Download Report

18 Chapter 1: Achievement Standard 90284 (Mathematics 2.1 )
1.7 Simultaneous linear/linear equations
Simultaneous linear/linear equations can be solved by
elimination or substitution. Both techniques should be
familiar.
Example T
At Level 2, you will be required to form simultaneous
equations from information given in a practical context.
Example U
Two identical washing machines and a drier sell for a
combined cost of $3 660.
The washing machine is $120 less than twice the cost of
the drier.
Solve using the elimination method.
3x – 5y = 11
2y = x – 4
Form and solve a pair of simultaneous equations to find
the cost of the washing machine.
Solution
Solution
Rearrange second equation to the same form as the first.
3x – 5y= 11
…(1)
Let x be the cost of a washing machine and y be the cost
of a drier.
–x + 2y= – 4
…(2)
From the information given:
2x + y = 3 660
Scale up equation (2) so that coefficients of x are
opposite amounts.
(2) × 3 = (3)
x = 2y – 120
–3x + 6y= –12
…(3)
Add to eliminate x:
Substitute x = 2y – 120 in the first equation.
2(2y – 120) + y= 3 660
y= –1
(1) + (3)
[combined cost is 2x + y]
4y – 240 + y= 3 660
[expanding]
Substitute y = –1 in either (1) or (2) to get x = 2.
The solution is (x,y) = (2,–1)
y= 780
x= 2 × 780 – 120 = 1 440
Graphics calculators can be used to solve
simultaneous linear/linear equations –
enter each equation in the form
ax + by = c
(some rearranging may be necessary first).
5y= 3 900 [rearranging and simplifying]
The washing machine costs $1 440.
Note: This problem was
solved using the
substitution method.
Exercise G: Simultaneous linear/linear equations
1. Solve using the elimination method. Give your answer as an ordered pair (x,y).
a. x – y = 5
b.
x +2y = –1
x + y = 5
3x +2y = 9
c. 2x – y = 3
3x + 2y = 22
SAMPL
E
PAGES
ONLY
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 18
3/5/10 3:25:44 PM
Draw straightforward non-linear graphs 39
Chapter 2: Achievement Standard 90285
Mathematics 2.2
Draw straightforward non-linear graphs
Externally assessed 3 credits
Graphs and their features
Features of graphs
Points on a graph can be found by substituting suitable
x-values into its equation to find the related y-values.
Example A
To sketch the graph whose equation is y = x3, a table of
values is drawn up.
x
–2
–1
−
1
2
0
1
2
1
2
y
–8
–1
−
1
8
0
1
8
1
8
When drawing a graph, choose suitable scales for
the axes so that key features, such as intercepts and
symmetry, can be shown clearly.
A function is said to be
• increasing if its y-values increase as the x-values
increase. For example, the function y = x3 is
increasing for all values of x.
• decreasing if its y-values decrease as the x-values
increase.
Example B
The graph of the parabola y = (x – 2)2 – 5 is drawn.
[for example if x = –2 then y = (–2)3 = –8]
y
The points are plotted as ordered pairs (–2,–8), (–1,–1),
etc, and joined as shown.
2
y
x
8
–5
6
The function y = (x – 2)2 – 5 is decreasing for x < 2 and
increasing for x > 2. At the point (2,–5) the graph is
neither increasing nor decreasing.
4
2
–2
–1
0
–2
–4
–6
–8
1
2
x
Graphics calculators enable graphs to be
drawn quickly and accurately – tables of points
on the graph can be automatically generated,
and features identified.
SAMPL
E
Note: In this Achievement Standard,P
it A
is unwise
GEtoSrely
solely on graphics calculators, as many problems involve
interpreting graphs and/or writing down
the equations of
ON
LYskill.
graphs which have been drawn – not a calculator
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 39
3/5/10 3:25:50 PM
Find and use straightforward derivatives and integrals 107
AS 90286: Practice test
c. The slant height of a cone is 6 cm.
1. a. The point P(–2,4) lies on the curve
g(x) = 2x3 + 5x2 – x – 1.
i. Find the gradient of the curve at P.
6 cm
h
r
ii. The curve has the same gradient at the
point Q as it does at the point P.
Find the coordinates of Q.
What is the maximum volume the cone can
have?
1
(Note: Volume of a cone = πr2h)
3
b. Water flows from a spout, making an arch in the
shape of a parabola.
water
spout
The height of the water (in centimetres) can be
2
modelled by the function h = 6x – x2, where
3
x is the horizontal distance of the water from
the spout (in centimetres).
Use calculus to find the maximum height the
water reaches.
SAMPL
E
PAGES
ONLY
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 107
3/5/10 3:26:20 PM
108 Chapter 3: Achievement Standard 90286 (Mathematics 2.3 )
2. a. The graph shows the function
b. A water trough has a parabolic cross-section, as
shown.
f(x) = x(x + 3)(5 – x)
= –x3 + 2x2 + 15x
5m
y
40
1.6 m
30
1.2 m
20
10
–4
–2
2
4
6
The trough is 1.6 m wide, 1.2 m deep at its
deepest point and 5 m long.
