
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