document stringlengths 121 3.99k | embedding listlengths 384 384 |
|---|---|
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... | [
-0.0056667146272957325,
0.017399175092577934,
0.054401978850364685,
-0.08554059267044067,
-0.049488749355077744,
0.00790510792285204,
-0.049339354038238525,
-0.042600229382514954,
-0.020310375839471817,
-0.05890442430973053,
0.08088645339012146,
0.01900116167962551,
0.00555709283798933,
0.... |
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° 3... | [
-0.018168989568948746,
-0.02913554385304451,
0.030376529321074486,
-0.045857980847358704,
-0.02104325033724308,
0.015303206630051136,
-0.04847551882266998,
-0.04402092844247818,
-0.06751479208469391,
-0.028053006157279015,
0.04054361209273338,
-0.060537394136190414,
0.016511879861354828,
0... |
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 wh... | [
-0.013774562627077103,
0.021954413503408432,
0.020321208983659744,
-0.04575701057910919,
0.015789175406098366,
0.0010870278347283602,
-0.009663901291787624,
0.011556120589375496,
-0.036655619740486145,
-0.022408388555049896,
0.1343853771686554,
-0.0396350733935833,
-0.031283553689718246,
0... |
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°, −... | [
-0.03717261552810669,
0.014893529936671257,
0.03597526624798775,
-0.05714454874396324,
0.014444485306739807,
-0.02217547968029976,
-0.02995366044342518,
0.0334157720208168,
-0.04551146551966667,
-0.0794769823551178,
0.0876179039478302,
-0.062026966363191605,
0.006378043908625841,
0.0585075... |
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... | [
-0.03422199934720993,
0.00399892870336771,
-0.014300178736448288,
-0.09377742558717728,
-0.020009230822324753,
0.026830697432160378,
-0.05524986609816551,
0.016045426949858665,
-0.022537661716341972,
0.0614582784473896,
0.029121579602360725,
0.016780393198132515,
0.03480800613760948,
0.083... |
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 =... | [
-0.07932998985052109,
-0.0006660960498265922,
0.05859466642141342,
-0.0403473861515522,
-0.025980662554502487,
-0.03418593853712082,
-0.011075307615101337,
0.017638063058257103,
-0.028226107358932495,
-0.09302019327878952,
0.07445651292800903,
-0.0727209821343422,
0.031033378094434738,
0.0... |
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 w... | [
-0.04000534489750862,
0.09055008739233017,
0.027374915778636932,
-0.0579155832529068,
-0.0694705992937088,
0.024977032095193863,
-0.03142930194735527,
0.009051090106368065,
-0.0506678931415081,
-0.05797279253602028,
0.031152019277215004,
0.018195537850260735,
0.03437882289290428,
0.0857954... |
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 =... | [
0.029671935364603996,
0.04465069621801376,
0.013624254614114761,
-0.058124542236328125,
-0.0636674240231514,
0.04257951304316521,
-0.006685344502329826,
0.04903851076960564,
-0.07820148766040802,
-0.00649023475125432,
0.03346290439367294,
-0.10012113302946091,
-0.021702218800783157,
0.0129... |
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 gr... | [
-0.07980819791555405,
0.017875507473945618,
0.0011870450107380748,
-0.03173457458615303,
-0.10074963420629501,
0.036235492676496506,
0.027277668938040733,
0.0524633452296257,
-0.016905250027775764,
-0.04520963504910469,
0.05341743677854538,
0.01338943187147379,
0.02765970677137375,
0.05362... |
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 versio... | [
-0.029902011156082153,
0.03220852091908455,
0.0222860649228096,
-0.05496789887547493,
-0.030511924996972084,
0.05075828358530998,
-0.02130059525370598,
-0.008144357241690159,
-0.027840109542012215,
-0.073392353951931,
0.045010436326265335,
-0.054834604263305664,
0.0185847245156765,
0.02106... |
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
● Solv... | [
-0.017683004960417747,
0.058107081800699234,
0.0052863904275000095,
