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Transcript of C4 revision guide
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Pure C4
Revision Notes
March 2013
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C4 14/04/13 1
Contents
Core
4
1
Algebra ................................................................................................................. 3
Partial fractions ........................................................................................................................................................ 3
2
Coordinate Geometry .......................................................................................... 5
Parametric equations ................................................................................................................................................ 5
Conversion from parametric to Cartesian form ...................................................................................................................... 6
Area under curve given parametrically................................................................................................................................... 7
3
Sequences and series ......................................................................................... 8
Binomial series (1 +x)n for anyn.......................................................................................................................... 8
4 Differentiation .................................................................................................... 10
Relationship between and
......................................................................................................................... 10
Implicit differentiation ........................................................................................................................................... 10
Parametric differentiation ...................................................................................................................................... 11Exponential functions, a
x...................................................................................................................................... 11
Related rates of change .......................................................................................................................................... 12
Forming differential equations ............................................................................................................................... 12
5 Integration .......................................................................................................... 13
Integrals of ex and
.......................................... .................................................................................................. 13
Standard integrals ................................................................................................................................................... 13
Integration using trigonometric identities .............................................................................................................. 13
Integration by reverse chain rule ......................................................................................................................... 14
Integrals of tanxand cotx .................................................................................................................................................... 15
Integrals of secxand cosecx ................................................................................................................................................ 15
Integration using partial fractions .......................................................................................................................... 16
Integration by substitution, indefinite .................................................................................................................... 16
Integration by substitution, definite ....................................................................................................................... 18
Choosing the substitution ...................................................................................................................................................... 18
Integration by parts................................................................................................................................................. 19
Area under curve .................................................................................................................................................... 20
Volume of revolution ............................................................................................................................................. 20
Volume of revolution about thexaxis ................................................................................................................................. 21
Volume of revolution about theyaxis ................................................................................................................................. 21
Parametric integration ............................................................................................................................................ 21
Differential equations ............................................................................................................................................. 22
Separating the variables ........................................................................................................................................................ 22
Exponential growth and decay ............................................................................................................................... 23
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C4 14/04/13 SDB2
6 Vectors ................................................................................................................ 24
Notation .................................................................................................................................................................. 24
Definitions, adding and subtracting, etcetera ........................................................................................................ 24
Parallel and non - parallel vectors......................................................................................................................................... 25
Modulus of a vector and unit vectors ................................................................................................................................... 26
Position vectors ..................................................................................................................................................................... 27
Ratios ..................................................................................................................................................................................... 27
Proving geometrical theorems .............................................................................................................................................. 28
Three dimensional vectors ..................................................................................................................................... 29
Length, modulus or magnitude of a vector ........................................................................................................................... 29
Distance between two points ................................................................................................................................................ 29
Scalar product ........................................................................................................................................................ 29
Perpendicular vectors ............... ............... .............. ............... ............... ............... .............. ................. .............. ................ ...... 30
Angle between vectors .......................................................................................................................................................... 31
Vector equation of a straight line .......................................................................................................................... 31
Intersection of two lines ......................................................................................................................................... 32
2 Dimensions ............. ................ ............... ............... ................ .............. ................ .............. ................ ............... ................ ... 32
3 Dimensions ............. ................ ............... ............... ................ .............. ................ .............. ................ ............... ................ ... 33
7
Appendix ............................................................................................................. 34
Binomial series (1 +x)
n
for any n proof ........................................................................................................... 34
Derivative of xq for q rational ................................................................................... ......................................... 34
for negative limits ....................................................................................... .............................................. 35Integration by substitution why it works ............................................................................................................ 36
Parametric integration ............................................................................................................................................ 36
Separating the variables why it works ................................................................................................................ 37
Index ............................................................................................................................. 38
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C4 14/04/13 3
1 Algebra
Partial fractions
1) You must start with a proper fraction: i.e. the degree of the numerator must be less than
the degree of the denominator.
If this is not the case you must firstdo long division to find quotient and remainder.
2) (a) Linear factors (not repeated)
........)().....)(.(
......+
bax
A
bax
(b) Linear repeated factors (squared)
........)()().....(.)(
......22
+
+
bax
B
bax
A
bax
(c) Quadratic factors)
........)().....)(.(
......22
++
+
+ bax
BAx
bax
or ........)().....)(.(
......22
+++
+
++ cbxax
BAx
cbxax
Example: Express)1)(1(
252
2
xx
xx
+
+ in partial fractions.
Solution: The degree of the numerator, 2, is less than the degree of the denominator, 3, so
we do not need long division and can write
)1)(1(
252
2
xx
xx
+
+
211 x
CBx
x
A
+
++
multiply both sides by (1x)(1+x2)
5 x+ 2x2 A(1 +x2) + (Bx+ C)(1 x)
5 1 + 2 = 2A A= 3 clever value!, put x= 1
5 =A+ C C= 2 easy value, put x= 0
2 =AB B= 1 equate coefficients of x2
)1)(1(
252
2
xx
xx
+
+
21
2
1
3
x
x
x +
++
.
