AQA – Further Pure 3 – Key points and Papers
Name: ……............................................................................... Tutor group: .............................
Key dates Further pure 3 exam:
10th June 2013 am
Term dates: Term 1: Monday 3 September 2012 - Wednesday 24 October 2012 (38 teaching days)
Term 4: Monday 18 February 2013 - Friday 22 March 2013 (25 teaching days)
Term 2: Monday 5 November 2012 - Friday 21 December 2012 (35 teaching days)
Term 5: Monday 8 April 2013 - Friday 24 May 2013 (34 teaching days)
Term 3: Monday 7 January 2013 - Friday 8 February 2013 (25 teaching days)
Term 6: Monday 3 June 2013 - Wednesday 24 July 2013 (38 teaching days)
AQA - FP3
2
AQA - FP3
Scheme of Assessment Mathematics Advanced Subsidiary (AS) Advanced Level (AS + A2) The Scheme of Assessment has a modular structure. The A Level award comprises four compulsory Core units, one optional Applied unit from the AS scheme of assessment, and one optional Applied unit either from the AS scheme of assessment or from the A2 scheme of assessment. For the written papers, each candidate will require a copy of the AQA Booklet of formulae and statistical tables issued for this specification. All the units count for 331/3% of the total AS marks 162/3% of the total A level marks Written Paper 1hour 30 minutes 75 marks Grading System
The AS qualifications will be graded on a five-point scale: A, B, C, D and E. The full A level qualifications will be graded on a six-point scale: A*, A, B, C, D and E. To be awarded an A* in Further Mathematics, candidates will need to achieve grade A on the full A level qualification and 90% of the maximum uniform mark on the aggregate of the best three of the A2 units which contributed towards Further Mathematics. For all qualifications, candidates who fail to reach the minimum standard for grade E will be recorded as U (unclassified) and will not receive a qualification certificate.
Further pure 3 subject content Series and limits Differential equations Polar coordinates
AQA - FP3
3
AQA - FP3
Further pure 3 specifications Candidates will be expected to be familiar with the knowledge, skills and understanding implicit in the modules Core 1, Core 2, Core 3, Core 4 and Further Pure 1. Candidates may use relevant formulae included in the formulae booklet without proof.
Series and limits Maclaurin series Expansions of ex, ln(1+ x), cos x and sin x, and (1+ x)n for rational values of n.
Knowledge and use, for k > 0, of lim x k e − x as x tends to infinity and lim x k ln x as x tends to zero. Improper integrals.
Use of the range of values of x for which these expansions are valid, as given in the formulae booklet, is expected to determine the range of values for which expansions of related functions are valid;
1+ x 2 x e.g ln ; (1 − 2 x) e . 1− x 1
e
E. g. ∫ x ln x dx ,
∫
0
∞
0
xe − x dx.
Candidates will be expected to show the limiting processes used
Use of series expansion to find limits.
x2 e x ex −1 Sin(3x ) 2+ x − 2 ; lim ; lim ; lim x →0 x 0 x 0 x 0 → → → x x cos(2 x) − 1 x
E.g.lim
Differential equations The concept of a differential equation and its order. Boundary values and initial conditions, general solutions and particular solutions.
The relationship of order to the number of arbitrary constants in the general solution will be expected.
Differential Equations. First Order
Analytical solution of first order linear differential equations of the dy + Py = Q where P and Q form dx are functions of x .
To include use of an integrating factor and solution by complementary function and particular integral.
Numerical method
Numerical methods for the solution of differential equations dy = f ( x, y ) . of the form dx Euler’s formula and extensions to second order methods for this first order differential equation.
Formulae to be used will be stated explicitly in questions, but candidates should be familiar with standard notation such as used in Euler’s formula yr +1 = yr + hf ( xr , yr ) , the formula yr +1 = yr −1 + 2hf ( xr , yr ) , 1 and the formula yr +1 = yr + (k1 + k2 ) 2 where k1 = hf ( xr , yr ) and k2 = hf ( xr + h, yr + k1 )
AQA - FP3
4
AQA - FP3
Differential equations .Second Order
Solution of differential equations d2y dy of the form a 2 + b + cy = 0, dx dx where a, b and c are integers, by using an auxiliary equation whose roots may be real or complex. Solution of equations of the form d2y dy a 2 + b + cy = f ( x) where a, dx dx b and c are integers by finding the complementary function and a particular integral. Solution of differential equations d2y dy + P + Qy = R of the form: 2 dx dx where P, Q, and R are functions of x. A substitution will always be given which reduces the differential equation to a form which can be directly solved using the other analytical methods in this specification or by separating variables.
Including repeated roots.
Finding particular integrals will be restricted to cases where f(x)is of the form ekx , cos(kx) , sin(kx) or a polynomial of degree at most 4, or a linear combination of any of the above.
Level of difficulty as indicated by:-
d2y − 2y = x use the substitution x = et dx 2 d 2 y dy − − 2 y = et to show that 2 dt dt (a) Given x
2
Hence find y in terms of t Hence find y in terms of x
d2y dy dy =0 (b) (1 − x ) 2 − 2 x use the substitution u = dx dx dx du 2 xu A = to show that and hence that u = where A is an 2 dx 1 − x 1 − x2 2
arbitrary constant. Hence find y in terms of x. Polar Coordinates
Relationship between polar and Cartesian coordinates.
The convention r > 0 will be used. The sketching of curves given by equations of the form r = f(θ ) may be required. Knowledge of the formula tan φ = r
dθ is not required. dr
Use of the formula area β 1 = ∫ r 2 dθ . α 2
AQA - FP3
5
AQA - FP3
The formulae booklet
AQA - FP3
6
AQA - FP3
AQA - FP3
7
AQA - FP3
AQA - FP3
8
AQA - FP3
AQA - FP3
9
AQA - FP3
Content Series and limits ........................................................................................................................ 11 Series............................................................................................................................................................................ 11 Limits............................................................................................................................................................................ 12
Differential equations ............................................................................................................... 17 Generalities and definitions .......................................................................................................................................... 17 First order linear differential equation .......................................................................................................................... 18 Auxiliary equations method. ......................................................................................................................................... 19 Second order linear differential equations .................................................................................................................... 24 Numerical methods to solve first order differential equation ........................................................................................ 29
Polar coordinates ...................................................................................................................... 33 Past papers ................................................................................................................................ 37
AQA - FP3
10
AQA - FP3
Series and limits Series MacLaurin's series The function f ( x) and all its derivatives exist at x = 0 The Maclaurin series for a function f ( x ) is given by: f ''(0) 2 f '''(0) 3 f ( r ) (0) r x + x + ... + x + ... 2! 3! r! where f ', f '', f ''',... denote the first, second, third,... derivatives of f , respectively. f ( x ) = f (0) + f '(0) x +
f ( r ) is the derivative of order r.