-0.012440606020390987,
-0.04061184078454971,
0.026969170197844505,
-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 ... | [
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 ta... | [
-0.009879173710942268,
0.0059527065604925156,
-0.000645646417979151,
-0.008359718136489391,
-0.058040034025907516,
0.048892684280872345,
-0.0035303516779094934,
-0.03263775259256363,
-0.05608223378658295,
-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° w... | [
0.008106395602226257,
-0.0381806343793869,
0.05605209618806839,
-0.028636878356337547,
-0.06365350633859634,
0.03993484750390053,
-0.02323935553431511,
-0.07498277723789215,
-0.06600544601678848,
-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 (18... | [
-0.03881235793232918,
0.02005922980606556,
-0.009691301733255386,
-0.05000996217131615,
0.013693052344024181,
-0.03706236556172371,
0.012718725949525833,
-0.02683405764400959,
-0.03285663574934006,
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... | [
-0.04603499919176102,
0.058071255683898926,
0.09010346978902817,
-0.0037512104026973248,
-0.07519226521253586,
0.052059050649404526,
0.008556120097637177,
-0.003354455344378948,
-0.09993486851453781,
-0.00739734061062336,
-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° ... | [
-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... | [
0.010279553011059761,
0.0428256057202816,
0.05937739089131355,
-0.01879449188709259,
-0.07244641333818436,
0.05109157785773277,
-0.027194788679480553,
-0.029518254101276398,
-0.10338125377893448,
-0.07087022811174393,
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 ... | [
-0.09999018162488937,
0.08629908412694931,
0.02746148779988289,
-0.07336703687906265,
-0.0650477334856987,
0.051825862377882004,
0.00005612675522570498,
-0.058786701411008835,
-0.06819798052310944,
-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
__... | [
-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 ... | [
-0.060138676315546036,
0.09009862691164017,
0.05405065789818764,
-0.034870605915784836,
-0.07936310023069382,
0.044792499393224716,
0.06418002396821976,
-0.07624496519565582,
-0.04474056512117386,
-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 =... | [
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…° ... | [
0.02935585379600525,
0.00958210788667202,
0.045546580106019974,
-0.016068216413259506,
-0.008288867771625519,
-0.015551134012639523,
-0.07473377883434296,
0.006624842062592506,
-0.06459261476993561,
0.04436689242720604,
0.08414311707019806,
-0.05507560819387436,
-0.023512504994869232,
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 ) .
E... | [
0.0164723452180624,
0.05864585191011429,
0.042534563690423965,
-0.04179215058684349,
-0.055367808789014816,
0.008968864567577839,
-0.01829376444220543,
-0.002721866127103567,
-0.06382793933153152,
-0.059466298669576645,
0.07010405510663986,
-0.06717757135629654,
-0.026195712387561798,
0.00... |
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... | [
-0.008709532208740711,
0.03711675852537155,
-0.02448982559144497,
-0.08097850531339645,
0.005475939251482487,
0.004336853511631489,
-0.05137117579579353,
0.03276880457997322,
-0.07737088948488235,
0.014721506275236607,
0.07798916101455688,
-0.060665313154459,
-0.0023066424764692783,
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... | [
-0.003957527689635754,
0.044421929866075516,
0.07039124518632889,
-0.03209157660603523,
-0.03311266005039215,
0.030366964638233185,
-0.05658510699868202,
-0.05084259435534477,
-0.06351561844348907,
-0.09476904571056366,
0.06298675388097763,
-0.06918511539697647,
0.012961742468178272,
0.018... |
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... | [
-0.0073067424818873405,
0.027874384075403214,
0.045753732323646545,
-0.0533931665122509,
0.0038575620856136084,
0.0325535349547863,
