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FP2 Revision Thread
10
Seeing as threads for most of the other modules have been made, I decided to make one for P5.
I'm going to post some notes on general techniques you can use, and maybe later I'd post some specific notes on a part or two of chapter 4 (slightly leaning towards the parabola and intrinsic coordinates) if time permits.
Feel free to add any notes of your own.
I'm going to post some notes on general techniques you can use, and maybe later I'd post some specific notes on a part or two of chapter 4 (slightly leaning towards the parabola and intrinsic coordinates) if time permits.
Feel free to add any notes of your own.
Scroll to see replies
Reply 1
dvs
OP
10
General Integration Techniques
- When an integrand involves sqrt(1-cosx) or sqrt(1+cosx), use the following identies:
cosx = 1 - 2sin²(x/2) = 2cos²(x/2) - 1
These should help you get rid of the square root.
e.g. sqrt(1-cosx) = sqrt[1-1+2sin²(x/2)] = sqrt(2) sin(x/2), which you can now integrate easily.
- For integrands of forms similar to:
1/(acosx + bsinx), 1/(a+bsinx) or 1/(a+bcosx) (This includes rsecx.)
The substitution t=tan(x/2) can make things simpler. Use the following identities to apply the substitution correctly:
sec²(x/2) = 1 + tan²(x/2) = 1 + t²
sinx = 2sin(x/2)cos(x/2) = 2tan(x/2)/sec²(x/2) = 2t/(1+t²)
cosx = cos²(x/2) - sin²(x/2) = (1 - tan²(x/2))/sec²(x/2) = (1-t²)/(1+t²)
e.g.
∫ 1/(1+2sinx) dx
= ∫ 1/[1+4t/(1+t²)] . 2/(1+t²) dt
= ∫ 2/(t²+4t+1) dt
= 2 ∫ 1/[(t+2)² - 3] dt
= 1/sqrt[3] ln|(t+2-sqrt[3])/(t+2+sqrt[3])|
Don't forget to back-substitute t=tan(x/2) in your final answer.
- Spotting "function and its derivative"-type integrands helps quite a bit.
f(x) . f'(x) dx = (1/n) [f(x)]n + C
f'(x)/f(x) dx = ln|f(x)| + C
e.g. ∫ x/(x²+1) dx = (1/2) ∫ 2x/(x²+1) dx = (1/2) ln|x²+1| + C, since x²+1 is the function and 2x is its derivative.
e.g. ∫ cos²x sinx dx = - ∫ cos²x(-sinx) dx = -(1/3)cos³x + C
- Integrands that contain fractions may require long divison. This happens when the numerator is greater than the denumerator.
e.g. x/(x-2) = 1 + 2/(x-2), which you can now integrate easily.
Reduction Formulae Techniques
- Adding zero (+1-1) might help.
e.g.
xn/(xn + 1)
= (xn + 1 - 1)/(xn + 1)
= (xn + 1)/(xn + 1) - 1/(xn + 1)
= 1 - 1/(xn + 1)
[link to see where this can be helpful.]
- Splitting up indices might make the integrand easier to work with.
[link containing a few examples.]
- Sometimes you have to use integration by parts twice.
e.g. ∫ xn sinx dx
First time: u=xn, dv/dx=sinx
Second time: u=xn-1, dv/dx=cosx
General Coordinate Geometry Techniques
- To find the locus of a point as some variable (e.g. t) varies:
1. Write down the coordinates of the point, e.g. (2t, at²)
2. Express the x-coord & y-coord seperately, e.g. x=2t, y=at²
3. Make the variable the subject of one of your equations (it helps if it's the simpler one), e.g. t=x/2
4. Plug this in the other equation to get the locus of the point's path, e.g. y=at²=a(x/2)² => 4y=ax²
Sometimes you might need to use some trig identities.
e.g. Find the locus of P (3cost, 5/sint) as t varies.
x = 3cost => cost = x/3
y = 5/sint => sint = 5/y
cos²t + sin²t = 1
(x/3)² + (5/y)² = 1
And simplify if need be.
e.g. Find the locus of Q (acos2u, bsinu) as u varies.
x = acos2u => cos2u = x/a
y = bsinu => sinu = y/b
cos2u = 1 - 2sin²u
x/a = 1 - 2(y/b)²
etc.
Use your imagination and make sure you know your identities.
- You don't always have to solve a quadratic equation to use its roots.
If an equation of the form x²+ax+b=0 has roots p and q, then p+q=-a and pq=b.
e.g. Suppose you have a question that's asking you for the coordinates of the midpoint of a line joining two points on a parabola.
Let's say that the equations of the line and the parabola are y=x+c and y²=ax, respectively.