Range of validity Some series are valid for all values of x ∈ ¡, but some series are valid for only some values of x. Refer to the formulae book to find the range of validity. For example:
ln(1 + x) = x −
x 2 x3 + − ... is valid for − 1 < x ≤ 1 2 3
Multiplying and composing Maclaurin's series f ( x) and g ( x) are two functions • The Maclaurin's series of the function f × g ( x) is the product the two maclaurin's series. • To obtain the Maclaurin's series of the function f ( g ( x)), substitute x in the Maclaurin's series f by the Maclaurin's series of g ( x). x2 x3 + ... and sin( x) = x − + ... 2 6 2 x x3 •The maclaurin's series of e x sin( x) = (1 + x + + ...)( x − + ...) 2 6 3 4 3 x x x3 x 4 x = x − + x 2 − + + ... = x − x 2 + − + ... 6 6 2 3 6 Examples : e x = 1 + x +
2
•The maclaurin's series of e
1 x3 x3 = 1 + x − + ... + x − + ... + .. 6 6 2 2 4 x x = 1 + x + − + ... 2 8
sin( x )
Maclaurin's series and limits If a function f is not defined when x = 0, we study the value of the function when x is very close to 0. If a value exists, it is called the limit of f ( x) when x tends to 0. • When the limit is not obvious, work out the Maclaurin's series of the function and substitute x by 0 in the series (if possible) to obtain the limit. Example: ex − 1 lim ( e x − 1) = 0 and lim( x) = 0. Not only the function f is not defined at x = 0, x →0 x →0 x but also its limit when x tends to 0 f ( x) =
0 1 x2 x can not be determined (" ").The Maclaurin's series of f is f ( x) = 1 + x + + ... − 1 = 1 + + ... x 0 2 2 0 So when x tends to 0, f ( x) tends to 1+ + ... = 1 or lim f ( x ) = 1 x →0 2 AQA - FP3
11
AQA - FP3
Limits Limits you have to know: You are allowed to use the following results without proof: •when x → ∞ , x k e− x → 0 for any real number k . •when x → 0 , x k ln( x) → 0 for k > 0.
Improper integrals The integral
∫
b
a
f ( x)dx is said IMPROPER if
a) the interval of integration is infinite, or
b) f ( x ) is not defined at one or both of the end points x = a and x = b.
Method To work out if an improper integral has a value or not (exists or not) 1) Replace "∞" or "a", the value where f is not defined, by a letter. "N" for example. 2) Integrate to find an expression in terms of "N". 3) Work out the limit of this expression when "N" tends to "∞" or "a". 4) If the limit exists then the improper integral has a value. If the limit is "∞", the improper integral does not exist. ∞
1 dxis an improper integral. 1 + x2 N 1 N Let's work out ∫ dx = [ Arc tan( x )]0 = Arctan(N) − Arctan(0) 0 1 + x2
Example: ∫
0
Arctan(0) = 0 and when N → ∞, Arctan(N) → conclusion : ∫
∞
0
AQA - FP3
1 dx exists and 1 + x2
∫
∞
0
π
2
.
1 π dx = 2 1+ x 2
12
AQA - FP3
AQA - FP3
13
AQA - FP3
2
4
(ii) Evaluate
f
1
³
4. (i) Show that
1 dx . (2 x 1)(3x 2)
§ 2a 1 · 2 lim ¨ ¸ = . a of 3a 2 © ¹ 3
§ sec x cos x · (ii) Hence find lim ¨ ¸. x o0 x2 © ¹
cos x 1
x x ...... , 2! 4! or otherwise, show that the first three non-zero terms in the expansion, in ascending powers of x, of sec x are x 2 5x 4 1 . 2 24
3. (i) By using the expansion
2. Find the terms in the series expansion of ln 3 x in ascending powers of x, up to and including the term in x3 and state the range of values of x for which this expansion is valid.
(ii) Hence, by using Maclaurin’s theorem, show that the first four terms in the S· § expansion, in ascending powers of x, of sin ¨ x ¸ are 6¹ © 1 3 1 2 3 3 x x x . 2 2 4 12
1. (i) Given that y
d3 y dy d 2 y S· § , and . sin ¨ x ¸ , find 2 dx dx 6¹ dx 3 ©
Chapter assessment
Series and Limits
x o0
(ii) Find lim x
x2 . (e 1)(1 cos x)
7. The function f is defined by f(x) =
1 . ex 1 (i) Use Maclaurin’s theorem to show that when f(x) is expanded in ascending powers of x: 1 1 (a) the first two terms are x 2 4 (b) the coefficient of x2 is zero.
(ii) Determine the range of values of x for which the series expansion for cos 6x ln(1 + 6x) is valid.
6. (i) Find the first four non-zero terms in the series expansion of cos 6x ln(1 + 6x) in ascending powers of x.
x of
5. Find lim (2 x)2 e x .
AQA - FP3
14
AQA - FP3
S
©
6¹
S cos §¨ x ·¸
S sin §¨ x ·¸ 6¹ ©
cos §¨ x ·¸ 6¹ ©
S
sin §¨ x ·¸ 6¹ ©
1 2
S f ƍƍƍ(0) = cos §¨ ·¸ ©6¹
3 2
3 S f ƍ(0) = cos §¨ ·¸ ©6¹ 2 1 S f Ǝ(0) = sin §¨ ·¸ 2 ©6¹
S f(0) = sin §¨ ·¸ ©6¹
sin §¨ x ·¸ 6¹ ©
S
S
¸ sin x cos cos x sin 6¹ 6 6
S·
and then using the series expansions for sin x and cos x that appear in the AQA formulae booklet.
writing sin ¨ x
§ ©
If the question had not stated the method to be used, this series expansion could have been obtained by
1 3 1 3 3 x x2 x +… 2 2 4 12
x2 x3 f(x) = f(0) + xf ƍ(0) + f ƍƍ(0) + f ƍƍƍ(0) + … Maclaurin’s series 2 6 S 1 3· § 3 · x2 § 1 · x3 § sin §¨ x ·¸ x¨ ¸ 2 ¨ 2 ¸ 6 ¨ 2 ¸ ... 6¹ 2 2 © © ¹ © ¹ © ¹
S
S (ii) Let y = f(x) = sin §¨ x ·¸ 6¹ ©
dx 3
dx 2 d3 y
dx d2 y
dy
1. (i) y
Solutions to Chapter assessment
Series and Limits
ln 3 3
x
18
x2
81
x3
1 2
x2
1 24
x4 ... 1
§ sec x cos x (ii) ¨ x2 ©
Question 3(a) includes the phrase ‘...or otherwise…’. This means that you could use other methods as an alternative to the series expansion for cos x. For example, this series for sec x could be found using Maclaurin’s theorem.
(1 + y)1 1 y + y2 … § x2 x4 · ¸ with y ¨ © 2 24 ¹
|1
|
6
x
2
x2
terms in x 6 and higher powers
Dividing both the numerator and denominator by the common factor. (This crucial step must be clearly shown)
terms in x 4 and higher powers
6
x4
x2 5 x4 · § x2 x4 · § ¨ 1 2 24 ... ¸ ¨ 1 2 24 ... ¸ · © ¹ © ¹ ¸| x2 ¹ x2
x 3
using the binomial expansion
with X
using series for ln(1 + X)
using log law
x2 x4 º ª | «1 .... 2 24 »¼ ¬ ° § x 2 x 4 · § x 2 x 4 · 2 °½ | ®1 ¨ ¸ ¨ ¸ ...¾ °¯ © 2 24 ¹ © 2 24 ¹ °¿ 2 4 4 x x x ½ | ®1 ...¾ 2 24 4 ¯ ¿ x2 5 x4 sec x | 1 2 24
3. (i) sec x |
i.e. 3 x d 3
x 3
...