0.000602048123255372,
0.05274200811982155,
-0.08520890772342682,
0.03153535723686218,
0.11659297347068787,
-0.02893052250146866,
-0.05469546467065811,
0.01379... |
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... | [
-0.027302518486976624,
0.06570399552583694,
0.05927257612347603,
-0.00813221000134945,
-0.052011944353580475,
0.022469691932201385,
-0.010977276600897312,
-0.12262283265590668,
-0.09709782898426056,
-0.015477281995117664,
0.011095714755356312,
-0.05618630722165108,
-0.028892675414681435,
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° = − ... | [
0.045183565467596054,
0.06420135498046875,
0.03413212299346924,
-0.005349040497094393,
-0.05555636063218117,
0.005887928418815136,
-0.02538461796939373,
-0.009451300837099552,
-0.06777816265821457,
0.02748837321996689,
0.030955836176872253,
-0.03495802357792854,
0.047869134694337845,
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 ... | [
-0.06931402534246445,
0.06071430444717407,
0.07691103219985962,
-0.029394913464784622,
-0.06703834980726242,
0.012365635484457016,
-0.017557699233293533,
-0.08935701847076416,
-0.06265708059072495,
-0.1074625551700592,
0.02144034579396248,
-0.05551455542445183,
0.008152100257575512,
0.0905... |
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 ... | [
-0.007498653139919043,
0.0666094720363617,
0.01585250534117222,
-0.0035379657056182623,
-0.011304820887744427,
0.024079596623778343,
-0.10670746117830276,
-0.03912610560655594,
-0.09547779709100723,
-0.0015183856012299657,
0.06147965043783188,
-0.07842094451189041,
-0.01717389188706875,
0.... |
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
− y... | [
0.03019065037369728,
0.04462366923689842,
0.0691971555352211,
0.008328454568982124,
-0.027961911633610725,
0.011232059448957443,
-0.06106271967291832,
-0.08211274445056915,
-0.09454295039176941,
-0.01187760941684246,
0.028349043801426888,
-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
= 4... | [
-0.03330410644412041,
0.04959657043218613,
-0.02925816923379898,
-0.04194098711013794,
0.007832659408450127,
0.021461129188537598,
0.004744387697428465,
-0.01155386958271265,
-0.01771898753941059,
-0.05502661317586899,
0.14055678248405457,
-0.08660637587308884,
0.024131054058670998,
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° ... | [
-0.03457292914390564,
0.019327063113451004,
0.023233668878674507,
0.00841791182756424,
-0.057557400315999985,
0.02296893112361431,
-0.015583604574203491,
-0.04273909330368042,
-0.08170225471258163,
-0.028272204101085663,
0.02820177935063839,
-0.055274754762649536,
0.021342337131500244,
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 i... | [
-0.009185361675918102,
0.09424050897359848,
-0.0024613174609839916,
0.01704336889088154,
0.04042156785726547,
0.04533959552645683,
-0.018993670120835304,
-0.055371589958667755,
-0.07095731049776077,
0.05170753598213196,
0.027492264285683632,
-0.05954413488507271,
0.03593229502439499,
-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... | [
0.003044706303626299,
0.05201176181435585,
-0.021641768515110016,
-0.014711277559399605,
0.03549176827073097,
0.04944372549653053,
-0.05338750407099724,
0.04168127104640007,
-0.09402193874120712,
-0.056846871972084045,
0.03402604162693024,
-0.0726269856095314,
0.06103646755218506,
0.013596... |
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 +... | [
-0.06631626188755035,
0.13035814464092255,
0.07419974356889725,
0.026210933923721313,
-0.019311651587486267,
0.07075368613004684,
-0.04782409220933914,
0.013677211478352547,
-0.08793962001800537,
0.0309456754475832,
-0.06936056911945343,
-0.11137841641902924,
-0.03532185032963753,
-0.01971... |
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 pyra... | [
0.03696100041270256,
0.03221121430397034,
-0.04707975685596466,
-0.07145076990127563,
-0.08015643060207367,