What is the maximum number of litres of water
the trough can hold?
x
–10
–20
i. Find the area between the graph of f and
the x-axis between x = 0 and x = 5.
ii. Find the total area between the graph of f
and the x-axis for –3 ≤ x ≤ 5.
SAMPL
E
PAGES
ONLY
ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 108
3/5/10 3:26:20 PM
114 Chapter 4: Achievement Standard 90287 (Mathematics 2.4 )
Gradient of a straight line
Example E
the gradient of a line is a measure of the steepness of
the line.
1. The gradient of the line passing through the points
(–9,3) and (–5,7) is
y −y
m= 7 − 3 [m = 2 1 ]
x 2 − x1
−5 − −9
= 1
• For a non-vertical straight line, the gradient is
constant.
• The gradient of a vertical line is undefined.
rise
between any
The gradient is often described as the
run
two points on the line.
2. To calculate the gradient of the line PQ shown,
6
y
The gradient of a straight line (non-vertical) passing
through A(x1,y1) and B(x2,y2) is
m=
y 2 − y1
x
5
x 2 − x1
y
y2
gradient of AB is
B
y 2 − y1
x 2 − x1
y2 – y1
y1
A
first identify two points on the line: (0,6) and (12,0).
Then use the gradient formula:
y −y
0−6
m=
[m = 2 1 ]
12 − 0
x 2 − x1
1
m= −
2
Collinear points lie on the same straight line.
x2 – x1
x1
12
x2
B
x
Note: The formula for the gradient is undefined when
x2– x1 = 0 (cannot divide by zero), ie when x1 = x2.
If two points have the same x-coordinates then the line
passing through them is vertical.
C
A
To prove that three points A, B and C are collinear it
is necessary to show that the gradient of AB and the
gradient of BC are the same (if two lines have the same
slope and share a point, then they are one and the
same line).
Exercise C: Gradient of a straight line
1. Find the gradient of each of the following lines, where defined (one line is vertical and has an undefined gradient).
a. PQ where P is (4,4) and Q is (10,7)
b. ST where S is (7,–6) and T is (–1,2)
c. CD where C is (3.6,4.1) and D is (4.1,5)
d. EF where E is (0,12) and F is (0,–2)
SAMPL
E
PAGES
ONLY
ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 114
3/5/10 3:26:22 PM
Use coordinate geometry methods 115
e. MN where M is (–3,–4) and N is (2,–6)
5. Find the gradient of the line joining the midpoint of
AB to the midpoint of CD, where A is (25,40),
B is (–17,13), C is (22,–9) and D is (16,2).
f. GH where G is (–9,–10) and H is (–7,–6)
2. The line AB passes through (
1 1
1 3
, 3 ) and ( −2 , 5 ).
2 4
2 4
Line CD passes through (4.75,3.6) and
(–0.6,–7.1).
Line EF passes through (256,458) and (587,1120).
Which pair of lines has the same gradient?
6. The lines GH and IJ have the same gradient, where
G is (11,–7), H is (2,5), I is (–5,–2), and J is (m,n).
a. Find a possible set of values for m and n, where
m and n are integers such that
i. both m and n are negative
ii. m is negative and n is positive.
3. Which line is steeper:
KL which passes through K(–3.5, –1) and L(0,6), or
MN which passes through M(4,–6) and N(7,1)?
iii. m is positive and n is negative.
4. Are the points I(–3,–5), J(1,5) and K(9,25) collinear?
b. Is it possible to find m and n where both are
positive? Explain why or why not.
SAMPL
E
PAGES
ONLY
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 115
3/5/10 3:26:22 PM
Solve trigonometry problems requiring modelling of practical situations 205
Radian measure
Angles can be measured in radians as well as in
degrees.
1
A quick rule for converting
between degrees and
radians is shown in the
diagram alongside.
×
degrees
1
–1
1
radians
×
1 radian
π
180
Example G
180
π
1. Convert 85° to radians, to 3 dp.
2. Convert 2.5 radians to degrees, to 1 dp.
3. Convert 135° to radians, in terms of π.
–1
Solution
In a circle of radius 1, the angle subtending an arc of
length 1 at the centre of the circle, is said to be of size
1 radian.
The circumference of a circle of radius 1 is 2π, so an
angle of 360° is equivalent to 2π radians.
Dividing by 2 gives the conversion relationship:
π radians = 180°
Using this relationship, angles can be converted between
degrees and radians. For example,
π
radians or 1.57 radians (3 sf).
90° =
2
180
1 radian = π degrees or 57.3°.
π
radians
180
= 1.484 radians
180
2. 2.5= 2.5 × π degrees
1. 85°= 85 ×
= 143.2°
3. 135°= 135 ×
π
radians
180
135π
radians
180
3π
radians
[simplifying]
=
4
Note: When finding a trigonometric function
of an angle in radians, the calculator must be
put in radian mode.