x ln 3 §¨ 1 ·¸ 3¹ © x ln 3 ln §¨ 1 ·¸ 3¹ © x 1 x 2 1 x 3 ln 3 §¨ ·¸ §¨ ·¸ ... 3 2©3¹ 3© 3¹
The expansion is valid for 1 d 1
2. ln 3 x
AQA - FP3
15
AQA - FP3
As xĺ 0,
f
· ¸ ¸ ¸ ¹
f
a
a
t of
lim (2 t )2 e t
t of
lim 4e t 4te t t 2 e t
t of
5. Using the substitution t x leads to t o f when x o f so lim (2 x )2 e x becomes lim (2 t )2 e t
§ 10 · ln ¨ ¸ © 9¹
2a 1 · ª § 3 ·º lim «ln §¨ ¸ ln ¨ ¸ » a of ¬ © 5 ¹¼ © 3a 2 ¹ 2a 1 · º ª § §3· ln «lim ¨ ¸ ln ¨ ¸ ©5 ¹ ¬ a of © 3a 2 ¹ »¼ 2· 3· § § ln ¨ ¸ ln ¨ ¸ ©3¹ ©5 ¹
ª § 2 x 1 ·º lim «ln ¨ ¸» a of ¬ © 3 x 2 ¹¼ 1
a of
lim >ln 2 x 1 ln(3 x 2)@ 1
1
³
3 · § 2 ¨ ¸ dx © 2 x 1 3x 2 ¹ a§ 2 3 · lim ³ ¨ ¸ dx a of 1 © 2 x 1 3x 2 ¹
§2 1 ¨ a lim ¨ 2 a of ¨3 a © 20 30 2 3
1 dx (2 x 1)(3 x 2)
x of
1
³
2a 1 · 4. (i) lim §¨ ¸ a of © 3a 2 ¹
(ii)
x2 · § lim ¨ 1 ... x o0 6 ¸¹ ©
x2 ĺ 0 and terms in higher powers of x also ĺ 0 6 § sec x cos x · So lim ¨ ¸ 1 x o0 © x2 ¹
§ sec x cos x · lim ¨ ¸ x o0 © x2 ¹
The interval of integration is infinite so replace the upper limit (f) by a and find the limit as a f. This and the next five lines show the limiting process used.
Using partial fractions
t of
0
1 18 x 2 54 x 4 ... 6 x 18 x 2 72 x 3 324 x 4 ...
f(x) = f(0) + xf ƍ(0) +
7. (i) f(x) =
x2 f ƍƍ(0) + … 2
1 1 f(0) = 2 ex 1 f(x) = (ex + 1)1 f ƍ(x) = 1(ex + 1)2 (ex) = ex (ex + 1)2 1 f ƍ (0) = 4 f Ǝ(x) = ex (ex + 1)2 + 2ex(ex + 1)3 (ex) 1 2 f Ǝ(0) = = 0 4 8
Maclaurin’s series
(product rule)
(chain rule)
The series expansion for ln(1 + 6x) is valid for 1 < 6x 1 ie
1 1 xd 6 6 The series expansion for cos 6x ln(1 + 6x) is valid only for those values of x 1 1 which satisfy both all values of x and x d . 6 6 1 1 So the required range of values is x d . 6 6
(ii) The series expansion for cos 6x is valid for all values of x
ln(1 + x)
Replace x by 6x in the series expansion for
Replace x by 6x in the series expansion for cos x
1 6 x 72 x 3 378 x 4 ... So cos6 x ln 1 6 x | 1 6 x 72 x 3 378 x 4
cos 6 x ln 1 6 x
(6 x )2 (6 x )3 (6 x )4 ... 2 3 4 2 3 4 6 x 18 x 72 x 324 x ...
(6 x )2 (6 x )4 ... 2! 4! 1 18 x 2 54 x 4 ... 1
0
ln 1 6 x (6 x )
6. (i) cos 6 x
x of
so lim (2 x )2 e x
As t o f , e t o 0 and, for k > 0, lim t k e t
1 1 x2 0 ... x §¨ ·¸ 2 © 4¹ 2
(ii)
1 1 x 2 4 2 the coefficient of x is zero.
x
x2
x
the first two terms are
16
1
1 2
1 2
§1 1 · ¨ x ... ¸ lim ¨ 2 4 2 ¸ x o0 ¨¨ 1 x ... ¸¸ 2 24 © ¹
using (i) 1 x2 u (e 1)(1 cos x ) (e 1) (1 cos x ) 1 1 x2 | §¨ x ... ·¸ u 2 x x4 § · ©2 4 ¹ ... ¸ 1 ¨1 2! 4! © ¹ §§ 1 2 1 3 ·· ¨ ¨ x x ... ¸ ¸ x2 2 4 © ¹¸ lim x lim ¨ x o 0 (e 1)(1 cos x ) x o0 ¨ § x 2 x4 · ¸ ¨ ¨ 2 24 ... ¸ ¸ ¹ ¹ © ©
(b)
(a)
So in the expansion of f(x)
f(x) =
The question does not say ‘Hence’ so you may prefer to use the expansion of ex from the formula booklet rather than using the results from part (i).
AQA - FP3
AQA - FP3
Differential equations Generalities and definitions Definitions • A differential equation is an equation involving the derivatives of a function. •The ORDER of a differential equation is the same as the highest order of derivation occuring in the equation. •A differential equation is linear if it is LINEAR in y and the derivative of y. (Any equation containing powers of y and/or its derivative or products of y and/or its derivatives are non-linear)
Solving differential equations •To solve a differential equation is to find all the functions satisfying the equation. All these solutions constitue a FAMILY of solutions. • Solutions that involve ARBRITRARY constants are called GENERAL SOLUTIONS. • A solution which contains NO arbritrary CONSTANT is called a PARTICULAR SOLUTION. • To work out a particular solution, you need initial/boundary conditions: y ( x0 ) = y0
Methods to solve first order differential equations •Method 1:Direct integration This method can be used if the differential equation can be written as dy By integrating both sides, you obtain = f ( x). dx y = ∫ f ( x )dx •Method 2 :Separating variables This method can be used if the differential equation can be written as dy g ( y ) = f ( x). By integrating both sides, you obtain dx
∫
g ( y )dy = ∫ f ( x )dx
•Method 3:Recognising the derivative of a product function This method can be used if the differential equation can be written as dy du y = f ( x), where u is a function of x. + dx dx d Re− write as ( u × y ) = f ( x) and integrate both sides: dx 1 u × y = ∫ f ( x )dx so y = ∫ f ( x)dx u u
AQA - FP3
17
AQA - FP3
First order linear differential equation Standard form A first order linear equation can be re-arrange in the form dy + P( x ) y = Q ( x ) where P ( x ) and Q( x ) are two functions. dx This form is called the STANDARD form the equation.
Integrating factors dy + P( x) y = Q ( x), dx we want to multiple both sides by a function I ( x),
Considering an equation
so that the left-hand side of the equation becomes the derivative of a product function. dy + IP × y = IQ dx
i.e I ×
with
dI = IP dx
Such a function is called an INTEGRATING FACTOR and I ( x) = e ∫
P ( x ) dx
Example : dy 1 − y = x 2 where x > 0 dx x
Find the general solution of the equation •The integrating factor I ( x) = e∫
1 − dx x
= e− ln( x ) = e
1 ln x
=
1 x
•Multiplying the equation by I ( x), it becomes d 1 × y = x and by integrating dx x 1 y = x3 + cx c ∈ ¡ 2
1 dy 1 − y=x this is x dx x 2 1 1 y = x2 + c x 2
Substitution The substitution to use will be given to you in the question. Use this substitution to transform the given differential equation into one which you can solve using either of the known methods: (direct integration, separating variables, integrating factors) Example: 1 dy + xy = xy 2 into a diff. eq in z and x. to transform the diff. eq. y dx b)Solve the new differential equation. a) Use the substitution z =
c) Find y in terms of x. Solution : z =
1 y
1 1 dz dy and =− 2 z dx z dx 1 dz x x dz − 2 + = 2 − xz = − x z dx z z dx
so y =
after substitution, we have
An integrating factor is I ( x ) = e ∫ e
e AQA - FP3
x2 − 2
−
x2 2
dz − xe dx
x2 − 2
z = ∫ − xe
z = − xe
−
x2 2
−
x2 2
=e
−
x2 2
x x − d −2 e z = − xe 2 dx
x2 − 2
dx = e
− xdx
2
+c
2
x2
z = 1 + ce 2 this gives y =
1 x2 2
ce + 1 18
AQA - FP3
Auxiliary equations method. Definitions a
dy + by = f ( x) dx
with a, b ∈ ¡
dy + by = 0. dx The general solution of the reduced equation is called The COMPLEMENTARY FUNCTION
• The REDUCED equation is a
• A PARTICULAR INTEGRAL satisfies the equation a
dy + by = f ( x) dx
dy + by = f ( x) is dx the sum of the complementary function and the particular integral yG = yP + yC
•The general solution of a
Solving first order linear differential equations dy + by = f ( x ) is a differential equation where a and b are real numbers dx dy The reduced eqaution is a + by = 0 dx •The AUXILIARY equation associated with this equation is aλ + b = 0 a
•The complementary function is : y = Ceλ x where λ is solution to aλ + b = 0. •Finding the particular integral: ⊗ if f ( x) is a polynomial then y P is also a polynomial of the same degree ⊗ if f ( x) = ACos (kx ) + BSin(kx ) then yP = aCos (kx) + bSin(kx) a and b to be worked out. ⊗ if f ( x) = Ae then yP = ae if k ≠ λ kx
kx
y P = axe kx if k = λ
where a is to be worked out
•The general solution is yG = yP + yC
AQA - FP3
19
AQA - FP3
AQA - FP3
20
AQA - FP3
dy 2y dx 3 12 x 2 .