0.016630632802844048,
0.00223378068767488,
-0.0013065810780972242,
-0.08953610062599182,
-0.05020107328891754,
0.059420812875032425,
-0.05412749946117401,
-0.024717174470424652,
0.02... |
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 2... | [
0.025005022063851357,
-0.01579180732369423,
-0.008918183855712414,
-0.08755729347467422,
-0.03894795849919319,
0.005673043895512819,
-0.09762173146009445,
0.06928750872612,
-0.10424664616584778,
0.10550964623689651,
0.025021549314260483,
-0.09188618510961533,
-0.026333849877119064,
0.01788... |
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. Sma... | [
0.04062908515334129,
-0.01759522221982479,
-0.004290326964110136,
-0.07569000124931335,
-0.09903721511363983,
0.003453226061537862,
-0.040559522807598114,
-0.023300280794501305,
-0.04176913946866989,
0.07260175049304962,
0.02293035015463829,
-0.03371328487992287,
0.006657063961029053,
-0.0... |
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 ... | [
0.001238883240148425,
0.03219946473836899,
-0.013150774873793125,
-0.02002209611237049,
-0.03340228646993637,
0.020176496356725693,
-0.05468110367655754,
-0.03823411837220192,
-0.06301996856927872,
0.07297851890325546,
0.06581316143274307,
-0.058116428554058075,
-0.00007100145739968866,
0.... |
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... | [
0.04612082988023758,
-0.0026882158126682043,
0.002365644322708249,
-0.05062064155936241,
-0.031101375818252563,
0.02188156172633171,
-0.06443235278129578,
-0.08068928122520447,
0.007016538642346859,
0.0946122258901596,
-0.00004867401730734855,
-0.0813499391078949,
-0.019492262974381447,
-0... |
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... | [
-0.03834836557507515,
0.004759977571666241,
-0.007363012991845608,
-0.11487159878015518,
0.008037004619836807,
0.03798384964466095,
-0.023217132315039635,
-0.031469084322452545,
-0.13457566499710083,
0.08900336921215057,
0.029158461838960648,
0.023197636008262634,
-0.0024151981342583895,
-... |
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
... | [
-0.008913731202483177,
0.031700726598501205,
0.025224531069397926,
-0.064027339220047,
0.02078738808631897,
0.0627160295844078,
-0.03858461603522301,
0.03497915342450142,
-0.08117462694644928,
0.05999080091714859,
0.06569579988718033,
-0.1261410266160965,
-0.05735474079847336,
0.0476069673... |
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.
Mu... | [
-0.07015806436538696,
-0.015085387974977493,
-0.047300633043050766,
-0.09736853837966919,
0.040446650236845016,
0.025401080027222633,
-0.021294299513101578,
-0.04768820106983185,
-0.0850469097495079,
0.11032646894454956,
-0.03383564576506615,
-0.08823936432600021,
-0.003998155239969492,
-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 +... | [
0.0037805314641445875,
-0.004121602047234774,
0.048771969974040985,
-0.11100907623767853,
-0.041454240679740906,
-0.002287721261382103,
-0.05985318496823311,
-0.047410592436790466,
-0.11219077557325363,
0.11581569910049438,
-0.061649806797504425,
-0.0658707395195961,
0.0243300162255764,
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... | [
-0.07565922290086746,
0.035143688321113586,
0.020206985995173454,
-0.08580482751131058,
0.01210551243275404,
-0.03998926281929016,
-0.013960635289549828,
0.044045835733413696,
-0.07563894987106323,
0.07908333837985992,
0.04063819721341133,
-0.04949714615941048,
-0.0810493603348732,
-0.0121... |
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... | [
0.022144107148051262,
-0.04686625301837921,
0.001691351761110127,
-0.10809078067541122,
-0.0787372812628746,
0.0134972482919693,
-0.04992883652448654,
-0.024995610117912292,
-0.06005128100514412,
0.0695619136095047,
0.008152863010764122,
-0.04183708131313324,
-0.009290764108300209,
0.06397... |
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.