N.B. You can put in any value for x, so you can always find as many equations as youneed to solve for A, B, C, D . . . .
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C4 14/04/13 SDB4
Example: Express2
2
)3)(12(
227
+
xx
xx in partial fractions.
Solution: The degree of the numerator, 2, is less than the degree of the denominator,3, so we do not need long division and can write
2
2
)3)(12(
227
+
xx
xx 3)3(12 2
+
+
x
C
x
B
x
A multiply by denominator
x2 7x+ 22 A(x 3)2 + B(2x 1) + C(2x 1)(x 3)
9 21 + 22 = 5B B= 2 clever value, put x= 3
7/2+ 22 = 2.52A A= 3 clever value, put x=
22 = 9AB + 3C C= 1 easy value, put x= 0
2
2
)3)(12(
227
+
xx
xx
3
1
)3(
2
12
32
+
xxx
Example: Express9
392
23
+
x
xxx in partial fractions.
Solution: Firstly the degree of the numerator is not lessthan the degree of thedenominator so we must divide top by bottom.
x2 9 ) x3 + x2 9x 3 ( x + 1
x3 9x
x2 3
x2 96
9
392
23
+
x
xxx = x+ 1 +
9
62
Factorise to give x2 9 = (x 3)(x+ 3) and write
9
62
33)3)(3(
6
++
+
x
B
x
A
xx multiplying by denominator
6 A(x+ 3) + B(x 3)
6 = 6A A= 1 clever value, put x= 3
6 = 6B B= 1 clever value, put x= 3
9
392
23
+
x
xxx = x+ 1 +
3
1
3
1
+
xx
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C4 14/04/13 5
2 Coordinate Geometry
Parametric equations
If we definexandyin terms of a single variable (the letters tor are often used) then thisvariable is called a parameter: we then have the parametric equation of a curve.
Example: y= t2 3, x= 2 + t is the parametric equation of a curve. Find
(i) the points where the curve meets thex-axis,
(ii) the points of intersection of the curve with the line y= 2x+ 1.
Solution:
(i) The curve meets thex-axis wheny= 0 t2= 3 t= 3
curve meets thex-axis at (2 3, 0) and (2 + 3, 0).
(ii) Substitute forxandyin the equation of the line
t2
3 = 2(2 + t) + 1 t2 2t 8 = 0 (t 4)(t+ 2) = 0
t = 2 or 4
the points of intersection are (0, 1) and (6, 13).
Example: Find whether the curvesy= t2 2, x= 2t+ 3 and y= t+ 3, x= 3t 4 intersect.
If they do give the point of intersection, otherwise give reasons why they do notintersect.
Solution: If they intersect there must be a value of t which makes their x-coordinatesequal, so find this value of t
2t+ 3 = 3t 4 t= 7
But for this value of t the y-coordinates are 47 and 10.
So we cannot find a value of t to give equal x-and equal y-coordinates and thereforethe curves do not intersect.
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C4 14/04/13 SDB6
Conversion from parametric to Cartesian form
Eliminate the parameter (tor or ) to form an equation betweenxandyonly.
Example: Find the Cartesian equation of the curve given by y= t2 3, x= t+ 2.
Solution: x= t+ 2 t=x 2, and y= t2 3
y= (x 2)2 3,
which is the Cartesian equation of a parabola with vertex at (2, 3)
With trigonometric parametric equations the formulae
sin2t + cos
2t = 1 and sec
2t tan
2t = 1
will often be useful.
Example: Find the Cartesian equation of the curve given by
y= 3sin t+ 2, x= 3cos t 1.
Solution: Re-arranging we have
3
1cosand,
3
2sin
+=
=
xt
yt , which together with sin
2t + cos
2t = 1
13
1
3
222
=
++
xy
(x+ 1)2 + (y 2)2 = 9
which is the Cartesian equation of a circle with centre (1, 2) and radius 3.
Example: Find the Cartesian equation of the curve given byy= 3tan t, x= 4sec t.
Hence sketch the curve.
Solution: Re-arranging we have4
secand,3
tanx
ty
t == ,
which together with sec2t tan
2t = 1
134 2
2
2
2
= yx
which is the standard equation of a hyperbolawith centre (0, 0)
andx-intercepts (4, 0), (4, 0).
6 4 2 2 4 6
4
2
2
4 y
x
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C4 14/04/13 7
Area under curve given parametrical ly
We know that the area between a curve and thex-axis is given by A = ydx
ydx
dA= .
But, from the chain ruledtdx
dxdA
dtdA =
dtdxy
dtdA =
Integrating with respect to t
A = dtdtdx
y .
Example: Find the area between the curve y= t2 1, x= t
3+ t, the x-axis and the linesx= 0 and x= 2.
Solution: The area is A= .
A = we should write limits for t, notx
Firstly we need to find y anddt
dx in terms of t.
y= t2 1 and
dt
dx= 3t
2+ 1.
