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192
Chapter 9
9.5 Graphs of sine, cosine and tangent
■ The graphs of sine, cosine and tangent are periodic. They repeat themselves after a certain
inter
val.
You need to be able to draw the graphs for a given range of angles.
■ The graph of y = sin θ:
• repeats ever
y 360° and cro
sses the x-axis at …, −180°, 0, 180°, 360°, …
• has a maximum value of 1 and a minimum value of −1.
y
θ 90°y = sin θ
–90° –180° 180° 270° 360° 450° 540°01
1
2
–1–12
■ The graph of y = cos θ:
• repeats ever
y 360° and cro
sses the x-axis at …, −90°, 90°, 270°, 450°, …
• has a maximum value of 1 and a minimum value of −1.
y
θ 90° –90° –180° 180° 270° 360° 450° 540°01
12
–1–12y = co s θsin θ = −1 when θ = −90°, 270°, etc.sin θ = 1 when θ = 90°, 450°, etc.
sin θ = 0 when
θ = −180°, 0°, 180°, 360°, 540°, etc.
cos θ = 1 when θ = 0°, 360°, etc.
cos θ = 0 when
θ = −90°, 90°, 270°, 450, etc.
cos θ = −1 when θ = −180°, 180°, 540°, etc. | [
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193
Trigonometric ratios
■ The graph of y = tan θ:
• repeats ever
y 180° and cro
sses the x-axis at … −180°, 0°, 180°, 360°, …
• has no maximum or minimum value
•
has ver
tical as
ymptotes at x = −90°, x = 90°, x = 270°, …
y
θ –180° –120° –60° 60° 30° 120° 180° 240° 300° 360° –150° –90° –30° 90° 150° 210° 270° 330°0y = tan θ
tan θ = 0 when θ = 0°, 180°, 360°, etc.tan θ does not
have maximum and minimum points but approaches negative or positive
infinity as the curve approaches the asymptotes at −90°, 90°, 270°, etc. tan θ is undefined
for these values of θ.
Example 11
a Sketch the graph of y = cos θ in the interval −360° < θ < 360°.
b i Sketch the gra
ph of y = sin x in the interva
l −180° < x < 270 °
ii sin (−
30°) = − 0.5. Use your gra
ph to determine two further values of x for which sin x =
− 0.5.
a y
y = cos θ
θ90° 180° 270° 360° –90°
–11
–180° –270° –360°OThe axes are θ and y. The curve meets the θ-axis at θ = ±270° and θ = ±90°. The curve crosses the y-axis at (0, 1). | [
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194
Chapter 9
1 Sketch the gra
ph of y = cos θ in the interva
l −180° < θ < 180°.
2 Sketch the gra
ph of y = tan θ in the interva
l −180° < θ < 180°.
3 Sketch the gra
ph of y = sin θ in the interva
l −90° < θ < 270°.
4 a cos 30° = √ __
3 ___ 2 Use your graph in question 1 to find another value of θ for which cos θ = √ __
3 ___ 2
b tan 60° = √ __
3 . Use your graph in question 2 to find other values of θ for which:
i tan θ =
√ __
3 ii tan θ = − √ __
3
c sin 45° = 1 ___ √ __
2 Use your graph in question 3 to find other values of θ for which:
i sin θ =
1 ___ √ __
2 ii sin θ = − 1 ___ √ __
2 Exercise 9F
Example 12
Sketch on separate sets of axes the graphs of:
a y =
3 sin x, 0 <
x < 360°
b y =
−tan θ, −
180° < θ < 180°b i
y
y = sin x
x
–11
–180° –90° 90° 180° 270° O
ii Using the symmetry of the graph:
sin
(−150°) = − 0.5
sin 210° = − 0.5
x = −150° or 210°
9.6 Transforming trigonometric graphs
You can use your knowledge of transforming graphs to
transform the graphs of trigonometric functions. You need to be able to apply
tra
nslations and stretches to graphs
of trigonometric functions.
← Chapter 4LinksThe line x = −90° is a line of
symmetry.
The line x = 90° is a line of symmetry. You could also find this value by working out sin (180° − (−30°)). | [
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195
Trigonometric ratios
a y
y = 3 sin x ×3
x
–33
90° 180° 270° 360°
×3O
b y y = –t/a.ss01n θ
O x –180° –90° 90° 180°y = 3f(x) is a vertical stretch of the graph
y = f(x) with scale factor 3. The intercepts on the x-axis remain unchanged, and the graph has a maximum point at (90°, 3) and a minimum point at (270°, −3).
y = −f(x) is a reflection of the graph y = f(x) in the
x-axis. So this graph is a reflection of the graph
y = tan x in the x -axis.
Example 13
Sketch on separate sets of axes the graphs of:
a y =
−1 + sin x, 0 <
x < 360° b y =
1 _ 2 + cos x, 0 < x < 360°
a y
y = –1 + sin x
–1x
–2+1
90° 180° 270° 360°O
b y
y = + cos x
–x 360° 270° 180° 90°11
1
21
2
1
21
21
2
OOy = f(x) − 1 is a translation of the graph
y = f(x) by vector ( 0 −1 ) .
The graph of y = sin x is translated by 1 unit in
the negative y-direction.
y = f(x) + 1 _ 2 is a translation of the graph
y
= f(x) by vector ( 0
1 _ 2 ) .
The graph of y = cos x is translated by 1 _ 2 unit in
the positive
y-direction. | [
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196
Chapter 9
Example 14
Sketch on separate sets of axes the graphs of:
a y =
tan (θ
+ 45°), 0 < θ < 360° b y =
cos (θ
− 90°), −360° < θ < 360°
y = f(θ + 45°) is a translation of
the graph y = f(θ ) by vector ( −45° 0 ) .
Remember to translate any
asymptotes as well.
y = f(θ − 90°) is a translation of
the graph y = f(θ ) by vector ( 90° 0 ) .a
y
Oy = t/a.ss01n (θ + /four.ss015°)
/four.ss015°
θ1
/four.ss015° 135° 225° 315° 360°
b y
θ O 90°90°
–90° –180° –270° –360° 180° 270° 360°1
–1y = cos (θ – 90°)
The graph of y = cos θ is translated
by 90° to the right. Note that this
is exactly the same curve as
y = sin θ , so another property is
that cos (θ − 90°) = sin θ .The graph of y = tan θ is translated
by 45° to the left. The asymptotes are now at θ = 45° and θ = 225°.
The curve meets the y -axis where
θ = 0 so y = 1.
Example 15
Sketch on separate sets of axes the graphs of:
a y =
sin 2x
, 0 < x < 360° b y =
cos θ __ 3 , −540° < θ < 540° c y = tan (− x), −360° < x < 360°
a y
Oy = sin 2x
x90° 180° 270° 360°1
–1y = f(2x) is a horizontal stretch of the graph
y = f(x) with scale factor 1 _ 2 .
The graph of y = sin x is str etched horizontally
with scale factor 1 _ 2
The period is now 180° and tw
o complete ‘waves’
are seen in the interval 0 < x < 360°. | [
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197
Trigonometric ratios
b y
y = cos 1
–1θ180° –180° 360° –360° –5/four.ss010° 5/four.ss010°θ
3
O
c y
xy = t/a.ss01n (–x)
O 360° 180° –360° –180°y = f( 1 _ 3 θ ) is a horizontal stretch of the graph
y
= f(θ ) with scale factor 3.
y = f(− x) is a reflection of the graph y = f(x) in
the y-axis.The graph of y = cos θ is str etched horizontally
with scale factor 3. The period of cos θ __ 3 is 1080°
and only one complete wave is seen while
−540 < θ < 540°. The curve crosses the θ-axis at
θ = ±270°.
The graph of y = tan (−x) is refl ected in the y-axis.
In this case the asymptotes are all vertical so they remain unchanged.
1 Write down i the maximum value, and ii the minimum value, of the following expressions,
and in each case giv
e the smallest positive (or zero) value of x for which it occurs.
a cos x b 4 sin x c cos (−
x)
d 3 + sin x e −sin x f sin 3x
2 Sketch, on the same set of ax
es, in the interval 0 < θ < 360°, the graphs of cos θ and cos 3θ.
3 Sketch, on separa
te sets of axes, the graphs of the following, in the interval 0 < θ < 360°.
Give the coordinates of points of intersection with the axes, and of maximum and minimum points where appropriate.
a
y =
−cos θ b y =
1 _ 3 sin θ c y = sin 1 _ 3 θ d y = tan (θ − 45°)
4 Sketch, on separa
te sets of axes, the graphs of the following, in the interval −180° < θ < 180°.
Give the coordinates of points of intersection with the axes, and of maximum and minimum
points where appropriate.
a y =
−2 sin θ b y =
tan (θ
+ 180°) c y =
cos 4θ d y =
sin (−
θ )
5 Sketch, on separa
te sets of axes, the graphs of the following in the interval −360° < θ < 360°.
In each case give the periodicity of the function.a
y =
sin 1 _ 2 θ b y = − 1 _ 2 cos θ c y = tan (θ − 90°) d y = tan 2θExercise 9G Plot transformations of
tri
gonometric graphs using GeoGebra.Online | [
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198
Chapter 9
6 a By considering the graphs of
the functions, or otherwise, verify that:
i cos θ =
cos (−
θ )
ii sin θ =
−sin (−
θ )
iii sin (θ
− 90°) = −cos θ.
b Use the results in a ii
and iii to show that sin (90° −
θ ) = cos θ.
c In Example 14 you saw tha
t cos (θ
− 90°) = sin θ.
Use this result with part
a i to show that cos (90° −
θ ) = sin θ.
7 The graph sho
ws the curve y
θ 90° 180° 270° 360° –90°
–11
–180° –270° –360°0
y = cos (x
+ 30°), −360° < x < 360°.
a Write down the coor
dinates of
the points where the curve
crosses the x-axis. (2 marks)
b Find the coordinates of
the
point where the curve crosses the y-axis. (1 mark)
8 The graph sho
ws the curve with equation y
θ 120° 300° –60°
–11
–240°0
y = sin (x + k), −360° < x < 360°,
where k is a constant.
a Find one possible va
lue
for k. (2 marks)
b Is there more than one possib
le
answer to part a? Give a reason for your answer.
(2 marks)
9 The variation in the depth of
water in a rock pool can be modelled using the function
y = sin (30t
)°, where t is the time in hours and 0 < t < 6.
a Sketch the function for the gi
ven interval. (2 marks)
b If t = 0 r
epresents midday, during what times will the rock pool be at least half full? (3 marks)P
E
E/P
E/P
Give non-exact answers to 3 significant figures.
1 Triangle ABC
has area 10 cm2. AB = 6 cm, BC = 8 cm and ∠ABC is obtuse. Find:
a the size of
∠ABC
b the length of AC
2 In each triangle below
, find the size of x and the area of the triangle.
40°x
80°xx2.4 cm6 cm
5 cm5 cm
3 cm1.2 cm
3 cmabcMixed Exercise 9 | [
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199
Trigonometric ratios
3 The sides of a triangle are 3 cm, 5 cm and 7 cm respectiv
ely. Show that the largest angle is 120°,
and find the area of the triangle.
4 In each of the figures be
low calculate the total area.
ab
A A
D DC
CB
8.2 cm10.4 cm
4.8 cm3.9 cm
75°100°
30.6°B
2.4 cm
5 In △ABC, AB = 10 cm, BC = a √ __
3 cm, AC = 5 √ ___ 13 cm and ∠ ABC = 150°. Calculate:
a the value of
a
b the exact area of
△ABC.
6 In a triangle, the largest side has length 2 cm and one of the other sides has length √ __
2 cm.
Given tha
t the area of the triangle is 1 cm2, show that the triangle is right-angled and isosceles.
7 The three points A
, B and C, with coordinates A(0, 1), B(3, 4) and C (1, 3) respectiv
ely, are
joined to form a triangle.
a Show that cos ∠ACB
= − 4 _ 5 (5 marks)
b Calculate the ar
ea of △ABC. (2 marks)
8 The longest side of a triangle has length (2x
− 1) cm. The other sides have lengths (
x − 1) cm
and (x
+ 1) cm. Given tha
t the largest angle is 120°, work out
a the value of
x (5 marks)
b the area of the triangle
. (3 marks)
9 A park is in the shape of
a triangle ABC as shown.
110°
1.2 km1.4 km
C
ABN
A park keeper walks due north from his hut at A until he
reaches point B. He then walks on a bearing of 110° to point C.
a Find how far he is from his hut w
hen at point C.
Give your answer in km to 3 s.f. (3 marks)
b Work out the bearing of
the hut from point C.
Give your answer to the nearest degree. (3 marks)
c Work out the ar
ea of the park. (3 marks)
10 A windmill has four identical triangular sails made fr
om wood. If each triangle has sides of
length 12 m, 15 m and 20 m, work out the tota
l area of wood needed. (5 marks)
11 Two points
, A and B are on level ground. A church tower at point C has an angle of elevation
from A of 15° and an angle of elevation from B of 32°. A and B are both on the same side of
C, and A, B and C lie on the same straight line. The distance AB = 75 m.
Find the height of the chur
ch tower. (4 marks)P
P
E/P
E/P
E/P
E/P
E/P | [
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200
Chapter 9
12 Describe geometrically the tr
ansformations which map:
a the graph of
y = tan x onto the gra
ph of tan 1 _ 2 x
b the graph of
y = tan 1 _ 2 x onto the gra ph of 3 + tan 1 _ 2 x
c the graph of
y = cos x onto the gra
ph of −cos x
d the graph of
y = sin (x
− 10) onto the graph of sin (x
+ 10).
13 a Sketch on the same set of ax
es, in the interval 0 < x < 180°, the graphs of y = tan (x
− 45°)
and y = −2 cos x, sho
wing the coordinates of points of intersection with the axes. (6 marks)
b Deduce the number of solutions of
the equation tan (x
− 45°) + 2 cos x =
0, in the interval
0 < x < 180°. (2 marks)
14 The diagram sho
ws part of the graph of y = f(x). y
x 0 pC q
AD
B
120°
It crosses the x-axis at A(120°, 0) and B( p, 0).
It crosses the
y-axis at C (0, q) and has a maxim
um
value at D, as shown.
Given that f(x) = sin (x
+ k), where k > 0, write down
a the value of
p (1 mark)
b the coordinates of
D (1 mark)
c the smallest va
lue of k (1 mark)
d the value of
q. (1 mark)
15 Consider the function f(x)
= sin px, p
∈ ℝ, 0 < x < 360°.
The closest point to the origin that the graph of f(x) crosses the x-axis has x-coordinate 36°.
a Determine the va
lue of p and sketch the graph of y = f(x). (5 marks)
b Write down the period of
f(x). (1 mark)
16 The graph be
low shows y = sin θ, 0 <
θ < 360°, with one y
θ01
–190°α180° 270° 360°
value of θ (θ
= α) marked on the axis.
a Copy the gra
ph and mark on the θ -axis the positions
of 180° − α, 180° + α, and 360° − α.
b Verify that:
sin α = sin (180° − α) = −sin (180° + α) = −sin (360° − α).
17 a Sketch on separa
te sets of axes the graphs of y = cos θ (0 <
θ < 360°) and
y = tan θ (0 <
θ < 360°), and on each θ-axis mark the point ( α, 0) as in question 16.
b Verify that:i
cos α = −cos (180° − α) = −cos (180° + α) = cos (360° − α)
ii tan α = −tan (180° − α) = tan (180° + α) = −tan (360° − α)
18 A series of sand dunes has a cross-section w
hich can be modelled using a sine curve of the form
y = sin (60x
)° where x is the length of the series of dunes in metres.
a Draw the gr
aph of y = sin (60x
)° for 0 < x < 24°. (3 marks)
b Write down the n
umber of sand dunes in this model. (1 mark)
c Give one r
eason why this may not be a realistic model. (1 mark)E/P
E
E/P
E/P | [
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201
Trigonometric ratios
1 This version of the cosine rule is used to find a missing
side if you kno
w two sides and the angle between them:
a2 = b2 + c2 − 2bc cos A
2 This version of the c
osine rule is used to find an angle if
you know all three sides:
cos A =
b2 + c2 − a2 __________ 2bc
3 This version of the sine rul
e is used to find the length of a missing side:
a _____ sin A = b _____ sin B = c _____ sin C
4 This version of the sine rul
e is used to find a missing angle:
sin A _____ a = sin B _____ b = sin C _____ c
5 The sine rule sometimes produc
es two possible solutions for a
b bcA
BC1 C2
missing angle:
sin θ =
sin (180° −
θ )
6 Area of a t
riangle = 1 _ 2 ab sin C.
7 The graphs of sine, c
osine and tangent are periodic. They repeat themselves after a certain
interval.
• The graph of y = sin θ: repeats ever
y 360° and crosses the x -axis at …, − 180°, 0, 180°, 360°, …
• has a maximum value of 1 and a minimum value of −1.
•
The graph of y = cos θ: repeats ever
y 360° and crosses the x -axis at …, − 90°, 90°, 270°, 450°, …
• has a maximum value of 1 and a minimum value of −1
•
The graph of y = tan θ: repeats ever
y 180° and crosses the x -axis at … − 180°, 0°, 180°, 360°, …
• has no maximum or minimum value
•
has ver
tical as
ymptotes at x
= −90°, x = 90°, x = 270°, …A
CB
ac
bSummary of key pointsIn this diagram AB = BC = CD = DE = 1 m.
B1 m1 m
1 m1 mA
CD E
Prove that ∠AEB + ∠ADB = ∠ACB .Challenge
Try drawing triangles ADB and
AEB back to back.
BA
BA
DEHint | [
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202
After completing this chapter you should be able to:
● Calculat
e the sine, cosine and tangent of any angle → pages 203–208
● Know the exact trigonometric ratios for 30°, 45° and 60° → pages 208–209
● Know and use the relationships tan θ ; sin θ _____ cos θ
and sin2 θ + cos2 θ ; 1 → pages 209–213
● Solve simple trigonometric equations of the forms sin θ = k,
cos θ = k and tan θ = k → pages 213–217
● Solve more complicated trigonometric equations of the
forms sin nθ = k and sin (θ ± α) = k and equivalent equations
involving cos and tan → pages 217–219
● Solve trigonometric equations that produce quadratics → pages 219–222Objectives
1 a Sketch the graph of y = sin x for 0 < x < 540°.
b How many solutions ar
e there to the equation
sin x =
0.6 in the range 0 < x < 540°?
c Given that sin−1(0.6) = 36.9° (to 3 s.f.), write
down three other solutions to the equation sin
x =
0.6. ← Section 9.5
2 Work out the marked angles in these triangles.a
θ 16.3 cm8.7 cm bθ6.1 cm
20 cm
← GCSE Mathematics
3 Solve the following equations.
a 2x
– 7 = 15 b 3x
+ 5 = 7x – 4
c sin x =
–0.7 ← GCSE Mathematics
4 Solve the following equations.a
x2 – 4x + 3 = 0 b x2 + 8x – 9 = 0
c 2x2 – 3x – 7 = 0 ← Section 2.1Prior knowledge check
Trigonometric equations can be used
to model many real-life situations such as the rise and fall of the tides or the angle of elevation of the sun at different times of the day.Trigonometric
identities and equations 10 | [
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-0.06948476284742355,
0.0856902003288269,
-0.14321191608905792,
-0.02322051115334034,
0.06849046796560287,
-0.043340858072042465,
-0.0355055145919323,
0.06... |
203Trigonometric identities and equations
10.1 Angles in all four quadr ants
You can use a unit circle with its centre at the origin
to help you understand the trigonometric ratios.
■ For a point P(x, y) on a unit circle such that
OP makes an angle θ with the positive x-axis:
• cos θ = x = x-coordinate of P
• sin θ = y = y-coordinate of P
• tan θ = y __ x = gr adient of OP
You can use these definitions to find the values of sine, cosine and tangent for any angle θ. You always
measure positive angles θ anticlockwise from the positive x-axis.
1(x,y)
θP
xOy
yx
Use GeoGebra to explore the
va
lues of sin θ, cos θ and tan θ for any
angle θ in a unit circle.Online
You can also use these definitions to generate the graphs of y = sin θ and y = cos θ.
Py
y = sin θ
45° 90° 180° 270° 360°1
–10O (–1, 0)45°
(1, 0)
(0, –1)θ(0, 1)
y
y = cos θ45°
90°
180°
270°
360°1 –1 0
θ The point P corresponding to an angle θ
is the same as the point P corresponding to an
angle θ + 360°. This shows you that the graphs
of y = sin θ and y = cos θ are periodic with period
360°. ← Section 9.5LinksTo plot y = sin θ, read off
the y-coordinates as P moves around the circle.
To plot y = cos θ, read off the
x-coordinates as P moves around the circle.1P
x Oy
x(x,y)
yθ
A unit
circ
le is a circle with
a radius of 1 unit.Notation
When θ is obtuse,
cos θ is negative
because the x-coordinate of P is negative. | [
0.005260057281702757,
0.05481402575969696,
0.03412600979208946,
0.0145505890250206,
-0.10791610181331635,
-0.0027307935524731874,
0.013563727028667927,
-0.02047872729599476,
-0.07544013112783432,
0.00233823386952281,
0.08433511108160019,
-0.04725604131817818,
-0.060158610343933105,
0.04963... |
204
Chapter 10
Example 1
Write down the values of:
a sin 90° b sin 180° c sin 270°
d cos 180° e cos (−
90)° f cos 450°
a sin 90° = 1
b sin 18
0° = 0
c sin 27
0° = −1
d cos 18
0° = −1
e cos (−90
°) = 0
f cos 45
0° = 0
Example 2
Write down the values of:a
tan 45° b tan 135° c tan 225°
d tan (−
45°) e tan 180° f tan 90°
a tan 45° = 1
b tan 135°
= −1
c tan 22
5° = 1
d tan (−45°
) = tan 31
5° = −1
e tan 18
0° = 0
f tan 90
° = undefinedWhen θ = 45°, the coordinates of OP are
( 1 ___
√ __
2 , 1 ___
√ __
2 ) so the gradient of OP is 1.
O xy
( , )1
2
45°12
When θ = –45° the gradient of OP is –1.
When θ = 180°, P has coordinates (–1, 0) so the
gradient of OP = 0 __ 1 = 0.
When θ = 90°, P has coordinates (0, 1) so the
gradient of OP = 1 __ 0 . This is undefined since you
cannot divide by zero. tan θ is undefined when θ = 270° or any
other odd multiple of 90°. These values of θ
correspond to the asymptotes on the graph of
y = tan θ. ←Section 9.5LinksThe y-coordinate is 1 when θ = 90°.
O xy
(0, 1)
(0, –1)90°
-90°
If θ is negative, then measure clockwise from the
positive x-axis.
An angle of −90° is equivalent to a positive angle
of 270°. The x-coordinate is 0 when θ = −90° or
270°. | [
-0.009879173710942268,
0.0059527065604925156,
-0.000645646417979151,
-0.008359718136489391,
-0.058040034025907516,
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-0.004918475169688463,
0.054447077214717865,
-0.008084431290626526,
0.05877421796321869... |
205Trigonometric identities and equations
The x-y plane is divided into quadrants:
y
xFirst
quadrantSecond
quadrant
Fourth
quadrantThird
quadrantO+90° –270°
+270°
–90°0
+360°–360° –180° +180°Angles may lie outside the range 0–360°, but they
will always lie in one of the four quadrants. For example, an angle of 600° would be equivalent to 600° – 360° = 240°, so it would lie in the third quadrant.
Example 3
Find the signs of sin θ, cos θ and tan θ in the second quadrant.
P(x, y)y
y
xO1
θ
x
As x is − ve and y is + ve in this quadrant
sin θ = +ve
cos θ = −ve
tan θ = +ve _____ −ve = −ve
So only sin θ is positive.Draw a circle, centre O and radius 1, with P(x, y)
on the circle in the second quadrant.In the second quadrant, θ is obtuse,
or 90° < θ < 180°.
Thi s diagram is often referred to as
a CAST diagram since the word is spelled out from the bottom right going anti-clockwise.Notationy90°
All
CosSin
Tan
270°0, 360° 180°x■ You can use the quadrants to determine whether each of the trigonometric ratios is positive
or negative.
For an angle θ in the
second quadrant, only
sin θ is positive.
For an angle θ in the
third quadrant, only tan θ
is positive.For an angle θ in the first quadrant,
sin θ , cos θ and tan θ are all positive.