Now we want to solve these two equations simultaneously to find the coordinates of those points.
(x+c)²=ax
x²+2xc+c²=ax
x²+(2c-a)x+c²=0
Finding the roots p & q of this equation will prove to be a horrendous task. So instead, use the fact that:
p + q = -(2c-a) = a-2c
Similarly, we can find r+t, where r & t are the y-intercepts of the points.
Now all we have to do is plug this into the formula for the midpoint:
[(p+q)/2, (r+t)/2]
And we're done.
- Know what form of coordinates you should use.
For example, dealing with a certain ellipse question you might find using coordinates of the form (p, q) to suit your needs perfectly; some other times, however, you're going to have to use the parametric coordinates (acost, bsint).
The same applies to all the other conics.
Unfortunately, I can't really say when you use the different forms. If the question specifically uses one of the forms, it's obviously a good idea to use that one in your solution. In other cases, just go with your gut and hope you get lucky!
I hope I haven't slipped up and made some silly mistakes. Please pm me if you find any.
- When an integrand involves sqrt(1-cosx) or sqrt(1+cosx), use the following identies:
cosx = 1 - 2sin²(x/2) = 2cos²(x/2) - 1
These should help you get rid of the square root.
e.g. sqrt(1-cosx) = sqrt[1-1+2sin²(x/2)] = sqrt(2) sin(x/2), which you can now integrate easily.
- For integrands of forms similar to:
1/(acosx + bsinx), 1/(a+bsinx) or 1/(a+bcosx) (This includes rsecx.)
The substitution t=tan(x/2) can make things simpler. Use the following identities to apply the substitution correctly:
sec²(x/2) = 1 + tan²(x/2) = 1 + t²
sinx = 2sin(x/2)cos(x/2) = 2tan(x/2)/sec²(x/2) = 2t/(1+t²)
cosx = cos²(x/2) - sin²(x/2) = (1 - tan²(x/2))/sec²(x/2) = (1-t²)/(1+t²)
e.g.
∫ 1/(1+2sinx) dx
= ∫ 1/[1+4t/(1+t²)] . 2/(1+t²) dt
= ∫ 2/(t²+4t+1) dt
= 2 ∫ 1/[(t+2)² - 3] dt
= 1/sqrt[3] ln|(t+2-sqrt[3])/(t+2+sqrt[3])|
Don't forget to back-substitute t=tan(x/2) in your final answer.
- Spotting "function and its derivative"-type integrands helps quite a bit.
f(x) . f'(x) dx = (1/n) [f(x)]n + C
f'(x)/f(x) dx = ln|f(x)| + C
e.g. ∫ x/(x²+1) dx = (1/2) ∫ 2x/(x²+1) dx = (1/2) ln|x²+1| + C, since x²+1 is the function and 2x is its derivative.
e.g. ∫ cos²x sinx dx = - ∫ cos²x(-sinx) dx = -(1/3)cos³x + C
- Integrands that contain fractions may require long divison. This happens when the numerator is greater than the denumerator.
e.g. x/(x-2) = 1 + 2/(x-2), which you can now integrate easily.
Reduction Formulae Techniques
- Adding zero (+1-1) might help.
e.g.
xn/(xn + 1)
= (xn + 1 - 1)/(xn + 1)
= (xn + 1)/(xn + 1) - 1/(xn + 1)
= 1 - 1/(xn + 1)
[link to see where this can be helpful.]
- Splitting up indices might make the integrand easier to work with.
[link containing a few examples.]
- Sometimes you have to use integration by parts twice.
e.g. ∫ xn sinx dx
First time: u=xn, dv/dx=sinx
Second time: u=xn-1, dv/dx=cosx
General Coordinate Geometry Techniques
- To find the locus of a point as some variable (e.g. t) varies:
1. Write down the coordinates of the point, e.g. (2t, at²)
2. Express the x-coord & y-coord seperately, e.g. x=2t, y=at²
3. Make the variable the subject of one of your equations (it helps if it's the simpler one), e.g. t=x/2
4. Plug this in the other equation to get the locus of the point's path, e.g. y=at²=a(x/2)² => 4y=ax²
Sometimes you might need to use some trig identities.
e.g. Find the locus of P (3cost, 5/sint) as t varies.
x = 3cost => cost = x/3
y = 5/sint => sint = 5/y
cos²t + sin²t = 1
(x/3)² + (5/y)² = 1
And simplify if need be.
e.g. Find the locus of Q (acos2u, bsinu) as u varies.
x = acos2u => cos2u = x/a
y = bsinu => sinu = y/b
cos2u = 1 - 2sin²u
x/a = 1 - 2(y/b)²
etc.