Secondly we are integrating with respect to t and so the limits of integration must be for
values of t.
x= 0 t= 0, and
x= 2 t3+ t = 2 t3+ t 2 = 0 (t 1)(t2+ t+ 2) = 0 t= 1.
so the limits for t are from 0 to 1
A =1
0
dxdt
dt = 1
0(t
2 1) (3t
2+ 1) dt
= 1
03t4 2t2 1 dt
=
1
0
35
32
53
ttt = 15
11
Note that in simple problems you may be able to eliminate t and find y dx in the usualmanner. However there will be some problems where this is difficult and the above techniquewill be better.
??
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3 Sequences and series
Binomial series (1 + x)n for anyn
(1 +x)n = ....!3
)2)(1(
!2
)1(1 32 +
+
++ x
nnnx
nnnx
This converges provided that |x| < 1.
Example: Expand (1 + 3x)2
, giving the first four terms, and state the values of x for
which the series is convergent.
Solution:
(1 + 3x)2
= 1 + (2) 3x +2 3( 2) ( 3) ( 2) ( 3) ( 4)(3 ) (3 ) ...
2! 3!x x
+ +
= 1 6x + 27x2
108x3
+ ...This series is convergent when |3x| < 1 |x| < 1/3.
Example: Use the previous example to find an approximation for2
1
0 9997.
Solution: Notice that2
1
0 9997 = 09997 2 = (1 + 3x)2 when x= 00001.
So writing x= 00001 in the expansion 1 6x + 27x2 108x3
09997 2 1 + 00006 + 000000027 + 0000000000108 = 1000600270108
The correct answer to 13 decimal places is 10006002701080 not bad eh?
Example: Expand 21
)4( x , giving all terms up to and including the term inx3, and state
for what values ofxthe series is convergent.
Solution: As the formula holds for (1+x)n we first re-write
2
1
)4( x =
2
1
2
1
414
x
=
2
1
412
x
and now we can use the formula
= 2 .....4!34!242
11
3
23
21
212
21
21
+
+
+
+
xxx
= 2 512644
32xxx
.
This expansion converges for 14
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C4 14/04/13 9
Example: Find the expansion of6
132
xx
x in ascending powers ofx up tox
2.
Solution: First write in partial fractions
2
1
3
2
6
132
++
=
xxxx
x, which must now be written as
1
2211
332
23
)1()1()1(2
1)1(3
2 +=
+
xx
xx
=
+
+
+
+
22
2!2
)2)(1(
2)1(1
2
1
3!2
)2)(1(
3)1(1
3
2 xxxx
=216
11
36
17
6
12
xx .
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C4 14/04/13 SDB10
4 Differentiation
Relationship between and
1
1
So y= 3x2
xdy
dxx
dx
dy
6
16 == .
Implicit differentiation
This is just the chain rule when we do not know explicitlywhatyis as a function of x.
Examples: The following examples use the chain rule (or implicit differentiation)
3 sin sin cos 5 10 5 using the product rule
( ) ( ) ( ) ( )
++=++=+
dx
dyxyxyx
dx
dyxyx
dx
d32333333
2222232
Example: Find the gradient of, and the equation of, the tangent to the curve
x2+y2 3xy = 1 at the point (1, 2).
Solution: Differentiating x2+y2 3xy= 1 with respect to x gives
2x+ 2ydx
dy
+
dx
dyxy 33 = 0
xy
xy
dx
dy
32
23
= 4=
dx
dy when x= 1 andy= 2.
Equation of the tangent is y 2 = 4(x 1) y= 4x 2.
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C4 14/04/13 11
Parametric d ifferentiation
Example: A curve has parametric equations x = t2+ t, y= t3 3t.
(i) Find the equation of the normal at the point where t= 2.
(ii) Find the points with zero gradient.
Solution: (i) When t = 2, x= 6 and y= 2.
33 2 = tdt
dyand 12 += t
dt
dx
when t= 2
Thus the gradient of the normal at the point (6, 2) is 5/9
and its equation is y 2 =5
/9 (x 6) 5x+ 9y= 48.
(ii) gradient = 0 when 012
33 2=
+
=t
t
dx
dy
3t2 3 = 0
t= 1
points with zero gradient are (0, 2) and (2, 2).
Exponential functions, ax
y= ax
lny = ln ax = xln a
aydx
dya
dx
dy
ylnln
1==
( ) aaadxd
xx ln=
3
32 1 95
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C4 14/04/13 SDB12
Related rates of change
We can use the chain rule to relate one rate of change to another.
Example: A spherical snowball is melting at a rate of 96 cm3s
1when its radius is 12 cm.
Find the rate at which its surface area is decreasing at that moment.
Solution: We know that V= 4/3r3 and that A= 4r2.
Using the chain rule we have
dt
drr
dt
dr
dr
dV
dt
dV== 24 , since 24 r
dr
dV=
dt
dV = 4r2
dt
dr
96 = 4 122 dt
dr
dt
dr =
6
1
Using the chain rule again
dt
drr
dt
dr
dr
dA
dt
dA== 8 , since r
dr
dA8=
12166
1128 == scm
dt
dA
Forming differential equations
Example: The mass of a radio-active substance at time t is decaying at a rate which isproportional to the mass present at time t. Find a differential equation connecting themass m and the time t.