For an angle θ in the fourth
quadrant, only cos θ is positive. | [
0.008106395602226257,
-0.0381806343793869,
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-0.028636878356337547,
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-0.032113201916217804,
0.024712922051548958,
-0.023540999740362167,
-0.0033954798709601164,
0.... |
206
Chapter 10
■ You can use these rules to find sin, cos or tan of any
positive or negative angle using the corresponding acute angle made with the
x-axis, θ.
A
CS
T360° – θ 180° + θ180° – θ
θ
θθθ
θy
x
cos (180° − θ ) = − cos θ
cos (180° + θ ) = − cos θ
cos (360° − θ ) = cos θtan (180° − θ ) = − tan θ
tan (180° + θ ) = tan θ
tan (360° − θ ) = − tan θsin (180° − θ ) = sin θ
sin (180° + θ ) = − sin θ
sin (360° − θ ) = − sin θ
Example 4
Express in terms of trigonometric ratios of acute angles:
a sin (−
100°) b cos 330° c tan 500°
a
80°A
CS
T80°
–100°y
x O
P
The acute angle made with the x -axis is 80°.
In the third quadrant only tan is + ve,
so sin is − ve.
So sin (−10
0)° = − sin 80
°
b
+330°30°y
O xA
CS
TP
The acute angle made with the x -axis is 30°.
In the fourth quadrant only cos is + ve.
So cos 33
0° = +cos 30
°For each part, draw diagrams showing the
position of OP for the given angle and insert the acute angle that OP makes with the x-axis. | [
-0.03881235793232918,
0.02005922980606556,
-0.009691301733255386,
-0.05000996217131615,
0.013693052344024181,
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0.009826259687542915,
0.0524747371673584,
0.0030614235438406467,
0.059461161494255066,
0.070... |
207Trigonometric identities and equations
c
40°+500°y
O xA
CS
TP
The acute angle made with the x -axis is 40°.
In the second quadrant only sin is + ve.
So tan 50
0° = −tan 40
°
Exercise 10A
1 Draw diagrams to show the following angles. Mark in the acute angle that OP makes with the
x-axis
.
a −80° b 100° c 200° d 165° e −145°
f 225° g 280° h 330° i −160° j −280°
2 State the quadrant tha
t OP lies in when the angle that OP makes with the positive x-axis is:
a 400° b 115° c −210° d 255° e −100°
3 Without using a calcula
tor, write down the values of:
a sin (−
90°) b sin 450° c sin 540° d sin (−
450°) e cos (−
180°)
f cos (−
270°) g cos 270° h cos 810° i tan 360° j tan (−
180°)
4 Express the follo
wing in terms of trigonometric ratios of acute angles:
a sin 240° b sin (−
80°) c sin (−
200°) d sin 300° e sin 460°
f cos 110° g cos 260° h cos (−
50°) i cos (−
200°) j cos 545°
k tan 100° l tan 325° m tan (−
30°) n tan (−
175°) o tan 600°
5 Given tha
t θ is an acute angle, express in terms of sin θ:
a sin (−θ ) b sin (180° + θ ) c sin (360° − θ )
d sin (−
(180° + θ )) e sin (− 180° + θ ) f sin (− 360° + θ )
g sin (540° + θ ) h sin (720° − θ ) i sin (θ + 720°) The results
ob
tained in questions
5 and 6 are true for all
values of θ.Hint
6 Given tha
t θ is an acute angle, express in terms of cos θ or tan θ:
a cos (180° − θ ) b cos (180° + θ ) c cos (−θ ) d cos (−(180° − θ ))
e cos (θ − 360°) f cos (θ − 540°) g tan (−θ ) h tan (180° − θ )
i tan (180° + θ ) j tan (− 180° + θ ) k tan (540° − θ ) l tan (θ − 360°) | [
-0.04603499919176102,
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0.09010346978902817,
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-0.010111751034855843,
-0.04867052286863327,
0.033200256526470184,
0... |
208
Chapter 10
Draw a diagram showing the positions of θ and
180° – θ on the unit circle.Problem-solving
a Prove that sin (18 0° − θ ) = sin θ
b Prove that cos (−θ ) = cos θ
c Prove that tan (18 0° − θ ) = −t an θChallenge
10.2 Exact values of trigonometrical ratios
You can find sin, cos and tan of 30°, 45° and 60° exactly using triangles.
Consider an equilateral triangle ABC of side 2 units. Draw a perpendicular from A to meet BC at D.Apply the trigonometric ratios in the right-angled triangle ABD.
■ sin 30° = 1 __ 2 cos 30° = √ __
3 ___ 2 tan 30° = 1 ___
√ __
3 = √ __
3 ___ 3
sin 60° = √ __
3 ___ 2 cos 60° = 1 __ 2 tan 60° = √ __
3
Consider an iso
sceles right-angled triangle PQR with
PQ = RQ = 1 unit.
■ sin 45° = 1 ___
√ __
2 = √ __
2 ___ 2 cos 45° = 1 ___
√ __
2 = √ __
2 ___ 2 tan 45° = 130°
2
160°A
BCD3
BD = 1 unitAD = √ ______ 22 − 12 = √ __
3
PR = √ ______ 12 + 12 = √ __
2
11
RQP
245°
45°
Example 5
Find the exact value of sin (−210°).
O xy
–210°150°
sin (−210°) = sin (1 50°) = sin (3 0°) = 1 __ 2 sin (−210°) = sin (150°)
Use sin (180° − θ ) = sin θ | [
-0.051817361265420914,
0.04967747628688812,
0.03546268492937088,
0.004772591404616833,
-0.029259521514177322,
0.028434986248612404,
-0.056135814636945724,
0.021161822602152824,
-0.0642160028219223,
-0.04046836122870445,
0.05806143954396248,
-0.05493220314383507,
0.03557640314102173,
0.0205... |
209Trigonometric identities and equations
Exercise 10B
1 Express the following as trigonometric ratios of either 30°, 45° or 60°, and hence find their
exact v
alues.
a sin 135° b sin (−
60°) c sin 330° d sin 420° e sin (−
300°)
f cos 120° g cos 300° h cos 225° i cos (−
210°) j cos 495°
k tan 135° l tan (−
225°) m tan 210° n tan 300° o tan (−
120°)
The diagram shows an isosceles right-angled triangle ABC .
AE = DE = 1 unit. Angle ACD = 30°.
a Cal
culate the exact lengths of
i CE ii DC iii BC iv DB
b Sta
te the size of angle BCD .
c Hen
ce find exact values for
i sin 15° ii cos 15°45° 30°1
1A CEDBChallenge
10.3 Trigonometric identities
You can use the definitions of sin, cos and tan, together with
1P
Oy
x(x,y)
θ
Pythagoras’ theorem, to find two useful identities.
The unit circle has equation x2 + y2 = 1.
Since cos θ = x and sin θ = y, it follows that cos2θ + sin2θ = 1.
■ For all values of θ, sin2θ + cos2θ ≡ 1.
Since tan θ = y __ x it follows that tan θ = sin θ _____ cos θ
■ For all values of θ such that cos θ ≠ 0, tan θ ≡ sin θ _____ cos θ
You can use these two identities to simplify
trigonometrical expressions and complete proofs. The equation of a circle with radius r and
ce
ntre at the origin is x2 + y2 = r2. ← Section 6.2Links
tan θ is undefined when
the denominator = 0. This occurs
when cos θ = 0, so when θ = … – 90°,
90°, 270°, 450°, …Watch out The se results are called
trigonometric identities. You use the ≡ symbol instead of = to show that they are always true for all values of
θ (subject to any conditions given).Notation | [
0.010279553011059761,
0.0428256057202816,
0.05937739089131355,
-0.01879449188709259,
-0.07244641333818436,
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0.07313168048858643,
-0.04478728771209717,
-0.026703035458922386,
-0.015... |
210
Chapter 10
Example 6
Simplify the following expressions:
a sin2 3θ + cos2 3θ b 5 − 5 sin2 θ c sin 2θ __________
√ _________ 1 − sin2 2θ
a sin2 3θ + cos2 3θ = 1
b 5 −
5 sin2 θ = 5(1 − sin2 θ )
= 5 co
s2 θ
c sin 2θ ____________
√ __________ 1 − sin2 2θ = sin 2θ ________
√ _______ cos2 2θ
= sin 2θ ______ cos 2θ
= tan 2θsin2 θ + cos2 θ = 1, with θ replaced by 3 θ.
sin2 2θ + cos2 2θ = 1, so 1 − sin2 2θ = cos2 2θ.
Example 7
Prove that cos4 θ − sin4 θ ____________ cos2 θ ; 1 − tan2 θ
LHS ≡ cos4 θ − sin4 θ _____________ cos2 θ
≡ (cos2 θ + sin2 θ )(cos2 θ − sin2 θ ) ____________________________ cos2 θ
≡ (cos2 θ − sin2 θ ) ______________ cos2 θ
≡ cos2 θ ______ cos2 θ − sin2 θ ______ cos2 θ
≡ 1 − tan2 θ = RHSThe numerator can be factorised as the
‘difference of two squares’.
Divide through by cos2 θ and note that
sin2 θ _____ cos2 θ ≡ ( sin θ _____ cos θ ) 2
≡ tan2 θ.tan θ = sin θ _____ cos θ , so sin 2θ ______ cos 2θ = tan 2θ.Always look for factors.
sin2 θ + cos2 θ = 1, so 1 − sin2 θ = cos2 θ.
When you have to prove an identity like this you may quote the basic identities like ‘sin
2 + cos2 ≡ 1’.Problem-solving
To prove an identity, start from the left-hand side, and manipulate the expression until it matches the right-hand side.
← Sections 7.4, 7.5
Example 8
a Given that cos θ = − 3 __ 5 and that θ is reflex, find the value of sin θ.
b Given tha
t sin α = 2 __ 5 and that α is obtuse, find the exact value of cos α.sin2 θ + cos2 θ ≡ 1. | [
-0.09999018162488937,
0.08629908412694931,
0.02746148779988289,
-0.07336703687906265,
-0.0650477334856987,
0.051825862377882004,
0.00005612675522570498,
-0.058786701411008835,
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-0.012531319633126259,
0.03376102074980736,
0.05836615338921547,
0.029999587684869766,
0.024... |
211Trigonometric identities and equations
Example 9
Given that p = 3 cos θ, and that q = 2 sin θ, show that 4p2 + 9q2 = 36.
As p = 3 cos θ, and q = 2 sin θ,
cos θ = p __ 3 and sin θ = q __ 2
Using
sin2 θ + cos2 θ ≡ 1,
( q __ 2 ) 2
+ ( p __ 3 ) 2
= 1
so q2
___ 4 + p2
___ 9 = 1
∴ 4p2 + 9 q2 = 36You need to eliminate θ from the equations.
As you can find sin θ and cos θ in terms of p
and q , use the identity sin2 θ + cos2 θ ≡ 1.Problem-solvinga Since sin2 θ + cos2 θ ≡ 1,
sin2 θ = 1 − (− 3 __ 5 ) 2
= 1 − 9 ___ 25
= 16 ___ 25
So si
n θ = − 4 __ 5
b Using
sin2 α + cos2 α ≡ 1,
cos2 α = 1 − 4 ___ 25 = 21 ___ 25
As α is obtuse, cos α is negative
so cos α = − √ ___ 21 ____ 5 Obtuse angles lie in the second quadrant, and
have a negative cosine.
The question asks for the exact value so leave your answer as a surd.
Exercise 10C
1 Simplify each of the follo wing expressions:
a 1 − cos2 1 _ 2 θ b 5 sin2 3θ + 5 cos2 3θ c sin2 A − 1
d sin θ _____ tan θ e √ _________ 1 − cos2 x __________ cos x f √ __________ 1 − cos2 3A ___________
√ __________ 1 − sin2 3A
g (1 + sin x)2 + (1 − sin x)2 + 2 cos2 x h sin4 θ + sin2 θ cos2 θ
i sin4 θ + 2 sin2 θ cos2 θ + cos4 θ
2 Given that 2 sin θ = 3 cos θ, find the value of tan θ.
3 Given tha
t sin x cos y =
3 cos x sin y, expr
ess tan x in terms of
tan y.‘θ is reflex’ means θ is in the 3rd or 4th quadrants,
but as cos θ is negative, θ must be in the 3rd
quadrant. sin θ = ± 4 _ 5 but in the third quadrant
sin θ is negative. If yo u use your calculator to find
cos–1 (– 3 _ 5 ) , then the sine of the result, you will get
an i
ncorrect answer. This is because the cos–1
function on your calculator gives results between
0 and 180°.Watch out
Multiply both sides by 36. | [
-0.01568686030805111,
0.07753204554319382,
0.023014064878225327,
-0.030406074598431587,
-0.06038428470492363,
0.0497027225792408,
-0.04388625547289848,
-0.021755123510956764,
-0.09227098524570465,
-0.02307935431599617,
0.07817547023296356,
-0.010161170735955238,
-0.04360487312078476,
0.039... |
212
Chapter 10
4 Express in terms of
sin θ only:
a cos2 θ b tan2 θ c cos θ tan θ
d cos θ _____ tan θ e (cos θ − sin θ )(cos θ + sin θ )
5 Using the identities sin2 A + cos2 A ≡ 1 and/or tan A = sin A _____ cos A (cos A ≠ 0), prove that:
a (sin θ + cos θ )2 ≡ 1 + 2 sin θ cos θ b 1 _____ cos θ − cos θ ≡ sin θ tan θ
c tan x + 1 _____ tan x ≡ 1 _________ sin x cos x d cos2 A − sin2 A ≡ 2 cos2 A − 1 = 1 − 2 sin2 A
e (2 sin θ − cos θ )2 + (sin θ + 2 cos θ )2 ≡ 5 f 2 − (sin θ − cos θ )2 ≡ (sin θ + cos θ )2
g sin2 x cos2 y − cos2 x sin2 y ≡ sin2 x − sin2 y
6 Find, without using your calcula
tor, the values of:
a sin θ and cos θ, given that tan θ = 5 __ 12 and θ is acute.
b sin θ and cos θ, given that cos θ = − 3 _ 5 and θ is obtuse.
c cos θ and tan θ, given that sin θ = − 7 __ 25 and 270° < θ < 360°.
7 Given tha
t sin θ = 2 _ 3 and that θ is obtuse, find the exact value of: a cos θ b tan θ
8 Given that tan θ = − √ __
3 and that θ is reflex, find the exact value of: a sin θ b cos θ
9 Given that cos θ = 3 _ 4 and that θ is reflex, find the exact value of: a sin θ b tan θ
10 In each of the follo wing, eliminate θ to give an equation relating x and y:
a x =
sin θ, y = cos θ b x = sin θ, y = 2 cos θ
c x = sin θ, y = cos2 θ d x = sin θ, y = tan θ
e x = sin θ + cos θ, y = cos θ − sin θ In part e find expressions for x + y and x − y.Problem-solving
11 The diagram sho
ws the triangle ABC with AB = 12 cm,
ABC
10 cm 8 cm
12 cm
BC = 8 cm and AC =
10 cm.
a Show that cos B =
9 ___ 16 (3 marks)
b Hence find the exact va
lue of sin B. (2 marks)
Use the cosine rul e: a2 = b2 + c2 − 2bc cos A ← Section 9.1 Hint
12 The diagram sho
ws triangle PQR with PR = 8 cm, Q
RP6 cm
8 cm30°
QR = 6 cm and angle QPR
= 30°.
a Show that sin Q =
2 __ 3 (3 marks)
b Given tha
t Q is obtuse, find the exact value
of cos Q (2 marks)P
P
E/P
E/P | [
-0.060138676315546036,
0.09009862691164017,
0.05405065789818764,
-0.034870605915784836,
-0.07936310023069382,
0.044792499393224716,
0.06418002396821976,
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-0.019168876111507416,
0.023969756439328194,
-0.00858987681567669,
0.002607664791867137,
0.09... |
213Trigonometric identities and equations
Example 10
Find the solutions of the equation sin θ = 1 _ 2 in the interval 0 < θ < 360°.
Method 1
sin θ = 1 __ 2
So θ = 30°
A
CS
T30° 150°
30° 30°
So x = 30°
or x =
180° − 30° = 150°
Method 2
y
Oy =
θ 90° 180° 270° 360°1
2
sin θ = 1 __ 2 where the line y = 1 __ 2 cuts the curve.
Hen
ce θ = 30° or 150°You can check this by putting sin 150° in your
calculat
or.Putting 30° in the four positions shown gives the
angles 30°, 150°, 210° and 330° but sine is only positive in the 1st and 2nd quadrants.
Draw the graph of y = sin θ for the given interval.
■ When you use the inverse trigonometric functions on your calculator, the angle you get is
called the princip
al value.
Your calculator will give principal values in the following ranges:
cos−1 in the range 0 < θ < 180°
sin−1 in the range −90° < θ < 90°
tan−1 in the range −90° < θ < 90° Use the symmetry properties of the
y = sin θ graph. ← Sections 9.510.4 Simple trigonometric equations
You need to be able to solve simple trigonometric equations of the form sin θ = k and cos θ = k
(where −1 < k < 1) and tan θ = p (where p ∈ 핉) for given intervals of θ.
■ Solutions to sin θ = k and cos θ = k only exist
when −1 < k < 1.
■ Solutions to tan θ = p exist for all values of p. The graphs of y = sin θ and y = cos θ
have a maximum value of 1 and a minimum value of – 1.
The graph of y = tan
θ has no maximum or
minimum value. ← Section 9.5Links
The i nverse
trigonometric functions are also called arccos ,
arcsin and arctan .Notation | [
0.017883963882923126,
0.006526624783873558,
0.029291434213519096,
-0.01747271418571472,
-0.06323817372322083,
-0.011628393083810806,
-0.03729813173413277,
-0.0200732983648777,
-0.033409975469112396,
-0.05275741219520569,
0.004155786242336035,
-0.03762640804052353,
0.010529776103794575,
0.0... |
214
Chapter 10
Example 11
Solve, in the interval 0 < x < 360°, 5 sin x = −2.
Method 1
5 sin x =
−2
sin x =
−0.4
Principal value is x = − 23.6° (3 s.f.)
23.6° 23.6°A
CS
T
x = 203.6° (204° to 3 s.f.)
or x
= 336.4° (336° to 3 s.f.)
Method 2
Oy
x
–1
–2–90° 90° 180° 270° 360°1
sin−1(−0.4) = − 23.578…°
x = 203.578…° (204° to 3 s.f.)
or x = 336.421…° (336° to 3 s.f.)First rewrite in the form sin x = …
Sine is negative so you need to look in the 3rd
and 4th quadrants for your solutions.
You can now read off the solutions in the given interval.
Note that in this case, if
α = sin−1(−0.4), the
solutions are 180 − α and 360 + α.
Draw the graph of y = sin x starting fr om −90°
since the principal solution given by sin−1(−0.4) is
negative.
Use the symmetry properties of the y = sin θ graph.
Example 12
Solve, in the interval 0 , x < 360°, cos x = √ __
3 ___ 2
A student writes down the follo
wing working:
a The principal solution is correct but the
student has found a second solution in the second quadrant where cos is negative.cos
−1 ( √ __
3 ___ 2 ) = 30°
So x = 30° or x = 180° − 30° = 150°
a Identify the error made by the student.
b Write down the corr
ect answer.In your exam you might have to analyse student
working and identify errors. One strategy is to solve the problem yourself, then compare your working with the incorrect working that has been given.Problem-solving The p rincipal value will not always
be a solution to the equation.Watch out | [
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0.00... |
215Trigonometric identities and equations
b x = 30° from the calculator
A
CS
T30°
30°
x = 30° or 330°cos x is positive so you need to look in the 1st
and 4th quadr
ants.
Read off the solutions, in 0 , x < 360°, from your
diagram.
Note that these results are α and 360° − α
where α = cos−1 ( √ __
3 ___ 2 ) .
Example 13
Find the values of θ in the interval 0 < θ < 360° that satisfy the equation sin θ = √ __
3 cos θ.
sin θ = √ __
3 cos θ
So tan θ = √ __
3
tan−1( √ __
3 ) = 60°
A
CS
T60°
240°60°
θ = 60° or 240°Since cos θ = 0 does not satisfy the equation,
divide both sides by cos θ and use the identity
tan θ ≡ sin θ _____ cos θ
This is the principal solution.
Tangent is positive in the 1st and 3rd quadrants,
so insert the angle in the correct positions.
Exercise 10D
1 The diagram shows a sketch of y = tan x.
a Use your calcula
tor to find
the principal solution to
the equation tan x =
−2.
b Use the graph and y
our answer
to part a to find solutions to the equation tan
x =
−2 in the range 0 < x < 360°.
2 The diagram sho
ws a sketch of y = cos x.
a Use your calcula
tor to find the principal solution
to the equation cos x =
0.4.
b Use the graph and y
our answer to part a to find
solutions to the equation cos x =
±0.4 in the range
0 < x < 360°.OAy
x
–290° 180° 270° 360°2 The principal solution is
ma
rked A on the diagram.Hint
Oy
x
–190° 180° 270° 360°1You can use the identity tan θ ≡ sin θ _____ cos θ to solve equations. | [
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216
Chapter 10
3 Solve the follo
wing equations for θ, in the interval 0 < θ < 360°:
a sin θ = −1 b tan θ = √ __
3 c cos θ = 1 _ 2
d sin θ = sin 15° e cos θ = −cos 40° f tan θ = −1
g cos θ = 0 h sin θ = −0.766
4 Solve the follo
wing equations for θ, in the interval 0 < θ < 360°:
a 7 sin θ = 5 b 2 cos θ = − √ __
2 c 3 cos θ = −2 d 4 sin θ = −3
e 7 tan θ = 1 f 8 tan θ = 15 g 3 tan θ = −11 h 3 cos θ = √ __
5
5 Solve the follo
wing equations for θ, in the interval 0 < θ < 360°:
a √ __
3 sin θ = cos θ b sin θ + cos θ = 0 c 3 sin θ = 4 cos θ
d 2 sin θ − 3 cos θ = 0 e √ __
2 sin θ = 2 cos θ f √ __
5 sin θ + √ __
2 cos θ = 0
6 Solve the follo
wing equations for x, giving your answers to 3 significant figures where
appropriate, in the intervals indicated:
a sin x =
− √ __
3 ___ 2 , −180° < x < 540° b 2 sin x = −0.3, −180° < x < 180°
c cos x =
−0.809, −180° < x < 180° d cos x =
0.84, −360° < x < 0°
e tan x =
− √ __
3 ___ 3 , 0 < x < 720° f tan x = 2.90, 80° < x < 440°
7 A teacher asks two students
to solve the equa
tion
2 cos x
= 3 sin x
f
or −180° < x < 180°.
The attempts are shown:
a Identify the mistake made b
y Student A. (1 mark)
b Identify the mistake made b
y Student B and explain the effect it has on their
solution. (2 marks)
c Write down the corr
ect answers to the question. (1 mark)
8 a Sketch the gra
phs of y = 2 sin x and
y = cos x on the same set of ax
es (0 < x < 360°).
b Write down ho
w many solutions there are in the given range for the equation 2 sin x =
cos x.
c Solve the equation 2 sin x =
cos x alge
braically, giving your answers in exact form.
9 Find all the va
lues of θ, to
1 decimal place, in the
interval 0 < θ < 360° for
which tan2 θ = 9. (5 marks)
10 a Show that 4 sin2 x – 3 cos2 x = 2 can be written as 7 sin2 x = 5. (2 marks)
b Hence solve, f
or 0 < x < 360°, the equation 4 sin2 x – 3 cos2 x = 2.
Give your answers to 1 decimal place. (7 marks)
11 a Show that the equa
tion 2 sin2x + 5 cos2x = 1 can be written as 3 sin2x = 4. (2 marks)
b Use your result in part
a to explain why the equation
2 sin2 x + 5 cos2 x = 1 has no solutions. (1 marks) Give your answers
ex
actly where possible,
or round to 3 significant
figures.Hint
E/PStudent A:
tan x =
3 __ 2
x =
56.3° or x = − 123.7°Student B:
4 cos2x = 9 sin2x
4(1 − sin2x) = 9 sin2x
4 = 13 sin2x
sin x = ± √ ___
4 ___ 13 , x = ±33.7° or x = ±146.3°
E/P
When you take square roots of both sides of an equation you
need to consider both the positive and the negative square roots.Problem-solving
E/P
E/P | [
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0.0208... |
217Trigonometric identities and equations
a Let X = 3θ
So cos X° = 0 .766
As X = 3 θ,
then as 0 < θ < 360°
So 3 × 0 < X < 3 × 360°
So the interval for X is
0 < X < 1080°
X = 40.
