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Level 3, 2004

Calculus Differentiate and use derivatives to solve problems (90635)

Integrate functions and solve problems by integration, differential equations or numerical methods (90636)

Manipulate real and complex numbers, and solve equations (90638)

Sketch graphs and find equations of conic sections (90639)

National Statistics

Assessment Report

Assessment Schedule

NCEA Level 3 (Calculus) 2004 — page 2

Calculus Level 3, 2004 General Comments Candidates did well at the achievement level in all four assessment activities. The length of paper 90639 for conic sections seems to have affected the results at Merit level. The numbers who achieved Merit in 90635 Differentiation, 90636 Integration and 90638 Complex Numbers were greater than for 90639. Candidates found the excellence questions for integration and conic sections particularly challenging. Candidates needed good algebra skills for all of this paper. They also needed efficient checking systems to help themselves avoid making unnecessary arithmetic and algebraic errors. Calculus: Differentiate and use derivatives to solve problems (90635) National Statistics

Number of Percentage achieved

Results Not Achieved Achieved Merit Excellence

7933 31.3 42.5 22.6 3.5

Assessment Report General Comments

Most candidates demonstrated a high level of differentiation skills while completing this assessment activity. Those who achieved the higher grades did so because they had good algebra skills and a good understanding of related rates of change. Some users of the graphic calculator failed to show the derivatives needed to solve the differentiation problems. As indicated in the specifications and on the front page of the examination booklet, CAO achieved an N in all of the application problems. In Questions 2 and 8, many candidates ignored the reminders that they did not need to spend time to do the second derivative test. Often a lack of good algebra skills resulted in candidates being unable to complete the application of differentiation problems once they had arrived at the correct derivative. Specific Comments

1. Differentiating functions and using differentiation to solve problems. Most candidates were confident in differentiating the functions included in the exam. Errors resulted from candidates who failed to apply the chain rule fully, or at all. In Question 2, a significant number of candidates did not realise that they needed to solve the first derivative equal to zero to minimise the total cost of producing the calculators. Question 3 was only partially answered by many candidates. Either they located some, but not all of the turning points, or they failed to determine the nature of the turning points that they did find. A lack of y-


coordinates for the turning points identified was regarded as a minor error in this round of marking, but may not be in the future. 2. Demonstrating knowledge of concepts and techniques of differentiation. The candidates completed Questions 5, 6 and 7 with confidence and a high amount of success. Errors were mainly careless, involving partial application of the chain rule in question 5 and careless

rearrangement of the equation when making

dy

dxthe subject in Question seven.

3. Solving the merit level differentiation problems. Candidates were less successful in their efforts to complete Questions 4, 8 and 10. Those who were able to set up the model in Question 4 were generally able to successfully complete the question. The responses from candidates suggested that the related rates of change topic is an area that teachers need to work more with their students. Most candidates were able to differentiate the function in question 8. However many students were not able to then successfully solve the resulting equation to answer the problem. Few candidates did Question 10 well. Many candidates did not read the question carefully and interpreted the graph as if it represented the E(t) graph rather than the )(tE ! graph. Those candidates who realised that they were interpreting a gradient function still found it difficult to successfully answer the questions posed. 4. Solving problems involving a combination of differentiation techniques. The candidates who achieved Excellence demonstrated a good knowledge of the trigonometric functions included in the Level 3 Trigonometry standard, as well as a good knowledge of solids of revolutions included in the Level 3 Integration standard. Students were then able to successfully set up the models needed to solve the related rates sub-questions of Questions 9 and 11. Calculus: Integrate functions and solve problems by integration, differential equations or numerical methods (90636) National Statistics



7865 44.3 39.8 15.6 0.3


Many candidates did not realise that they are expected to show the integral used. This needs to be emphasised by teachers, and candidates need to read the instructions of the assessment activity more carefully. Candidates need to check their answers more carefully. Some gave negative answers to volume questions. They need to be reminded that answers can be checked by differentiation.