Y... | [
-0.029484575614333153,
-0.057661283761262894,
-0.00357726844958961,
-0.1309952735900879,
-0.025521982461214066,
0.04268478602170944,
-0.003951233811676502,
0.05411916598677635,
-0.05379980802536011,
0.08711190521717072,
0.07375721633434296,
-0.10739995539188385,
-0.06352902203798294,
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
ecto... | [
0.004277675878256559,
-0.011714023537933826,
0.01028799545019865,
-0.10400412976741791,
-0.015074500814080238,
-0.011863152496516705,
-0.05112283304333687,
0.04171120747923851,
-0.06422419846057892,
0.058954060077667236,
0.0646265298128128,
-0.11563097685575485,
-0.016037791967391968,
0.03... |
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 ... | [
-0.02276705391705036,
0.017447808757424355,
0.021190010011196136,
-0.06433277577161789,
-0.058741386979818344,
0.03146833926439285,
-0.06792069226503372,
0.02360422909259796,
-0.03908625990152359,
0.05815693363547325,
-0.01737956888973713,
-0.035319678485393524,
-0.018835216760635376,
-0.0... |
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... | [
-0.04516260698437691,
-0.010824974626302719,
-0.002312145894393325,
-0.07821246981620789,
0.057018447667360306,
-0.010846172459423542,
-0.023247655481100082,
0.008278094232082367,
-0.009302924387156963,
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 h... | [
0.012362947687506676,
0.025287562981247902,
-0.013545703142881393,
-0.11473420262336731,
0.0053300438448786736,
0.04620837792754173,
-0.05309167131781578,
-0.003060348564758897,
0.0062494827434420586,
0.08878674358129501,
0.05178995802998543,
-0.06017650291323662,
-0.04769153520464897,
0.0... |
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 midpoi... | [
-0.038875874131917953,
0.03227316215634346,
0.009734873659908772,
-0.06888306140899658,
0.005809166468679905,
0.0479385070502758,
-0.03518497943878174,
0.02204795740544796,
-0.09565571695566177,
0.04132601246237755,
-0.018182117491960526,
-0.11930627375841141,
-0.019465215504169464,
0.0300... |
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 respe... | [
-0.011685410514473915,
0.08683216571807861,
-0.0016839035088196397,
-0.060049328953027725,
-0.012255455367267132,
0.05335879698395729,
0.03509117290377617,
0.08266430348157883,
-0.10498149693012238,
-0.0088519137352705,
0.0316321924328804,
-0.014298641122877598,
0.006077084224671125,
-0.02... |
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 vertice... | [
0.01120834145694971,
0.051842089742422104,
-0.02010309509932995,
-0.05375223606824875,
-0.005393057595938444,
0.07344502955675125,
-0.04207450896501541,
0.04633847251534462,
-0.07775949686765671,
-0.0019129524007439613,
0.060169704258441925,
-0.13044196367263794,
-0.008673401549458504,
0.0... |
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 ... | [
0.02515784278512001,
-0.028131749480962753,
0.03349294513463974,
-0.09638581424951553,
-0.01962324231863022,
-0.06718854606151581,
-0.027514245361089706,
0.07450667768716812,
0.008953303098678589,
0.07578858733177185,
0.014329138211905956,
-0.04475840553641319,
-0.018840830773115158,
0.060... |
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 |... | [
0.053414054214954376,
0.03037256747484207,
0.036959208548069,
-0.0727737545967102,
-0.007749298587441444,
0.04671185836195946,
-0.08807694166898727,
-0.005215668585151434,
-0.06909162551164627,
0.0004133709880989045,
0.041935134679079056,
-0.06791659444570541,
-0.037248652428388596,
0.0027... |
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 minute... | [
0.03758266940712929,
0.009631291031837463,
0.016412869095802307,
-0.03480171412229538,
-0.002178119495511055,
-0.0010817316360771656,
0.00920837465673685,
-0.0812297910451889,
-0.08547723293304443,
0.06557045131921768,
0.11569410562515259,