0°, 320°, 400°, 680°, 760°, 1040°
i.e. 3θ = 40.0°, 320°, 400°, 680°, 760°, 1040°
So θ = 13.3°, 107°, 133°, 227°, 253°, 347°
b sin 2θ ______ cos 2θ = 1 __ 2 , so tan 2θ = 1 __ 2
Let
X = 2 θ
So tan X = 1 __ 2
As X
= 2θ, then as 0 < θ < 360°
The interval for X is 0 < X < 720°
yy = tan x
O X
–1
–290° 180° 270° 360° 450° 540° 630° 720°12
The principal solution for X is 26.565…°
Add multiples of 180°:X = 26.565…°, 206.565…°, 386.565…°, 566.565…°
θ = 13.3°, 103°, 193°, 283°A
CS
T40°
40°10.5 Harder trigonometric equations
You need to be able to solve equations of the form sin nθ = k, cos nθ = k and tan nθ = p.
Example 14
a Solve the equation cos 3θ = 0.766, in the interval 0 < θ < 360°.
b Solve the equation 2 sin 2θ = cos 2θ, in the interval 0 < θ < 360°.
If the r ange of values
for θ is 0 < θ < 360°, then the range
of values for 3 θ is 0 < 3θ < 1080°.Watch out
Remember X = 3 θ.The value of X from your calculator
is 40.0. You need to list all values in the 1st and 4th quadrants for three complete revolutions.Replace 3 θ by X and solve.
Use the identity for tan to rearrange the equation.
Draw a graph of tan X for this
int
erval.
Alternatively, you could use a CAST diagram as in part a .Let X = 2 θ, and double both values
to find the interval for X.
Convert your values of X back into
values of θ.
Round each answer to a sensible degree of accuracy at the end. | [
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218
Chapter 10
You need to be able to solve equations of the form sin (θ + α) = k, cos (θ + α) = k and tan (θ + α) = p.
Example 15
Solve the equation sin (x + 60°) = 0.3 in the interval 0 < x < 360°.
Let X = x + 60°
So sin X =
0.3
The interval for X is
0° + 60° < X < 360° + 60°
So 60° < X < 420°
100 Oy
X
–0.5
–1200 300 4000.51
The principal value for X is 17.45…°
X = 162.54…°, 377.45…°
Subtract 60° from each value:x = 102.54…°, 317.45…°Hence x = 102.5° or 317.5°Draw a sketch of the sin graph for the given
interval.
You could also use a CAST diagram to solve this problem.Adjust the interval by adding 60° to both values.
Exercise 10E
1 Find the values of θ, in the interval 0 < θ < 360°, for which:
a sin 4θ = 0 b cos 3θ = −1 c tan 2θ = 1
d cos 2θ = 1 _ 2 e tan 1 _ 2 θ = − 1 ___ √ __
3 f sin (−θ ) = 1 ___ √ __
2
2 Solve the follo
wing equations in the interval given:
a tan (45° − θ ) = − 1, 0 < θ < 360° b 2 sin (θ − 20°) = 1, 0 < θ < 360°
c tan (θ + 75°) = √ __
3 , 0 < θ < 360° d sin (θ − 10°) = − √ __
3 ___ 2 , 0 < θ < 360°
e cos (70° −
x) = 0.6, 0 < θ < 180°
3 Solve the follo
wing equations in the interval given:
a 3 sin 3θ = 2 cos 3θ, 0 < θ < 180°
b 4 sin (θ + 45°) = 5 cos (θ + 45°), 0 < θ < 450°
c 2 sin 2x
– 7 cos 2x
= 0, 0 < x < 180°
d √ __
3 sin (x – 60°) + cos(x – 60°) = 0, –180° < x < 180°This is not in the given interval so it does not
correspond to a solution of the equation. Use the symmetry of the sin graph to find other solutions. | [
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219Trigonometric identities and equations
Solve the equation sin(3 x − 45°) = 1 _ 2 in the interval 0 < x < 180°.Challenge
10.6 Equations and identities
You need to be able to solve quadratic equations in sin θ, cos θ or tan θ. This may give rise to two sets
of solutions.
5 sin 2x + 3 sin x – 2 = 0
(5 sin x
– 2)(sin x +
1) = 0
5 sin x
– 2 = 0 sin x +
1 = 0Setting each factor equal to zero produces two
linear equations in sin x.This is a quadratic equation in the form 5A
2 + 3A – 2 = 0 where A = sin x.
Factorise
Example 16
Solve for θ, in the interval 0 < x < 360°, the equations
a 2 cos2 θ − cos θ − 1 = 0 b sin2 (θ − 30°) = 1 _ 2
a 2 cos2 θ − cos θ − 1 = 0
So (2 co
s θ + 1)(cos θ − 1) = 0
So cos θ = − 1 __ 2 or cos θ = 1
cos θ = − 1 __ 2 so θ = 120°Compare with 2x2 − x − 1 = (2x + 1)(x − 1)
Set each factor equal to 0 to find two sets of
solutions.4 Solve for 0 < x < 180° the equations:
a sin(x
+ 20°) = 1 _ 2 (4 marks)
b cos 2x
= −0.8, giving your answers to 1 decimal place. (4 marks)
5 a Sketch for 0
< x < 360° the graph of y = sin (x
+ 60°) (2 marks)
b Write down the e
xact coordinates of the points where the graph meets the
coordinate axes. (3 marks)
c Solve, f
or 0 < x < 360°, the equation sin (x
+ 60°) = 0.55, giving your answers to
1 decimal place. (5 marks)
6 a Given tha
t 4 sin x =
3 cos x, write do
wn the value of tan x. (1 mark)
b Solve, f
or 0 < θ < 360°, 4 sin 2θ = 3 cos 2θ giving your answers to 1 decimal place. (5 marks)
7 The equation tan kx =
− 1 ___ √ __
3 , where k is a constant and k > 0, has a solution at x = 60°
a Find a possible va
lue of k. (3 marks)
b State, with justifica
tion, whether this is the only such possible value of k. (1 mark)E
E
E
E/P | [
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0... |
220
Chapter 10
A
CS
T60°
60°60°60°
θ = 120° or θ = 240°
y
O θ 90° 180° 270° 360°y = cos θ
Or cos θ = 1 so θ = 0 or 360°
So the solutions are
θ = 0°, 120°, 240°, 360°
b sin2 (θ − 30°) = 1 __ 2
sin (θ − 30°) = 1 ___ √ __
2
or s
in (θ − 30°) = − 1 ___ √ __
2
So θ − 30° = 45° or θ − 30° = − 45°
45°A
CS
T45° 45°
45°
So from sin (θ − 30°) = 1 ___ √ __
2
θ − 30° = 45°, 135°
a
nd from sin (θ − 30°) = − 1 ___ √ __
2
θ − 30° = 225°, 315°
S
o the solutions are: θ = 75°, 165°, 255°,
345°120° makes an angle of 60° with the horizontal.
But cosine is negative in the 2nd and 3rd quadrants so
θ = 120° or θ = 240°.
Sketch the graph of y = cos θ.
Use your calculator to find one solution for each equation.There are four solutions within the given interval.
Draw a diagram to find the quadrants where sine is positive and the quadrants where sine is negative.The solutions of x2 = k are x = ± √ __
k . | [
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0.032... |
221Trigonometric identities and equations
In some equations you may need to use the identity sin2 θ + cos2 θ ≡ 1.
Example 17
Find the values of x, in the interval −180° < x < 180°, satisfying the equation
2 cos2 x + 9 sin2 x = 3 sin2 x.
2 cos2 x + 9 sin x = 3 sin2 x
2(1
− sin2 x) + 9 sin x = 3 sin2 x
5 sin2 x − 9 sin x − 2 = 0
So (5 si
n x
+ 1)(sin x −
2) = 0
sin x = − 1 __ 5
11.5° 11.5°A
CS
T
The solutions are − 168.5° and − 11.5° (1 d.p.)As sin2 x + cos2 x ≡ 1, you are able to rewrite
cos2 x as (1 − sin2 x), and so form a quadratic
equation in sin x.
The factor (sin x – 2) do es not produce
any solutions, because sin x = 2 has n
o solutions.Watch out
Your calculator value of x is x = −11.5° (1 d.p.).
Insert into the CAST diagram.
The smallest angle in the interval, in the 3rd
quadrant, is (−180 + 11.5) = −168.5°; there are no values between 0 and 180°.
Exercise 10F
1 Solve for θ, in the interval 0 < θ < 360°, the following equations.
Give your answers to 3 significant figures where they are not exact.
a 4 cos2 θ = 1 b 2 sin2 θ − 1 = 0 c 3 sin2 θ + sin θ = 0
d tan2 θ − 2 tan θ − 10 = 0 e 2 cos2 θ − 5 cos θ + 2 = 0 f sin2 θ − 2 sin θ − 1 = 0
g tan2 2θ = 3
2 Solve for θ, in the interval −180° < θ < 180°, the following equations.
Give your answers to 3 significant figures where they are not exact.a
sin2 2θ = 1 b tan2 θ = 2 tan θ
c cos θ (cos θ − 2) = 1 d 4 sin θ = tan θ
3 Solve for θ, in the interval 0 < θ < 180°, the following equations.
Give your answers to 3 significant figures where they are not exact.a
4 (sin2 θ − cos θ ) = 3 − 2 cos θ b 2 sin2 θ = 3(1 − cos θ ) c 4 cos2 θ − 5 sin θ − 5 = 0
4 Solve for θ, in the interval −180° < θ < 180°, the following equations.
Give your answers to 3 significant figures where they are not exact.a
5 sin2 θ = 4 cos2 θ b tan θ = cos θ In part c , on ly one factor leads to valid solutions. Hint | [
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222
Chapter 10
1 Solve the equation cos2 3θ – cos 3θ = 2 in the interval −180° < θ < 180°.
2 Sol
ve the equation tan2 (θ – 45°) = 1 in the interval 0 < θ < 360°.Challenge5 Find all the solutions, in the interv
al 0 < x < 360°, to the equation 8 sin2 x + 6 cos x – 9 = 0
giving each solution to one decimal place. (6 marks)
6 Find, for 0 <
x < 360°, all the solutions of sin2 x + 1 = 7 _ 2 cos2 x giving each solution
to one decima
l place. (6 marks)
7 Show that the equa
tion 2 cos2 x + cos x – 6 = 0 has no solutions. (3 marks)
8 a Show that the equa
tion cos2 x = 2 – sin x can
be written as sin2 x – sin x + 1 = 0. (2 marks)
b Hence show that the equa
tion cos2 x = 2 – sin x
has no solutions. (3 marks)
9 tan2 x – 2 tan x – 4 = 0
a Show that tan x =
p ± √ __
q where p and q are numbers to be found. (3 marks)
b Hence solve the equation tan2 x – 2 tan x – 4 = 0 in the interval 0 < x < 540°. (5 marks)E
E
E/P
E/P
If you have to answer a question involving the
number of solutions to a quadratic equation, see if you can make use of the discriminant.Problem-solving
E/P
1 Write each of the following as a trigonometric ratio of an acute angle:
a cos 237° b sin 312° c tan 190°
2 Without using your ca
lculator, work out the values of:
a cos 270° b sin 225° c cos 180° d tan 240° e tan 135°
3 Given tha
t angle A is obtuse and cos A =
− √ ___
7 ___ 11 , show that tan A = −2 √ __
7 _____ 7
4 Given tha
t angle B is obtuse and tan B =
+ √ ___ 21 ____ 2 , find the exact value of: a sin B b cos B
5 Simplify the following e
xpressions:
a cos4 θ − sin4 θ b sin2 3θ − sin2 3θ cos2 3θ
c cos4 θ + 2 sin2 θ cos2 θ + sin4 θ
6 a Given that 2 (sin x + 2 cos x) = sin x + 5 cos x, find the e xact value of tan x.
b Given tha
t sin x cos y +
3 cos x sin y =
2 sin x sin y −
4 cos x cos y, expr
ess tan y in terms
of
tan x.
7 Prov
e that, for all values of θ :
a (1 + sin θ )2 + cos2 θ ≡ 2(1 + sin θ ) b cos4 θ + sin2 θ ≡ sin4 θ + cos2 θP
P
PMixed exercise 10 | [
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223Trigonometric identities and equations
8 Without attempting to solv
e them, state how many solutions the following equations have in
the interval 0 < θ < 360°. Give a brief reason for your answer.
a 2 sin θ = 3 b sin θ = − cos θ
c 2 sin θ + 3 cos θ + 6 = 0 d tan θ + 1 _____ tan θ = 0
9 a Factorise 4xy
− y2 + 4x − y. (2 marks)
b Solve the equation 4 sin θ cos θ − cos2 θ + 4 sin θ − cos θ = 0, in the interval
0 < θ < 360°. (5 marks)
10 a Express 4 cos 3θ − sin (90° − 3θ ) as a single trigonometric function. (1 mark)
b Hence solve
4 cos 3θ − sin (90° − 3θ ) = 2 in the interva l 0 < θ < 360°.
Give your answers to 3 significant figures. (3 marks)
11 Given
that 2 sin 2θ = cos 2θ:
a Show that tan 2θ = 0.5. (1 mark)
b Hence find the values of
θ, to one decimal place, in the interval 0 < θ < 360°
for which 2 sin 2θ = cos 2θ. (4 marks)
12 Find all the va
lues of θ in the interval 0 < θ < 360° for which:
a cos (θ + 75°) = 0.5,
b sin 2θ = 0.7, giving your answers to one decimal place.
13 Find the values of
x in the interval 0 < x < 270° which satisfy the equation
cos 2x
+ 0.5 ___________ 1 – cos 2x = 2 (6 marks)
14 Find, in degrees, the v
alues of θ in the interval 0 < θ < 360° for which
2 cos2 θ – cos θ – 1 = sin2 θ
Give your answers to 1 decimal place, where appropriate. (6 marks)
15 A teacher asks one of his students to solve the equa
tion 2 sin 3x
= 1 for –360° < x < 360°.
The attempt is shown below:
sin 3x = 1 _ 2
3x =
30°
x = 10°
Additional solution at 180° − 10° = 170°
a Identify two mistak es made by the student. (2 marks)
b Solve the equation. (2 marks)
16 a Sketch the gra
phs of y = 3 sin x and
y = 2 cos x on the same set of
axes (0 < x < 360°).
b Write down ho
w many solutions there are in the given range for the equation 3 sin x = 2 cos x.
c Solve the equation 3 sin x =
2 cos x a
lgebraically, giving your answers to one decimal place.P
E
E
E/P
E
E
E/P | [
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-0.058388106524944305,
0.013207442127168179,
0.0596... |
224
Chapter 10
17 The diagram sho
ws the triangle ABC with AB = 11 cm,
BC
= 6 cm and AC =
7 cm.
a Find the exact va
lue of cos B, giving y
our answer in
simplest form. (3 marks)
b Hence find the exact va
lue of sin B. (2 marks)
18 The diagram sho
ws triangle PQR with PR = 6 cm, QR
= 5 cm
and angle QPR
= 45°.
a Show that sin Q =
3 √ __
2 ____ 5 (3 marks)
b Given tha
t Q is obtuse, find the exact value of
cos Q. (2 marks)
19 a Show that the equa
tion 3 sin2 x – cos2 x = 2 can be written as 4 sin2 x = 3. (2 marks)
b Hence solve the equation 3 sin2 x – cos2 x = 2 in the interval –180° < x < 180°,
giving your answers to 1 decimal place. (7 marks)
20 Find all the solutions to the equation 3 cos2 x + 1 = 4 sin x in the interv al
–360° < x < 360°, giving your answers to 1 decimal place. (6 marks) E
ABC
7 cm 6 cm
11 cm
E/P
45°
RPQ
5 cm
6 cm
E/P
E
Solve the equation tan4 x – 3 tan2 x + 2 = 0 in the interval 0 < x < 360°.Challenge
1 For a point P(x, y) on a unit circle such that OP
1P
x Oy
x(x,y)
yθ
makes an angle θ with the positive x-axis:
• cos θ = x = x-coordinate of P
• sin θ = y = y-coordinate of P
• tan θ = y __ x = gradient of OP
2 You can use the quadrants t
o determine whether each of the trigonometric ratios is positive or
negative.
For an angle θ in the first quadrant,
sin θ, cos θ and tan θ are all positive.For an angle θ in the
second quadrant, only
sin θ is positive.
For an angle θ in the
third quadrant, only tan θ
is positive.For an angle θ in the fourth
quadrant, only cos θ is positive.y90°
All
CosSin
Tan
270°0, 360° 180°xSummary of key points | [
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0.0258... |
225Trigonometric identities and equations
3 You can use these rules to find sin, cos or tan of any positive or negative angle using the
corr
esponding acute angle made with the x-axis, θ.
A
CS
T360° – θ 180° + θ180° – θ
θ
θθθ
θy
x
4 The trigonometric ratios of 30°, 45° and 60° have exact forms, given below:
sin 30° = 1 __ 2 cos 30° = √ __
3 ___ 2 tan 30° = 1 ___
√ __
3 = √ __
3 ___ 3
sin 45° = 1 ___
√ __
2 = √ __
2 ___ 2 cos 45° = 1 ___
√ __
2 = √ __
2 ___ 2 tan 45° = 1
sin 60° = √ __
3 ___ 2 cos 60° = 1 __ 2 tan 60° = √ __
3
5 For all values o
f θ, sin2 θ + cos2 θ ; 1
6 For all values o
f θ such that cos θ ≠ 0, tan θ ; sin θ _____ cos θ
7 • Solutions to sin θ = k and cos θ = k only exist when −1 < k < 1
• Solutions to tan θ = p exist for all values of p.
8 When you use the inv
erse trigonometric functions on your calculator, the angle you get is
called the principal value.
9 Your calculat
or will give principal values in the following ranges:
• cos−1 in the range 0 < θ < 180°
• sin−1 in the range −90° < θ < 90°
• tan−1 in the range −90° < θ < 90°cos (180° − θ ) = − cos θ
cos (180° + θ ) = − cos θ
cos (360° − θ ) = cos θtan (180° − θ ) = − tan θ
tan (180° + θ ) = tan θ
tan (360° − θ ) = − tan θsin (180° − θ ) = sin θ
sin (180° + θ ) = − sin θ
sin (360° − θ ) = − sin θ | [
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0.0... |
226Review exercise2
1 Find the equation of the line w hich passes
through the points A(−2, 8) and B(4, 6), in
the form ax + by + c = 0. (3 marks)
← Section 5.2
2 The line l passes thr ough the point (9, −4)
and has gradient 1 _ 3 . Find an equation for l,
in the form ax + by + c = 0, where a, b and
c are integers. (3 marks)
← Section 5.2
3 The points A(0, 3), B(k, 5) and C(10, 2k),
where k is a constant, lie on the same
straight line. Find the two possible values of k.
(5 marks)
← Section 5.1
4 The scatter graph shows the height, h cm,
and inseam leg measurement,
l cm, of six
adults. A line of
best fit has been added to
the scatter graph.
1501551601651701751801856870727476788082Inseam leg measurement (cm)Height (cm)
a Use two points on the scatter graph to calcula
te the gradient of the
line. (2 marks)
b Use your answ
er to part a to write a
linear model relating height and inseam in the form l = kh, where k is a constant to be found.
(1 mark)
c Comment on the validity of
your model
for small values of h. (1 mark)
← Section 5.5E
E
E/p
E/p5 The line l1 has equation y = 3x − 6.
The line l2 is perpendicular to l1 and passes
through the point (6, 2).
a Find an equation for
l2 in the form
y = mx + c, where m and c are
constants (3 marks)
The lines l1 and l2 intersect at the point C.
b Use algebr
a to find the coordinates of C.
(2 marks)
The lines l1 and l2 cross the x-axis at the
points A and B respectively.
c Calculate the e
xact area of triangle
ABC. (4 marks)
← Sections 5.3, 5.4
6 The lines y = 2 x and 5y + x − 33 = 0
intersect at the point P. Find the distance
of the point from the origin O, giving your answer as a surd in its simplest form.
(4 marks)
← Sections 5.2, 5.4
7 The perpendicular bisector of the line segment joining (5, 8) and (7, −4) crosses the
x-axis at the point Q. Find the
coordinates of Q. (4 marks)
← Section 6.1
8 The circle C has centre (−3, 8) and passes through the point (0, 9). Find an equa
tion
for C. (4 marks)
← Section 6.2
9 a Show that x2 + y2 − 6x + 2y − 10 = 0
can be written in the form (x − a)
2 + (y − b)2 = r2, where a, b and r
are numbers to be found. (2 marks)
b Hence write down the centre and r
adius of the circle with equation
x2 + y2 − 6x + 2y − 10 = 0. (2 marks)
← Section 6.2E
E
E/p
E
E/p | [
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-0.03... |
227
Review exercise 2
10 The line 3x + y = 14 intersects the cir
cle
(x − 2)2 + (y − 3)2 = 5 at the points A
and B.
a Find the coordinates of
A and
B. (4 marks)
b Determine the length of the chor
d
AB. (2 marks)
← Section 6.3
11 The line with equation y = 3x − 2 does
not intersect the circle with centre (0, 0)
and radius r. Find the range of possible values of r.
(8 marks)
← Section 6.3
12 The circle C has centre (1, 5) and passes through the point
P (4, −2). Find:
a an equation for the cir
cle C. (4 marks)
b an equation for the tangent to the cir
cle at P. (3 marks)
← Section 6.4
13 The points A(2, 1), B(6, 5) and C(8, 3) lie
on a circle.
a Show that ∠ABC = 90°. (2 marks)
b Deduce a geometrical property of
the
line segment AC . (1 mark)
c Hence find the equation of the
cir
cle. (4 marks)
← Section 6.5
14 2 x 2 + 20x + 42 _______________ 224x + 4 x 2 − 4 x 3 = x + a ________ bx(x + c)
where
a, b and c are constants. Work out
the values of a, b and c. (4 marks)
← Section 7.1
15 a Show that (2x − 1) is a factor of
2x3 − 7x2 − 17x + 10. (2 marks)
b Factorise 2x3 − 7x2 − 17x + 10
completely. (4 marks)
c Hence, or otherwise, sk
etch the graph
of y = 2x3 − 7x2 − 17x + 10, labelling
any intersections with the coordinate axes clearly.
(2 marks)
← Section 7.3E/p
E/p
E/p
E/p
E/p
E16 f(x) = 3 x3 + x2 − 38x + c
Given that f(3) = 0,
a find the value of
c, (2 marks)
b factorise f(x) complete
ly, (4 marks)
← Section 7.3
17 g(x) = x3 − 13x + 12
a Use the factor theorem to show tha
t
(x − 3) is a factor of g(x). (2 marks)
b Factorise g(x
) completely. (4 marks)
← Section 7.3
18 a It is claimed that the follo wing
inequality is true for all real numbers a
and b. Use a counter-example to show that the claim is false:
a
2 + b2 < (a + b)2 (2 marks)
b Specify conditions on a and
b that
make this inequality true. Prove your result.
(4 marks)
← Section 7.5
19 a Use proof b y exhaustion to prove that
for all prime numbers p, 3 < p < 20,
p2 is one greater than a multiple
of 24. (2 marks)
b Find a counterexample tha
t disproves
the statement ‘ All numbers which are one greater than a multiple of 24 are the squares of prime numbers.’
(2 marks)
← Sections 7.5
20 a Show that x2 + y2 − 10x − 8y + 32 = 0
can be written in the form (x − a)
2 + (y − b)2 = r2, where a, b and r
are numbers to be found. (2 marks)
b Circle C
has equation x2 + y2 − 10x −
8y + 32 = 0 and circle D has equation x
2 + y2 = 9. Calculate the distance
between the centre of circle C and the centre of circle D.
(3 marks)
c Using your answ
er to part b, or
otherwise, prove that circles C and D do not touch.
(2 marks)
← Sections 6.4, 7.5E/p
E
E/p
E/p
E/p | [
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228
Review exercise 2
21 a Expand (1 − 2x)10 in ascending
powers of x up to and including the
term in x3. (3 marks)
b Use your answ
er to part a to
evaluate (0.98)10 correct to 3 decimal
places. (1 mark)
← Sectio n 8.5
22 If x is so small tha t terms of x3 and
higher can be ignored,
(2 − x)(1 + 2x)5 ≈ a + bx + cx2.
Find the values of the constants a, b and c.