Candidates often did not read the questions with sufficient care, especially with regard to initial conditions and which axis the shapes were being rotated around to form the solid of revolution. Candidates need to take more care as to which order the limits of an integral should be written.

Specific Comments

1. Integrating functions. Candidates found it difficult to integrate composite functions successfully. Many found it difficult to find the correct coefficients when integrating. 2. Solving problems by integration, differential equations or numerical methods. In Question 2, candidates found it difficult to set up the appropriate integral to find the shaded area. Many used volume and others used numerical methods. Others found the correct integral and then were not able to correctly substitute into it.

In Question 3, candidates did not read the correct y-values off the graph and substituted x-values into the Simpson’s Rule formula. Others were not able to correctly find the h value for their formula.

In Question 4, many candidates did not evaluate the constant of integration. 3. Finding integrals and using integration to solve problems. Candidates found Question 5 quite challenging and did not choose an appropriate substitution. Several integrated a product by integrating individual terms. Others partially substituted and then integrated the x and u parts separately.

In Question 6, some candidates found an integral expression and then stopped. Others integrated the function around the x-axis instead of the y-axis as required.

In Question 7, candidates treated the graph as a straight line or a rectangle.

In Question 8, several candidates wrote down the general solution directly without integration. The instruction for all candidates was that they had to show the results of any integration needed to solve the problems. 4. Using a variety of integration techniques to solve problem(s). The excellence explanatory notes state that “problems may include finding volumes of solids of revolution formed by rotating around an axis parallel to the x or y axis”. Therefore, it was interesting to note that very few students demonstrated knowledge of how to set up an integral around a line parallel with the x-axis in this exam.

Many candidates who could set up the appropriate integral did not know how to integrate sin

2(1

4

x) .

Calculus: Manipulate real and complex numbers, and solve equations (90638) National Statistics



7853 39.0 40.9 16.2 3.8



Candidates using a graphic calculator had a significant advantage when completing this assessment activity. Questions 2(a) and 3 could be keyed into the graphic calculator for a quick, correct answer. Specific Comments 1. Manipulating real and complex numbers. In Question 1(a)(i), some candidates did not know how to multiply complex numbers in rectangular form.

In Question 1(a)(ii), some candidates found it difficult to deal with the negative imaginary part when converting to polar form.

In Question 1(c), some candidates did not know how to multiply 2 with 2 or how to simplify to get a final answer. 2. Solving equations. In Question 2(b), some candidates did not know how to expand a perfect square.

Candidates found the writing of logarithm symbols difficult, especially when the base was important and stood out, eg log3 written log3 in Question 2(c). Some candidates also treated the log symbol itself as a variable/number which could itself be cancelled.

In Question 2(c), some candidates did not know how to deal with an expression that had a linear factor as the numerator. 3. Solving Merit level equations.

Many candidates were not able to expand, collect terms and factorise a quadratic expression in Question 3. Few remembered to check for invalid solutions.

Candidates found it difficult to use De Moivre’s theorem to find the square roots of 5 – 12i in Question 4.

Candidates also found it difficult to use the factor theorem effectively and divide polynomials accurately when solving the problem posed in Question 5. 4. Solving problems involving real and complex numbers. The number of candidates who attempted Question 6 seemed to be significantly low. Those who did showed that candidates needed more practice at designing an appropriate sequence of statements for a “show” or “prove question”. Calculus: Sketch graphs and find equations of conic sections (90639) National Statistics



7604 43.8 43.7 11.8 0.6



The overall length of the examination paper may have resulted in the better students not having sufficient time to complete the conic sections assessment activity to the best of their ability. There was confusion among the candidates as to the degree of accuracy required by the instruction to sketch a graph. Many did not understand the importance of symmetry and shape, or the need for intercepts and asymptotes to be placed accurately on the grid. Some candidates demonstrated a lack of familiarity with the information provided on the formulae sheet. Good candidates sketched graphs for the contextual problems to help clarify the problem needing to be solved. This practice needed to be adopted by more candidates. Candidates needed to take more care to avoid the simple errors, both arithmetic and algebraic, that were common in the scripts marked. Specific Comments 1. Sketching graphs of conic sections. Because of the overall poor quality of responses, candidates were not penalised for poor quality sketches in this first NCEA Level 3 Calculus paper. It will be expected that in future candidates should draw smooth curves, with no sharp corners, that show the characteristics properties of the relevant conic: hyperbolas need to approach asymptotes, circles need to be circular and ellipses need to be symmetrical.