-0.09322608262300491,
-0.04467717930674553,
-0.012... |
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... | [
-0.0034278398379683495,
0.02331121265888214,
-0.016362184658646584,
-0.032197169959545135,
-0.0033362354151904583,
0.01996871456503868,
0.027930954471230507,
-0.006654248107224703,
-0.03476482257246971,
0.08206915110349655,
0.1400941014289856,
-0.10520898550748825,
-0.03174063190817833,
0.... |
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... | [
0.004687814973294735,
0.027991794049739838,
0.03238503634929657,
0.025806495919823647,
0.04265283793210983,
-0.05387207120656967,
-0.006289628800004721,
-0.005950962658971548,
-0.03913070261478424,
0.06271026283502579,
0.03849753364920616,
-0.030436357483267784,
0.019827311858534813,
-0.03... |
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 d... | [
-0.002372327959164977,
0.026118125766515732,
0.04655126854777336,
-0.08171463012695312,
0.008176189847290516,
0.025616036728024483,
-0.03467079624533653,
0.015207808464765549,
-0.1137067973613739,
0.05999048426747322,
0.046292826533317566,
-0.08624127507209778,
0.0019370658555999398,
0.015... |
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 ⟶ A... | [
-0.038080666214227676,
0.044777534902095795,
-0.04153566062450409,
-0.06327030062675476,
-0.019678298383951187,
0.01159821730107069,
-0.031110072508454323,
0.014363858848810196,
-0.07755477726459503,
0.042681433260440826,
0.01681922934949398,
-0.031385388225317,
0.056949373334646225,
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 → page... | [
0.0009528724476695061,
0.043320611119270325,
-0.017280912026762962,
-0.0033084091264754534,
0.010023534297943115,
0.04627092927694321,
-0.09116639196872711,
0.024461882188916206,
-0.05848410725593567,
0.0470445342361927,
0.06751221418380737,
-0.04018401727080345,
-0.05456769838929176,
0.01... |
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 t... | [
-0.02279318869113922,
0.08728340268135071,
0.028825532644987106,
-0.019523276016116142,
-0.05104001984000206,
0.07803194224834442,
0.006411666050553322,
-0.010502159595489502,
0.03607599064707756,
0.06873584538698196,
0.09363330900669098,
0.05671586096286774,
-0.0032731725368648767,
0.0287... |
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 thi... | [
0.0758010596036911,
0.047043755650520325,
-0.0001327585632679984,
-0.07190245389938354,
-0.06863921880722046,
0.04398687556385994,
-0.005878149531781673,
0.022327812388539314,
-0.020699361339211464,
-0.03698641061782837,
0.04195931553840637,
-0.06439541280269623,
0.02832835726439953,
-0.03... |
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... | [
-0.028354132547974586,
0.0745006650686264,
0.008716126903891563,
-0.07967345416545868,
-0.04001491516828537,
0.08400769531726837,
0.004016008693724871,
0.06275711953639984,
-0.0638917088508606,
0.02093605324625969,
0.06198914349079132,
0.0005971722421236336,
0.01704634726047516,
-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 tang... | [
0.05790619179606438,
0.08613777160644531,
0.010358507744967937,
-0.05873865261673927,
-0.03958775848150253,
0.06029767915606499,
0.01735677197575569,
-0.04800715297460556,
-0.0004080428334418684,
0.031791508197784424,
0.10342828929424286,
-0.033854905515909195,
-0.01622486487030983,
0.0054... |
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... | [
0.03444387763738632,
0.10872754454612732,
-0.004970457870513201,
-0.04066607728600502,
-0.04129614308476448,
0.0393415130674839,
-0.018656805157661438,
0.02222255803644657,
0.039297886192798615,
0.042143020778894424,
0.10130821168422699,
0.017751256003975868,
-0.005604993551969528,
0.03823... |
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 ... | [
0.0046971701085567474,
0.05172157287597656,
0.013381105847656727,
-0.05180977284908295,