(5 marks)
← Section 8.4
23 The coefficient of x in the binomial
expansion of (2 − 4x)q, where q is a
positive integer, is −32q. Find the value of q.
(4 marks)
← Section 8.4
24 The diagram shows triangle ABC, with AB =
√ __
5 cm, ∠ABC = 45° and
∠BCA = 30°. Find the exact length
of AC. (3 marks)
5 cm
Not to scale 45° 30°
B CA
← Section 9.2
25 The diagram shows triangle ABC, with
AB = 5 cm, BC = (2
x − 3) cm,
CA = (
x + 1) cm and ∠ABC = 60°.
(x + 1) cm(2x – 3) cm 5 cm
Not to scale60°
C AB
a Show that x satisfies the equation x
2 − 8x + 16 = 0. (3 marks)
b Find the value of
x. (1 mark)
c Calculate the ar
ea of the triangle,
giving your answer to 3 significant figures.
(2 marks)
← Section 9.4E
E/p
E/p
E
E/p26 Ship B is 8 km, on a bearing of 030°,
from ship
A.
Ship C is 12 km, on a bearing of 140°,
from ship
B.
a Calculate the distance of
ship C from
ship A. (4 marks)
b Calculate the bearing of
ship C from
ship A. (3 marks)
← Section 9.4
27 The triangle ABC has v ertices A(−2, 4),
B(6, 10) and C(16, 10).
a Prov
e that ABC is an isosceles
triangle. (2 marks)
b Calculate the siz
e of ∠ABC . (3 marks)
← Sections 5.4, 9.4
28 The diagram shows ΔABC.
Calcula
te the area of ΔABC. (6 marks)
3.5 cm4.3 cm
8.6 cm40°B
AD C
← Section 9.4
29 The circle C has centre (5, 2) and radius 5. The points X
(1, −1), Y (10, 2) and
Z (8, k) lie on the circle, where k is a positive integer.
a
Write down the equa
tion of the circle.
(2 marks)
b Calculate the v
alue of k. (1 mark)
c Show that cos ∠XYZ = √ __
2 ___
10 (5 marks)
← Sections 6.2, 9.4
30 a On the same set of axes , in the interval
0 < x < 360°, sketch the graphs of
y = tan (x
− 90°) and y = sin x. Labe
l
clearly any points at which the graphs cross the coordinate axes.
(5 marks)
b Hence write down the number of
solutions of the equation tan
(x
− 90°) = sin x in the interva
l
0 < x < 360°. (1 mark)
← Section 9.6E/p
E/p
E/p
E/p
E | [
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229
Review exercise 2
31 The graph sho
ws the curve
y = sin (x + 45°), −360° < x < 360°.
y = sin(x + 45°)
Oy
x
a Write down the coordinates of each
point wher
e the curve crosses the
x-axis. (2 marks)
b Write down the coor
dinates of the
point where the curve crosses the y-axis.
(1 mark)
← Section 9.6
32 A pyramid has four triangular faces and a square base
. All the edges of the pyramid
are the same length, s cm. Show that the total surface area of the pyramid is (
√ __
3 + 1)s2 cm2. (3 marks)
← Sections 9.4, 10.2
33 a Given that sin θ = cos θ, find the value
of tan θ. (1 mark)
b Find the values of
θ in the interval
0 < θ < 360° for which
sin θ = cos θ. (2 marks)
← Sections 10.3, 10.4
34 Find all the values of x in the interval 0 <
x < 360° for which 3 tan2x = 1.
(4 marks)
← Section 10.4
35 Find all the values of θ in the interva
l 0 < θ < 360° for which
2 sin (θ
− 30°) = √ __
3 . (4 marks)
← Section 10.5
36 a Show that the equation
2 cos2 x = 4 − 5 sin x ma y be written
as 2 sin2 x − 5 sin x + 2 = 0. (2 marks)
b Hence solve, f
or 0 < x < 360°, the
equation 2 cos2 x = 4 − 5 sin x.
(4 marks)E
E/p
E
E
E
E37 Find all of the solutions in the interval 0 <
x < 360° of 2 tan2 x − 4 = 5 tan x
giving each solution, in degrees
, to one
decimal place. (6 marks)
← Section 10.6
38 Find all of the solutions in the interval 0 <
x < 360° of 5 sin2 x = 6(1 − cos x)
giving each solution, in degrees, to one decimal place.
(7 marks)
← Section 10.6
39 Prove that cos2 x (tan2 x + 1) = 1 for all
values of x where cos x and tan x are
defined. (4 marks)
← Sections 7.4, 10.3E
E
E/p
1 The diagram shows
a sq
uare ABCD on
a set of coordinate axes. The square intersects the x -axis
at the points B and S, and the equation
of the line which passes through B and C is y = 3 x − 12.
a
Cal
culate the area of the square.
b Fin
d the coordinates of S .
← Sections 5.2, 5.4
2 Prove that the circle ( x + 4)2 + (y – 5)2 = 82
lies completely inside the circle x
2 + y2 + 8x – 10 y = 59.
← Sections 1.5, 6.2
3 Prove that for all positive integers n and k ,
( n k ) + ( n k + 1 ) = ( n + 1 k + 1 ) .
← Sections 7.4, 8.2
4 Solve for 0° < x < 360° the equation
2 sin3 x – sin x + 1 = co s2 x.
← Section 10.6B S
CDA
Oy
xChallenge | [
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23011Vectors
After completing this chapter you should be able to:
● Use vectors in t
wo dimensions → pages 231–235
● Use column vectors and carry out arithmetic operations
on vect
ors → pages 235–238
● Calculate the magnitude and direction of a vector → pages 239–242
● Understand and use position vect ors → pages 242–244
● Use vectors to solve geometric problems → pages 244–247
● Understand vector magnit ude and use vectors in speed
and distance calculations → pages 248–251
● Use vectors to solve problems in context → pages 248–251Objectives
1 Write the column vector for
AB
CD
the translation of shape
a A to
B
b A to
C
c A to
D ← GCSE Mathematics
2 P divides the line AB in the ratio AP : PB = 7 : 2.
A
PB
Find:
a AP ___ AB b PB ___ AB c AP ___ PB ← GCSE Mathematics
3 Find x to one decimal place.
a b
79138
5
187
370°30°40° 110°
xxxx
c d
79138
5
187
370°30°40° 110°
xxxx
← Sections 9.1, 9.2Prior knowledge check
Pilots use vector addition to work
out the resultant vector for their speed and heading when a plane encounters a strong cross-wind. Engineers also use vectors to work out the resultant forces acting on structures in construction. | [
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231Vectors
11.1 Vectors
A vector has both magnitude and direction.
You can represent a vector using a directed line segment.
This is vector ⟶ PQ . It starts
at P and finishes at Q.This is vector ⟶ QP . It starts
at Q and finishes at P.Q
PQ
P
The direction of the arrow shows the direction of the vector. Small (lower case)
letters are also used to represent vectors. In print, the small letter will be in bold type. In writing, you should underline the small letter to show it is a vector: a or
a ~
■ If ⟶ PQ = ⟶ RS then the line segments PQ and RS are equal in length
and are parallel.
■ ⟶ AB = − ⟶ BA as the line segment AB is equal
in length, parallel and in the opposite
direction to BA.
You can add two vectors together using the triangle law for vector addition.
■ Triangle la
w for
vector addition:
⟶ AB + ⟶ BC = ⟶ AC
If ⟶ AB = a, ⟶ BC = b and ⟶ AC = c, then a + b = ca
R
SQP
B
AB
–aa
A
B
C
A cb a The res ultant is the vector sum of
two or more vectors.
⟶ AB + ⟶ BC + ⟶ CD = ⟶ AD Notation
ADC
B
Example 1
The diagram shows vectors a, b and c.
Draw a diagram to illustrate the vector addition
a + b + c.
c/a.ss01
/a.ss01 + b
/a.ss01 + b + cb
First use the triangle law for a + b, then use it
again for ( a + b ) + c .
The resultant goes from the start of a to the end of c.c ba
Explore vector addition using
GeoGe
bra.Online | [
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232
Chapter 11
■ Subtracting a vect
or is
equivalent to ‘adding a
negative vector’: a − b = a + (−b)
If you travel from P to Q, then back from Q to P, you are back where you started, so your displacement is zero.
■
Adding the vect
ors ⟶ PQ and ⟶ QP gives Q
P ⟶ QP = − ⟶ PQ .
So ⟶ PQ + ⟶ QP = ⟶ PQ − ⟶ PQ = 0.Hint
the zero v
ector 0: ⟶ PQ + ⟶ QP = 0ba–b
a – ba To subtract b, you reverse
the d
irection of b then add.Hint
Example 2
In the diagram, ⟶ QP = a , ⟶ QR = b , ⟶ QS = c and ⟶ RT = d .
Find in terms of a, b, c and d:
a ⟶ PS b ⟶ RP
c ⟶ PT d ⟶ TS RQ
P
TSca
b
dYou can multiply a vector by a scalar (or number).
a
3a
1
2ab
–2b
12 – b
■ Any vector parallel to the vector a may be Real n umbers are examples of
scalars . They have magnitude but no direction.Notation
written as λa, where λ is a non-zero scalar. If the number is
positive (≠ 1) the
n
ew vector has a
different length but the same
direction.If the number is negative (≠ − 1)
the new vector has a different length and the opposite
direction.
a ⟶ PS = ⟶ PQ + ⟶ QS = −a + c
= c − a
b ⟶ RP = ⟶ RQ + ⟶ QP = −b + a
= a − b
c ⟶ PT = ⟶ PR + ⟶ RT = (b − a) + d
= b + d − a
d ⟶ TS = ⟶ TR + ⟶ RS = −d + ( ⟶ RQ + ⟶ QS )
= −d + (−b + c)
= c − b − dAdd vectors using △ PQS .
Add vectors using △ RQP .
Add vectors using △ PRT .
Use ⟶ PR = − ⟶ RP = −(a − b) = b − a .
Add vectors using △ TRS and △ RQS . | [
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233Vectors
Example 4
Show that the vectors 6a + 8b and 9a + 12b are parallel.
9a + 12 b = 3 __ 2 (6a + 8 b)
∴ the vectors are parallel.Here λ = 3 _ 2 Thi s is called the parallelogram law
for vector addition.NotationExample 3
ABCD is a parallelogram. ⟶ AB = a , ⟶ AD = b . Find ⟶ AC .
ADC
B ab
⟶ AC = ⟶ AB + ⟶ BC
⟶ BC = ⟶ AD = b
So ⟶ AC = a + bUsing the triangle law for addition of vectors.
AD and BC are opposite sides of a parallelogram
so they are parallel and equal in magnitude.
Example 5
In triangle ABC, ⟶ AB = a and ⟶ AC = b .
P is the midpoint of AB.
Q divides AC in the ratio 3 : 2.
Write in terms of
a and b:
a ⟶ BC b ⟶ AP c ⟶ AQ d ⟶ PQ BQ
CPA
a ⟶ BC = ⟶ BA + ⟶ AC
= − ⟶ AB + ⟶ AC
⟶ BC = b − a
b ⟶ AP = 1 _ 2 ⟶ AB = 1 _ 2 a
c ⟶ AQ = 3 _ 5 ⟶ AC = 3 _ 5 b
d ⟶ PQ = ⟶ PA + ⟶ AQ
= − ⟶ AP + ⟶ AQ
= 3 _ 5 b − 1 _ 2 a ⟶ BA = − ⟶ AB
AP is the l ine segment between A
and P , whereas ⟶ AP is the vector from A to P .Watch outAP = 1 _ 2 AB so ⟶ AP = 1 _ 2 a
Q divides AC in the ratio 3 : 2 so A Q = 3 _ 5 AC.
Going from P to Q is the same as going from P to
A, then from A to Q . | [
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234
Chapter 11
Exercise 11A
1 The diagram shows the vectors a, b, c and d.
Dra
w a diagram to illustrate these vectors:
a a +
c b −b
c c −
d d b +
c + d
e a −
2b f 2c
+ 3d
g a
+ b + c + d
2 ACGI is a squar
e, B is the midpoint of AC , F is the midpoint
AIEBCG
HF
Db
d
of CG, H is the midpoint of GI, D is the midpoint of AI.
⟶ AB = b and ⟶ AD = d. Find, in terms of b and d:
a ⟶ AC b ⟶ BE c ⟶ HG d ⟶ DF
e ⟶ AE f ⟶ DH g ⟶ HB h ⟶ FE
i ⟶ AH j → BI k → EI l ⟶ FB
3 OACB is a par
allelogram. M, Q, N and P are
OPBNCQA
MD
m
p
the midpoints of OA , AC , BC and OB
respectively.
Vectors p and m are equal to ⟶ OP and ⟶ OM
respecti
vely. Express in terms of p and m.
a ⟶ OA b ⟶ OB c ⟶ BN d ⟶ DQ
e ⟶ OD f ⟶ MQ g ⟶ OQ h ⟶ AD
i ⟶ CD j ⟶ AP k ⟶ BM l ⟶ NO
4 In the diagram, ⟶ PQ = a , ⟶ QS = b , ⟶ SR = c and ⟶ PT = d .
Find in terms of a, b, c and d:
a ⟶ QT b ⟶ PR
c ⟶ TS d ⟶ TR
5 In the triangle PQR,
PQ = 2a and QR = 2b.
The midpoint of PR is M. Find, in terms of a and b:
a ⟶ PR b ⟶ PM c ⟶ QM
6 ABCD is a tra
pezium with AB parallel to DC and DC = 3AB.
M divides DC such that DM : MC =
2 : 1. ⟶ AB = a and ⟶ BC = b.
Find, in terms of
a and b:
a ⟶ AM b ⟶ BD c ⟶ MB d ⟶ DA ab
dc
Q
P
RSTa
d
cb
P
Draw a sketch to show the
information given in the question.Problem-solving | [
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235Vectors
7 OABC is a par
allelogram. ⟶ OA = a and ⟶ OC = b.
The point P divides OB in the ratio 5:3.
Find, in terms of a and b:a
⟶ OB b ⟶ OP c ⟶ AP
8 State with a reason w
hether each of these vectors is parallel to the vector a − 3b:
a 2a
− 6b b 4a
− 12b c a +
3b d 3b
− a e 9b
− 3a f 1 _ 2 a − 2 _ 3 b
9 In triangle ABC, ⟶ AB = a and ⟶ AC = b .
P is the midpoint of AB and Q is the midpoint of AC .
a Write in terms of
a and b:
i ⟶ BC ii ⟶ AP iii ⟶ AQ iv ⟶ PQ
b Show that
PQ is parallel to BC.
10 OABC is a quadrila
teral. ⟶ OA = a, ⟶ OC = 3 b and ⟶ OB = a + 2b.
a Find, in terms of
a and b:
i ⟶ AB ii ⟶ CB
b Show that
AB is parallel to OC.
11 The vectors 2a
+ kb and 5a + 3b are parallel. Find the value of k.A B
P
C Oa
b
P
BQ
CPA
P AB
CO
P
11.2 Representing vectors
A vector can be described by its change in position or displacement relative to the x- and y-axes.
a
4
3 a = ( 3 4 ) where 3 is the change in the x-direction
and 4 is the change in the y-direction.
This is called column vector form.
■ To multiply a column v
ector by a scalar, multiply each component by the scalar: λ ( p q ) = ( λp λq )
■ To add tw
o column vectors, add the x-components and the y-components: ( p q ) + ( r s ) = ( p + r q + s ) The t op number
is the x -component and
the bottom number is the
y-component.Notation
Example 6
a = ( 2 6 ) and b = ( 3 −1 )
Find a 1 _ 3 a b a + b c 2a − 3b | [
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236
Chapter 11
a 1 __ 3 a = ( 2 __ 3
2 )
b a +
b = ( 2 6 ) + ( 3 −1 ) = ( 5 5 )
c 2a − 3b = 2 ( 2 6 ) − 3 ( 3 −1 )
= ( 4 12 ) − ( 9 −3 ) = ( 4 − 9 12 + 3 ) = ( −5 15 ) Both of the components are divided by 3.
Add the x-components and the y-components.
Multiply each of the vectors by the scalars then
subtract the x- and y-components.
You can use unit vectors to represent vectors in two dimensions.
■ A unit vector is a v
ector of length 1. The unit vectors
(0, 1)
(1, 0) ij
Oy
x
along the x - and y -axes are usually denoted by
i and j respectively.
• i = ( 1 0 ) j = ( 0 1 )
■ You can write an
y two-dimensional vector in the form pi + qj.
By the triangle law of addition:
5i5i + 2j
2j
AC
B ⟶ AC = ⟶ AB + ⟶ BC
= 5i
+ 2j
You can also write this as a column vector: 5i + 2j = ( 5 2 )
■ For any t
wo-dimensional vector: ( p q ) = pi + qj
Example 7
a 1 __ 2 a = 1 __ 2 (3i − 4j) = 1.5i − 2j
b a +
b = 3i − 4j + 2i + 7j
= (3 + 2)i + (− 4 + 7)
j = 5i + 3j
c 3a
− 2b = 3(3i − 4j) − 2(2i + 7j)
= 9i − 12j − (4i + 14j)
= (9 − 4)i + (−12 − 14)j = 5i − 26ja = 3i − 4j, b = 2i + 7j
Find a 1 _ 2 a b a + b c 3a − 2b
Divide the i component and the j component by 2.
Add the i components and the j components.
Multiply each of the vectors by the scalar then
subtract the i and j components. | [
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-0... |
237Vectors
Example 8
a Draw a diagram to represent the vector −3i + j
b Write this as a column vector
.
a
–3i–3i + j
j
b −3i + j = ( −3 1 ) 3 units in the direction of the unit vector −i and
1 unit in the direction of the unit vector j.
Example 9
Given that a = 2i + 5j, b = 12i − 10j and c = −3i + 9j, find a + b + c, using column vector notation in
your working.
a + b + c = ( 2 5 ) + ( 12 −10 ) + ( −3 9 ) = ( 11 4 ) Add the numbers in the top line to get 11
(the x-component), and the bottom line to get 4 (the y-component). This is 11i + 4j.
Exercise 11B
1 These vectors are drawn on a grid of unit squares. v1
v2
v5v3v4
v6
Express the vectors v1, v2, v3, v4, v5 and v6 in:
(i) i, j notation
(ii) column vector for
mExample 10
Given a = 5i + 2j and b = 3i − 4j,
find 2a − b in terms of i and j.
2a = 2 ( 5 2 ) = ( 10 4 )
2a
− b = ( 10 4 ) − ( 3 −4 ) = ( 10 − 3 4 − (−4) ) = ( 7 8 )
2a
− b = 7i + 8jTo find the column vector for vector 2a multiply
the i and j components of vector a by 2.
To find the column vector for 2a − b subtract the components of vector b from those of vector 2a.
Remember to give your answer in terms of i and j. Explore this solution as a vector
dia
gram on a coordinate grid using GeoGebra.Online | [
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0.... |
238
Chapter 11
2 Given tha
t a = 2i + 3j and b = 4i − j, find these vectors in terms of i and j.
a 4a b 1 _ 2 a c −b d 2b + a
e 3a
− 2b f b −
3a g 4b
− a h 2a
− 3b
3 Given tha
t a = ( 9 7 ) , b = ( 11 −3 ) and c = ( −8 −1 ) find:
a 5a b − 1 _ 2 c c a + b + c d 2a − b + c
e 2b
+ 2c − 3a f 1 _ 2 a + 1 _ 2 b
4 Given tha
t a = 2i + 5j and b = 3i − j, find:
a λ if a + λb is parallel to the vector i b μ if μa + b is parallel to the vector j
5 Given tha
t c = 3i + 4j and d = i − 2j, find:
a λ if c + λd is parallel to i + j b μ if μc + d is parallel to i + 3j
c s if
c − sd is parallel to 2i + j d t if
d − tc is parallel to −2i + 3j
6 In triangle ABC, ⟶ AB = 4 i + 3j and ⟶ AC = 5i + 2j.
AB
C
Find BC.
(2 marks)
7 OABC is a par
allelogram. AB
P
C O
P divides AC in the ratio 3 : 2. ⟶ OA = 2 i + 4j, ⟶ OC = 7 i.
Find in i, j format and column vector format:
a ⟶ AC b ⟶ OP c ⟶ AP
8 a =
( j 3 ) , b = ( 10 k ) , c = ( 2 5 )
Given tha
t b − 2a = c, find the values of j and k.
(2 marks)
9 a =
( p −q ) , b = ( q p ) , c = ( 7 4 )
Given tha
t a + 2b = c, find the values of p and q. (2 marks)
10 The resultant of the v
ectors a = 3i − 2j and b = pi − 2pj is parallel to the vector c = 2i − 3j.
Find:
a the value of
p (4 marks)
b the resultant of v
ectors a and b. (1 mark)P
P
E
P
E/P
E/P
E/PYou can consider b – 2 a = c as two linear
equations. One for the x -components
and one for the y -components.Problem-solving | [
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239Vectors
11.3 Magnitude and direction
You can use Pythagoras’ theorem to calculate the magnitude of a vector.
■ For the vect
or a = xi + yj = ( x y ) ,
the magnitude of the vector is given by:
|a | = √ ______ x2 + y2
You need to be abl
e to find a unit vector
in the direction of a given vector.
■ A unit vector in the dir
ection of a is a ___ |a|
If |a|
= 5 then a unit vector in the direction
of a is a __ 5 . a
a
5 You u se straight lines on either side of
the vector:
|a| = |xi + y j| = | ( x y ) |Notation
A unit vector is any vector with
magnitude 1.
A unit vector in the direction of a is sometimes written as a
^.Notation
Example 11
Given that a = 3i + 4j and b = −2i − 4j:
a find |a|
b find a unit vector in the direction of
a
c find the exact va
lue of |2a + b |
a a = ( 3 4 )
|a| = √ ________ 32 + 42
|a| = √ ___ 25 = 5
b a unit v
ector is a ____ |a| = 3i +
4j _______ 5
= 1 __ 5 (3i + 4j) or ( 0.6 0.8 )
c 2a
+ b = 2 ( 3 4 ) + ( –2 –4 ) = ( 6 – 2 8 – 4 ) = ( 4 4 )
|2a +
b| = √ ________ 42 + 42 = √ ___ 32 = 4 √ __
2 Unless specified in the question it is acceptable
to give your answer in i, j form or column vector form.
You need to give an exact answer, so leave your answer in surd form:
√ ___ 32 = √ ______ 16 × 2 = 4 √ __
2 ← Section 1.5It is often quicker and easier to convert from i, j
form to column vector form for calculations.
Using Pythagoras.
a4
3 Explore the magnitude of a vector
us
ing GeoGebra.Online | [
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240
Chapter 11
You can define a vector by giving its magnitude, and the angle between the vector and one of the
coordinate axes. This is called magnitude-direction form.
θ4i + 5j
Oy
x
tan θ = 5 __ 4
θ = tan−1 ( 5 __ 4 ) = 51.3° (3 s.f.)Identify the angle that you need to find.
A diagram always helps.
You have a right-angled triangle with base 4 units and height 5 units, so use trigonometry.Example 12
Find the angle between the vector 4i + 5j
and the positive x-axis.This might be referred to as the angle between
the vector and i.
Example 13
Vector a has magnitude 10 and makes an angle of 30° with j.
Find a in i, j and column vector format.30° a
Oy
x
60°30°10
Oy
xxy
cos 60° = x ___ 10 x = 10 co s 60 ° = 5
sin 60
° = y ____ 10 y = 10 si n 60 ° = 5 √ __
3
a = 5 i + 5 √ __
3 j or a = ( 5 5 √ __
3 ) The d irection of a vector can be
given relative to either the positive x -axis (the i
direction) or the positive y -axis (or the j direction).Watch outUse trigonometry to find the lengths of the x-
and y-components for vector a.