4
S
.
(b) Solve this differential equation, given that y = 20 when x = 3.
5. (a) Show that x 2 is an integrating factor for the first-order differential equation 1 dy 2 y 6( x 3 9) 2 dx x
(b) Hence find the general solution of this first order differential equation.
4. (a) Find a particular integral of the differential equation dy 4 y 12 17 cos x dx
given that y = 1 when x
3. By using an integrating factor, find the solution of the differential equation dy 2 y tan x sin 2 x dx
(b) Hence, write down the general solution of
2. (a) Find the complementary function and a particular integral of the differential dy 2 y 3 12 x 2 . equation dx
1. (a) Show that xe 2 x is an integrating factor for the first-order differential equation dy § 2 x 1 · 2 x ¨ ¸y e . dx © x ¹ dy § 2 x 1 · 2 x (b) Hence, find the general solution of ¨ ¸y e . dx © x ¹
Chapter assessment
Introduction to differential equations
7. Solve the differential equation dy 2 x 6 y x 2 ln x dx 1 given that y when x = 1. 25 x>0
(b) Find the general solution of this differential equation.
6. (a) Find the value of the constant k for which kxe 2 x is a particular integral of the differential equation dy 2 y 4e 2 x dx
AQA - FP3
21
AQA - FP3
1
x
³ 2 x
³
dx
dx
e 2 x u e ln x
e 2 x ln x
e
e
2
x2
dx
0 or
dx
dy
2y
The GS of
2 y is y Ae 2 x dx dy so the CF of 2 y 3 12 x 2 is y Ae 2 x . dx Since f(x) = 3 12 x 2 try a PI of the form y = ax2 + bx + c dy Substituting this into 2 y 3 12 x 2 gives dx 2 ax b 2(ax 2 bx c ) 3 12 x 2 Equating coefficients of x2 gives 2a = 12 a = 6 2a 2b = 0 Equating coefficients of x gives b= 6 b 2c = 3 Equating constant terms gives c = 1.5 so a particular integral is y = 6x2 + 6x + 1.5
dy
2y
A
dy
1 2 x A 2 x xe e 2 x
³ x dx
2. (a) Reduced equation
y
yxe
2x
x
§ 2x 1 · 2 x 2 x xe 2 x ¨ ¸ y xe e dx © x ¹ dy (e 2 x 2 xe 2 x )y x dx
dy
d y( xe 2 x ) dx
xe 2 x
(b) xe 2 x
e2 x u x so xe 2 x is an integrating factor.
1. (a) Integrating factor I
2 x 1
Solutions to chapter assessment
Introduction to differential equations
dx
2 tan x dx
³(sec
2
sec 2 x
x 1) dx
, y=1
4
1
A
S
1§ S · ¨1 A ¸ 2© 4 ¹ 1
Solving c + 4b = 17 and 4c b = 0 simultaneously gives b = 4, c = 1 so a particular integral is y = 3 + 4cos x + sin x
4. (a) Since f(x) = 12 + 17cos x try a PI of the form y a b cos x c sin x dy Substituting this into 4 y 12 17 cos x gives dx b sin x c cos x 4(a b cos x c sin x ) 12 17 cos x Equating constant terms gives 4a = 12 a=3 c + 4b = 17 Equating coefficients of cos x gives Equating coefficients of sin x gives 4c b = 0
So y
3 12 x 2 is
sin 2 x sec 2 x
S· § cos x ¨ tan x x 1 ¸ 4¹ © 2
4
S
2
e ln(sec x )
tan x x A cos 2 x tan x x A
When x
y
2y
e 2 ln sec x
e³
2 y tan x sec 2 x
I
d sin 2 x ( y sec 2 x ) dx cos 2 x 2 y sec x ³ tan 2 x dx
sec 2 x
dy
dx
dy
Ae 2 x 6 X 2 6 X 1.5
3. Integrating factor
y
(b) General solution of
AQA - FP3
22
AQA - FP3
dx
dy
4y
0 or
dx
dy
4 y
dx dy
dy
2 xy
x2 u
x
2
dx
2
x
3
9 A
3 2
y
4 3
x
9 108
3 2
(k e 2 x
k
Since the answer is given in the question all these steps should be shown.
Instead of finding the integral by inspection you may prefer to use the substitution u = (x3 + 9)
4
4e 2 x
4 u 63 A 3 108
2 y 4e 2 x gives dx ke 2 x kx 2e 2 x ) 2kxe 2 x 4e 2 x
dy
A
180 = 180
1
3 4 108 x 3 9 2 2 x 3x 2
kxe 2 x
y
yx
3
Substituting this into
6. (a) PI
So
2
When x = 3, y = 20
4 3
1 2
x 2 u 6( x 3 9)
x2
e ln x
6 x 2( x 3 9)2
y
2
³x
e 2 ln x
e
dx 1 d ( yx 2 ) 6 x 2( x 3 9)2 dx 1 yx 2 ³ 6 x 2( x 3 9)2 dx
x2
(b) x 2
5. (a) Integrating factor I
The GS of
4 y is y Ae 4 x dx dy so the CF of 4 y 12 17 cos x is y Ae 4 x . dx General solution is y Ae 4 x 3 4cos x sin x
dy
(b) Reduced equation
x
y
dx 3
dy
2y 0 or dx
dy 2y
dx
dx dy 3
x
3 3x 2y
x3 u
x3 u
3
1 x ln x 2
x3
e ln x
1 4 x ln x 2
y
dx
e 3 ln x
e
3
³x
4 xe 2 x
gives
3 is not a constant we cannot x
dx
dy
Using integration by parts
use the complementary function, particular integral method
Since
x 2 ln x to a form with unitary coefficient of
1 x ln x 2
6y
d 1 4 ( yx 3 ) x ln x dx 2 1 yx 3 ³ x 4 ln x dx 2 1 5 1 1 1 1 x u ln x ³ x 5 u u dx 5 2 5 2 x 1 5 1 x ln x ³ x 4 dx 10 10 1 5 1 5 x ln x x A 10 50
x
x
dy 3
Integrating factor I
dx
dy
7. Rearranging 2 x
dy
dx
dy
2 y is y Ae 2 x dx dy so the CF of 2 y 4e 2 x is y Ae 2 x dx dy 2 y 4e 2 x is y Particular integral of dx General solution is y Ae 2 x 4 xe 2 x The GS of
(b) Reduced equation
AQA - FP3
23
AQA - FP3
So y
1 25
1 25
A
1 2 1 2 3 x ln x x 10 50 50 x 3
When x = 1, y
3 50
0
1 A 50
Second order linear differential equations Definitions a
dy d2y + b + cy = f ( x ) 2 dx dx
with a, b, c ∈ ¡
d2y dy + b + cy = 0. 2 dx dx The general solution of the reduced equation is called The COMPLEMENTARY FUNCTION
• The REDUCED equation is a
• A PARTICULAR INTEGRAL satisfies the equation a
d2y dy + b + cy = f ( x ) 2 dx dx
dy + by = f ( x) is dx the sum of the complementary function and the particular integral
•The general solution of a yG = y P + yC
Solving second order linear differential equations d2y dy + b + cy = f ( x) is a differential equation where a, b and c are real numbers 2 dx dx d2y dy The reduced equation is a 2 + b + cy = 0 dx dx •The AUXILIARY equation associated with this equation is aλ 2 + bλ + c = 0 a
The auxiliary equation is a quadratic equation, three cases are possible: Case1 : aλ 2 + bλ + c = 0 has two distinct solutions λ1 and λ2 The complementary function is y = C1e λ1 x + C2 eλ2 x
C1 , C2 ∈ ¡
Case 2 : aλ 2 + bλ + c = 0 has equal/repeated root λ0 The complementary function is y = (C1 x + C2 )eλ0 x
C1 , C2 ∈ ¡
Case 3: aλ 2 + bλ + c = 0 has two conjugate complex solutions λ1 = p + iq and λ2 = p − iq The complementary function is y = e px ( C1Cos (qx) + C2 Sin(qx) )
C1 , C2 ∈ ¡
•Finding the particular integral: ⊗ if f ( x) is a polynomial then y P is also a polynomial of the same degree ⊗ if f ( x) = ACos (kx) + BSin( kx ) then yP = aCos (kx) + bSin(kx) a and b to be worked out. ⊗ if f ( x) = Ae then yP = ae if k ≠ λ kx
kx
or yP = axe kx if k = λ1 or λ2
where a is to be worked out
or y p = ax e if k = λ0 (the repeated root.) 2 kx
• The general solution is yG = yP + yC
Substitution d2y dy + P ( x) + Q ( x ) y = R ( x) is a differential equation 2 dx dx where P,Q and R are functions of x. Note: this equation is written in its standard form. These equations are solved using substitution. The substitution to use will be given in the question.