Solution: Remember thatdt
dm means the rate at which the mass is increasingso in this
case we must consider the rate of decay as a negative increase
dt
dm m
dt
dm = km, where k is the (positive) constant of proportionality.
1 ln m =
kt2 + lnA
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C4 14/04/13 13
5 Integration
Integrals of ex and
+= cedxe xx
+= cxdxx
||ln1
for a further treatment of this result, see the appendix
Example: Find dxx
xx +
2
3 3
Solution: dxx
xx +
2
33
= + dx
xx
3 = x2 + 3 lnx + c.
Standard integrals
x must be in RADIANS when integrating trigonometric functions.
f(x) dxxf )( f(x) dxxf )(
xn
1
1
+
+
n
xn
sinx cosx
x1 ln x cosx sinx
e
x
e
x
secxtanx secxsec2x tanx
cosecxcotx cosecx
cosec2x cotx
Integration us ing trigonometric identities
Example: Find dxx2
cot .
Solution: cot2x = cosec
2x 1
= dxxdxx 1coseccot 22
= cotx x + c.
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C4 14/04/13 SDB14
Example: Find .sin 2 dxx
Solution: sin2x = (1 cos 2x)
= dxxdxx )2cos1(sin 212 = x sin 2x + c.
You cannotchangex to 3xin the above result to find dxx3sin2
. see next example
Example: Find dxx3sin 2 .
Solution: sin23x = (1 cos 2 3x) = (1 cos 6x)
= dxxdxx )6cos1(3sin 212 = x
1/12sin 6x + c.
Example: Find sin 3xcos 5x dx.
Solution: Using the formula 2 sinAcosB = sin(A+B) + sin(AB)
sin 3xcos 5x dx
= 1/2sin 8x + sin (2x) dx = 1/2sin 8x sin 2x dx= 1/16cos 8x + cos 2x + c.
Integration by reverse chain rule
Some integrals which are not standard functions can be integrated by thinking of the chainrule for differentiation.
Example: Find dxxx 3cos3sin 4 .
Solution: dxxx 3cos3sin 4
If we think of u= sin 3x, then the integrand looks like dx
du
u
4
if we ignore theconstants, which would integrate to give
1/5u
5
so we differentiate u5 = sin
53x
to give ( ) ( ) xxxxxdx
d3cos3sin153cos33sin53sin 445 ==
which is 15 times what we want and so
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C4 14/04/13 15
dxxx 3cos3sin 4 = cx +3sin 5151
Example: Find dxxx
)32( 2
Solution: dxxx
)32( 2
If we think of u= (2x2 3), then the integrand looks like
dx
du
u
1if we ignore the
constants, which would integrate to ln u
so we differentiate ln u = ln 2x2 3
to give ( )2 2 21 4
ln 2 3 42 3 (2 3)
d xx x
dx x x = =
which is 4 times what we want and so
dx
x
x
)32( 2 = ln 2x2 3 + c.
In general cxfdxxf
xf+= )(ln)(
)('
Integrals of tan xand cot x
tanx dx = dxxx
cos
sin =
dx
x
x
cos
sin,
and we now have cxfdxxf
xf+=
)(ln
)(
)('
tanx dx = ln cosx + c
tanx dx = ln secx + c
cotxcan be integrated by a similar method to give
cotx dx = ln sinx + c
Integrals of sec xand cosec x
secxdx = dxxx
xxxdx
xx
xxx +
+=
++
tansec
tansecsec
tansec
)tan(secsec 2
The top is now the derivative of the bottom
and we have cxfdxxf
xf+= )(ln)(
)('
secxdx = ln secx+ tanx + c
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C4 14/04/13 SDB16
and similarly
cosecxdx = ln cosecx+ cotx + c
Integration using partial fractions
For use with algebraic fractions where the denominator factorises.
Example: Find
+dx
xx
x
2
6
2
Solution: First express2
62 +xx
x in partial fractions.
2
62 +xx
x
21)2)(1(
6
++
+ x
B
x
A
xx
x
6x A(x+ 2) + B(x 1).
put x = 1 A= 2,
put x = 2 B= 4
++=+ dxxxdxxx
x
2
4
1
2
2
62
= 2 ln |x 1| + 4 ln |x+ 2| + c.
Integration by substitution, indefinite
(i) Use the given substitution involving x and u, (or find a suitable substitution).
(ii) Find eitherdx
du or
du
dx, whichever is easier and re-arrange to find dx in terms
of du, i.e dx= ... du(iii) Use the substitution in (i) to make the integrand a function of u, and use your
answer to (ii) to replace dx by ...... du.
(iv) Simplify and integrate the function of u.
(v) Use the substitution in (i) to write your answer in terms of x.
Example: Find dxxx 53 2 using the substitution u= 3x2 5.