Exercise 11C
1 Find the magnitude of each of these vectors.
a 3i
+ 4j b 6i
− 8j c 5i
+ 12j d 2i
+ 4j
e 3i
− 5j f 4i
+ 7j g −3i
+ 5j h −4i
− j | [
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0.146... |
241Vectors
2 a =
2i + 3j, b = 3i − 4j and c = 5i − j. Find the exact value of the magnitude of:
a a +
b b 2a
− c c 3b
− 2c
3 For each of the f
ollowing vectors, find the unit vector in the same direction.
a a =
4i + 3j b b =
5i − 12j c c =
−7i + 24j d d =
i − 3j
4 Find the angle that each of these v
ectors makes with the positive x-axis.
a 3i
+ 4j b 6i
− 8j c 5i
+ 12j d 2i
+ 4j
5 Find the angle that each of these v
ectors makes with j.
a 3i
− 5j b 4i
+ 7j c −3i
+ 5j d −4i
− j
6 Write these vectors in
i, j and column vector form.
a b c d
60°45° 15
Oy
x20° 8
Oy
x Oy
x25°
205
Oy
x
7 Draw a sketch for each vector and work out the exact value of its magnitude and the angle it
mak
es with the positive x-axis to one decimal place.
a 3i
+ 4j b 2i
− j c −5i
+ 2j
8 Given tha
t |2i − kj | =
2 √ ___ 10 , find the exact va lue of k. (3 marks)
9 Vector a
= pi + qj has magnitude 10 and makes
an angle θ with the positive x-axis where
sin θ = 3 _ 5 . Find the possible va lues of p and q.
(4 marks)
10 In triangle ABC, ⟶ AB = 4 i + 3j, ⟶ AC = 6 i − 4j. B
A
Ca Find the angle between ⟶ AB and i.
b Find the angle between ⟶ AC and i.
c Hence find the size of
∠BAC , in degrees, to one decimal place.
11 In triangle PQR, ⟶ PQ = 4 i + j, ⟶ PR = 6 i − 8j. Q
P
Ra Find the size of ∠QPR, in degrees,
to one decimal place. (5 marks)
b Find the area of triangle
PQR. (2 marks)E/P
E/P
E/P The area of a
tri
angle is 1 _ 2 ab sin θ.
← Section 9.3Hint
θa
bMake sure you consider all the possible cases.Problem-solving | [
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242
Chapter 11
In the diagram below ⟶ AB = pi + q j
and ⟶ AD = ri + s j.
ABCD is a parallelogram.
Prove that the area of ABCD is ps − qr .Challenge
ABC
DDraw the parallelogram
on a coordinate grid, and choose a position for the origin that will simplify your calculations.Problem-solving
11.4 Position vectors
You need to be able to use vectors to describe the position of a point in two dimensions.
Position vectors are vectors giving the position of a point, relative to a fixed origin.
The position vector of a point A is the vector ⟶ OA , where O is the origin.
Oy
A
xIf ⟶ OA = ai + bj then the position vector of A is ( a b ) .
■ In general
, a point P with coordinates ( p, q) has a position vector
⟶ OP = pi + qj = ( p q ) .
■ ⟶ AB = ⟶ OB − ⟶ OA , where ⟶ OA and
Oy
BA
x ⟶ OB are the position vectors of
A and B respectively. Use the triangle law:
⟶ AB = ⟶ AO + ⟶ OB = − ⟶ OA + ⟶ OB
So ⟶ AB = ⟶ OB − ⟶ OA
← Sec tion 11.1Link
Example 14
The points A and B in the diagram have coordinates (3, 4)
Oy
x2 4 6 8 10 12246
A
B
and (11, 2) respectively.
Find, in terms of i and j:
a the position vector of
A b the position vector of
B
c the vector ⟶ AB
a ⟶ OA = 3i + 4j
b ⟶ OB = 11i + 2j
c ⟶ AB = ⟶ OB − ⟶ OA
= (
11i + 2j) − (3i + 4j) = 8i − 2jIn column vector form this is ( 3 4 ) .
In column vector form this is ( 11 2 ) .
In column vector form this is ( 8 −2 ) . | [
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243Vectors
Example 15
⟶ OA = 5 i − 2j and ⟶ AB = 3 i + 4j. Find:
a the position vector of
B
b the exact va
lue of | ⟶ OB | in simplified surd for m.
x
ABy
O
a ⟶ OA = ( 5 −2 ) and ⟶ AB = ( 3 4 )
⟶ OB = ⟶ OA + ⟶ AB = ( 5 −2 ) + ( 3 4 ) = ( 8 2 )
b | ⟶ OB | = √ ________ 82 + 22 = √ _______ 64 + 4 = √ ___ 68 = 2 √ ___ 17 It is usually quicker to use column vector form for
calculations.
√ ___ 68 = √ ______ 4 × 17 = 2 √ ___ 17 in simplified surd form.In i, j form the answer is 8 i + 2 j.
Exercise 11D
1 The points A , B and C ha ve coordinates (3, − 1), (4, 5) and (− 2, 6) respectively, and O is the origin.
Find, in terms of i and j:
a i the position vectors of
A, B and C ii ⟶ AB iii ⟶ AC
b Find, in surd for
m: i | ⟶ OC | ii | ⟶ AB | iii | ⟶ AC |
2 ⟶ OP = 4i − 3j, ⟶ OQ = 3i + 2j
a Find ⟶ PQ
b Find, in surd for
m: i | ⟶ OP | ii | ⟶ OQ | iii | ⟶ PQ |
3 ⟶ OQ = 4i − 3j, ⟶ PQ = 5i + 6j
a Find ⟶ OP
b Find, in surd for
m: i | ⟶ OP | ii | ⟶ OQ | iii | ⟶ PQ |
4 OABCDE is a r
egular hexagon. The points A and B have position vectors a and b respectively,
where O is the origin.
Find, in terms of a and b, the position vectors ofa
C b D c E.P | [
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0.07442537695169449,
0.11658822745084763,
-0.15025849640369415,
-0.03664077818393707,
0... |
244
Chapter 11
5 The position vectors of 3 v
ertices of a parallelogram
Use a sketch to check that you
have considered all the possible positions for the fourth vertex.Problem-solving
are ( 4 2 ) , ( 3 5 ) and ( 8 6 ) .
Find the possible position vectors of the fourth vertex.
6 Given tha
t the point A has position vector 4i − 5j and the point B has position vector 6i + 3j,
a find the vector ⟶ AB . (2 marks)
b find | ⟶ AB | giving your answer as a simplified surd. (2 marks)
7 The point A lies on the cir
cle with equation x2 + y2 = 9. Given that ⟶ OA = 2ki + kj,
find the exact value of k. (3 marks)P
E
E/P
The point B lies on the line with equation 2 y = 12 − 3 x. Given that | OB| = √ ___ 13 ,
find p
ossible expressions for ⟶ OB in the form p i + qj.Challenge
11.5 Solving geometric probl ems
You need to be able to use vectors to solve geometric problems and to find the position vector of a
point that divides a line segment in a given ratio.
■ If the point P divides the line segment
AB in the ratio λ : μ, then A
B
AP : PB = λ : P
O m ⟶ OP = ⟶ OA + λ ______ λ + μ ⟶ AB
= ⟶ OA + λ ______ λ + μ ( ⟶ OB − ⟶ OA )
Example 16
In the diagram the points A and B have A
OBP
ba
position vectors a and b respectively
(referred to the origin O). The point P divides AB in the ratio 1
: 2.
Find the position vector of
P.
⟶ OP = ⟶ OA + 1 __ 3 ⟶ AB
= ⟶ OA + 1 __ 3 ( ⟶ OB − ⟶ OA )
= 2 __ 3 ⟶ OA + 1 __ 3 ⟶ OB
= 2 __ 3 a + 1 __ 3 b
Give your final answer in terms of a and b.There are 3 parts in the ratio in total, so P is 1 _ 3 of
the wa
y along the line segment AB.
Rewrite ⟶ AB in terms of the position vectors for A
and B
. | [
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245Vectors
Example 17
OABC is a parallelogram. P is the point where A
OCB
P
the diagonals OB and AC intersect.
The vectors a and c are equal to ⟶ OA and ⟶ OC
respecti
vely.
Prove that the diagonals bisect each other.
If the diagonals bisect each other, then P
must be the midpoint of OB and the midpoint
of AC .
From the diagram,
⟶ OB = ⟶ OC + ⟶ CB = c + a
and ⟶ AC = ⟶ AO + ⟶ OC .
= − ⟶ OA + ⟶ OC = −a + c
P lies on OB ⇒ ⟶ OP = λ(c + a)
P lies on AC ⇒ ⟶ OP = ⟶ OA + ⟶ AP
= a
+ μ(−a + c)
⇒ λ(c + a) = a + μ(−a + c)
⇒ λ = 1 − μ and λ = μ
⇒ λ = μ = 1 __ 2 , so P is the midpoint of both
diagonals, so the diagonals bisect each
other.If P is halfway along the line segment then it
must be the midpoint.The two expressions for ⟶ OP must be equal.Express ⟶ OB and ⟶ AC in terms of a and c.
Use the fact that P lies on both diagonals to find
two different routes from O to P, giving two
different forms of ⟶ OP .
Form and solve a pair of simultaneous equations
by equating the coefficients of a and c.
Example 18
In triangle ABC, ⟶ AB = 3 i − 2j and ⟶ AC = i − 5j,
Find the exact size of
∠BAC in degrees.
CBAWork out what information you would need to
find the angle. You could:
● find the lengths of all three sides then use the co
sine rule
● convert ⟶ AB and ⟶ AC to magnitude-direction
for
m
The working here shows the first method.Problem-solving Use GeoGebra to show that
di
agonals of a parallelogram bisect each other.OnlineYou can solve geometric problems by comparing coefficients on both sides of an equation:
■ If a and
b are two non-parallel vectors and pa + qb = ra + sb then p = r and q = s. | [
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246
Chapter 11
Exercise 11E
1 In the diagram, ⟶ WX = a, ⟶ WY = b and
X
WZ
Y
c ba ⟶ WZ = c. It is given tha t ⟶ XY = ⟶ YZ .
Prov
e that a + c = 2b.
2 OAB is a triangle.
P, Q and R are the midpoints
O RBQ
PA
of OA, AB and OB respectively.
OP and OR are equal to p and r respectively.
a Find i ⟶ OB ii ⟶ PQ
b Hence prov
e that triangle PAQ is similar to triangle OAB .
3 OAB is a triangle. ⟶ OA = a and ⟶ OB = b.
MNA
B O
The point M divides OA in the ratio 2 : 1.
MN is par
allel to OB.
a Express the vector ⟶ ON in terms of a and b .
b Show that
AN : NB =
1 : 2P
P
P ⟶ BC = ⟶ AC − ⟶ AB = ( 1 −5 ) − ( 3 −2 ) = ( −2 −3 )
| ⟶ AB | = √ __________ 32 + (−2)2 = √ ___ 13
| ⟶ AC | = √ _________ 12 + (−5)2 = √ ___ 26
| ⟶ BC | = √ ____________ (−2)2 + (−3)2 = √ ___ 13
cos ∠BAC =
| ⟶ AB |2 + | ⟶ AC |2 − | ⟶ BC |2 _______________________
2 × | ⟶ AB | × | ⟶ AC |
= 13 + 26
− 13 ______________
2 × √ ___ 13 × √ ___ 26 = 26 ______
26 √ __
2 = 1 ___
√ __
2
∠BAC =
cos−1 ( 1 ___
√ __
2 ) = 45°Use the triangle law to find ⟶ BC .
Leave your answers in surd form.
cos A = b2 + c2 − a2 __________ 2bc ← Sec tion 9.1
Check your answer by entering the
ve
ctors directly into your calculator.Online | [
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247Vectors
4 OABC is a squar
e. M is the midpoint of OA , and Q divides BC A B
Q
C OMP
in the ratio 1 : 3.
AC and MQ
meet at P .
a If ⟶ OA = a and ⟶ OC = c, express ⟶ OP in terms of a and c .
b Show that
P divides AC in the ratio 2 : 3.
5 In triangle ABC the position v
ectors of the vertices A, B and C are ( 5 8 ) , ( 4 3 ) and ( 7 6 ) . Find:
a | ⟶ AB | b | ⟶ AC | c | ⟶ BC |
d the size of
∠BAC , ∠ABC and ∠ACB to the nearest degree.
6 OPQ is a triangle.
O QP
RS
a
b2 ⟶ PR = ⟶ RQ and 3 ⟶ OR = ⟶ OS
⟶ OP = a and ⟶ OQ = b.
a Show that ⟶ OS = 2 a + b.
b Point T
is added to the diagram such
that ⟶ OT = −b.
Prove that points T, P and S lie on a straight line.P
P
OPQR is a parallelogram.
N is the midpoint of PQ and M is the midpoint of QR .
⟶ OP = a and ⟶ OR = b. The lines ON and OM intersect the diagonal PR at points
X and Y respectively.
a Exp
lain why ⟶ PX = −j a + j b, w here j is a constant.
b Sho
w that ⟶ PX = (k − 1) a + 1 _ 2 kb, where k is a constant.
c Exp
lain why the values of j and k must satisfy these simultaneous equations:
k − 1 = − j
1 _ 2 k = j
d Hen
ce find the values of j and k .
e Ded
uce that the lines ON and OM divide the diagonal PR into 3 equal parts.Challenge
ORQ
MN P
X
Ya
bTo show that T , P and S lie
on the same straight line
you need to show that any
two of the vectors ⟶ TP , ⟶ TS
or ⟶ PS are parallel.Problem-solving | [
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248
Chapter 11
11.6 Modelling with vectors
You need to be able to use vectors to solve problems in context.
In mechanics, vector quantities have both magnitude and direction. Here are three examples:●
vel
ocity
● displacement
● for
ce
You can also refer to the magnitude of these vectors. The magnitude of a vector is a scalar quantity
− it has size but no direction:
● speed is the magnitude o
f the velocity vector
● distance in a st
raight line between A and B is the magnitude of the displacement vector ⟶ AB
When modelling with vectors in mechanics
, it is common to use the unit vector j to represent North
and the unit vector i to represent East.
Example 19
A girl walks 2 km due east from a fixed point O to A, and then 3 km due south from A to B. Find:
a the total distance tra
velled
b the position vector of
B relative to O
c | ⟶ OB |
d the bearing of B
from O.
a The distance the girl has walked is
2 km +
3 km =
5 km
b Rep
resenting the girl’s journey on a diagram:
θO A
B3 km2 kmN
⟶ OB = (2 i − 3 j) km
c | ⟶ OB | = √ ________ 22 + 32 = √ ___ 13 = 3.61 km (3 s.f.)
d tan θ = 3 __ 2
θ = 56.3°
The bearing of B from O is
56.3° + 90° = 146.3° = 146°Note that the distance of B from O is not the
same as the distance the girl has walked.
Remember to include the units with your answer.
⟶ OB is the length of the line segment OB in th e
diagram and represents the girl’s distance from
the starting point.j represents North, so 3 km south is written as
–3
j km.
A three-figure bearing is always measured clockwise from north. | [
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249Vectors
Example 20
In an orienteering exercise, a cadet leaves the starting point O and walks 15 km on a bearing of
120° to reach
A, the first checkpoint. From A he walks 9 km on a bearing of 240° to the second
checkpoint, at B. From B he returns directly to O.
Find:a
the position vector of
A relative to O
b | ⟶ OB |
c the bearing of B
from O
d the position vector of
B relative to O.
a
θO
A
B9 km15 km120°
240°N
N
The position vector of A relative to O is ⟶ OA .
AO30°15 cos 30°
15 km15 sin 30°
⟶ OA = (15 cos 30 °i + 15 si n 30 °j) km
=
(13.0i − 7.5 j) km
b
θO
A
B9 km15 km
240°60°30°
60°N
N
| ⟶ OB |2 = 152 + 92 − 2 × 15 × 9 × cos 60°
= 171
| ⟶ OB | = √ ____ 171 = 1 3.1 km (3 s.f.)Start by drawing a diagram.
Draw a right angled triangle to work out the
lengths of the i and j components for the position vector of A relative to O.
| ⟶ OB | is the length of OB in triangle OAB .
Use the co
sine rule in triangle OAB .∠OAB = 360° − (240° + 60°) = 60° | [
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250
Chapter 11
Exercise 11F
1 Find the speed of a particle moving with these v elocities:
a (3i + 4j) m s−1 b (24i − 7j) km h−1
c (5i + 2j) m s−1 d (−7i + 4j) cm s−1
2 Find the distance moved b y a particle which travels for:
a 5 hours at ve
locity (8i + 6j) km h−1
b 10 seconds at ve locity (5i − j) m s−1
c 45 minutes at velocity (6i + 2j) km h−1
d 2 minutes at velocity (− 4i − 7j) cm s−1.
3 Find the speed and the distance trav
elled by a particle moving in a straight line with:
a velocity (
−3i + 4j) m s−1 for 15 seconds b velocity (2 i + 5j) m s−1 for 3 seconds
c velocity (5
i − 2j) km h−1 for 3 hours d velocity (12 i − 5j) km h−1 for 30 minutes.
4 A particle P is acce
lerating at a constant speed.
When t = 0, P has velocity u = (2i + 3j) m s−1
and at time t = 5 s, P
has velocity v = (16i − 5j) m s−1.
The acceleration vector of the particle is given by the formula: a = v −
u _____ t
Find the accelera
tion of P in terms of i and j. Speed is the magnitude of
th
e velocity vector.Hint
Find the speed in each case
the
n use:
Distance travelled = speed × timeHint
The units of acceleration
wi
ll be m/s2 or m s−2.Hintc sin θ ____ 9 = sin 60° _______
√ ____ 171
sin θ = 9 × sin 60° ___________
√ ____ 171 = 0.
596…
θ = 36.6° = 37° (3 s.f.)
The bearing of B from O = 120 + 37
= 157°
d
BO13.1 cos 67°
13.1 sin 67°
13.1 km67°N N
⟶ OB = (5.1i − 12. 1j ) kmUse the sine rule to work out θ.
157° − 90° = 67°
Draw a right angled triangle to work out the
lengths of the i and j components for the position vector of B relative to O. | [
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251Vectors
5 A particle P of
mass m = 0.3 kg moves under the action of
a single constant force F newtons.
The acceleration of P is a = (5i + 7j) m s−2.
a Find the angle between the acceler
ation and i. (2 marks)
Force
, mass and acceleration are related by the formula F = ma.
b Find the magnitude of
F. (3 marks)
6 Two f
orces, F1 and F2, are given by the vectors F1 = (3i − 4j) N and F2 = ( pi + qj) N.
The resultant f
orce, R = F1 + F2 acts in a direction which is parallel to the vector (2i − j).
a Find the angle between R
and the vector i. (2 marks)
b Show that
p + 2q = 5. (3 marks)
c Given tha
t p = 1, find the magnitude of R. (3 marks)
7 The diagram sho
ws a sketch of a field in the shape of a triangle ABC. B
A
CGiven ⟶ AB = 30 i + 40j metres and ⟶ AC = 40 i − 60j metres,
a find ⟶ BC (2 marks)
b find the size of
∠BAC , in degrees, to one decimal place (4 marks)
c find the area of the fie
ld in square metres. (3 marks)
8 A boat has a position vector of
(2i + j) km and a buoy has a position v
ector of (6i − 4j) km,
re
lative to a fixed origin O.
a Find the distance of the boat fr
om the buoy.
b Find the bearing of the boat fr
om the buoy.
The boat travels with constant velocity (8i − 10j) km/h.
c Verify that the boa
t is travelling directly towards the buoy
d Find the speed of the boat.
e Work out ho
w long it will take the boat to reach the buoy.E
E/P
E/P
P
Draw a sketch showing the
initial positions of the boat, the buoy and the origin.Problem-solving
1 Two f
orces F1 and F2 act on a particle.
F1 = −3i + 7j newtons
F2 = i − j newtons
The resultant force R acting on the particle is given by R = F1 + F2.
a Calculate the ma
gnitude of R in newtons. (3 marks)
b Calculate
, to the nearest degree, the angle between the line of action of R and the
vector j. (2 marks)EMixed exercise 11 | [
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252
Chapter 11
2 A small boat
S, drifting in the sea, is modelled as a particle moving in a straight line at constant
speed. When first sighted at 09:00, S is at a point with position vector (−2i − 4j) km rela
tive to a
fixed origin O, where i and j are unit vectors due east and due north respectively. At 09:40, S is
at the point with position vector (4i − 6j) km.
a Calculate the bearing on w
hich S is drifting.
b Find the speed of S
.
3 A football pla
yer kicks a ball from point A on a flat football field. The motion of the ball is
modelled as that of a particle travelling with constant velocity (4i + 9j) m s−1.
a Find the speed of the ball.
b Find the distance of the ball fr
om A after 6 seconds.
c Comment on the validity of
this model for large values of t.
4 ABCD is a tra
pezium with AB parallel to DC and DC = 4AB.
M divides DC such that DM : MC =
3 : 2, ⟶ AB = a and ⟶ BC = b.
Find, in terms of a and b:
a ⟶ AM b ⟶ BD c ⟶ MB d ⟶ DA
5 The vectors 5a
+ kb and 8a + 2b are parallel. Find the value of k. (3 marks)
6 Given tha
t a = ( 7 4 ) , b = ( 10 −2 ) and c = ( −5 −3 ) find:
a a +
b + c b a −
2b + c c 2a
+ 2b − 3c
7 In triangle ABC, ⟶ AB = 3 i + 5j and ⟶ AC = 6 i + 3j, find: B
ACa ⟶ BC (2 marks)
b ∠BAC (4 marks)
c the area of the triangle
. (2 marks)
8 The resultant of the v
ectors a = 4i − 3j and b = 2 pi
− pj is parallel to the vector
c = 2i − 3j. Find:
a the value of
p (3 marks)
b the resultant of v
ectors a and b. (1 mark)
9 For each of the f
ollowing vectors, find
i a unit vector in the same direction
ii the angle the vector mak
es with i
a a =
8i + 15j b b =
24i − 7j c c =
−9i + 40j d d =
3i − 2jP
P
P
E/P
E
E/P | [
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253Vectors
10 The vector a
= pi + qj, where p and q are positive constants, is such that |a| = 15.
Given that a makes an angle of 55° with i, find the values of p and q.
11 Given tha
t |3i − kj | = 3 √ __
5 , find the value of k. (3 marks)
12 OAB is a triangle
. ⟶ OA = a and ⟶ OB = b. The point M divides OA in the ratio 3 : 2.
MN is par
allel to OB.
a Express the vector ⟶ ON in terms of a and b. (4 marks)
MNA
B Ob Find vector ⟶ MN . (2 marks)
c Show that
AN : NB =
2 : 3. (2 marks)
13 Two f
orces, F1 and F2, are given by the vectors F1 = (4i − 5j) N and F2 = ( pi + qj) N.
The resultant f
orce, R = F1 + F2 acts in a direction which is parallel to the vector (3i − j)
a Find the angle between R
and the vector i. (3 marks)
b Show that
p + 3q = 11. (4 marks)
c Given tha
t p = 2, find the magnitude of R. (2 marks)
14 A particle P is acce
lerating at a constant speed. When t = 0, P has velocity u = (3i + 4j) m s−1
and at time t = 2 s, P
has velocity v = (15i − 3j) m s−1.
The acceleration vector of the particle is given by the formula: a = v −
u _____ t
Find the magnitude of the acce
leration of P. (3 marks)P
E/P
E/P
E/P
E
The point B lies on the line with equation 3 y = 15 − 5 x.
Given that | ⟶ OB | = √ ___ 34 ____ 2 , find two possible expressions for ⟶ OB In the form p i + qj.Challenge | [
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254
Chapter 11
1 If ⟶ PQ = ⟶ RS then the line segments PQ and RS are equal in length and are parallel.
2 ⟶ AB = − ⟶ BA as the line segment AB is equal in length, parallel and in the opposite direction
to
BA.
3 Triangle la
w for vector addition: ⟶ AB + ⟶ BC = ⟶ AC
If ⟶ AB = a, ⟶ BC = b and ⟶ AC = c, then a + b = c
4 Subtracting a v
ector is equivalent to ‘adding a negative vector’: a − b = a + (−b)
5 Adding the vect
ors ⟶ PQ and ⟶ QP gives the zero vector 0: ⟶ PQ + ⟶ QP = 0.
6 Any vect
or parallel to the vector a may be written as λa, where λ is a non-zero scalar.
7 To multiply a column v
ector by a scalar, multiply each component by the scalar: λ ( p q ) = ( λp λq )
8 To add tw
o column vectors, add the x-components and the y-components ( p q ) + ( r s ) = ( p + r q + s )
9 A unit vector is a v
ector of length 1. The unit vectors along the x- and y-axes are usually
denoted by i and j respectively. i = ( 1 0 ) j = ( 0 1 )
10 For any t
wo-dimensional vector: ( p q ) = pi + qj
11 For the vect
or a = xi + yj = ( x y ) , the magnitude of the vector is given by: |a| = √ ______ x2 + y2
12 A unit vector in the dir
ection of a is a ___ |a|
13 In general
, a point P with coordinates ( p, q
) has position vector:
⟶ OP = pi + qj = ( p q )
14 ⟶ AB = ⟶ OB − ⟶ OA , where ⟶ OA and ⟶ OB are the position vectors of A and B respectively.