Use your imagination and make sure you know your identities.
- You don't always have to solve a quadratic equation to use its roots.
If an equation of the form x²+ax+b=0 has roots p and q, then p+q=-a and pq=b.
e.g. Suppose you have a question that's asking you for the coordinates of the midpoint of a line joining two points on a parabola.
Let's say that the equations of the line and the parabola are y=x+c and y²=ax, respectively.
Now we want to solve these two equations simultaneously to find the coordinates of those points.
(x+c)²=ax
x²+2xc+c²=ax
x²+(2c-a)x+c²=0
Finding the roots p & q of this equation will prove to be a horrendous task. So instead, use the fact that:
p + q = -(2c-a) = a-2c
Similarly, we can find r+t, where r & t are the y-intercepts of the points.
Now all we have to do is plug this into the formula for the midpoint:
[(p+q)/2, (r+t)/2]
And we're done.
- Know what form of coordinates you should use.
For example, dealing with a certain ellipse question you might find using coordinates of the form (p, q) to suit your needs perfectly; some other times, however, you're going to have to use the parametric coordinates (acost, bsint).
The same applies to all the other conics.
Unfortunately, I can't really say when you use the different forms. If the question specifically uses one of the forms, it's obviously a good idea to use that one in your solution. In other cases, just go with your gut and hope you get lucky!
I hope I haven't slipped up and made some silly mistakes. Please pm me if you find any.
Reply 2
hyperbolic functions:
coshx = ½(ex+e-x)
sinhx = ½(ex-e-x)
and you can derive all other functions from those two functions.
for intergrating or differentiating, if were unable to find the identity to get to standard integrals, just convert the hyp. function into the exp. form, and integrate or differentiate, as integratin e is not that hard.
i hope i helped
coshx = ½(ex+e-x)
sinhx = ½(ex-e-x)
and you can derive all other functions from those two functions.
for intergrating or differentiating, if were unable to find the identity to get to standard integrals, just convert the hyp. function into the exp. form, and integrate or differentiate, as integratin e is not that hard.
i hope i helped
Reply 3
15
I'll revive this thread with a copy and paste from one of my other posts, as it isn't long until the exam.
Intrinsic coordinates:
1: Converting intrinsic form
to cartesian form [y].
Differentiate to give (ds/dw)=f'(w) so ds=f'(w) dw
(dy/ds)=sinw
∫1 dy = ∫sinw ds
y=∫sinw.f'(w) dw
Integrate to give a relationship between y and w. Remember not to neglect the arbitrary constant.
(dx/ds)=cosw
∫1 dx = ∫cosw ds
x=∫cosw.f'(w) dw
Integrate to give a relationship between x and w. Remember not to neglect the arbitrary constant.
Eliminate the parameter 'w' to give the cartesian relationship between y and x. Typically you'll use a trigonometric identity or make w the subject in the two equations to eliminate w.
2: Converting cartesian form [y] to intrinsic form
y=f(x) so (dy/dx)=f'(x)=tanw
s=∫[1+(dy/dx)^2]^0.5 dx with lower limit a(where arc length is measured from) and upper limit x.
s=∫[1+(f(x))^2]^0.5 dx with lower limit a and upper limit x.
Perform the integration to find the relation s=h(x) and eliminate x using (dy/dx)=tanw, to form a relation involving only s and w.
Tou can convert from x=f(y) or parametric equations x=f(t), y=g(t) in a similar manner using the appropriate arc length formula.
Intrinsic coordinates:
1: Converting intrinsic form
to cartesian form [y].Differentiate to give (ds/dw)=f'(w) so ds=f'(w) dw
(dy/ds)=sinw
∫1 dy = ∫sinw ds
y=∫sinw.f'(w) dw
Integrate to give a relationship between y and w. Remember not to neglect the arbitrary constant.
(dx/ds)=cosw
∫1 dx = ∫cosw ds
x=∫cosw.f'(w) dw
Integrate to give a relationship between x and w. Remember not to neglect the arbitrary constant.
Eliminate the parameter 'w' to give the cartesian relationship between y and x. Typically you'll use a trigonometric identity or make w the subject in the two equations to eliminate w.
2: Converting cartesian form [y] to intrinsic form
y=f(x) so (dy/dx)=f'(x)=tanw
s=∫[1+(dy/dx)^2]^0.5 dx with lower limit a(where arc length is measured from) and upper limit x.
s=∫[1+(f(x))^2]^0.5 dx with lower limit a and upper limit x.
Perform the integration to find the relation s=h(x) and eliminate x using (dy/dx)=tanw, to form a relation involving only s and w.