Solution: (i) u= 3x2 5
(ii) x
dudxxdx
du
66 ==
(iii) We can see that there an x will cancel, and 3 5 dxxx 53 2 = x
duux
6= du
u
6
(iv) = duu 21
6
1 = c
u+
23
2
3
6
1
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C4 14/04/13 17
(v) = cx
+
9
)53( 23
2
Example: Find + dxx 211
using the substitution x = tan u.
Solution: (i) x = tan u.
(ii) udu
dx 2sec= dx = sec2u du.
(iii) + dxx 211
= + duuu2
2sec
tan1
1
(iv) = duu
u2
2
sec
sec since 1 + tan2 u = sec2u
= += cudu
(v) = tan1
x + c.
Example: Find
dxx
x
4
3
2 using the substitution u
2=x
2 4.
Solution: (i) u2=x
2 4.
(ii) We know that ( ) dxdu
uudx
d
22
= so differentiating gives
dux
udxx
dx
duu == 22 .
(iii) We can see that an x will cancel and ux = 42 so
dxx
x
4
3
2 = dux
u
u
x3
(iv) = += cudu 33
(v) = cx + 43 2
A justification of this technique is given in the appendix.
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C4 14/04/13 SDB18
Integration by substitution, definite
If the integral has limits then proceed as before but remember to change the limits from
values of x to the corresponding values of u.Add (ii) (a) Change limits from x to u, and
new (v) Put in limits for u.
Example: Find 6
223 dxxx using the substitution u= 3x 2.
Solution: (i) u= 3x 2.
(ii) 3=dx
du
3
dudx =
(ii) (a) Change limits from x to u
x= 2 u= 3 2 2 = 4, and x = 6 u= 3 6 2 = 16
(iii) 6
223 dxxx =
+164
2
1
33
2 duu
u
(iv) = +16
4
2
1
2
3
91 2 duuu
=
16
42
3
2
3
2
5
2
5
91 2
+
uu
(v) = [ ]64102434
52
91 + [ ]832
34
52
91 + = 524 to 3 S.F.
Choosing the substitution
In general put u equal to the awkward bit but there are some special cases where this willnot help.
1
put u =x
2+ 1
put u = x 2 2 5 put u = 2x+ 5 or u2 = 2x+ 5 4 put u or u2 = 4 x2 only ifnis ODD
put x = 2 sin u only ifnis EVEN (or zero)
this makes 4 = 4cos = 2 cos u
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C4 14/04/13 19
There are many more possibilities use your imagination!!
Integration by parts
The product rule for differentiation is
To integrate by parts
(i) choose u anddx
dv
(ii) find v and
dx
du
(iii) substitute in formula and integrate.
Example: Find dxxx sin
Solution: (i) Choose u=x, because it disappears when differentiated
and choose xdx
dvsin=
(ii) u=x and1=dx
du
=== xdxxvxdxdv
cossinsin
(iii) = dxdxdu
vuvdxdx
dvu
dxxx sin = xcosx dxx)cos(1= xcosx + sinx + c.
Example: Find
dxxln
Solution: (i) It does not look like a product,dx
dvu , but if we take u= lnx and
1=dx
dv then
dx
dvu = lnx 1 = lnx
(ii) u= lnx xdx
du 1= and 1=
dx
dv v =x
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(iii) = dxxxxdxx1
ln1ln
= xlnx x + c.
Area under curve
We found in Core 2 that the area under the curve is written
as the integral .We can consider the area as approximately the sum of therectangles shown.
If each rectangle has width x and if the heights of therectangles are y1,y2, ...,yn
then the area of the rectangles is approximately the areaunder the curve
bb
aa
y x y dx
and as x 0 we haveb b
aa
y x y dx
This last result is true for any function y.
Volume of revolution
If the curve of y=f(x) is rotated about the xaxis then the volume of the shape formed can
be found by considering many slices each of width x: one slice is shown.
The volume of this slice (a disc) is approximately y2x
Sum of volumes of all slices from a to b 2b
a
y x
x
x +x
a b
y = f (x)
y
x
y
a b
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and as x0 we have (using the result aboveb b
aa
y x y dx )
2 2b b
aa
y x y dx .
Volume of revolution about the xaxis
Volume when y=f(x), between x= a and x= b, is rotated about the xaxis
is V= b
adxy 2 .
Volume of revolution about the yaxis
Volume when y=f(x) , between y= c and y= d, is rotated about the yaxis
is V= d
cdyx
2 .
Volume of rotation about they-axis is not in the syllabus but is included for completeness.
Parametric integration
When x and y are given in parametric form we can find integrals using the techniques inintegration by substitution.
think of cancelling the dts
See the appendix for a justification of this result.
Example: If x= tan t and y= sin t, find the area under the curve from t= 0 to t =
/4.
Solution: The area = dxy for some limits on x = dtdtdx
y .
We know that y = sin t, and also that
x= tan t tdt
dx 2sec=
area = dxy =dx
y dtdt
24 4
0 0sin sec tan sect t dt t t dt
= =
= [ ] 40sec
t = 12
To find a volume of revolutionwe need dxy2 and we proceed as above writing
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Example: The curve shown has parametric equations
x= t21, y= t3.