15 If the point P divides the line segment
AB in the ratio λ : μ, then A
B
AP : PB = λ : P
O m ⟶ OP = ⟶ OA + λ _____ λ + μ ⟶ AB
= ⟶ OA + λ _____ λ + μ ( ⟶ OB − ⟶ OA )
16 If a and
b are two non-parallel vectors and pa + qb = ra + sb then p = r and q = sSummary of key points | [
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0.000... |
255
Differentiation
After completing this chapter you should be able to:
● Find the derivative, f
9(x) or dy ___ dx , of a simple function → pages 259–268
● Use the derivative t o solve problems involving gradients, tangents
and normals → pages 268–270
● Identify increasing and decreasing functions → pages 270–271
● Find the second order deriv ative, f 0(x ) or d 2 y ____ d x 2 , of a
simple function → pages 271–272
● Find stationary points of functions and det ermine their
nature → pages 273–276
● Sketch the gradient function of a given function → pages 277–278
● Model real-life situations with differentiation → pages 279–281Objectives
1 Find the gradients of these lines .
1
–5y
xab c
64y
x(6, 6)
4y
xO
OO
← Sec tion 5.1
2 Write each of these expressions in the
form
xn where n is a positive or negative
real number.
a x3 × x7 b 3 √ __
x2 c x 2 × x 3 ______ x 6
d √ ___
x 2 ___
√ __
x ← Sections 1.1, 1.4
3 Find the equation of the str aight line that
passes through:
a (0, −2) and (6, 1) b (3, 7) and (9, 4)
c (10, 5) and (−2, 8) ← Section 5.2
4 Find the equation of the perpendicular to
the line
y = 2x − 5 at the point (2, 1).
← Section 5.3Prior knowledge check
Differentiation is part of calculus, one of the most powerful tools in mathematics. You will use differentiation in mechanics to model rates of change, such as speed and acceleration.
→ Exercise 12K Q512 | [
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256
Chapter 12
12.1 Gradients of curves
The gradient of a curve is constantly changing. You can use a tangent to find the gradient of a curve
at any point on the curve. The tangent to a curve at a point A is the straight line that just touches the curve at A.
■ The gradient of a curve at a given point is defined as the gradient of the tangent to the curv
e at that point.
The diagram shows the curve with equation y = x2.
The tangent, T, to the curve at the point A(1, 1) is shown. Point A is joined to point P by the chord AP.
a
Calculate the gr
adient of the tangent, T.
b Calculate the gr
adient of the chord AP when P
has coordinates:
i (2, 4)
ii (1.5, 2.25)
iii (1.1, 1.21)
iv (1.01, 1.0201)
v (1 +
h, (1 + h)2)
c Comment on the rela
tionship between your
answers to parts a and b.Example 1
y
x
–1–0.5 0.5 1 1.52 2.5 3 –1 –1.51
O23456
–2–3
–4AP
T
(1, 1)y = x2y
x
–0.5–0.5 0.5y =
x3 – 2x + 1
1 1.5 2 2.5 –1 –1.50.511.522.5
–1
–1.5OThe tangent to the curve at (1, 0) has gradient 1, so
the gradient of the curve at the point (1, 0) is equal to 1.
The tangent just touches the curve at (1, 0).
It does not cut the curve at this point, although it may cut the curve at another point. | [
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257Differentiation
This time (x1, y1) is (1, 1) and (x2, y2) is (1.5, 2.25).Use the formula for the gradient of a straight line
between points ( x1, y1) and ( x2, y2). ← Section 5.1
This point is closer to (1, 1) than (1.1, 1.21) is.
This gradient is closer to 2.
This becomes h(2 + h) _______ h
You can use this formula to confirm the answers
to questions i to iv. For example, when h = 0.5, (1 + h, (1 + h)
2) = (1.5, 2.25) and the gradient of
the chord is 2 + 0.5 = 2.5.
As h gets closer to zero, 2 + h gets closer to 2, so the gradient of the chord gets closer to the gradient of the tangent.The points used are (1, 1) and (2, 3).a Gradient of tangent = y2 − y1 _______ x2 − x1
= 3 − 1 ______ 2 − 1
= 2
b i Gra
dient of chord joining (1, 1) to (
2, 4)
= 4 − 1 ______ 2 − 1
= 3
ii Gra
dient of the chord joining (1, 1) to
(
1.5, 2.25)
= 2.2
5 − 1 _________ 1.5 − 1
= 1 .2
5 ____ 0.5
= 2.5
iii Gra
dient of the chord joining (1, 1) to
(
1.1, 1.21)
= 1.21
− 1 _________ 1.1 − 1
= 0.21 _____ 0.1
= 2.1
iv Gra
dient of the chord joining (1, 1) to
(
1.01, 1.0201)
= 1.02
01 − 1 ___________ 1.01 − 1
= 0.02
01 ________ 0.01
= 2.01
v Gra
dient of the chord joining (1, 1) to
(
1 + h, (1 + h )2)
= (1
+ h)2 − 1 ____________ (1 + h) − 1
= 1 + 2
h + h2 − 1 ________________ 1 + h − 1
= 2h
+ h2 ________ h
= 2 +
h
c As P
gets closer to A , the gradient of the
chord AP gets closer to the gradient of
the tangent at A .h is a constant.
(1 + h)2 = (1 + h)(1 + h) = 1 + 2h + h2 Explore the gradient of the chord
AP us
ing GeoGebra.Online | [
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258
Chapter 12
1 The diagram sho
ws the curve with equation y = x2 − 2x. y
O x–1 1234 –2
–11234
–2a Copy and complete this table showing estimates for
the gradient of
the curve.
x-coordinate −1 0 1 2 3
Estimate for gradient of curve
b Write a hypothesis a
bout the gradient of the curve at
the point where x = p.
c Test your h
ypothesis by estimating the gradient of
the graph at the point (1.5, −0.75).
2 The diagram sho
ws the curve with equation y = √ ______ 1 − x2 .
The point A has coor
dinates (0.6, 0.8).
The points B, C and D lie on the curve with x-coordinates 0.7, 0.8 and 0.9 respectively.
xy
O–0.2 –0.4 –0.6 –0.8 –1.0 0.2 0.4 0.6 0.8 1.0
–0.20.20.40.60.81.0
A
B
C
Dy = 1 – x2
a Verify that point A lies on the curve.
b Use a ruler to estimate the gradient of
the curve at point A.
c Find the gradient of the line segments:
i AD
ii AC
iii AB
d Comment on the rela
tionship between your answers to parts b and c. Place a ruler on the graph
to a
pproximate each tangent.HintExercise 12A
Use algebra for part c . Hint | [
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-0.0467884... |
259Differentiation
3 F is the point with coordina
tes (3, 9) on the curve with equation y = x2.
a Find the gradients of the chor
ds joining the point F to the points with coordinates:
i (4, 16) ii (3.5, 12.25) iii (3.1, 9.61)
iv (3.01, 9.0601) v (3 +
h, (3 + h)2)
b What do y
ou deduce about the gradient of the tangent at the point (3, 9)?
4 G is the point with coordina
tes (4, 16) on the curve with equation y = x2.
a Find the gradients of the chor
ds joining the point G to the points with coordinates:
i (5, 25) ii (4.5, 20.25) iii (4.1, 16.81)
iv (4.01, 16.0801) v (4 +
h, (4 + h)2)
b What do y
ou deduce about the gradient of the tangent at the point (4, 16)?
y = f(x)
xy
OAB
You can formalise this approach by letting the x-coordinate of A be x0 and the x-coordinate of B be
x0 + h. Consider what happens to the gradient of AB as h gets smaller.
y = f(x)
xy
A
x0 x0 + hB
O12.2 Finding the derivative
You can use algebra to find the exact gradient of a curve at a given point. This diagram shows two
points, A and B, that lie on the curve with equation y = f(x).
As point B moves closer to point A the gradient of chord AB
gets closer to the gradient of the tangent to the curve at A.
h rep resents a
small change in the value
of x. You can also use dx to
represent this small change. It
is pronounced ‘delta x ’.NotationPoint B has coordinates (x0 + h, f(x0 + h)).
Point A has coordinates (x0, f(x0)). | [
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260
Chapter 12
The vertical distance from A to B is f(x0 + h) − f(x0).
AB
f(x 0 + h) – f(x 0)
h
The horizontal distance is x0 + h − x0 = h.
So the gradient of AB is f ( x 0 + h) − f ( x 0 ) ______________ h
As h g
ets smaller, the gradient of AB gets closer to the gradient of the tangent to the curve at A.
This means that the gradient of the curve at A is the limit of this expression as the value of h
tends to 0.
You can use this to define the gradient function.
■ The gradient function, or derivative, of the curve y = f(x)
is written as f9( x) or dy ___ dx .
f 9 (x
) =
lim
h → 0 f(x + h) − f(x) _____________ h
The gradient function can be used to find the gr
adient of
the curve for any value of x.
Using this rule to find the derivative is called differentiating from first principles. lim
h → 0 means ‘the
limit as h tends to 0’. You can’t
evaluate the expression when h = 0, but as h gets smaller the expression gets closer to a fixed (or limiting ) value.Notation
The point A with coordinates ( 4, 16 ) lies on the curve with equation
y = x2.
At point A the curve has gradient g.
a Show that
g = lim
h → 0 (8 + h) .
b Deduce the value of
g.Example 2
a g = lim
h → 0 f(4 + h) − f(4) _____________ h
= lim
h → 0 (4 + h)2 − 42 ____________ h
= lim
h → 0 16 + 8h + h2 − 16 ___________________ h
= lim
h → 0 8h + h2 ________ h
= lim
h → 0 (8 + h)
b g =
8Use the definition of the derivative with x = 4.
As h → 0 the limiting value is 8, so the gradient at
point A is 8.The 16 and the −16 cancel, and you can cancel h in the fraction.The function is f(x ) = x2. Remember to square
everything inside the brackets. ← Section 2.3 | [
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261Differentiation
Prove, from first principles, that the derivative of x3 is 3x2.Example 3
f(x) = x3
f9(x) = lim
h → 0 f(x + h) − f( x) ____________ h
= lim
h → 0 (x + h)3 − (x)3 _____________ h
= lim
h → 0 x3 + 3 x2h + 3 xh2 + h3 − x3 _________________________ h
= lim
h → 0 3x2h + 3 xh2 + h3 _________________ h
= lim
h → 0 h(3x2 + 3 xh + h2) _________________ h
= lim
h → 0 (3x2 + 3 xh + h2)
As h → 0, 3 xh → 0 and h2 → 0.
So f9(x) = 3 x2(x + h)3 = (x + h)(x + h)2
= (x + h)(x2 + 2hx + h2)
which expands to give x3 + 3x2h + 3xh2 + h3‘From first principles’ means that you have to use
the definition of the derivative. You are starting your proof with a known definition, so this is an example of a proof by deduction.
Any terms containing h, h2, h3, etc will have a
limiting value of 0 as h → 0.
1 For the function f(x) = x2, use the definition of the derivative to show that:
a f9
(2) = 4 b f9
(−3) = −6 c f9
(0) = 0 d f9
(50) = 100
2 f(x
) = x2
a Show that f 9 (x) = lim
h → 0 (2x + h) . b Hence deduce that f9 (x) = 2x.
3 The point A with coor
dinates (−2, −8) lies on the curve with equation y = x3.
At point A the curve has gradient g.
a Show that g =
lim
h → 0 (12 − 6h + h 2 ) . b Deduce the value of g.
4 The point A with coor
dinates (−1, 4) lies on the curve with
equation y = x3 − 5x.
The point B also lies on the curve and has x -coordinate (− 1 + h).
a Show that the gr
adient of the line segment AB is given
by h2 − 3h − 2.
b Deduce the gradient of the curv
e at point A.
5 Prov
e, from first principles, that the derivative of 6x is 6. (3 marks)
6 Prov
e, from first principles, that the derivative of 4x2 is 8x. (4 marks)
7 f(x
) = ax2, where a is a constant. Prove, from first principles, that f9(x) = 2ax. (4 marks)P
Draw a sketch showing
points A and B and the
chord between them.Problem-solving
E/P
E/P
E/PExercise 12BDifferentiation
Factorise the numerator. | [
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-0.061... |
262
Chapter 12
Find the derivative, f 9(x
), when f(x) equals:
a x6 b x 1 _ 2 c x−2 d x2 × x3 e x __ x5 Example 412.3 Differentiating x n
You can use the definition of the derivative to find an expression for the derivative of x n where n is
any number. This is called differentiation.
■ For all real values of n, and for a constant a :
● If f(x) = x n then f 9 (x) = n x n − 1
If y = x n then dy ___ dx = nx n − 1
● If f(x) = ax n then f 9 (x) = anx n − 1
If y = ax n then dy ___ dx = anx n − 1
a f(x) = x6
So f9(x) = 6x5
b f(x) = x 1 __ 2
So f9(x) =
1 __ 2 x − 1 __ 2
= 1 ____ 2 √ __
x
c f(x) =
x−2
So f9(x) = −2x−3
= − 2 __ x3
d f(x) = x2 × x3
= x5
So f9(x) = 5x4You can leave your answer in this form or write it
as a fraction.
You need to write the function in the form x n
before you can use the rule.
x2 × x3 = x 2 + 3
= x5
f9(x) and dy ___ dx both represent the
derivative. You usually use dy ___ dx
whe
n an expression is given in
the form y = …Notation
Multiply by the power, then subtract 1 from the
power:
6 × x 6 − 1 = 6x 5
The new power is 1 _ 2 − 1 = − 1 _ 2
x − 1 _ 2 = 1 ___
√ __
x
← Section 1.4f(x) = 1 __ x
a Giv
en that f 9 (x) = lim
h → 0 f (x + h) − f (x) ______________ h , show that f 9 (x) = lim
h → 0 −1 _______ x 2 + xh
b Ded
uce that f 9 (x) = − 1 __ x 2 Challenge | [
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-0.0256216... |
263Differentiation
e f(x) = x ÷ x5
= x−4
So f9(x) = −4x−5
= − 4 __ x5
Find dy ___ dx when y equals:
a 7x3 b −4 x 1 _ 2 c 3x−2 d 8 x 7 ____ 3x e √ ____ 36 x 3 Example 5
1 Find f 9(x ) given that f(x) equals:
a x7 b x8 c x4 d x 1 _ 3 e x 1 _ 4 f 3 √ __
x
g x−3 h x−4 i 1 __ x 2 j 1 __ x 5 k 1 ___ √ __
x l 1 ___ 3 √ __
x
m x3 × x6 n x2 × x3 o x × x2 p x 2 __ x 4 q x 3 __ x 2 r x 6 __ x 3
2 Find dy ___ dx given that y equals:
a 3x2 b 6x9 c 1 _ 2 x 4 d 20 x 1 _ 4 e 6 x 5 _ 4
f 10x−1 g 4 x 6 ____ 2 x 3 h x ____ 8 x 5 i − 2 ___ √ __
x j √ _________ 5 x 4 × 10x _________ 2 x 2 Exercise 12CUse the laws of indices to simplify the fraction:
x1 ÷ x5 = x 1 − 5 = x −4
Use the rule for differentiating ax n with a = 7 and
n = 3. Multiply by 3 then subtract 1 from the power.a dy ___ dx = 7 × 3x3 − 1 = 21x2
b dy ___ dx = −4 × 1 __ 2 x − 1 __ 2 = −2 x − 1 __ 2 = − 2 ___ √ __
x
c dy ___ dx = 3 × −2x−3 = −6x−3 = − 6 ___ x 3
d y =
8 __ 3 x 6
dy ___ dx = 6 × 8 __ 3 x 5 = 16 x 5
e y =
√ ___ 36 × √ ___ x 3 = 6 × (x3 ) 1 __ 2 = 6 x 3 __ 2
dy ___ dx = 6 × 3 __ 2 x 1 __ 2 = 9 x 1 __ 2 = 9 √ __
x Simplify the number part as much as possible.This is the same as differentiating x 1 _ 2 then
multiplying the result by −4.
Write the expression in the form ax n. Remember
a can be any number, including fractions.
3 _ 2 − 1 = 1 _ 2
Make sure that the
fu
nctions are in the form x n
before you differentiate.Hint | [
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264
Chapter 12
3 Find the gradient of the curv
e with equation y = 3 √ __
x at the point where:
a x =
4 b x =
9
c x =
1 _ 4 d x = 9 __ 16
4 Given tha
t 2y2 − x3 = 0 and y > 0, find dy ___ dx (2 marks)
12.4 Differentiating quadratics
You can differentiate a function with more than one term by differentiating the terms one-at-a-time.
The highest power of x in a quadratic function is x 2, so the highest power of x in its derivative will
be x.
You can find this expression for dy ___ dx by differentiating each of the terms one-at-a-time:
ax2Differentiate 2ax1 = 2ax bx = bx1Differentiate 1bx0 = bc Differentiate
The quadratic term tells
you the slope of thegradient function.An x term differentiatesto give a constant.Constant termsdisappear whenyou differentiate. 0E/P
The derivative is a straight line with
gra
dient 2 a. It crosses the x -axis once, at the
point where the quadratic curve has zero
gradient. This is the turning point of the quadratic curve.
← Section 5.1Links
Find dy ___ dx given that y equals:
a x2 + 3x b 8x − 7 c 4x2 − 3x + 5Example 6
a y = x2 + 3 x
So dy ___ dx = 2x + 3
b y =
8x − 7
So dy ___ dx = 8
c y =
4x2 − 3 x + 5
So dy ___ dx = 8 x − 3Differentiate the terms one-at-a-time.
The constant term disappears when you
differentiate. The line y = −7 would have zero gradient.
The derivative is 2ax + b = 2 × 4x − 3 = 8x − 3.4x2 − 3x + 5 is a quadratic expression with
a = 4, b = −3 and c = 5.Try rearranging unfamiliar equations into a form you recognise.Problem-solving
■ For the quadratic curve with equation
y = ax2 + bx + c, the derivative is given by
dy ___ dx = 2ax + b | [
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265Differentiation
1 Find dy ___ dx when y equals:
a 2x2 − 6x + 3 b 1 _ 2 x2 + 12x c 4x2 − 6
d 8x2 + 7x + 12 e 5 + 4x − 5x2
2 Find the gradient of the curv e with equation:
a y = 3
x2 at the point (2, 12) b y = x2 + 4x at the point (1, 5)
c y = 2
x2 − x − 1 at the point (2, 5) d y = 1 _ 2 x2 + 3 __ 2 x at the point (1, 2)
e y = 3
− x2 at the point (1, 2) f y = 4 − 2x2 at the point (−1, 2)
3 Find the y-coor
dinate and the value of the gradient at the point P with x-coordinate 1 on the
curve with equation y = 3 + 2x − x2.
4 Find the coordinates of
the point on the curve with equation y = x2 + 5x − 4 where the
gradient is 3.Let f(x) = 4x2 − 8x + 3.
a Find the gradient of
y = f(x) at the point ( 1 _ 2 , 0) .
b Find the coordinates of
the point on the graph of y = f(x) where the gradient is 8.
c Find the gradient of y
= f(x) at the points where the curve meets the line y = 4x − 5.Example 7
Exercise 12Da As y = 4x2 − 8 x + 3
dy ___ dx = f9(x) = 8x − 8 + 0
So f9 ( 1 __ 2 ) = −4
b dy ___ dx = f9(x) = 8x − 8 = 8
So x = 2
So y = f(2) = 3The point where the gradient is 8 is (2, 3).
c
4x2 − 8 x + 3 = 4 x − 5
4x2 − 12 x + 8 = 0
x2 − 3 x + 2 = 0
(x − 2)( x − 1) = 0
So x = 1 or x = 2At x = 1, the gradient is 0.At x = 2, the gradient is 8, as in part b .Differentiate to find the gradient function. Then
substitute the x -coordinate value to obtain the
gradient.
Put the gradient function equal to 8. Then solve the equation you have obtained to give the value of x .
Substitute this value of x into f(x) to give the value of y and interpret your answer in words.
To find the points of intersection, set the equation of the curve equal to the equation of the line. Solve the resulting quadratic equation to find the x-coordinates of the points of intersection.
← Section 4.4
Substitute the values of x into f9(x) = 8x − 8 to give the gradients at the specified points.
Use your calculator to check
so
lutions to quadratic equations quickly.Online | [
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266
Chapter 12
5 Find the gradients of the curv
e y = x2 − 5x + 10 at the points A and B where the curve meets
the line y = 4.
6 Find the gradients of the curv
e y = 2x2 at the points C and D where the curve meets the line
y = x + 3.
7 f(x
) = x2 − 2x − 8
a Sketch the gra
ph of y = f(x).
b On the same set of axes
, sketch the graph of y = f9(x).
c Explain why the
x-coordinate of the turning point of y = f(x) is the same as the x-coordinate
of the point where the graph of y = f9(x) crosses the x-axis.
12.5 Differentiating functions with two or more terms
You can use the rule for differentiating ax n to differentiate functions with two or more terms.
You need to be able to rearrange each term into the form ax n, where a is a constant and n is a real
number. Then you can differentiate the terms one-at-a-time.
■ If y = f(x) ± g( x), then dy ___ dx = f9(x) ± g9( x).P
P
P
Find dy ___ dx given that y equals:
a 4x3 + 2x b x3 + x2 − x 1 _ 2 c 1 _ 3 x 1 _ 2 + 4 x 2 Example 8
a y = 4 x3 + 2x
So dy ___ dx = 12x2 + 2
b y = x3 + x2 − x 1 __ 2
So dy ___ dx = 3x2 + 2x − 1 __ 2 x − 1 __ 2
c y = 1 __ 3 x 1 __ 2 + 4x2
So dy ___ dx = 1 __ 3 × 1 __ 2 x − 1 __ 2 + 8x
= 1 __ 6 x − 1 __ 2 + 8xDifferentiate the terms one-at-a-time.
Be careful with the third term. You multiply the
term by 1 _ 2 and then reduce the power by 1 to
get − 1 _ 2
Check that each term is in the form axn before
differentiating. | [
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267Differentiation
1 Differentia
te:
a x 4 + x−1 b 2x5 + 3x−2 c 6 x 3 _ 2 + 2 x − 1 _ 2 + 4
2 Find the gradient of the curv
e with equation y = f(x) at the point A where:
a f(x
) = x3 − 3x + 2 and A is at (−1, 4) b f(x ) = 3x2 + 2x−1 and A is at (2, 13)
3 Find the point or points on the curve with equation
y = f(x), where the gradient is zero:
a f(x
) = x2 − 5x b f(x ) = x3 − 9x2 + 24x − 20
c f(x
) = x 3 _ 2 − 6x + 1 d f(x ) = x−1 + 4x
4 Differentia
te:
a 2 √ __
x b 3 __ x2 c 1 ___ 3x3 d 1 _ 3 x3(x − 2)
e 2 __ x3 + √ __
x f 3 √ __
x + 1 ___ 2x g 2x +
3 ______ x h 3x2 − 6 _______ x
i 2x3 + 3x ________ √ __
x j x(x2 − x + 2) k 3x2(x2 + 2x) l (3x − 2) (4x + 1 __ x ) Exercise 12Ea Let y = 1 ____ 4 √ __
x
= 1 __ 4 x − 1 __ 2
The
refore dy ___ dx = − 1 __ 8 x − 3 __ 2
b Le
t y = x3(3x + 1)
= 3x4 + x3
Therefore dy ___ dx = 12 x3 + 3x2
= 3x2(4x + 1)
c Le
t y = x − 2 ______ x2
= 1 __ x − 2 __ x2
= x−1 − 2x−2
Therefore dy ___ dx = −x−2 + 4x−3
= − 1 __ x2 + 4 __ x3
= 4 −
x ______ x3 Differentiate:
a 1 ____ 4 √ __
x b x3(3x + 1) c x − 2 _____ x2
Multiply out the brackets to give a polynomial
function.Example 9
Use the laws of indices to write the expression in the form ax
n.
1 ____
4 √ __
x = 1 __ 4 × 1 ___
√ __
x = 1 __ 4 × 1 ___
x 1 _ 2 = 1 __ 4 x − 1 _ 2
Express the single fraction as two separate
fractions, and simplify: x __ x2 = 1 __ x Differentiate each term.
Write each term in the form ax n then
differentiate.