In Question 1, some candidates found it difficult to draw asymptotes for the hyperbolas with the correct gradient.

In Question 2, some candidates found it difficult to complete the square to write the circle in an appropriate form to be able to sketch. 2. Finding equations of conic sections. Candidates demonstrated a lack of accuracy in their efforts to write equations of conic sections. Omissions and simple errors were commonplace in Questions 4(b) and 4(c) especially. Some candidates did not show an understanding of the connection between the transformations of the conic and its algebraic form. 3. Solving problems involving conic sections. Good candidates understood and could apply parametric differentiation of trigonometric functions to find the equation of the tangent. They had strong algebraic skills, especially with regard to solving equations, and were able to draw appropriate models for contextual problems. Weaker candidates used models other than the parabola for Question 6, when the instructions of the

question did state that the motion was parabolic. Modelling the situation with the axy 42= form of the

parabola did complicate the problem. Students need to be reminded that when problem-solving, they can

also use the BAxy +=2 version from Level 2.

In Question 7, some candidates inappropriately applied the distance formula when attempting to solve the problem. Some candidates managed to complete the computational part of this question, but either failed to answer the question of whether the comet would effect the earth or answered the question incorrectly.


4. Solving the excellence level conic section problem. Candidates found the last question challenging. The few candidates who attempted it typically made the mistake of substituting the point (1,3) into the gradient expression for the tangent to the hyperbola despite being told that (1,3) did not lie on the hyperbola.

NCEA Level 3 Calculus (90635) 2004 – page 1 Assessment Schedule Calculus: Differentiate and use derivatives to solve problems (90635)

Achievement Criteria

No. Evidence Code Judgement Sufficiency

Ach

ieve

me

nt

Differentiate functions and use differentiation to solve problems.

1(a)

1(b)

1(c)

2

3

dy

dx= 5(6x ! 7)(3x

2! 7x)

4

dy

dx=

20x3

5x4+1

dy

dx= 7 sec 7x tan 7x

dC

dx=!6 000 000

x2

+ 0.3

Solve 0.3 x2 = 6 000 000

x = 4472.14

Number of Calculators = 4472 or 4473

!f (x) = 16x " 4x3

Solve 4x(4 " x2 ) = 0

x = 0, ±2

!!f (x) = 16 " 12x2

At (0,0)

!!f (0) > 0 Minimum.

At (±2,16 )

!!f (±2) < 0 Both maximum.

A1

A1

A1

A2

A2

Or equivalent.

Or equivalent.

Or equivalent. Must find derivative.

Accept unrounded answer. Must find derivative. Coordinates of all 3 turning points found.

However ignore minor errors such as: no y-coords found incorrect y-coords

due to an arithmetic error.

Nature of all 3 turning points identified. Accept CAO with derivative.

Achievement: four of code A with at least one A1 and one A2.

NCEA Level 3 Calculus (90635) 2004 – page 2



Mer

it

Demonstrate knowledge of concepts and techniques of differentiation.

Solve differentiation problems.

5

6(a) 6(b) 6(c)

7

4

8

10(a)

10(b)

10(c)

dy

dx= (cos x) 3x ! 2 + (sin x)

1

2(3x ! 2)

!1

2 " 3

1 0 Not differentiable when x = –2, 0.