Tou can convert from x=f(y) or parametric equations x=f(t), y=g(t) in a similar manner using the appropriate arc length formula.
Reply 4
Your summary was FANTASTIC - it really helped :-) Thanks !
Are you sure you didn't mean p+q = -b/a and pq = c/a ?
dvs
If an equation of the form x²+ax+b=0 has roots p and q, then p+q=-a and pq=b
If an equation of the form x²+ax+b=0 has roots p and q, then p+q=-a and pq=b
Are you sure you didn't mean p+q = -b/a and pq = c/a ?
Reply 5
dvs
OP
10
danguetta
Are you sure you didn't mean p+q = -b/a and pq = c/a ?
Yup. It would be p+q=-b/a and pq=c/a if the equation was ax²+bx+c=0, but the equation I used in my post was x²+ax+b=0.
I'm glad you found it helpful.
Reply 6
dvs
Yup. It would be p+q=-b/a and pq=c/a if the equation was ax²+bx+c=0, but the equation I used in my post was x²+ax+b=0.
I'm glad you found it helpful.
I'm glad you found it helpful.
Ahhh, clever
Thanks,
Daniel
Reply 7
12
Another technique to check the equation of the tangent of a conics: split up one element into 2 elements
Use x² = x.x, y² = y.y and x = (x/2) + (x/2), y = (y/2) + (y/2)
If the tangent passes through P(xo, yo) in the conic curve, then you just substitue xo, yo in one of the elements you've splitted up. The others keep the same.
E.g : An ellipse equation : x²/a² + y²/b² = 1
-> x.x/a² + y.y/b² = 1.
The tangent passes through P(xo, yo) in the ellipse is:
x.xo/a² + y.yo/b² = 1
or a parabola: y² = 4ax, i.e y.y = 2a(x + x)
The tangent passes through P(xo, yo) in the parabola isL
y.yo = 2a(x + xo).
But for rectangular hyperbola, xy = c², you must split xy = xy/2 + xy/2,
Tangent is: xo.y/2 + yo.x/2 = c².
I know it's easy part ... and maybe this technique is silly, but I just want to share
Hope all of you do well on Monday.
Use x² = x.x, y² = y.y and x = (x/2) + (x/2), y = (y/2) + (y/2)
If the tangent passes through P(xo, yo) in the conic curve, then you just substitue xo, yo in one of the elements you've splitted up. The others keep the same.
E.g : An ellipse equation : x²/a² + y²/b² = 1
-> x.x/a² + y.y/b² = 1.
The tangent passes through P(xo, yo) in the ellipse is:
x.xo/a² + y.yo/b² = 1
or a parabola: y² = 4ax, i.e y.y = 2a(x + x)
The tangent passes through P(xo, yo) in the parabola isL
y.yo = 2a(x + xo).
But for rectangular hyperbola, xy = c², you must split xy = xy/2 + xy/2,
Tangent is: xo.y/2 + yo.x/2 = c².
I know it's easy part ... and maybe this technique is silly, but I just want to share
Hope all of you do well on Monday.
Reply 8
I don't get this conics thing! It's not edexcel is it ? Cos I haven't HEARD of conics...
Reply 9
15
lesser weevil
I don't get this conics thing! It's not edexcel is it ? Cos I haven't HEARD of conics...
It's where the parabola, ellipse and hyperbola come from.
http://mathworld.wolfram.com/ConicSection.html
http://www.math2.org/math/algebra/conics.htm
Reply 10
OH my gosh!!! the exam is tomorrow!!!Reply 11
13
I've had this thread reopened for those people taking FP2 in june. No more introduction is needed further than what dvs gave last year.
Reply 12
13
ssmoose
I've had this thread reopened for those people taking FP2 in june. No more introduction is needed further than what dvs gave last year.
Great
Plus we've started off with some good notes from last year. Thanks everyone for writing these!
Reply 13
13
Just need to find that P4 one now...
Reply 14
13
Yeah. I asked Acaila to re-open the P6 thread, but I couldn't find the P4 thread from last June. Is there one from January?
Reply 15
13
Okay, I've found the FP1 thread from last June and have asked the mods to reopen it. I've also asked for the titles of these threads to be changed to FP1, FP2 and FP3 respectively.
Reply 16
13
Thread renamed "FP2"
Reply 17
How do you integrate arcosh(x)^2 ?
Reply 18
=gabriel=
How do you integrate arcosh(x)^2 ?
have you tried Integration by Parts? u=arcosh(x) and dv=arcosh(x)
or did you mean arcosh(x2) not [arcosh(x)]2..
Reply 19
TIA
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