AQA - FP3
24
AQA - FP3
AQA - FP3
25
AQA - FP3
2 2
du 1 u dx x
3x .
(a) Show that the substitution u =
[3]
[3]
[3]
[7]
[5]
[3]
[4]
dy transforms this differential equation into dx
dy = 2 when x = 0, find y in terms of x. dx
4. A differential equation is given by d2 y dy x 2 = 3x2 dx dx
(b) Given that y = 14 and
(a) Find p(x).
d2 y dy 4 5y = 12ex has general solution dx dx 2 y = Aex + Be5x + p(x) where p(x) is a particular integral satisfying the given differential equation.
4
.
3. The differential equation
when x =
S
(b) Hence find y in terms of x given that y = 2 when x = 0 and y
2. (a) Find the general solution of the differential equation d2 y + 4y = 8 3sin x. dx 2
(b) Find the general solution of the differential equation d2 y dy +4 + 3y = 9x2 dx dx 2
1. (a) Find the general solution of the differential equation d2 y dy +4 + 3y = 0 dx dx 2
Chapter assessment
Second order differential equations
Given that v xy , show that dy 1 dv v dx x dx x 2 d 2 y d 2 v 2 dv 2v . and x 2 dx dx 2 x dx x 2
(iii) Hence find the general solution of d2 y dy x 2 (2 x) (1 2 x) y dx dx
into
d2 y dy x 2 (2 x) (1 2 x) y dx dx d 2 v dv 2v e2 x . dx 2 dx
(ii) Hence show that the substitution v
5. (i)
[5]
[2]
[4]
e2 x .
e2 x
Total 50 marks
[8]
[3]
xy transforms the differential equation
(c) Hence find the general solution of the differential equation dy d2 y x 2 = 3x2 dx dx giving your answer in the form y = g(x).
(b) Find the general solution of du 1 u 3x dx x giving your answer in the form u = f(x).
AQA - FP3
26
AQA - FP3
Ae 3 x Be x
y
The general solution is y
Ae 3 x Be x
yC yP
dx
3y
a=3 b = 8 c 26 3
4
dy
Ae 3 x Be x
2 3x 2 8 x 8 3
Equating coefficients of x2: 3a = 9 Equating coefficients of x: 8a + 3b = 0 Equating constant terms: 2a + 4b+ 3c = 0 The particular integral is y p 3 x 2 8 x 263
d y
2a
2ax b
ax 2 bx c
dx 2
d2 y
dy 4 3y 0 dx 2 dx gives 2a + 4(2ax + b) + 3(ax2 + bx + c) = 9x2
Substituting into
2
dx 2
dx d2 y
dy
For a particular integral try yP
yc
(b) From part (a), the complementary function of
1. (a) The auxiliary equation is
k 2 4k 3 0 (k 3)(k 1) 0 k 3 or k 1 d2 y dy 4 3y 0 y The general solution of dx 2 dx
Solutions to chapter assessment
9 x 2 is
b sin x c cos x
a b cos x c sin x
8 3 sin x
So y
When x
2 2
2 2 3 2 sin 2 x 2 sin x 2
S , y 4
1 B2 2
(b) y A cos 2 x B sin 2 x 2 sin x When x = 0, y = 2 2=A+2
is y
d2 y
2 sin x
4a = 8 3b = 0 3c = 3
4 y 8 3 sin x dx 2 A cos 2 x B sin 2 x 2 sin x
The general solution of
A particular integral is y p
Equating constant terms: Equating coefficients of cos x: Equating coefficients of sin x:
3 2 2
Since the complementary function does not contain terms of the form a (constant) or terms of the form sin x or cos x the ‘obvious’ particular integral, y = a + b cosx + c sinx is appropriate.
A=0 3 B 2
a=2 b=0 c = 1
b cos x c sin x dx 2 Substitute into the differential equation: gives b cos x c sin x 4(a b cos x c sin x ) 8 3 sin x
dx d2 y
dy
For a particular integral try yP
k2 4 0 k r2i d2 y The complementary function of 4y dx 2 is yc A cos 2 x B sin 2 x
2. (a) The auxiliary equation is
AQA - FP3
27
AQA - FP3
x
or
So
4. (a) u
4 = A + 5 B
14 = A + B
d2 y
du dx dy
dx 2
d2 y
x
3x 2
A = 11, B = 3 11e x 3e 5 x 2 xe x
12e x
Ae x 5 Be 5 x 2e x 2 xe x
Ae x Be 5 x 2 x e x
du u 3 x 2 becomes x dx dx 2 d x du 1 u 3 x as required. dx x
dx
dy
So y
dx When x = 0, y = 14 dy 2 When x = 0, dx
dy
(b) General solution is y
6ae 12e a 2 So p( x ) 2 xe x
x
Substitute into
gives axe x
ae x axe x axe x
ae x axe x
dy 4 5 y 12e x dx 2 dx 2ae x 4(ae x axe x ) 5 axe x
d2 y
dx 2
dx d2 yP
dyP
3. (a) Since Ae x is part of the complementary function, for a particular integral try yP axe x
5/7 x 3 Cx 2 B ,
where B and C are arbitrary constants.