6x ! 8ydy

dx= 0

dy

dx =

3x

4 y

x2+ y

2= 25 5.0

d

d!=

t

y

y = (25 ! x

2 ) When x = 3

dy

dx=

!x

(25 ! x2 )

dx

dt=

25 ! 32

!3" –0.5

dx

dt=

dx

dy!

dy

dt

dx

dt=

2

3ms

!1

OR

0d

d22 =+

x

yyx 5.0

d

d!=

t

y

y

x

x

y!=

d

d When x = 3, y = 4

dx

dt=

dx

dy!

dy

dt

3

4

d

d!=

t

x × –0.5

=2

3ms

!1

!C (t) =[(t + 3)2

" 2] # [2t " 2(t + 3)]

(t + 3)4

=2(3! t)

(t +3)3, t +3" 0

C′(t) = 0 2(3 – t) = 0 t = 3 days OR

2)3(2)( !+= tttC

23 )3(2)3(2.2)( !!+++!=" ttttC

=6! 2t

(t +3)3, t +3" 0

C′(t) = 0 6 – 2t = 0 t = 3 days The price is not changing. The price is increasing at the greatest rate of increase. The price is decreasing.

A1 M1

M1

A1 M1

A1 A2 M2

A1 M1

A2 M2

A2 M2

Or equivalent.

Accept two of 6(a), 6(b), 6(c) for Q6.

Must have

x

y

d

das subject

Or equivalent.

Correct x

y

d

d

found. Or equivalent. Units not required.

Accept

dx

dt= !

2

3ms

!1

Must find derivative. Or equivalent. Units not required.

Only one M grade available from Q8 ( M1 or M2). Or equivalent. Accept two of 10(a), 10(b), 10(c) for Q10.

Merit: Achievement plus two of code M1 and two of code M2 or six of code M.




Exc

elle

nce

Solve problem(s) involving a combination of differentiation techniques.

9

11(a)

11(b)

d!

dt=

"

540radians min

#1

x = 150 cot!

dx

d! = #150 cosec

2!

dx

dt

= dx

d!$

d!

dt

When ! = "

6,

dx

dt

= #10"

9

or #3.49 m min-1

V = ! (400" y2 )dy

20"h

20#

= ! 400y "y3

3

#

$%%

&

'((

20"h

20

!!

"

#

$$

%

&'=3

32

20h

h(

3

32

20h

h!

! "=

240

d

dhh

h

V!! "=

t

V

V

h

t

h

d

d.

d

d

d

d=

2

40

80

hh !! "=

When h = 5,

!175

80

d

d=

t

h

= 0.1455 cm s–1

A2 M2 E

A2 M2 E

Or equivalent.

Units not required.

Accept 3.49.

Accept a minor error.

Or equivalent.

Units not required.

Accept one error.

Excellence: Merit plus two of code E.

θ

150

x


Assessment Schedule Calculus: Integrate functions and solve problems by integration, differential equations and numerical methods (90636)

Achievement

Criteria No. Evidence Code Judgement Sufficiency

Ach

ieve

me

nt

Integrate functions and solve problems by integration, differential equations or numerical methods.

1(a)

1(b)

1(c)

2

3

4

! 4

x+ 3ln x + C

(3t!1)6

18+ C

tan 4x

4+ C

Area = e2x

dx

!1

1.5

"

=

1

2(e3

! e!2 )

= )135.0085.20(2

1!

= 9.975

Distance

=

0.5

310 + 10 + 4(20 + 5) + 2(20)!" #$

=80

3= 26.6

.

km

Distance s = sin (

t

2 !

1

2) dt"

= !2cos (

t

2!

1

2) + C

at t = 1, s = 0 so 0 = !2 + C C = 2

Distance s = !2cos (

t

2!

1

2) + 2 m

A1

A1

A1

A2

A2

(A1)

A2

Or equivalent.

Accept without absolute value sign. Or equivalent. Or equivalent.

Or equivalent. Units not required. Or equivalent. Units not required.

Or equivalent.

Units not required.

Achievement: two of code A1 and two of code A2.




Mer

it

Find integrals and use integration to solve problems.

5

6

7

x 1 + x! dx let u = x + 1

= (u32!u

12

)du"

=2

5u

52!

2

3u

32+ C

=2

5(x + 1)

52!

2

3(x + 1)

32+ C

V = ! x

2dy"

V = ! ( y " 4)23

4

h

# dy

= !3

5( y " 4)

53

$

%&

'

()

4

h

=3!