The section betweenx= 0 andx= 8 above thex-axis
is rotated about thex-axis through 2radians. Findthe volume generated.
Solution: V = 8
0
2 dxy .
x= 0 t= 1 and x= 8 t= 3,
but the curve is above thex-axis y= t3> 0 t> 0, t = +1, or 3
also y = t3, x= t
21 t
dt
dx2=
V = ==3
1
73
1
62 22 dttdtttdtdt
dxy
=
38
1
28
t
= ( )134
8
Differential equations
Separating the variables
Example: Solve the differential equation xyydx
dy+= 3 .
Solution: xyydx
dy+= 3 = y(3 +x)
We first cheat by separating the xs and ys onto different sides of the equation.
dxxdyy
)3(1 += and then put in the integral signs
dxxdyy += 31
lny = 3x+ 1/2x2 + c.
See the appendix for a justification of this technique.
5 10
10
20
x
y
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Exponential growth and decay
Example: A radio-active substance decays at a rate which is proportional to the mass of thesubstance present. Initially 25 grams are present and after 8 hours the mass hasdecreased to 20 grams. Find the mass after 1 day.
Solution: Let m grams be the mass of the substance at time t.
dt
dmis the rate of increaseof m so, since the mass is decreasing,
dt
dm
=
km
kdt
dm
m=
1
= dtkdmm1
ln m = kt+ ln A
ln = kt ktm Ae= .
When t= 0, m= 25 A= 25
m = 25ekt.
When t = 8, m = 20
20 = 25e8k e8k = 08
8k = ln 08 k = 0027892943
So when t= 24, m = 25e24 0027892943
= 128.
Answer 128 grams after 1 day.
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6 Vectors
Notation
The book and exam papers like writing vectors in the form
a = 3i 4j + 7k.
It is allowed, and sensible, to re-write vectors in column form
i.e. a = 3i 4j + 7k =
7
4
3
.
Definit ions, adding and subtracting, etceteraA vector has both magnitude (length) and direction. If you always think of a vector as atranslation you will not go far wrong.
Directed line segments
The vector is the vector fromA toB,(or the translation which takes A to B).
This is sometimes called the
displacement vector from A to B.
Vectors in co-ordinate form
Vectors can also be thought of as column vectors,
thus in the diagram
=
3
7AB .
Negative vectors
is the 'opposite' of and so =
3
7.
Adding and subtracting vectors
(i) Using a diagram
Geometrically this can be done using a triangle (or a parallelogram):
Adding:
A
B
7
3
A
B
a
b
a+bb
a
a+b
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The sum of two vectors is called the resultant of those vectors.
Subtracting:
(ii) Using coordinates
+
+=
+
db
ca
d
c
b
a and
=
db
ca
d
c
b
a.
Parallel and non - parallel vectors
Parallel vectors
Two vectors are parallel if they have the same direction
one is a multiple of the other.
Example: Which two of the following vectors are parallel?
.1
2,
2
4,
3
6
Solution: Notice that
=
2
4
2
3
3
6 and so
3
6 is parallel to
2
4
but
1
2 is not a multiple of
2
4 and so cannot be parallel to the other two vectors.
Example: Find a vector of length 15 in the direction of
34
.
Solution:a =
3
4 has length a= a = 534 22 =+
and so the required vector of length 15 = 3 5 is 3a= 3
3
4 =
9
12.
b
aab
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Non-parallel vectors
If a and b are not parallel and if a + b = a + b, then
a a = b b ( )a = ( )b
but a and b are not parallel and one cannot be a multiple of the other
( ) = 0 = ( )
= and = .
Example: If a and b are not parallel and if
b+ 2a + b = a + 3b 5a, find the values of and .
Solution: Since a and b are not parallel, the coefficients of a and b must balance out
2 = 5 = 7 and 1 + = 3 = 2.
Modulus of a vector and unit vectors
Modulus
The modulusof a vector is its magnitude or length.
If 73 then the modulus of 7 3 58Or, if c = 35 then the modulus of c = c = c= 3 5 34
Unit vectors
A unit vectoris one with length 1.
Example: Find a unit vector in the direction of
5
12.
Solution: a =
5
12 has length a= a = 13512 22 =+ ,
and so the required unitvector is 131 a =
=
135
1312
131
5
12.
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Position vectors
If A is the point (1, 4) then the position vector of A is the vector from the origin to A,
usually written as = a=
4
1.
For two points A and B the position vectors are
= a and = bTo find the vector go from AO Bgiving
= a+b = b a
Ratios
Example: A,B are the points (2, 1) and (4, 7). Mlies onAB in the ratio 1 : 3. Find thecoordinates of M.
Solution : 26 = 26 0 51 5 21 0 51 5 2 52 5 Mis (25, 25)
x
B
A
O
b
a
ba
O
B
A
M
1
3
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Proving geometrical theorems
Example: In a triangle OBClet M and N be the midpoints of OB and OC.
Prove that BC= 2MN and that BC is parallel to MN.