y
This can also be written as
axe x 2ae x
So
is e
³ dx
dx
dy gives
yx
dy
x3
dx
dy
2
A x 2 B
3 x 2 Ax
d2v dv 2v dx 2 d x
e2 x
d2v 2 dv 2v v § 1 dv v · (2 x )¨ 2 ¸ (1 2 x ) e2 x dx 2 x d x x 2 x © x dx x ¹ d2v 2 dv 2v v 2 dv 2v dv v 2 2v e 2 x dx 2 x d x x x dx x 2 dx x x
dx dy v x x dx v 1 dv dx x dx x 2 dy dv yx dx dx d2 y dy d2 y d2v dy dy x x 2 2 2 dx dx dx dx dx dx 2 2 2 d y dv § 1 dv v · 2 ¸ x 2¨ dx 2 dx 2 © x dx x ¹ 2 2 d y d v 2 dv 2v x 2 dx d x 2 x dx x 2
dv dx dv dx dy
xy
Integrating gives y
5. (i) v
(ii)
1
x
1 du 1 u x dx x 2
I
du 1 u 3x dx x 1 1 e ln x e ln x x d §u · 3 ¨ ¸ 3 dx © x ¹ u 3x A x u 3 x 2 Ax
An integrating factor for
(c) Replacing u by
(b)
AQA - FP3
28
AQA - FP3
0
a
e2 x
xy y
v x
General solution for original equation is v
v
Particular integral is v 31 xe 2 x General solution is v Ae 2 x Be x 31 xe 2 x
1 3
3ae 2 x
A e 2 x Be x 1 2 x 3e . x
dv ae 2 x 2axe 2 x dx d2v 2ae 2 x 2ae 2 x 4axe 2 x dx 2 4ae 2 x 4axe 2 x Substituting into differential equation 4ae 2 x 4axe 2 x (ae 2 x 2axe 2 x ) 2axe 2 x e 2 x
k 2 or k 1 Complementary function: v Ae 2 x Be x Particular integral has form v axe 2 x
(k 2)(k 1) 0
d2v dv 2v e 2 x dx 2 d x Auxiliary equation: k 2 k 2
(iii)
Numerical methods to solve first order differential equation In this chapter, we want to solve equations which can be written dy = f ( x, y ) and y ( x0 ) = y0 dx There are three methods to solve numerically this equation.
Formulae to be used will be stated explicitly in questions. Knowing P0 ( x0 , y0 ), we work out P1 then P2 then P3 etc.
Euler's formula To work out Pr +1 , we consider that the gradient of the line Pr Pr +1 is (approx.) equal to the gradient at Pr . This gives: yr +1 = yr + hf ( xr , yr )
The mid-point formula We consider that the gradient of the line Pr −1 Pr +1 is (approx.) equal to the gradient at Pr : This gives
yr +1 = yr −1 + 2hf ( xr , yr )
The improved Euler's formula We consider that the gradient of the line Pr Pr +1 is ( approx.) the mean of the gradient at Pr and the gradient at Pr +1. This gives :
h yr +1 = yr + f ( xr , yr ) + f ( xr +1 , yr*+1 ) 2 with yr*+1 = yr + hf ( xr , yr )
Or as it is given in the exam question: 1 yr +1 = yr + ( k1 + k 2 ) 2 where k1 = hf ( xr , yr ) and k2 = hf ( xr + h, yr + k1 ) Possible layout for your workings out:
AQA - FP3
29
AQA - FP3
AQA - FP3
30
AQA - FP3
[4]
f(x, y) = x 2 y 2 y (1) = 0.5
(a) Use the Euler formula y r 1 y r h f xr , y r with h = 0.1 to obtain an approximation to y (1.1) giving your answer to four
where and
3. The function y(x) satisfies the differential equation dy f ( x, y ) dx
(iii) Use the formula y r 1 y r 1 2h f x r , y r with the same value of h to find y 3 , giving your answer to three decimal places. [5]
1.2156 , [4]
The use of the Euler formula y r 1 y r h f xr , y r with y 0 y (1) 1 gives y1 y (1 h) 1.1 . Determine the value of h that has been used. [4]
(ii) Show that, with this value of h, use of the same Euler formula gives y2 correct to four decimal places.
(i)
2. The function y(x) satisfies the differential equation dy x2 y2 dx
Show that the Euler formula y r 1 y r h f xr , y r with h = 0.2, gives y (0.2) = 1.656 correct to three decimal places.
1. The function y(x) satisfies the differential equation dy f ( x, y ) dx where f(x, y) = sin(x2 + y2) and y (0) = 1.5
Chapter assessment
Numerical methods for the solution of first order differential equations
[5]
Use the formula given in part (b)(i), together with your value for y(1.1) obtained in part (b)(i), to obtain an approximation to y(1.2), giving your answer to three decimal places.
Total 40 marks
(iii) Hence calculate, to two decimal places, the percentage error in the value of y (0.4) obtained in part (i). [2]
dy ( x y ) 2 is y tan( x A) x . dx Given that 0 < A < S, use the boundary condition y(0) = 1 to find the exact value of A. [2]
Use the formula y r 1 y r 1 2h f x r , y r with h = 0.2 and y(0.2) = 1.3085 to find an approximation to y(0.4).
(ii) The general solution of the differential equation
(i)
[5]
[5]
[4]
k1 h f x r , y r where k 2 h f x r h, y r k1 and with h = 0.1 to obtain a further approximation to y(1.1)
Use the formula yr 1 yr 12 k1 k2
4. The function y(x) satisfies the differential equation dy ( x y) 2 dx and also the condition y(0) = 1.
(ii)
(b) (i)
decimal places.
AQA - FP3
31
AQA - FP3
0,
y 0 h sin( x 0 y 0 )
y1
y0
y1
1.369 (to 3 d.p.)
y3
yr h f x r , y r
1.1, h
1.368774...
1.1 2 u 0.05 1.1 2 1.215625 2
y 1 2h x 22 y 22
x 2 y2
y1
x 2 y2
0.05 and f( x , y )
yr 1 2 h f x r , y r
x2
1.215625
1.1 0.05(1.05 2 1.1 2 )
y 1 h x 12 y 12
1.215625 ,
with r = 2 gives
y2
yr 1
1.1,
1.2156 (to 4 d.p.)
Using
(iii) y 1
y2
yr 1
1.1, h
y2
y1
with r = 1 gives
1.05 ,
2h
0.05
0.1
1 h(1 2 1 2 )
1.1
0.05 and f(x, y) = f( x , y )
y0 h x 0 y0
h
1.1
2
2
yr h f x r , y r
1, f( x , y )
Using
(ii) x 1
y3
1,
Using yr 1
2. (i) x 0
x 2 y 2 and y 1
1.5 0.2 sin(0 2 1.5 2 ) 1.6556...
2
2
0.2, f( x , y ) sin( x 2 y 2 ) and y(0.2)
yr h f x r , y r
1.5 , h
yr 1
y0
y (0.2) = 1.656 (to 3 d.p.)
Using
1. x 0
Solutions to chapter assessment
(ii)
(b) (i)
y0
0.5 , h
0.1, f( x , y )
0.618836...
k2
0.618836, h
0.1 1.2 2 0.745048 2 0.1412479...
0.1 f 1.2, 0.745048
h f x 0 h , y 0 k1
0.1 1.1 2 0.618836 2 0.126212...
hf 1.1,0.6188836
hf x 0 , y 0 k1
y0 1.1,
x0
x 2 y2
x 2 y 2 and y(1.1) y 1
0.1, f( x , y )
0.5 21 (0.111803 0.125869)
y 0 21 k1 k2
0.125869...
0.1 1.1 2 0.611803 2
0.1 u f(1.1,0.611803)
h f x 0 h , y 0 k1
0.111803...