5(h " 4)

53

x

0

k

! dx = x

k

9

! dx

2

3x

3

2"

#$

%

&'

0

k

=2

3x

3

2"

#$

%

&'

k

9

2

3k

3

2=

2

3( 27 )

2

3k

3

2

4

3k

3

2=18

k = 5.67

OR

x

0

k

! dx =1

2x

0

9

! dx = 9

93

2

0

2

3

=!!

"

#

$$

%

&k

x

93

22

3

=k

k = 5.67

(A1)

(A2) M

(A1, A2) M

(A1)

(A2) M

(A1)

(A2) M

Or equivalent (must be a function of x).

Or equivalent.

Or equivalent.

Merit: Achievement plus three of code M or four of code M.




Mer

it

8

dN

dt= kN

dn

N! = k dt!

CktN +=ln

N = Ae

kt

at t = 0, N = 50 000

A = 50 000

and N = 50 000ekt

at t = 30, N = 90 000

90 000 = 50 000e30k

ln(9

5) = 30k

k = 0.0196

thus N = 50 000e0.0196t

(A1)

(A2) M

If proved.

Or equivalent.

Exc

elle

nce

Use a variety of integration techniques to solve problems(s).

9 Volume found by rotating

y = sin(

1

4x) around

y = a is

Vol =

! y2

0

4!

" dx

= ! (sin1

4x " a)2 dx

0

4!

#

=! (sin1

4x)2 " 2asin

1

4x + a2

$

%&

'

()dx

0

4!

#

=!1

2 "

cos1

2x

2" 2asin

1

4x + a2

$

%

&&&&

'

(

))))

dx

0

4!

#

= !1

2x " sin

1

2x + 8acos

1

4x + a2

x

$

%&

'

()

0

4!

= ! 2! " 8a + a2

4!#

$%&

'(" ! 8a#$ &'

= 2!2"16a! + 4a

2!

2

Thus volume of remaining plastic is

= 2!2"16a! + 4a

2!

2" !(0.5)

24!

= !2"16a! + 4a

2!

2 cm

3

(A1)

(A2, M) E

Allow a minor error. Or equivalent.

Units not required.

Excellence: Merit plus code E.

NCEA Level 3 Calculus (90638) 2004 – page 1 Assessment Schedule Calculus: Manipulate real and complex numbers and solve equations (90638)

Achievement


Ach

ieve

me

nt

Manipulate real and complex numbers, and solve equations.

1(a)(i)

1(a)(ii)

1(b)

1(c)

1(d)

2(a)

2(b)

2(c)

wz = (1+ 3i)(2 + i)

= !1+ 7i

z = 2 ! i

z = 5 cis (!26.6° )

= 2.236 " cis (!0.464)

w

z= 0.6 cis 0.8

6 + 2

2 ! 2

"2 + 2

2 + 2

= 14 + 8 2

2

= 7 + 4 2

( 2 cis!

3)12

= 212

cis4!

= 4096

32x!1

= 4

(2x ! 1) log 3 = log 4

2x ! 1 = log 4

log 3

x = 1.131

(x + 3)2= 2x + 2

x2+ 4x + 7 = 0

x = !4 ± !12

2

x = !2 ± 3 i

OR x = !2 ± 1.732 i

log3(

x

x ! 2) = 2

x

x ! 2 = 32

x = 9x ! 18

x = 2.25

A1

A1

A1

A1

A1

A2

A2

A2

Accept CAO.

No alternative.

Accept CAO. Or equivalent / in any polar form. Or equivalent.

Accept at this point. Do not accept CAO. Accept CAO. Accept 4096+0i.

Accept CAO 3 sf. Accept any correct rounding to at least 2 sf with working. Or equivalent. Both solutions. Accept CAO.

Or equivalent.

Achievement: three of code A1 and two of code A2.


Achievement


Mer

it

Solve more complicated equations.