You can write the answer as a single fraction with
denominator x 3. | [
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268
Chapter 12
5 Find the gradient of the curv
e with equation y = f(x) at the point A where:
a f(x
) = x(x + 1) and A is at (0, 0) b f(x
) = 2x −
6 ______ x2 and A is a t (3, 0)
c f(x
) = 1 ___ √ __
x and A is at ( 1 _ 4 , 2) d f(x ) = 3x − 4 __ x2 and A is a t (2, 5)
6 f(x ) =
12 ____ p √ __
x + x , where p is a real constant and x > 0.
Given that f9(2) = 3, find p, giving your answer in the form a √ __
2 where a is a
rational number. (4 marks)
7 f(x
) = (2 − x)9
a Find the first 3 terms, in ascending po wers of x, of the
binomial expansion of f(x), giving each term in its
simplest form.
b If x
is small, so that x2 and higher powers can be ignored,
show that f9(x) < 9216x − 2304.E/P
P
Use the binomial
ex
pansion with a = 2, b = − x
and n = 9. ← Sec tion 8.3Hint
12.6 Gradients, tangents and normals
You can use the derivative to find the equation of the tangent to a curve at a given point. On the curve with equation y = f(x), the gradient of the tangent at a point A with x-coordinate a will be f9(a).
■ The tangent to the curve y = f(x) at the
point with coordinates ( a, f(a)) has
equation
y − f(a) = f9(a)(x − a) The equation of a straight line with
gra
dient m that passes through the point ( x1, y1)
is y − y1 = m (x − x1). ← Section 5.2Links
The normal to a curve at point A is the straight line through A which is perpendicular to the tangent
to the curve at A. The gradient of the normal will be − 1 ____ f 9 (a )
■ The normal to the curv e y = f(x) at the point with
coordinates ( a, f(a)) has equation
y − f(a ) = − 1 _____ f 9 (a ) (x − a ) y = f(x)Normal
at A
Tangent
at
A A
xy
O | [
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269Differentiation
Find the equation of the tangent to the curve
y = x3 − 3x2 + 2x − 1 at the point (3, 5).Example 10
Find the equation of the normal to the curve with equation y = 8 − 3
√ __
x at the point where x = 4.Example 11
1 Find the equation of the tangent to the curv e:
a y = x2 − 7x + 10 at the point (2, 0) b y = x + 1 __ x at the point (2, 2 1 _ 2 )
c y = 4 √ __
x at the point (9, 12) d y = 2x − 1 ______ x at the point (1, 1)
e y = 2
x3 + 6x + 10 at the point (−1, 2) f y = x2 − 7 __ x2 at the point (1, −6)
2 Find the equation of the nor
mal to the curve:
a y = x2 − 5x at the point (6, 6) b y = x2 − 8 ___ √ __
x at the point (4, 12)
3 Find the coordinates of
the point where the tangent to the curve y = x2 + 1 at the point (2, 5)
meets the normal to the same curve at the point (1, 2).PExercise 12Fy = x3 − 3x2 + 2x − 1
dy ___ dx = 3x2 − 6x + 2
When x = 3, the gradient is 11.
So the equation of the tangent at (3, 5) is
y − 5 = 11( x − 3)
y = 11 x − 28First differentiate to determine the gradient
function.
Then substitute for x to calculate the value of the gradient of the curve and of the tangent when x = 3.
You can now use the line equation and simplify.
y = 8 − 3 √ __
x
= 8 −
3 x 1 __ 2
dy ___ dx = − 3 __ 2 x − 1 __ 2
Whe
n x = 4, y = 2 and gradient of curve
and of tangent = − 3 __ 4
So gr
adient of normal is 4 __ 3 .
Equa
tion of normal is
y − 2 = 4 __ 3 (x − 4 )
3y − 6 = 4 x − 16
3y − 4 x + 10 = 0Write each term in the form ax n and differentiate
to obtain the gradient function, which you can use
to find the gradient at any point.
Find the y-coordinate when x = 4 by substituting
into the equation of the curve and calculating 8 − 3
√ __
4 = 8 − 6 = 2.
Find the gradient of the curve, by calculating
dy ___ dx = − 3 __ 2 (4 ) − 1 _ 2 = − 3 __ 2 × 1 __ 2 = − 3 __ 4
Gradient of normal
= − 1 _______________ gradient of curve
= − 1 ____
(− 3 _ 4 ) = 4 __ 3
Simplify by multiplying both sides b
y 3 and
collecting terms.
Explore the tangent and normal to
the c
urve using GeoGebra.Online | [
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270
Chapter 12
4 Find the equations of the nor
mals to the curve y = x + x3 at the points (0, 0) and (1, 2), and
find the coordinates of the point where these normals meet.
5 For f(x
) = 12 − 4x + 2x2, find the equations of the tangent and the normal at the point
where x = −1 on the curve with equation y = f(x).
6 The point P with
x-coordinate 1 _ 2 lies on the
curve with equa
tion y = 2x2.
The normal to the curve at P intersects the
curve at points P and Q. Find the coordinates of Q.
(6 marks)P
P
E/P
Draw a sketch showing the curve, the point P and
the normal. This will help you check that your
answer makes sense.Problem-solving
12.7 Increasing and decreasing functions
You can use the derivative to determine whether a function is increasing or decreasing on a
given interval.
■ The function f( x) is increasing on the interval [ a, b] if f9( x) > 0 for all values of x such that
a < x < b.
■ The function f( x) is decreasing on the interval [ a, b] if f9( x) < 0 for all values of x such that
a < x < b.
The inter val [a, b]
is the set of all real numbers,
x, that satisfy a < x < b .Notationy = x3 + xy = x4 – 2x2
(–1, –1)OO xy
xy
(1, –1)
The function f(x) = x3 + x is
increasing for all real values of x.The function f(x) = x 4 − 2x2 is
increasing on the interval [−1, 0]
and decreasing on the interval [0, 1].
Show that the function f(x) = x3 + 24x + 3 is
increasing for all real values of x.Example 12
f(x) = x3 + 24 x + 3
f9(x) = 3 x2 + 24
x2 > 0 for all real values of x
So 3 x2 + 24 > 0 fo r all real values of x .
So f( x) is increasing for all real values of x .First differentiate to obtain the gradient function.
State that the condition for an increasing function
is met. In fact f9 (x) > 24 for all real values of x .The line L is a tangent to the curve with equation y = 4 x
2 + 1. L cuts the y -axis at (0, − 8) and has a
positive gradient. Find the equation of L in the form y = mx + c.Challenge Use the discriminant to find the value of m
wh
en the line just touches the curve. ← Section 2.5Hint | [
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271Differentiation
1 Find the values of
x for which f(x) is an increasing function, given that f(x) equals:
a 3x2 + 8x + 2 b 4x − 3x2 c 5 − 8x − 2x2 d 2x3 − 15x2 + 36x
e 3 +
3x − 3x2 + x3 f 5x3 + 12x g x4 + 2x2 h x4 − 8x3
2 Find the values of x for which f(x) is a decreasing function, given that f(x) equals:
a x2 − 9x b 5x − x2 c 4 − 2x − x2 d 2x3 − 3x2 − 12x
e 1 −
27x + x3 f x + 25 ___ x g x 1 _ 2 + 9 x − 1 _ 2 h x2(x + 3)
3 Show that the function f(
x) = 4 − x (2x2 + 3) is decreasing for all x ∈ R . (3 marks)
4 a Given tha
t the function f(x) = x2 + px is increasing on the interval [−1, 1], find
one possible value for p. (2 marks)
b State with justification w
hether this is the only possible value for p. (1 mark)E/P
E/P
12.8 Second order derivatives
You can find the rate of change of the gradient function by differentiating a function twice.
= 15x2dy
dx= 30xd2y
dx2 Differentiate y = 5x3Differentiate
This is the rate of change of the gradient
function. It is called the second order derivative. It can also be written as f0(x).This is the gradient function. It describes the rate of change of the function with respect to x.Exercise 12GFind the interval on which the function
f(x) = x3 + 3x2 − 9x is decreasing.Example 13
f(x) = x3 + 3 x2 − 9 x
f9(x) = 3 x2 + 6 x − 9
If f9(x) < 0 then 3 x2 + 6 x − 9 < 0
So 3(x2 + 2x − 3) < 0
3(x
+ 3)( x − 1) < 0
So −3 <
x < 1
So f( x) is decreasing on the interval [ −3, 1].Write the answer clearly.Find f9(x) and put this expression < 0.
The d erivative is also called the
first order derivative or first derivative .
The second order derivative is sometimes
just called the second derivative .Notation ■ Differentiating a function y = f(x)
twice gives you the second order
derivative, f 0(x) or d 2 y ____ d x 2 Explore increasing and decreasing
fun
ctions using GeoGebra.OnlineSolve the inequality by considering the three
regions x < −3, −3 < x < 1 and x > 1, or by sketching the curve with equation
y = 3(x + 3)(x − 1)
← Section 3.5 | [
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272
Chapter 12
Given that y = 3x5 + 4 __ x2 find:
a dy ___ dx b d2y ___ dx2
a y = 3 x5 + 4 __ x2
= 3x5 + 4x−2
So dy ___ dx = 15 x4 − 8x−3
= 15x4 − 8 __ x3
b d2y ____ dx2 = 60 x3 + 24 x−4
= 60 x3 + 24 ___ x4 Express the fraction as a negative power of x .
Differentiate once to get the first order derivative.
Differentiate a second time to get the second
order derivative.Example 14
a f(x) = 3 √ __
x + 1 ____ 2 √ __
x
= 3 x 1 __ 2 + 1 __ 2 x − 1 __ 2
f9(x) = 3 __ 2 x − 1 __ 2 − 1 __ 4 x − 3 __ 2
b f0(x) = − 3 __ 4 x − 3 __ 2 + 3 __ 8 x − 5 __ 2 Given that f(x) = 3 √ __
x + 1 ____ 2 √ __
x , find:
a f9
(x) b f 0(
x)Example 15
1 Find dy ___ dx and d2y ___ dx2 when y equals:
a 12x2 + 3x + 8 b 15x + 6 + 3 __ x c 9 √ __
x − 3 __ x2 d (5x + 4)(3x − 2) e 3x + 8 ______ x2
2 The displacement of a particle in metres a
t
time t seconds is modelled by the function
f(t)
= t 2 + 2 _____ √ _ t
The accelera
tion of the particle in m s−2 is the second derivative of this function.
Find an expression for the acceleration of the particle at time t seconds.
3 Given tha
t y = (2x − 3)3, find the value of x when d 2 y ____ d x 2 = 0.
4 f(x
) = px 3 − 3px 2 + x 2 − 4
When x = 2, f 0(
x) = −1. Find the value of p.P
PExercise 12HDon’t rewrite your expression for f9(x ) as a
fraction. It will be easier to differentiate again if
you leave it in this form.
The velocity of the particle will be f9 (t) and
its ac
celeration will be f 0(t).
→ Statistics and Mechanics Year 2, Section 6.2LinksThe coefficient for the second term is
(− 3 __ 2 ) × (− 1 __ 4 ) = + 3 __ 8
The new power is − 3 __ 2 − 1 = − 5 __ 2
When you differentiate with
respect to x , you treat any
other letters as constants.Problem-solving | [
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273Differentiation
12.9 Stationary points
A stationary point on a curve is any point where the curve has gradient zero. You can determine
whether a stationary point is a local maximum, a local minimum or a point of inflection by looking at the gradient of the curve on either side.
Oy
A
BxPoint A is a local
maximum.
The origin
is a point of inflection. Point B is a local minimum. Point A is c alled
a local maximum because it
is not the largest value the
function can take. It is just the largest value in that immediate vicinity.Notation
a Find the coordinates of the stationary point on the curve with equation y = x 4 − 32x.
b By considering points on either side of the stationary point, deter
mine whether it is a local
maximum, a local minimum or a point of inflection.Example 16
a y = x4 − 32 x
dy ___ dx = 4 x3 − 32
Let dy ___ dx = 0
T
hen 4 x3 − 32 = 0
4x3 = 32
x3 = 8
x = 2
So y = 24 − 32 × 2
= 16 − 64
= −48
So (2, − 48) is a stationary point.Differentiate and let dy ___ dx = 0.
Solve the equation to find the value of x.
Substitute the value of x into the original
equation to find the value of y.■ Any point on the curve y = f(x) where f9 (x) = 0 is called a stationary point. For a small positive
value h:
Type of stationary point f9(x 2 h) f9( x) f9( x 1 h)
Local maximum Positive 0 Negative
Local minimum Negative 0 Positive
Point of inflectionNegative 0 Negative
Positive 0 Positive The p lural of
maximum is maxima and the
plural of minimum is minima .Notation | [
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274
Chapter 12
b Now consider the gradient on either side
of (
2, −48).
Value
of xx = 1.9 x = 2 x = 2.1
Gradient−4.56
which is − ve05.04 which
is +ve
Shape of curve
From the shape of the curve, the point (2, − 48)
is a local minimum point.Make a table where you consider a value of x slightly
less than 2 and a value of x slightly greater than 2.
Calculate the gradient for each of these values of
x close to the stationary point.
Deduce the shape of the curve.
In some cases you can use the second derivative, f 0(x), to determine the nature of a stationary point.
f 0(x) t ells you the rate of change of the gradient function. When f 9(x) = 0
and f
0(x) > 0 th
e gradient is
increasing from a negative
value to a positive value, so the stationary point is a minimum .Hint
a Find the coordinates of
the stationary points on the curve with equation
y =
2x3 − 15x2 + 24x + 6
b Find d 2 y ____ d x 2 and use it to determine the nature of the stationary points.Example 17
a y = 2x3 − 15 x2 + 24 x + 6
dy ___ dx = 6 x2 − 30 x + 24
Putting 6 x2 − 30 x + 24 = 0
6(x − 4)( x − 1) = 0
So x =
4 or x = 1
When x = 1,
y = 2 − 15 + 24 + 6 = 17When x = 4, y = 2
× 64 − 15 × 16 + 24 × 4 + 6
=
−10
So the stationary points are at (1, 17) and (4, − 10).Differentiate and put the derivative equal to zero.
Solve the equation to obtain the values of x for
the stationary points.
Substitute x = 1 and x = 4 into the original equation of the curve to obtain the values of y which correspond to these values.■ If a function f( x) has a stationary point when x = a, then:
● if f 0(a) . 0, the point is a local minimum
● if f 0(a) , 0, the point is a local maximum
If f 0(a) = 0, the point could be a local minimum, a local
maximum or a point of inflection. You will need to look at
points on either side to determine its nature. Explore the solution using
GeoGe
bra.Online | [
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275Differentiation
b d2y ____ dx2 = 12 x − 30
When x = 1, d2y ____ dx2 = −18 which is , 0
So (1, 17) i
s a local maximum point.
When x = 4, d2y ____ dx2 = 1 8 which is . 0
So (4, −10) i
s a local minimum point.Differentiate again to obtain the second
derivative.
Substitute x = 1 and x = 4 into the second derivative expression. If the second derivative is negative then the point is a local maximum point. If it is positive then the point is a local minimum point.
a The curve with equation y = 1 __ x + 27 x 3 has stationary points at x = ±a. Find the value of a.
b Sketch the gra
ph of y = 1 __ x + 27 x 3 .Example 18
a y = x−1 + 27 x3
dy ___ dx = −x−2 + 81 x2 = − 1 __ x 2 + 81 x 2
Whe
n dy ___ dx = 0:
− 1 __ x 2 + 81 x 2 = 0
81 x 2 = 1 __ x 2
81 x 4 = 1
x 4 = 1 ___ 81
x =
± 1 __ 3
So a
= 1 __ 3
b d 2 y ____ d x 2 = 2 x −3 + 162x = 2 __ x 3 + 162x
Whe
n x = − 1 __ 3 , y = 1 _____
(− 1 __ 3 ) + 27 (− 1 __ 3 ) 3 = −4
and d 2 y ____ d x 2 = 2 ______
(− 1 __ 3 ) 3 + 162 (− 1 __ 3 ) = −108
which is negative.
So the curve has a local maximum at (− 1 __ 3 , −4) .
When x = 1 __ 3 ,
y =
1 ____
( 1 __ 3 ) + 27 ( 1 __ 3 ) 3 = 4
and
d 2 y ____ d x 2 = 2 _____
( 1 __ 3 ) 3 + 162 ( 1 __ 3 ) = 108
which is positive.Write 1 __ x as x −1 to differentiate.
You need to consider the positive and negative roots:
(− 1 _ 3 ) 4 = (− 1 _ 3 ) × (− 1 _ 3 ) × (− 1 _ 3 ) × (− 1 _ 3 ) = 1 __ 81 Set dy ___ dx = 0 to determine the x-coordinates of the
stationary points.
To sketch the curve, you need to find the
coordinates of the stationary points and determine their natures. Differentiate your expression for
dy ___ dx to find d 2 y ____ d x 2
Substitute x = − 1 _ 3 and x = 1 _ 3 into the equation
of the cur
ve to find the y-coordinates of the
stationary points.
Check your solution using your
ca
lculator.Online | [
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0.058406829833984375,
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-0.021... |
276
Chapter 12
So the curve has a local minimum at ( 1 __ 3 , 4) .
The curve has an asymptote at x = 0.
As x → ∞ , y → ∞ .
As x → − ∞, y → − ∞.
1
31
x
13/four.ss01
–/four.ss01–
xy
Oy = + 27 x3
1 Find the least value of the following functions:
a f(x
) = x2 − 12x + 8 b f(x ) = x2 − 8x − 1 c f(x ) = 5x2 + 2x
2 Find the greatest v
alue of the following functions:
a f(x
) = 10 − 5x2 b f(x) = 3 + 2x − x2 c f(x) = (6 + x)(1 − x)
3 Find the coordinates of
the points where the gradient is zero
on the curves with the given equations. Establish whether these
points are local maximum points, local minimum points or points of inflection in each case.
a y =
4x2 + 6x b y = 9 + x − x2 c y = x3 − x2 − x + 1
d y =
x (x2 − 4x − 3) e y = x + 1 __ x f y = x2 + 54 ___ x
g y =
x − 3 √ __
x h y = x 1 _ 2 (x − 6) i y = x 4 − 12x2
4 Sketch the curves with equations given in question 3 parts a, b, c and d, labelling any stationary
points with their coordina
tes.
5 By considering the gradient on either side of the sta
tionary point on the curve
y = x3 − 3x2 + 3x, show that this point is a point of inflection.
Sketch the curve y = x3 − 3x2 + 3x.
6 Find the maximum va
lue and hence the range of values for the function f(x) = 27 − 2x4.
7 f(x)
= x 4 + 3x3 − 5x2 − 3x + 1
a Find the coordinates of
the stationary points of f(x),
and determine the nature of each.
b Sketch the gra
ph of y = f(x).P
P
PExercise 12I For each part of
qu
estions 1 and 2 :
● Fin
d f9(x).
● Set f 9
(x) = 0 and solve
to find the value of x
at the stationary point.
● Fin
d the corresponding
value of f( x).Hint
Use the factor theorem
wi
th small positive integer
values of x to find one factor of f9(x).
← Section 7.2Hint 1 __ x → ± ∞ as x → 0 so x = 0 is an asymptote of
the curve.
Mark the coordinates of the stationary points on
your sketch, and label the curve with its equation.
You could check dy ___ dx at specific points to help with
your sketch:
● When x
= 1 _ 4 , dy ___ dx = −10.9375 which is negative.
● When x
= 1, dy ___ dx = 80 which is positive. | [
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0.0172... |
277Differentiation
12.10 Sketching gradient functions
You can use the features of a given function to sketch the
corresponding gradient function. This table shows you features of the graph of a function, y = f(x), and the graph of its gradient function, y = f9(x), at corresponding values of x.
y = f (x) y = f9(x)
Maximum or minimum Cuts the x-axis
Point of inflection Touches the x-axis
Positive gradient Above the x-axis
Negative gradient Below the x-axis
Vertical asymptote Vertical asymptote
Horizontal asymptote Horizontal asymptote at the x-axisy = f(x)
y = f
/acute.sc(x)O xy
O xy
The diagram shows the curve with equation y = f(x). The curve has stationary points at (−1, 4) and (1, 0), and cuts the x-axis at (−3, 0).
Sketch the gradient function, y = f
9(
x), showing the
coordinates of any points where the curve cuts or meets
the x-axis.Example 19
y = f(x)
1(–1, 4)
–3 O xy
O –11y = f/caron.alt( x)
xy
Ignore any points wher e the curve
y = f(x) cuts the x-axis. These will not tell you
anything about the features of the graph of y = f9(x).Watch outx y = f( x) y = f9( x)
x , −1
Positive gradient Above x-axis
x = −1 Maximum Cuts x-axis
−1 , x , 1 Negativ
e gradient Below x-axis
x = 1 Minimum Cuts x-axis
x > 1 Positive gradient Above x-axis
Use GeoGebra to explore the key
fe
atures linking y = f(x) and y = f9(x).Online | [
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278
Chapter 12
The diagram shows the curve with equation y = f(x). The curve
has an asymptote at y = −2 and a turning point at (−3, −8). It cuts the x-axis at (−10, 0).
a
Sketch the gra
ph of y = f 9(
x).
b State the equation of
the asymptote of y = f 9(
x).Example 20
y = f(x)xy
O
(–3, –8)–2–10
1 For each graph given, sketch the graph of the corresponding gradient function on a separate set
of ax
es. Show the coordinates of any points where the curve cuts or meets the x-axis, and give
the equations of any asymptotes.a
xy
–11 O 8(–9, 12)(6, 15) b
y = 10
xy
O c
x = –7(4, 3)
xy
O
d
y = 3
–2 xy
6
O e
x = 6xy
O f
y = 4
y =
–4xy
O
2 f(x) = (x + 1)(x − 4)2
a Sketch the graph of y = f(x).
b On a separate set of
axes, sketch the graph of y = f 9(x
).
c Show that f 9(x
) = (x − 4)(3x − 2).
d Use the deriva
tive to determine the exact coordinates of the
points where the gradient function cuts the coordinate axes.PExercise 12Ja
Oy
x–3y = f/caron.alt(x)
b y =
0Draw your sketch on a separate set of axes. The
graph of y = f9(x) will have the same horizontal scale but will have a different vertical scale.
You don’t have enough information to work out the coordinates of the y-intercept, or the local maximum, of the graph of the gradient function.
The graph of y = f(x) is a smooth curve so the
graph of y = f9(x) will also be a smooth curve.
If y = f(x) has any horizontal asymptotes then the graph of y = f9(x) will have an asymptote at the x-axis.
This is an x3 graph
with a positive coefficient of x
3. ← Section 4.1Hint | [
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279Differentiation
12.11 Modelling with differentiation
You can think of dy ___ dx as small change in
y _______________ small change in x . It represents the rate of change of y with respect to x.
If you replace y and x with variables that represent real-life quantities, you can use the derivative to
model lots of real-life situations involving rates of change.
VThe volume of water in this water butt is constantly changing over time.
If V represents the volume of water in the water butt in litres, and t represents the time in seconds, then you could model V as a function of t.
If V = f(t) then
dV ___ dt = f9(t) would represent the rate of change of volume
with respect to time. The units of dV ___ dt would be litres per second.
V = 4 __ 3 π r3
dV ___ dr = 4π r2
When r = 5, dV ___ dr = 4π × 52
= 314 (3 s.f.)
So the rate of change is 314 cm3 per cm.Given that the volume, V cm3, of an expanding sphere is related to its radius, r cm, by the for mula
V = 4 _ 3 p r3, find the rate of change of volume with respect to radius at the instant when the radius
is 5 cm.
Substitute r = 5.
Interpret the answer with units.Example 21
Differentiate V with respect to r. Remember that
π is a constant.
A large tank in the shape of a cuboid is to be made from 54 m2 of sheet metal. The tank has a
horizontal base and no top. The height of the tank is x metres. Two opposite vertical faces are
squares.
a Show that the v
olume, V m3, of the tank is given by V = 18x − 2 _ 3 x3
b Given that x can vary, use differentiation to find the maximum or minimum value of V.
c Justify that the v
alue of V you have found is a maximum.Example 22 | [
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280
Chapter 12
Rearrange to find x.
x is a length so use the positive solution.