0.1 u 1 2 0.5 2
h f x 0 , y0
1,
0.5 0.1 1 2 0.5 2 0.611803...
y 0 h x 02 y02
y(1.1) 0.6188 (to 4 d.p.)
y1
k2
k1
x0
y(1.1) 0.6118 (to 4 d.p.)
x 2 y 2 and y(1.1) y 1
yr h f x r , y r
0.1, f( x , y )
y1
0.5 , h
with r = 0 gives
y0 yr 1
1,
Using
3. (a) x 0
AQA - FP3
32
AQA - FP3
0,
tan( x A ) x
©
S· ¸ x
4¹ S § y(0.4) tan ¨ 0.4 ·¸ 0.4 4¹ © 2.0649627...
y( x ) tan ¨§ x
Percentage error
2.0649927... 1.9102289 u 100 2.0649927... 7.4933... Percentage error = 7.49 (to 2 d.p.)
(iii) Using the exact solution
4
S
1 0.4(0.2 1.3085 )2 1.9102289
y 0 2 u 0.2 u x 1 y 1 2
1.3085 and y(0.4)
yr 1 2 h f x r , y r
y1
When x = 0, y = 1 so 1 = tan A A
(ii) y
y(0.4) 1.910 (to 3 d.p.)
yr 1
y2
Using
0.2,
with r = 1 gives
y0 1, x 1 f( x , y ) ( x y )2
4. (i) x 0
0.752565...
0.618836 21 (0.126212 0.141248)
y 0 21 k1 k2
y(1.2) 0.753 (to 3 d.p.)
y1
y2
Polar coordinates Definitions A point M can be placed in a set of two axis using CARTESIAN coordinates M ( x, y ) This position can also be determied by the distance from the origin O or POLE and the angle made by the line OM with the positive x-axis or INITIAL LINE The POLAR coordiantes of M(r,θ ) In order to have unicity in the polar coordinates of a given point, we will use the following conventions: r > 0 and − π < θ ≤ π or 0 ≤ θ < 2π
Curve in Polar coordinates In cartesian coordinates, an explicit equation of a curve will be given as y = f ( x). In polar coordinates, an explicit equation of a curve will be given as r = f (θ ).
o
Examples : r = 2Sinθ , r = e−2θ , r = 3,...
Conversions A point M has cartesian coordinates M ( x, y ) and polar coordinates M(r,θ ) Using the pythagoras' theorem and trigonometry, we have r 2 = x2 + y 2 θ = x rCos and y y = rSinθ Tanθ = x y y Note :If x > 0, θ = ArcTan( ) but if x < 0, θ = ArcTan( ) ± π x x
Area bounded by a polar curve The are bounded by a polar curve and the ray θ =α and θ =β is
A=
1 β 2 r dθ 2 ∫α
IMPORTANT : This formula is valid only if r > 0 for α ≤ θ ≤ β .
AQA - FP3
33
AQA - FP3
AQA - FP3
34
AQA - FP3
and the point B on C is the point where
Given that P is the point on C where T = D , show that the area of the region bounded by the curve C and the lines OA and OP is 2 1 S 2D
2S . The point O is the pole.
4
S
(AQA Jan2003 MAP5)
(ii) Hence find the value of D for which OP bisects the area between the curve C and the lines OA and OB.
(i)
T
(b) The point A on C is the point where T
(a) Sketch C.
4. The polar equation of a curve C is 1 S , d T d 2S . r T 4
3. A straight line through the pole O meets the curve with polar equation 4 r 2 sin T at the points A and B. 1 1 Show that 1. OA OB
2. The curve whose polar equation is r 5 2 cos T , S T d S and the line whose polar equation is r = 3sec T intersect at the points A and B. Find the polar coordinates of A and B.
1. A curve has polar equation r
6 . 1 sin T Find its cartesian equation in the form y = f(x).
Chapter assessment
Polar coordinates
Initial line Find the polar coordinates of K, the point of intersection of the two curves.
R
eT and part of the circle r = 2.
(AQA Jan2004 MAP5)
(ii) Find the area of the shaded region R, between the curves and the initial line, giving your answer in the form pln2 + q where p and q are rational numbers.
(i)
O
K
(b) The diagram shows a sketch of part of the curve r
5. (a) A curve C has polar equation r e kT 1 where k 0 and 0 d T d S . 2 The points P and Q on C have polar coordinates r1 , T1 and r2 , T 2 respectively, where T 2 ! T1 . Show that the area A bounded by C and the lines OP and OQ, where O is the pole, is given by 1 2 2 A r2 r1 . 4k
AQA - FP3
35
AQA - FP3
2.
1.
6y
x y
36 12 y y
3
x
12
2
36 x 2
2
0
5 2 cos T and r
6
r
3
S
1 2
,
r
5 2cos T
substituting cos T
The polar coordinates of the points of intersection A and B S S· are §¨ 6, ·¸ and §¨ 6, ¸. 3¹ 3¹ © ©
r
T
In the interval S T d S , when cos T
and
3 sec T
rearranged before squaring
using r 2 x 2 y 2 and y r sin T
(2 cos T 1)(cos T 3) 0 cos T 21 , since cos T 3 is not possible for real T.
2 cos 2 T 5 cos T 3
cos T(5 2 cos T ) 3
At the points of intersection of r 5 2 cos T 3 sec T
y
12 y
2
2
x 2 y 2 (6 y )2
x 2 y2
r r sin T 6 x 2 y2 y 6
r
6 1 sin T r 1 sin T 6
Solutions to chapter assessment
Polar coordinates
1 2
into
4.
3.
(a)
OA
1
1
T
4 1 2 sin D OA
,
4
S
d T d 2S .
2 2
S
S
S 4 4
S
S
1
S
3S 2 2 3S
1 2S
2S
d T d 2S .
4 2 sin D
O Initial line
4 S 1 The curve starts at the point §¨ , ·¸ and ends at the point §¨ , 2S ·¸ . © 2S ¹ ©S 4¹
r
T
As T increases, r decreases.
4
S
or you could use T (S D ) at B
2 sin D 4
Initial line
A
4 2 sin(S D ) 1 2 sin D 4 OB 2 sin D 2 sin D 4 1 4 4 4
S D , r = OB so OB
D
r is never equal to 0 in the interval
r
OB
1
When T
O
S D at B
D , r = OA so OA
4 2 sin T
D at A then T
When T
r
If T
AQA - FP3
36
AQA - FP3
1
D
T2
dT
1 2
2
1
T2
³T
S
e 2 kT dT
2
S
1 2D
eT 2 eT T ln 2 Coordinates of K are (2, ln 2)
2
S
1 . 4S
e kT u e kT
e kT
1 u 2 2 u ln 2 2 = 2ln 2
e 2 kT2
(ii) Area of sector of circle bounded by initial line and OK is
and r2 2
Points (r1, T1) and (r2, T2) lie on the curve r e kT so r12 e 2 kT1
r2
Since OP bisects the area bounded by OA, OB and curve C, the area bounded by OA, OP and curve C is half that bounded by OA, OB and curve C.