3

4

5

Squaring gives

x2+ 2x + 1 = 6 ! 2x

x2+ 4x ! 5 = 0

(x + 5)(x ! 1) = 0

x = !5, x = 1

x = !5 is not valid.

Solution is x = 1.

Let square root be a + bi.

(a + bi)2 = 5 ! 12i

a2 ! b2 + 2abi = 5 ! 12i

a2! b2 = 5, ab = !6

a = ± 3, b = ! 2

ie square roots are

3! 2i and ! 3+ 2i

OR

z2= 5 ! 12i

z = (13 cis (!67.4°))

12

= 13 cis (!33.7°

), 13 cis146.3°

OR

z = (13 cis (!1.176))

12

= 3.61 cis (!0.588), 3.61 cis 2.55

z = 2 ! i is another root

(z ! 2 ! i)(z !2 + i)(z ! a)

= z3! 6z

2+ 13z + k

(z2! 4z + 5)(z ! a) = z3

! 6z2 + 13z + k

Equating coefficients a = 2

ie k = !10,

ie other 2 roots are

z2= 2 ! i

z3= 2.

A2 M

A2 M

OR

A1 A2 M

OR

A1 A2 M

A2 M

If equation reached, must have both sols. and indicate that x = 1. Accept CAO ie x = 1. Must have both square roots.

Accept CAO.

Or equivalent. Accept answer in rectangular or polar form (RAD or DEG). Accept CAO. No alternative (must be 2 roots and not factors).

Merit: Achievement plus two of code M or three of code M.




Exc

elle

nce

Solve problem(s) involving real or complex numbers.

6(a)

6(b)

6(c)

z3= 1

z3! 1 = 0

(z ! 1)(z2+ z + 1) = 0

If w is one of the complex roots then

w2+ w+1= 0.

All other correct methods are acceptable, eg finding a value of w and substituting into the equation to show the equation is true.

1

w2+ w

4=

1

w2 (1 + w

2 )

= 1

w2 (!w)

from (a)

= !1

w3

= !1 as w3= 1

"#$

%&'

.

1+w = !w2 using (a)

= ! w2

= ! w using initial condition

= !w.

A2 M E

A1 M E

A1 M E

The information for 6(c) may not be used to obtain 6(a). Or equivalent. The information for 6(c) may not be used to obtain 6(b). If substitution is used, the statement must be shown for 1 complex value of w.

Or equivalent.

Or equivalent.

Excellence: Merit plus two of code E.

NB There are many equivalent ways of obtaining correct answers for Q6. Since Q6 says that w is one complex solution instead of a complex solution, it is acceptable to substitute just one complex root in 6(a), 6(b) and 6(c).


Assessment Schedule Calculus: Sketch graphs and find equations of conic sections (90639)

Achievement


Sketch graphs of conic sections.

1

x2

4 !

y2

9= 1

–2

–4

–6

4 6

4

6

2

A1

Hyperbola drawn: – asymptotes correctly drawn as

y = ±

3

2x

– vertices at (2, 0) and (–2, 0).

Both branches of hyperbola shown. Curve approaches but does not cross asymptotes.

Achievement: two of code A1 and two of code A2.

2

x2+ y

2! 4x + 6 y + 9 = 0

(x ! 2)2+ ( y + 3)

2= 4

0

1 2 3 4

–2

–4

yx

A1

Circle drawn: – centred at (2, –3) – y-intercept (0, –3)

3

x = 5cos! , y = 3sin!

x2

25+

y2

9= 1

x

y

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

0

2

A1

Ellipse drawn: – centred at (0, 0) – intercepts at

(5, 0) (–5,0) (0,– 3) (0, 3)

Find equations of conic sections from given information.

4(a)

(x ! 2)2

42+

( y+1)2

32=1

OR

(x ! 2)2

16+

( y+1)2

9=1

OR x = 4cos! + 2

y = 3sin! "1

A2

Or equivalent.

Ach

ieve

me

nt




4(b)

x2

82!

y2

42=1

OR

x2

64!

y2

16=1

OR x = 8sec!

y = 4 tan!

A2

Or equivalent.