Find the second derivative of V.Rearrange to find y in terms of x.a Let the length of the tank be y metres.
yx
x
Total area, A = 2x2 + 3xy
So 54 = 2x2 + 3xy
y = 54 − 2x2 ________ 3x
But V = x2y
So V = x2 ( 54 − 2x2 ________ 3x )
= x __ 3 (54 − 2x2)
So V = 18x
− 2 __ 3 x3
b dV ___ dx = 18 − 2x2
Put dV ___ dx = 0
0 = 18 −
2x2
So x2 = 9
x = −
3 or 3
But x is a length so x = 3
When x = 3, V = 18 ×
3 − 2 _ 3 × 33
= 54 − 18
= 36
V
= 36 is a maximum or minimum
value of V.
c d2V ____ dx2 = −4x
When x = 3, d2V ____ dx2 = −4 × 3 = −12
This is negative, so V = 36 is the maximum
value of V.You don’t know the length of the tank. Write it as
y metres to simplify your working. You could also draw a sketch to help you find the correct expressions for the surface area and volume of the tank.Problem-solving
Draw a sketch.
d2V ____ dx2 < 0 so V = 36 is a maximum.Substitute the expression for y into the equation.
Simplify.
Differentiate V with respect to x and put dV ___ dx = 0.
Substitute the value of x into the expression for V . | [
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281Differentiation
1 Find dθ ___ dt where θ = t2 − 3t.
2 Find dA ___ dr where A = 2pr.
3 Given tha
t r = 12 ___ t , find the value of dr __ dt when t = 3.
4 The surface area, A cm2, of an expanding sphere of radius r cm is given by A = 4pr2. Find the
rate of change of the area with respect to the radius at the instant when the radius is 6 cm.
5 The displacement, s metres
, of a car from a fixed point at time t seconds is given by s = t2 + 8t.
Find the rate of change of the displacement with respect to time at the instant when t = 5.
6 A rectangular garden is fenced on thr
ee sides, and the house forms the fourth side of the
rectangle.
a Given tha
t the total length of the fence is 80 m, show that the ar
ea, A , of the garden is given by
the formula A = y(80 − 2y), where y is the distance from the house to the end of the garden.
b Given tha
t the area is a maximum for this length of fence, find the dimensions of the
enclosed garden, and the area which is enclosed.
7 A closed cylinder has total surface area equa
l to 600p.
a Show that the v
olume, V cm3, of this cylinder is given by the formula V = 300pr − pr3,
where r cm is the radius of the cylinder
.
b Find the maximum volume of
such a cylinder.
8 A sector of a circle has ar
ea 100 cm2.
a Show that the perimeter of
this sector is given by the formula
P = 2r + 200 ____ r , r . √ ____
100 ____ p
b Find the minimum va
lue for the perimeter.
9 A shape consists of a r
ectangular base with a semicircular top, as shown.
a Given tha
t the perimeter of the shape is 40 cm, show that its
ar
ea, A cm2, is given by the formula
A = 40r − 2r2 − pr2 ___ 2
where
r cm is the radius of the semicir
cle. (2 marks)
b Hence find the maximum va
lue for the area of the shape. (4 marks)
10 The shape shown is a wir
e frame in the form of a large
rectangle split by parallel lengths of wire into 12 smaller
equal-sized rectangles.
a Given tha
t the total length of wire used to complete
the whole frame is 1512 mm, show that the ar
ea of
the whole shape, A mm2, is given by the formula
A = 1296x − 108x2 _____ 7
where x mm is the width of one of the sma
ller rectangles. (4 marks)
b Hence find the maximum area w
hich can be enclosed in this way. (4 marks)P
P
P
N
r cmM
O
E/P
r cm
E/P
y mm
x mmExercise 12K | [
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-0.0... |
282
Chapter 12
1 Prov
e, from first principles, that the derivative of 10x2 is 20x. (4 marks)
2 The point A with coor
dinates (1, 4) lies on the curve with equation y = x3 + 3x.
The point B also lies on the curve and has x-coordinate (1 + δ x ).
a Show that the gr
adient of the line segment AB is given by ( δx )2 + 3δx + 6.
b Deduce the gradient of the curv
e at point A.
3 A curve is giv
en by the equation y = 3x2 + 3 + 1 __ x2 , where x . 0. At the points A , B and
C on the curve, x = 1, 2 and 3 respectively. Find the gradient of the curve at A, B and C.
4 Calculate the
x-coordinates of the points on the curve with equation y = 7x2 − x3
at which the gradient is equal to 16. (4 marks)
5 Find the x-coor
dinates of the two points on the curve with equation y = x3 − 11x + 1
where the gradient is 1. Find the corresponding y-coordinates.
6 The function f is defined by f(
x) = x + 9 __ x , x [ R, x Þ 0.
a Find f9(
x). (2 marks)
b Solve f9
(x) = 0. (2 marks)
7 Given tha
t
y = 3 √ __
x − 4 ___ √ __
x , x . 0,
find dy ___ dx (3 marks)
8 A curve has equation
y = 12 x 1 _ 2 − x 3 _ 2 .
a Show that dy ___ dx = 3 __ 2 x − 1 _ 2 (4 − x). (2 marks)
b Find the coordinates of
the point on the curve where the gradient is zero. (2 marks)
9 a Expand ( x 3 _ 2 − 1)( x − 1 _ 2 + 1). (2 marks)
b A curve has equation
y = ( x 3 _ 2 − 1)( x − 1 _ 2 + 1), x . 0. Find dy ___ dx (2 marks)
c Use your answ
er to part b to calculate the gradient of the curve at the point
where x = 4. (1 mark)
10 Differentiate with r
espect to x:
2x3 + √ __
x + x2 + 2x _______ x2 (3 marks)
11 The curve with equation
y = ax2 + bx + c passes through the point (1, 2). The gradient
of the curve is zero at the point (2, 1). Find the values of a, b and c. (5 marks)E/P
P
E
E
E
E/P
E
E
E/PMixed exercise 12 | [
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283Differentiation
12 A curve C
has equation y = x3 − 5x2 + 5x + 2.
a Find dy ___ dx in terms of x. (2 marks)
b The points P and
Q lie on C. The gradient of C at both P and Q is 2.
The x-coordinate of P is 3.
i Find the x-coor
dinate of Q. (3 marks)
ii Find an equation for the tangent to
C at P, giving your answer in the form
y = mx + c, where m and c are constants. (3 marks)
iii If this tangent intersects the coordina
te axes at the points R and S, find the
length of RS, giving your answer as a surd. (3 marks)
13 A curve has equation
y = 8 __ x − x + 3x 2, x > 0. Find the equations of the tangent and the
normal to the curve at the point where x = 2.
14 The normals to the curv
e 2y = 3x3 − 7x2 + 4x, at the points O (0, 0) and A(1, 0),
meet at the point N
.
a Find the coordinates of
N. (7 marks)
b Calculate the ar
ea of triangle OAN . (3 marks)
15 A curve C
has equation y = x3 − 2x2 − 4x − 1 and cuts the y-axis at a point P.
The line L is a tangent to the curve at P, and cuts the curve at the point Q.
Show that the distance PQ is 2 √ ___ 17 . (7 marks)
16 Given tha
t y = x 3 _ 2 + 48 ___ x , x . 0
a find the value of
x and the value of y when dy ___ dx = 0. (5 marks)
b show that the v
alue of y which you found in part a is a minimum. (2 marks)
17 A curve has equation
y = x3 − 5x2 + 7x − 14. Determine, by calculation, the coordinates
of the stationary points of the curve.
18 The function f, defined for
x [ R, x . 0, is such that:
f
9(x) = x2 − 2 + 1 __ x2
a Find the value of
f
0 (x) a
t x = 4. (4 marks)
b Prov
e that f is an increasing function. (3 marks)
19 A curve has equation
y = x3 − 6x2 + 9x. Find the coordinates of its local maximum. (4 marks)
20 f(x
) = 3x4 − 8x3 − 6x2 + 24x + 20
a Find the coordinates of the stationary points of f(x), and determine the nature
of each of them.
b Sketch the gra
ph of y = f(x).E/P
E/P
E/P
E
E/P
E | [
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284
Chapter 12
21 The diagram sho
ws part of the curve with equation
y = f(x), where:
f(x) = 200 − 250 ____ x − x2, x . 0
The curve cuts the x
-axis at the points A and C.
The point B is the maximum point of the curve.
a Find f9(
x). (3 marks)
b Use your answ
er to part a to calculate the
coordinates of B. (4 marks)
22 The diagram sho
ws the part of the curve with
equation y = 5 − 1 _ 2 x2 for which y > 0.
The point P(x, y) lies on the curve and O is the origin.
a Show that
OP 2 = 1 _ 4 x4 − 4x2 + 25. (3 marks)
Taking f(
x) = 1 _ 4 x4 − 4x2 + 25:
b Find the values of
x for which f9(x) = 0. (4 marks)
c Hence, or otherwise, find the minim
um distance
from O to the curve, showing that your answer is
a minimum. (4 marks)
23 The diagram sho
ws part of the curve with
equation y = 3 + 5x + x2 − x3. The curve
touches the x-axis at A and crosses the x-axis at C. The points A and B are stationary points on the curve.
a
Show that
C has coordinates (3, 0). (1 mark)
b Using calculus and showing a
ll your working,
find the coordinates of A and B. (5 marks)
24 The motion of a damped spring is modelled using
this gr
aph.
On a separate graph, sketch the gradient function for this model. Choose suitable labels and units for each axis, and indicate the coordinates of any points where the gradient function crosses the horizontal axis.
25
The volume,
V cm3, of a tin of radius r cm is given b y the formula V = p(40r − r2 − r3).
Find the positive value of r for which dV ___ dr = 0, and find the value of V which
corresponds to this value of r.
26 The total surface area,
A cm2, of a cylinder with a fixed volume of 1000 cm3 is given
by the formula A = 2px2 + 2000 _____ x , where x cm is the radius. Show that when the rate
of change of the area with respect to the radius is zero, x3 = 500 ____ p E
ACOB
xy
E/P
O xy
P(x, y)
E B
C AO xy
P
0.50 2.1 1.2
Time (seconds)Displacement (cm)
P | [
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285Differentiation
27 A wire is bent into the plane shape
ABCDE as shown. Shape
ABDE is a rectangle and BCD is a semicircle with diameter BD.
The area of the region enclosed by the wire is R m2, AE = x metres,
and AB = ED = y metres. The total length of the wire is 2 m.
a Find an expression f
or y in terms of x. (3 marks)
b Prov
e that R = x __ 8 (8 − 4x − px). (4 marks)
Given tha
t x can vary, using calculus and showing your working:
c find the maximum va
lue of R. (Y ou do not have to prove that the value
you obtain is a maximum.) (5 marks)
28 A cylindrical biscuit tin has a close-fitting lid which o
verlaps the tin
by 1 cm, as shown. The radii of
the tin and the lid are both x cm.
The tin and the lid ar
e made from a thin sheet of metal of area
80p cm2 and there is no wastage. The volume of the tin is V cm3.
a Show that
V = p(40x − x2 − x3). (5 marks)
Given tha
t x can vary:
b use differentiation to find the positi
ve value of x for
which V is stationary. (3 marks)
c Prov
e that this value of x gives a maximum value of V . (2 marks)
d Find this maximum va
lue of V. (1 mark)
e Determine the percenta
ge of the sheet metal used in the
lid when V is a maximum. (2 marks)
29 The diagram sho
ws an open tank for storing water,
ABCDEF. The sides ABFE and CDEF are rectangles.
The triangular ends ADE and BCF are isosceles, and /AED = /BFC = 90°. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = x metres:a
show that the ar
ea of triangle ADE is 1 _ 4 x2 m2 (3 marks)
Given a
lso that the capacity of the container is 4000 m3 and that the total area of the two
triangular and two rectangular sides of the container is S m2:
b show that
S = x2
__ 2 + 16 000 √ __
2 ________ x (4 marks)
Given tha
t x can vary:
c use calculus to find the minimum v
alue of S. (6 marks)
d justify that the va
lue of S you have found is a minimum. (2 marks)E/P B
DA
EC
E/P
1 cmx cm
x cmLid
Tin
E
EFC B
A D
a Find the first four terms in the binomial expansion of ( x + h)7, in ascending powers of h .
b Hen
ce prove, from first principles, that the derivative of x7 is 7x6.Challenge | [
0.03797698765993118,
0.0666431412100792,
0.0018668669508770108,
0.049065206199884415,
0.009178305976092815,
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0.10970685631036758,
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-0.03361240029335022,
-0.053399... |
286
Chapter 12
1 The gradient of a curve at a given point is defined as the gradient of the tangent to the
cur
ve at that point.
2 The gradient function
, or derivative, of the curve y = f(x) is written as f9(x) or dy ___ dx
f 9 (x)
= lim
h → 0 f (x + h) − f(x) ____________ h
The gradient function can be used to find the gr
adient of the curve for any value of x.
3 For all real v
alues of n, and for a constant a:
● If f(x
) = x n then f 9 (x) = n x n − 1 ● If f(x) = ax n then f 9 (x) = anx n − 1
● If y = x n then dy ___ dx = nx n − 1 ● If y = ax n then dy ___ dx = anx n − 1
4 For the quadratic curve with equation y = ax2 + bx + c, the derivative is given by
dy ___ dx = 2ax + b
5 If y
= f(x) ± g(x), then dy ___ dx = f9(x) ± g9(x).
6 The tangent to the cur
ve y = f(x) at the point with coordinates (a, f(a)) has equation
y − f(a) = f9(a)(x − a)
7 The normal to the curv
e y = f(x) at the point with coordinates (a, f(a)) has equation
y − f(a ) = − 1 ____ f 9 (a ) (x − a )
8 ● The function f(x) is
increasing on the interval [a, b] if f9( x) > 0 for all values of x such that
a , x , b.
● The function f(x) is
decreasing on the interval [a, b] if f9(x) < 0 for all values of x such that
a , x , b.
9 Differentiating a function
y = f(x) twice gives you the second order derivative, f 0(x
) or d 2 y ____ d x 2
10 Any point on the curv
e y = f(x) where f9(x) = 0 is called a stationary point. For a small positive
value h:
11 If a function f(x) ha
s a stationary point when x = a, then:
● if f 0(
a) . 0, the point is a local minimum
● if f 0(
a) , 0, the point is a local maximum.
If
f 0(
a) = 0, the point could be a local minimum, a local maximum or a point of inflection.
You will need to look at points on either side to determine its nature.Type of stationary point f9(x − h) f9(x) f9(x + h)
Local maximum Positive 0 Negative
Local minimum Negative 0 Positive
Point of inflectionNegative 0 Negative
Positive 0 PositiveSummary of key points | [
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... |
287
Integration
After completing this unit you should be able to:
● Find y giv
en dy ___ dx for xn → pages 288–290
● Integrate polynomials → pages 290–293
● Find f(x) , given f ′(x ) and a point on the curve → pages 293–295
● Evaluate a definite integral → pages 295–297
● Find the area bounded by a cur ve and the x-axis → pages 297–302
● Find areas bounded by curves and straight lines → pages 302–306Objectives
1 Simplify these expressions
a x 3 ___
√ __
x b √ __
x × 2 x 3 ________ x 2
c x 3 − x ______
√ __
x d √ __
x + 4 x 3 ________ x 2
← Sections 1.1, 1.4
2 Find dy ___ dx when y equals
a 2 x 3 + 3x − 5 b 1 _ 2 x 2 − x
c x 2 (x + 1) d x − x 5 ______ x 2 ← Section 12.5
3 Sketch the curves with the following
equations:
a y =
(x + 1)(x − 3)
b y =
(x + 1) 2 (x + 5) ← Chapter 4Prior knowledge check
Integration is the opposite of differentiation.
It is used to calculate areas of surfaces, volumes of irregular shapes and areas under curves. In mechanics, integration can be used to calculate the area under a velocity-time graph to find distance travelled.
→ Exercise 13D Q813 | [
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-0.04713195934891701,
-0... |
288
Chapter 13
13.1 Integrating xn
Integration is the reverse process of differentiation:
xn
xnFunction Gradient Function
xn + 1
n + 1multiply by the powe r subtract one fr om the powe r
divide by the new powe r add one to the powe rnxn – 1
Constant terms disappear when you differentiate. This means that when you differentiate functions that
only differ in the constant term, they will all differentiate to give the same function. To allow for this, you need to add a constant of integration at the end of a function when you integrate.
Differentiate Integratey = x2 + 5
y = x2y = x2 + c
y = x2 – 19= 2xdy
dx
■ If dy ___ dx = xn, then y = 1 __ n + 1 x n + 1 + c, n ≠ −1.
■ If f ′(x) = xn, then f( x) = 1 __ n + 1 x n + 1 + c, n ≠ −1.Differentiating xn
Integrating xn
You cannot use this rule if n = − 1
be
cause 1 _____ n + 1 = 1 __ 0 and so is not defined.
You will learn how to integrate the function
x−1 in Year 2. → Year 2, Section 11.2Links
Example 1
Example 2Find y for the following:
a dy ___ dx = x4 b dy ___ dx = x−5
Find f(x) for the following:
a f ′(
x) = 3 x 1 _ 2 b f ′( x) = 3a y = x5 ___ 5 + c
b y =
x−4 ___ −4 + c = − 1 __ 4 x−4 + cThis is the constant of integration.
Use y = 1 _____ n + 1 xn + 1 + c with n = 4.
Don’t forget to add c .
Remember, adding 1 to the power gives − 5 + 1 = −4.
Divide by the new power (−4) and add c. | [
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-0.027... |
289
Integration
a f(x) = 3 × x 3 __ 2 ___
3 __ 2 +
c = 2 x 3 __ 2 + c
b f ′(x) =
3 = 3 x0
So f( x) = 3 × x 1 ___ 1 + c = 3 x + c
You can integrate a function in the form kxn by integrating xn and multiplying the integral by k.
■ If dy ___ dx = kxn, then y = k _____ n + 1 x n + 1 + c, n ≠ −1.
■ Using function notation, if f ′(x) = kxn,
then f( x) = k _____ n + 1 x n + 1 + c, n ≠ −1.
■ When integrating polynomials
, apply the rule of integration separately to each term. You d on’t need to multiply the
constant term ( c) by k . Watch out
Example 3
Given dy ___ dx = 6x + 2x−3 − 3 x 1 _ 2 , find y.
y = 6x2 ___ 2 + 2 ___ −2 x−2 − 3 __
3 __ 2 x 3 __ 2 + c
= 3x2 − x−2 − 2 x 3 __ 2 + cApply the rule of integration to each term of the
expression and add c.
Now simplify each term and remember to add c.x0 = 1, so 3 can be written as 3 x0.Remember 3 ÷ 3 _ 2 = 3 × 2 _ 3 = 2
Simplif
y your answer.
Exercise 13A
1 Find an expression f or y when dy ___ dx is the following:
a x5 b 10x4 c −x−2 d −4x−3 e x 2 _ 3 f 4 x 1 _ 2
g −2x6 h x − 1 _ 2 i 5 x − 3 _ 2 j 6 x 1 _ 3 k 36x11 l −14x−8
m −3 x − 2 _ 3 n −5 o 6x p 2x−0.4
2 Find y when dy ___ dx is given by the following expressions. In each case simplify your answer.
a x3 − 3 _ 2 x − 1 _ 2 − 6x−2 b 4x3 + x − 2 _ 3 − x−2 c 4 − 12 x −4 + 2 x − 1 _ 2
d 5 x 2 _ 3 − 10x4 + x −3 e − 4 _ 3 x − 4 _ 3 − 3 + 8x f 5x4 − x − 3 _ 2 − 12 x −5
3 Find f(x) w
hen f ′(x
) is given by the following expressions. In each case simplify your answer.
a 12x +
3 _ 2 x − 3 _ 2 + 5 b 6x5 + 6 x −7 − 1 _ 6 x − 7 _ 6 c 1 _ 2 x − 1 _ 2 − 1 _ 2 x − 3 _ 2
d 10x4 + 8x −3 e 2 x − 1 _ 3 + 4 x − 5 _ 3 f 9x2 + 4 x −3 + 1 _ 4 x − 1 _ 2
4 Find y gi
ven that dy ___ dx = (2x + 3)2. (4 marks) E/P
Start by expanding the brackets.Problem-solving | [
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0.010454344563186169,
-0... |
290
Chapter 13
Find y when dy ___ dx = (2 √ __
x − x2) ( 3 + x _____ x5 ) Challenge
13.2 Indefinite integr als
You can use the symbol ∫ to represent the process of integration.
■ ∫f ′(x)dx = f(x) + c
You can write the process of integrating xn as follows:
∫xn dx = xn + 1 _____ n + 1 + c, n ≠ −1
The elongated S
means integrate.The expression to
be integrated.The d x tells you to
integrate with respect to x.
When you are integrating a polynomial function, you can integrate the terms one at a time.
■ ∫(f(x) + g( x))dx = ∫f(x)dx + ∫g(x)dx This p rocess is
called indefinite integration .
You will learn about definite
integration later in this chapter.Notation
The d x tells you to integrate with respect to the
variable x , so any other letters must be treated as
constants.5 Find f(x) gi ven that f ′(x ) = 3 x −2 + 6 x 1 _ 2 + x − 4 . (4 marks) E
Example 4
Find:
a ∫( x 1 _ 2 + 2x3) dx b ∫( x − 3 _ 2 + 2) dx c ∫( p2 x −2 + q) dx d ∫(4t2 + 6) dt
a ∫( x 1 __ 2 + 2x3)dx = x 3 __ 2 ___
3 __ 2 + 2x4 ____ 4 + c
= 2 __ 3 x 3 __ 2 + 1 __ 2 x4 + c
b ∫( x − 3 __ 2 + 2)d x = x − 1 __ 2 ____
− 1 __ 2 + 2
x + c
= −2 x − 1 __ 2 + 2x + c
c ∫( p2 x −2 + q)dx = p2
___ −1 x −1 + qx + c
= − p2 x −1 + qx + c
d ∫(4t2 + 6)d t = 4t3 ____ 3 + 6 t + cFirst apply the rule term by term.
Simplify each term.
Remember − 3 __ 2 + 1 = − 1 __ 2 and the integral of the
constant 2 is 2 x.
The d t tells you that this time you must integrate
with respect to t.
Use the rule for integrating x n but replace x with t :
If dy ___ dt = kt n, then y = k _____ n + 1 tn + 1 + c, n ≠ −1. | [
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-0.021164093166589737,
0.06726480275392532,
-0.011... |
291
Integration
Example 5
Find:
a ∫ ( 2 __ x3 − 3 √ __
x ) dx b ∫x (x2 + 2 __ x ) dx c ∫ ((2x)2 + √ __
x + 5 ______ x2 ) dx
a ∫ ( 2 ___ x3 − 3 √ __
x ) dx
= ∫(2x −3 − 3 x 1 __ 2 )dx
= 2 ___ −2 x −2 − 3 __
3 __ 2 x 3 __ 2 + c
= −x −2 − 2 x 3 __ 2 + c
= − 1 ___ x2 − 2 √ ___
x3 + c
b ∫x (x2 + 2 __ x ) dx
= ∫(x3 + 2)d x
= x4 ___ 4 + 2x + c
c ∫ ((2x)2 + √ __
x + 5 _______ x2 ) dx
= ∫ (4x2 + x 1 __ 2 ___ x2 + 5 ___ x2 ) dx
= ∫(4x2 + x − 3 __ 2 + 5 x−2)dx
= 4 __ 3 x3 + x − 1 __ 2 ____
− 1 __ 2 + 5x−1 _____ −1 + c
= 4 __ 3 x3 − 2 x − 1 __ 2 − 5 x−1 + c
= 4 __ 3 x3 − 2 ___ √ __
x − 5 __ x + cFirst write each term in the form xn.
Apply the rule term by term.
Simplify each term.
Sometimes it is helpful to write the answer in the
same form as the question.Before you integrate, you need to ensure that each term of the expression is in the form kx n, where
k and n are real numbers.
First multiply out the bracket.
Then apply the rule to each term.
Simplify (2 x)2 and write √ __
x as x 1 _ 2 .
Write each term in the form xn.
Apply the rule term by term.
Finally simplify the answer.
Exercise 13B
1 Find the following integr als:
a ∫x3 dx b ∫x7 dx c ∫3x−4 dx d ∫5x2 dx
2 Find the following integr
als:
a ∫(x4 + 2x3)dx b ∫(2x3 − x2 + 5x)dx c ∫(5 x 3 _ 2 − 3x2)dx
3 Find the following integr
als:
a ∫(4x−2 + 3 x − 1 _ 2 )dx b ∫(6x−2 − x 1 _ 2 )dx c ∫(2 x − 3 _ 2 + x2 − x − 1 _ 2 )dx | [
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-0.... |
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