=
Exercise E: Radians and degrees conversions
1. Convert the following angles to radians. Give answers as exact multiples of π. (For example 90° =
a. 45°
b. 30°
c. 60°
d. 135°
e. 270°
f. 210°
g. 150°
h. 300°
π
radians.)
2
2. Convert to radians. Round answers to 4 dp.
a. 96°
b. 25°
c. 122°
d. 250°
e. 330°
f. 58.3°
g. 75.5°
h. 288.3°
3. Convert the following angles expressed in radians to degrees, to 1 dp.
SAMPL
E
PAGES
4.826
ONLY
a. 2
b. 1.96
c. 2.78
d. 0.07
e. 1.4
f. 0.86
g. 3.1892
h.
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 205
3/5/10 3:26:46 PM
Answers Chapter 9: Achievement Standard 90292 261
8. a. In ∆ACD sin A =
so h = b sin A
In ∆BCD sin B =
so h = a sin B
b. a sin B = b sin A
Exercise F: Arcs and
h
b
sectors
(pages 206–207)
h
a
1. a. i. 6 cm
Dividing both sides by ab and
cancelling gives
sin B
sin A
=
b
a
inverting and swapping gives
required result.
b. i. 2.66 cm ii. 4.66 cm2
c. i. 3.37 cm ii. 2.41 cm2
d. i. 16.9 cm ii. 108 cm
d. 65.6 m2
c. 19.7 cm
2. a. 78.5°
b. 4.6 m2
3. 12.9 ha
4. 13.7 m
5. 5 700 m2
6. 180 m2
7. 9.27 m
Exercise E: Radians and
degrees conversions
(page 205)
π
π
1. a. b.
4
6
3π
π
c.
d.
4
3
7π
3π
e.
f.
6
2
5π
g.
h. 5π
6
3
2. a. 1.6755
b. 0.4363
d. 7.4 cm (1 dp)
ii. 1.85 m2
c. i. 10.3 m
ii. 19.3 m2
Exercise A: Solving
d. i. 49.3 mm ii. 222 mm2
4. a. 14.7 cm
b. 6.0 cm
b. i. 2.46 m
triangle
(pages 203–204)
c. 17.2 m
2
2. a. i. 12.6 cm ii. 101 cm2
2
b. 90.4°
Chapter 9:
Achievement
Standard 90292
Exercise D: Area of a
1. a. 132.9 square units
ii. 15 cm2
3. a. 8 cm
2
2. a. 118 km/h at 062°
trigonometric equations
b. 40 cm2
using the sine graph
(pages 218–219)
b. 88.0 cm2
1. a. 65°, 115°
b. 245°, 295°
d. 209°, 331°
c. 28 cm
c. 38.7 cm
5. a. 1 radian
1
RS
c.
RS or
2
2
b. 15.9 cm
c. 29°, 151°
2. a. 1.5, 1.6416
Exercise G: Compound
b. 4.6416, 4.7832
c. 0.65, 2.4916
d. 3.7916, 5.6332
circular shapes and
3. a. i. 45.0°, 60.0°
segments
(pages 208–210)
1. a. 15.4 cm2
b. 1.21 m2
d. 82.5 cm2
c. 36.6 mm2
b. 0.0671 m2
d. 93.3 cm2
c. 5.99 m2
3. a. 21.3 cm2
b. 27.1 cm
4. a. 0.515 cm2
b. 6.66 cm
5 a. 44.6 cm2
b. 46.0 cm
6. a. 62.4 cm
2
ii. 240.0°, 300.0°
4. a. 0.6025, 2.5391
b. 3.8726, 5.5522
5. a. 56.0°, 124.0°, 416.0°, 484.0°
b. 236.0°, 304.0°, 596.0°,
664.0°
6. a. 1.000, 2.142
b. α = 0.65 radians
b. i. 45.0°, 135.0°
2. a. 126 cm2
ii. 60.0°
b. –1.000, –2.142
7. b. y = –0.6
7. 19.3 cm (3 sf)
b. sin x = – 0.6
c. –2.498, –0.6435, 5.640
2
c. 2.1293
d. 4.3633
AS 90291: Typical
e. 5.7596
f. 1.0175
g. 1.3177
h. 5.0318
assessment tasks
(pages 214–215)
3. a. 114.6°
b. 112.3°
1. a. i. 1 393 m2
c. 159.3° d. 4.0°
e. 80.2°
f. 49.3°
g. 182.7°
h. 276.5°
ii. 55.7 m
iii. 90.2°
b. i. 108.2 m
iii. 4 980 m
2
ii. 173.7 m
SAMPL
E
PAGES
ONLY
© ESA Publications (NZ) Ltd, Freephone 0800-372 266
Y12 Mat LW KP Book.indb 261
3/5/10 3:27:12 PM