4(c)

( y ! 2)2

= 8(x !1)

OR y =± 8(x !1) + 2

OR x = 2t2+1

y = 4t + 2

A2

Or equivalent.

Solve problems involving conic sections.

5

( 1169

22

=!yx )

dx

d!= 3sec! tan!

dy

d!= 4sec

2!

dy

dx=

4sec!

3tan! or

dy

dx=

16x

9y

dy

dx=

8

3

When 6

!" = ,

464.3

3

6

6sec3 ===

!x

309.2

3

4

6tan4 ===

!y

Equation of tangent:

!"

#$%

&'='3

6

3

8

3

4xy

OR 0363338 =!! yx

OR 343

8!= xy

OR 078.2038 =!! yx

A2

M

Cartesian equation of ellipse. Correct expression

for

dy

dx

Accept alternative methods. Allow one minor arithmetic error. Equation of tangent. Or equivalent.

Merit: Achievement plus two of code M or three of code M.

Mer

it


6

Parabola with turning point (0, 8):

y = Ax2+ 8

(–12, 2) is a point on the curve.

Model of parabola: y = !

1

24x

2+ 8

Line through (0, 0) with m = 2 is y = 2x. Substituting and rearranging gives

x2+ 48x !192 = 0

Point of intersection is (3.7128, 7.4256). Distance from (0, 0) to point = 8.30 m. (2dp) Alternative method: An ellipse is also an acceptable model. Ellipse centred at (0, 0) gives:

x2

122+

y2

62=1 and

y = 2x ! 2

Substituting and rearranging gives

612x2!1152x ! 4600 = 0

Point of intersection is (3.842, 5.684). Distance from (0, –2) to point = 8.59 m. (2dp)

A2

M

Accept alternative methods. Allow one minor error. Equation of parabola. Accept Accept if appropriate axes indicated on graph. Units not required.

7

Ellipse of form:

x2

2502+

y2

b2=1

Substituting the point (240, 20) gives

b

2=

250 000

49= 5102(4sf )

Ellipse is

x2

62 500+

49y2

250 000=1

When

x = 245, y =±14.21(2dp)

When the comet is at (245, –14.21), yes (the comet will affect the earth)

A2

M

Accept alternative methods. Allow one minor error.

Equation of ellipse. Or equivalent.

Consistent decision based on negative y –value required.

Mer

it

y

2= 24(x+8)

y

2= 24x


Achievement


Solve more difficult conic section problems.

8

Method 1:

cmxy += at (1,3) so c = 3!m

03648)49(

36)(49

3649

222

22

22

=!!!!

=+!

=!

ccmxxm

cmxx

yx

Discriminant = 0 gives

0)364)(49(464 2222=!!!! cmmc

057664129614464222222=!!++ mmccmc

08136494222222

=+!!+ mmccmc 081369

22=+! mc

Substituting mc != 3 ,

094)3( 22=+!+! mm

0622

=!+ mm

m = !1± 7 =1.65 or 3.65 (2dp)

The two possible tangents through (1,3) are:

y !3=1.65(x !1) and

y !3= !3.65( y !1)

OR

y =1.65x+1.35 and

y = !3.65y+6.65

A1

M

E

Suitable graph sketched as replacement for Q1 only. Equation formed by substituting the tangent equation into the equation for the hyperbola. Do not penalise minor errors.

Both equations given. Or equivalent.

Excellence: Merit plus code E.

Method 2:

dy

dx=

9x

4y=

y!3

x!1

yyxx 1249922!=!

361294922

=!=! yxyx

36129 =! yx

3634

349

2

2=!

"

#$%

&'' xx

861.4194.2 != orx

645.63545.1 !!= ory

The two possible tangents through (1,3) are:

194.21

3545.13

1

3

!

+=

!

!

x

y

and 861.41

645.63

1

3

+

+=

!

!

x

y

M

E

Do not penalise minor errors. Equation formed by substituting the tangent equation into the equation for the hyperbola. Both equations given. Or equivalent.

Exc

elle

nce

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