2S , area bounded by OA, OB and curve C is
1 1§2 1 · ¨ ¸ 2D 2 © S 4S ¹ 1 2 1 1 2D S S 8S 9 1 2D 8S 4S D 9
(ii) Putting D
1 2 kT T2 ªe º¼T 1 4k ¬ 1 2 kT2 e e 2kT1 4k 1 A = r 2 2 r 1 2 4k
A
D
4
S
(b)(i) At K, r = 2 and r
5.(a)
³
ª 1 º «¬ 2T »¼ S 4 1 2 2D S
1 2
Area bounded by OA, OP and curve C is
(b)(i) A
3 4
Using answer to part (a) with k = 1, r1 = 1 and r2 = 2
eT from (1, 0) to (2, ln2), initial line
Area of shaded region = 2 ln 2
Area bounded by the curve r 1 3 and OK is 2 2 1 2 = 4 4
Past papers
AQA - FP3
37
AQA - FP3
AQA - FP3
38
AQA - FP3
January 2008 AQA - FP3
39
AQA - FP3
AQA - FP3
40
AQA - FP3
AQA – Further pure 3 – Jan 2008 – Answers Question 1: dy = f ( x, y ) = x 2 − y 2 and dx
Exam report y (2) = 1
a ) y (2.1) = 1 + 0.1( 22 − 12 ) = 1.3
Numerical solutions of first order differential equations continues to be a good source of marks for all candidates. This was the best answered question on the paper. Although almost all candidates obtained the correct answer to part (a), some less able candidates showed a lack of understanding of the notation used in the given formula in part (b).
b) y (2.2) = y (2) + 2(0.1)( f (2.1, y (2.1))) =1 + 2 × 0.1( 2.12 − 1.32 ) = 1.544
Question 2: r = 1 + Tanθ
Exam report
π
P(1, 0) and Q( 3, ) 3 π 1 1 π a ) A = ∫ 3 (1 + Tanθ ) 2 dθ = ∫ 3 1 + Tan 2θ + 2Tanθ dθ 2 0 2 0 π
1⎞ 3 ⎡1 ⎤3 ⎛ 3 − ln ⎟⎟ − ( 0 − 0 ) = A = ⎢ Tanθ − ln Cosx ⎥ = ⎜⎜ + ln(2) 2⎠ 2 ⎣2 ⎦0 ⎝ 2 1 b) The area of the triangle OPQ is OP × OQ × SinPOQ 2 π 3 1 AreaOPQ = ×1× 3Sin = 2 3 4 The area shaded is therefore :
3 3 + ln(2) − 2 4
Question 3:
This question, which tested the areas of regions involving a curve whose equation was given in polar form, was relatively poorly answered. Full correct solutions were not often seen. In part (a), candidates generally wrote down the correct definite integral, then expanded (1+ tanθ)2 and integrated 2 tanθ correctly, but could not find a correct method to integrate 1+ tan2θ . Those who used the correct trigonometrical identity had no problem integrating the resulting sec2θ and completing the solution to reach the printed answer convincingly. It was disappointing to find a significant minority of candidates not attempting part (b) having failed to obtain the printed answer in part (a). Most of the other candidates found the correct lengths for OP and OQ but some then wrote down an incorrect formula for the area of triangle OPQ. Some others lost the final mark because they did not give the area of the triangle in an exact form anywhere in their working despite the form of the printed answer in part (a).
Exam report
2
d y dy + 4 + 5y = 5 2 dx dx The complementary function: Auxiliary equation: λ 2 + 4λ + 5 = 0 Discriminant:42 − 4 ×1× 5 = −4 = (2i ) 2 −4 + 2i λ1 = = −2 + i and λ2 = −2 − i 2 The complementary function is yc = e −2 x ( ACos( x) + BSin( x)) The particular integral y = a It is obvious that a = 1 the solution The general solution is : y = 1 + e−2 x ( ACosx + BSinx) b) When x = 0, y = 2 so 2 = 1 + A −2 x
A =1
y = 1 + e (Cosx + BSinx) dy = −2e−2 x (Cosx + BSinx) + e −2 x (− Sinx + B cos x) dx dy When x = 0, = 3 so 3 = −2 + B B=5 dx y = 1 + e −2 x (Cosx + 5Sinx) AQA - FP3 41
This question, which required candidates to solve a second order differential equation, was generally a good source of marks. It was disappointing to see some candidates trying to solve the auxiliary equation, m2 + 4m+ 5 = 0 , by factorisation. They obtained real solutions and this error was penalised heavily. Better candidates were able to write down the correct complementary function and find the particular integral but some wasted valuable time by starting with yp = ax2 + bx + c and showing that both a and b were zero. Candidates who were able to find the correct general solution in part (a) usually went on to apply the given boundary conditions correctly in their answers to part (b).
AQA - FP3
Question 4:
Exam report
∞
a ) ∫ xe −3 x dx is an improper integral because 1
the interval of integration is infinite 1 1 b) ∫ xe −3 x dx = − xe −3 x + ∫ e −3 x dx 3 3 1 1 = − xe −3 x − e −3 x + c 3 9 N 1 ⎡ 1 ⎤ c) ∫ xe −3 x dx = ⎢ − xe −3 x − e −3 x ⎥ 1 9 ⎣ 3 ⎦1 1 1 1 1 = − Ne −3 N − e −3 N + e −3 + e −3 3 9 3 9 −3 N −3 N lim Ne = 0 and lim e = 0 N
N →∞
∫
∞
1
Part (a) was generally not well answered with a significant minority either not attempting it or making a statement which they then contradicted in part (c). The method of integration by parts was understood with the great majority obtaining the correct answer to part (b). Although there continues to be an improvement in candidates’ solutions to the evaluation of an improper integral, there were still a significant minority who made no attempt to show the limiting process used.
N →∞
xe−3 x dx exists and
∫
∞
1
4 xe −3 x dx = e −3 9
Question 5:
Exam report
dy 4x + 2 y=x dx x + 1 An integrating factor is 4x
2x
2∫ 2 2 ∫ 2 dx I = e x +1 = e x +1 = e 2ln( x +1) = ( x 2 + 1) 2 The equation becomes: dy + 4 x( x 2 + 1) 2 y = x( x 2 + 1) 2 ( x 2 + 1) 2 dx d ( x 2 + 1) 2 y ) = x( x 2 + 1) 2 ( dx 1 ( x 2 + 1) 2 y = ∫ x( x 2 + 1) 2 dx = ∫ 2 x( x 2 + 1) 2 dx 2 1 1 ( x 2 + 1) 2 y = × ( x 2 + 1)3 + c 2 3 1 c y = ( x 2 + 1) + 2 6 ( x + 1) 2 1 5 When x = 0, y = 1 so 1 = + c so c = 6 6 ( x 2 + 1) + 5 y= 6( x 2 + 1) 2
Although many candidates were able to write down the integrating factor in terms of an integral, a significant minority could not then integrate
4x x +1 2
correctly. Those who found the
correct simplified integrating factor generally used it appropriately and either solved the resulting integral by a suitable substitution or, more frequently, just multiplied out and integrated x5 + 2x3 + x . Although some candidates failed to insert the constant of integration and so lost the final two marks, this was not a common error.
AQA - FP3
42
AQA - FP3
Question 6:
Exam report
r Sin 2θ = 8 2
a) r 2 × 2 Sinθ Cosθ = 8 rSinθ rCosθ = 4 4 x c) r = 2 Secθ and r 2 Sin 2θ = 8 xy = 4
y=
so (2 Secθ ) Sin2θ = 8 2
4 × 2 Sinθ Cosθ = 8 Cos 2θ Tanθ = 1 so θ = For θ =
π 4
, r = 2 Sec
π 4
π 4
=2 2
π
A(2 2, ) 4
Those candidates who replaced sin 2θ by 2sinθ cosθ generally obtained the correct Cartesian equation in part (a). The sketch of the curve C (rectangular hyperbola) required in part (b) was not answered as well as expected with many sketches consisting of closed loops. Candidates presented a variety of acceptable methods for part (c). Those who eliminated r were required to obtain a trigonometrical equation in a single angle before any mark was awarded. Usually candidates who had found the correct equation went on to obtain the correct exact values for the polar coordinates of A. The second most popular method involved working with the cartesian form of the equation of C. Candidates who recognised the cartesian form of the equation of the line equivalent to the polar form given in part (c) of the question generally had no difficulty getting the cartesian coordinates for A as (2, 2), but then a significant minority could make no further progress.
Question 7:
Exam report 2
3
x x + + ... 2 3 (2 x) 2 (2 x)3 + + ... ln(1 + 2 x) = (2 x) − 2 3 8 ln(1 + 2 x) = 2 x − 2 x 2 + x3 + ... 3 ii ) This is valid for − 1 < 2 x ≤ 1 1 1 −