Solution: Write the vectors as b, and as c.Then = = band = = c.To find , go fromMto Ousing b andthen from O toNusing c
= b + c = c bAlso, to find , go fromB to O using b andthen from O to C using c
= b +c=cb.But = b + c = (cb) = BC
BC is parallel to MN
andBC is twice as long as MN.
Example: P lies on OAin the ratio 2 : 1, and Qlies on OBin the ratio 2 : 1. Prove thatPQis parallel toABand that PQ=
2/3
AB.
Solution: Let a= , and b= ba a, band = 2/3b2/3a
=2/3 (b a)
= 2/3 PQis parallel toABand
PQ= 2/3AB.
B
C
O
c
b
O
P
A
Q
B
1
2
2
1
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Three dimensional vectors
Length, modulus or magnitude of a vector
The length, modulus or magnitude of the vector 1
2
3
a
a
a
=
is
= a 2 2 2`1 2 3
a a a a= = + + ,
a sort of three dimensional Pythagoras.
Distance between two points
To find the distance between A, (a1, a2, a3) and B, (b1, b2, b3) we need to find the length of
the vector . = b a 1 1 1 12 2 2 23 3 3 3
b a b a
b a b a
b a b a
= =
= AB = ( ) ( ) ( )2 2 21 1 2 2 3 3b a b a b a + + Scalar product
a.b = ab cos
where a and b are the lengths of a and b
and is the angle measured from a to b.
Note that (i) a . a = aa cos 0o = a
2
(ii) a.(b+c) = a.b + a.c.(iii) a . b = b . a since cos = cos ()
In co-ordinate form
a.b = 22112
1
2
1 . babab
b
a
a+=
= ab cos
or a.b = 332211
3
2
1
3
2
1
. bababab
b
b
a
a
a
++=
= ab cos .
a
b
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Perpendicular vectors
If a and b are perpendicular then = 90o and cos = 0
thus a perpendicular tob a.b = 0
and a.b = 0 eithera is perpendicular to b or a or b = 0.
Example: Find the values of so that a= 3i 2j+ 2k and
b= 2i+ j+ 6k are perpendicular.
Solution: Since a and b are perpendicular a.b = 0
23 2
2 0 6 2 12 0
2 6
. = + =
2= 9 = 3.
Example: Find a vector which is perpendicular to a,
2
1
1
, and b,
1
1
3
.
Solution: Let the vector c,
r
q
p
, be perpendicular to both a and b.
r
q
p
.
2
1
1
= 0 and
r
q
p
.
1
1
3
= 0
pq+ 2r = 0 and 3p+ q+ r = 0.
Adding these equations gives 4p+ 3r = 0.
Notice that there will never be a unique solution to these problems, so havingeliminated one variable, q, we find p in terms of r, and then find q in terms of r.
4
5
4
3 rq
rp =
=
c is any vector of the form
r
r
r
4
5
4
3
,
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and we choose a sensible value of r = 4 to give
4
5
3
.
Angle between vectors
Example: Find the angle between the vectors
= 4i 5j+ 2k and = i+ 2j 3k, to the nearest degree.Solution: First re-write as column vectors (if you want)
a =
2
5
4
and b=
3
2
1
a= a= 5345254 222 ==++ , b = b= 14321 222 =++
and a.b =
2
5
4
.
3
2
1
= 4 10 6 = 20
a.b = ab cos 20 = 3 5 14 cos
cos =703
20 = 0.796819
= 143o to the nearest degree.
Vector equation of a straight line
r=
z
y
x
is usually used as the position vector of a
general point,R.
In the diagram the line l passes through the point A
and is parallel to the vector b.
To go from OtoRfirst go toA, usinga, and thenfromAtoR using some multiple ofb.
l
O
a
R
b
A
r
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The equation of a straight line through the pointAand parallel to the vector b is
r = a+ b.
Example: Find the vector equation of the line through the pointsM, (2, 1, 4),
andN, (5, 3, 7).
Solution: We are looking for the line throughM (orN) which is parallel to the vector . = nm =
=
3
4
7
4
1
2
7
3
5
equation is r =
+
3
4
7
4
1
2
.
Example: Show that the pointP, (1, 7, 10), lies on the line
r =
+
3
2
1
4
3
1
.
Solution: Thexco-ord of P is 1 and of the line is 1
1 = 1 = 2.
In the equation of the line this gives y= 7 and z= 10
P, (1, 7, 10) does lie on the line.
Intersection of two lines
2 Dimensions
Example: Find the intersection of the lines
1l , r = 22 1 1 1
, and , 3 2 3 1
+ = + l r .
Solution: We are looking for values of and which give the same x and yco-ordinates on each line.
Equatingxco-ords 2 = 1 +
equatingyco-ords 3 + 2= 3
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Adding 5 + = 4 = 1 = 2
lines intersect at (3, 1).
3 Dimensions
This is similar to the method for 2 dimensions with one important difference you can notbecertain whether the lines intersect without checking.
You will always (or nearly always) be able to find values of and by equatingxcoordinates andycoordinates but the zcoordinates might or might not be equal andmust be checked.
Example: Investigate whether the lines
l1, r =
+
1
2
1
3
1
2
and l2, r =
+
1
3
1
5
1
3
intersect
and if they do find their point of intersection.
Solution: Ifthe lines intersect we can find values of and to give the same x,y and zcoordinates in each equation.
Equating xcoords 2 = 3 + , I
equating ycoords 1 + 2 = 1 + 3, II
equating zcoords 3 + = 5 + . III
2 I+ II 5 = 5 + 5 = 2, in I = 3.
We must now check to see if we get the same point for the values of and
In l1, = 3 gives the point (1, 7, 6);
in l2, = 2 gives the point (1, 7, 7).
Thexandyco-ords are equal (as expected!), but thezco-ordinates are different and sothe lines do notintersect.
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7 Appendix
Binomial series (1 + x)n for any n proof
Suppose that
f(x) = (1 +x)n = a+ bx+ cx2+ dx3+ ex4+
putx= 0, 1 = a
f(x) = n(1 +x)n 1= b+ 2cx+ 3dx2+ 4ex3+
put x= 0, n= b
f(x) = n(n1)(1 +x)n 2= 2c+ 3 2dx+ 4 3ex2+
put x= 0,
n(n
1) = 2c
! f(x) = n(n1)(n2)(1 +x)n3= 3 2d+ 4 3 2ex+ put x= 0, n(n1)(n2) = 3 2d
!
Continuing this process, we have
! and ! , etc.
giving f(x) = (1 +x)n
= 1 + nx+
! + ! + ! + ! + + ! +
Showing that this is convergent for x< 1, is more difficult!
Derivative of xq for q rational
Suppose that q is any rational number, , where r and s are integers, s 0.Then y= x
q = ys=xr
Differentiating with respect to x s = r xr1
= since q=
= qx q1 sinceys=xrand y= xq
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which follows the rule found for xn, where n is an integer.
for negative limits
We know that the area under any curve, fromx= atox= bis approximately
If the curve is above thex-axis, all theyvalues are positive, and if a< bthen all values of xare positive, and so the integral is positive.
1
ln|x|
ln ln
Example: Find
3
1
1dx
x.
Solution: The integral wanted is shown as A
in the diagram.
By symmetry A=A (Apositive)
and we need to decide whether the integral
is +A or A.
From x=
1 to x=
3, we are going inthe direction ofxdecreasing
all xare negative.
And the graph is below thex-axis,
theyvalues are negative, y x is positive
0 0 the integral is positive and equal toA.
The integral, A =A = 3
1
1dx
x= [ ] 31lnx = ln 3 ln 1 = ln 3
A=3
11 dxx
= ln 3
Notice that this is what we get if we write ln x in place of lnx
3
1
1dx
x= [ ] 31ln | |x
= ln 3 ln 1 = ln 3
As it will always be possible to use symmetry in this way, since we can never have onepositive and one negative limit (because there is a discontinuity atx= 0), it is correct to write
ln xfor the integral of 1/x.
x
, 0 a b
y
3 2 1 1 2 3
1
1
A
A'
y=1/x
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Separating the variables why it works
We show this with an example.
If y= 6y3
then
= 18y2
and so 18 18 6 Notice that we cancel the dx.
Example: Solve = x2secy
Solution: = x2secy
cosy
= x2
cos cos cancelling the dx siny = + c
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Index
Binomial expansion
convergence, 8
for any n, 8
proof - for any n, 34
Differential equations
exponential growth and decay, 23
forming, 12
separating the variables, 22
Differentiation
derivative of xq for q rational, 34
exponential functions, ax, 11
related rates of change, 12
Implicit differentiation, 10
Integration1/xfor negative limits, 35
by parts, 19
by substitution why it works, 36choosing the substitution, 18
ex, 13
lnx, 13
parametric, 21
parametric why it works, 36
reverse chain rule, 14
secxand cosecx, 15
separating variables why it works, 37
standard integrals, 13
substitution, definite, 18
substitution, indefinite, 16
tanxand cotx, 15
using partial fractions, 16
using trigonometric identities, 13
volume of revolution, 20
Parameters
area under curve, 7
equation of circle, 6
equation of hyperbola, 6
intersection of two curves, 5
parametric equations of curves, 5
parametric to Cartesian form, 6
trig functions, 6
Parametric differentiation, 11
Parametric integration, 36
areas, 21
volumes, 21
Partial fractions, 3
integration, 16
linear repeated factors, 4
linear unrepeated factors, 3
quadratic factors, 3
top heavy fractions, 4
Scalar product, 29
perpendicular vectors, 30
properties, 29
Vectors, 24
adding and subtracting, 24
angle between vectors, 31
displacement vector, 24distance between 2 points, 29
equation of a line, 31
intersection of two lines, 32
length of 3-D vector, 29
modulus, 26
non-parallel vectors, 26
parallel vectors, 25
position vector, 27
proving geometrical theorems, 28
ratios, 27
unit vector, 26