-0.05967383459210396,
0.03421381488442421,
0.013849047012627125,
0.007181062828749418,
-0.030647210776805878,
0.030145954340696335,
0.06259773671627045,
-0.08757885545492172,
0.03474423289299011,
-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 r... | [
-0.03476310521364212,
0.09939083456993103,
0.0402657650411129,
0.03485766798257828,
0.012130330316722393,
0.02812100574374199,
0.048979587852954865,
-0.014095032587647438,
0.04737777262926102,
0.045380085706710815,
0.06666013598442078,
-0.02306460402905941,
0.014071312732994556,
-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 √ __... | [
0.01373356394469738,
0.06604833900928497,
0.0330975241959095,
-0.018002688884735107,
0.00032862371881492436,
0.027202708646655083,
0.06947272270917892,
-0.014265363104641438,
-0.04618728905916214,
0.07320968806743622,
0.0692266896367073,
-0.1263413280248642,
0.022794941440224648,
-0.023612... |
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 b... | [
0.03908907622098923,
0.127048522233963,
0.022875163704156876,
-0.0013093098532408476,
0.011182256042957306,
0.016975929960608482,
-0.0346963070333004,
0.0055518182925879955,
-0.030499190092086792,
0.049966197460889816,
0.08024712651968002,
-0.0367974229156971,
0.022301601245999336,
0.00050... |
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 + ... | [
0.024645080789923668,
0.04479121044278145,
0.08906380087137222,
-0.06594657152891159,
-0.04466661438345909,
0.007266992703080177,
0.00783770065754652,
-0.03277609869837761,
0.009543223306536674,
0.059896986931562424,
0.03504301235079765,
-0.06637509167194366,
0.012991795316338539,
-0.03868... |
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
, s... | [
-0.032000329345464706,
0.10889829695224762,
0.01341982837766409,
-0.023537056520581245,
-0.0021177581511437893,
0.039925090968608856,
0.008774988353252411,
0.013910114765167236,
-0.017120052129030228,
0.06552106887102127,
0.03409064933657646,
-0.0706656277179718,
-0.003833746537566185,
-0.... |
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 equ... | [
0.011055950075387955,
0.08838456869125366,
-0.0019119130447506905,
-0.011773900128901005,
-0.009889899753034115,
0.04079638421535492,
-0.028773058205842972,
0.006010227371007204,
0.027708008885383606,
0.015253998339176178,
0.05280586704611778,
-0.06630959361791611,
-0.01915043592453003,
-0... |
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 __... | [
0.012095384299755096,
0.08793006837368011,
0.031102538108825684,
-0.03584098443388939,
0.014770867303013802,
0.06984058022499084,
-0.016980094835162163,
-0.019016703590750694,
0.016895517706871033,
0.06319601833820343,
0.09794049710035324,
-0.026831910014152527,
0.0038875332102179527,
0.01... |
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 (... | [
0.005873055662959814,
0.05656995624303818,
0.07041271030902863,
-0.04068222641944885,
-0.010473676025867462,
0.06818141788244247,
-0.01930898241698742,
-0.014217070303857327,
-0.004005999770015478,
0.0490807481110096,
0.08344332873821259,
-0.059313416481018066,
0.004980350844562054,
0.0093... |
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 T... | [
0.01874510571360588,
0.06435774266719818,
-0.051173143088817596,
-0.017296787351369858,
0.00439302995800972,
0.023425357416272163,
-0.059616297483444214,
0.012654739432036877,
-0.09157358855009079,
0.016284549608826637,
0.0789838433265686,
-0.03795613348484039,
0.009795049205422401,
-0.030... |
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 x... | [
0.04924332723021507,
0.10340869426727295,
0.040375933051109314,
-0.002396893687546253,
-0.0581330806016922,
0.023663945496082306,
-0.016507083550095558,
0.052913736552000046,
-0.05283796042203903,
0.0333092100918293,
0.0677429586648941,
-0.06862161308526993,
0.015798619017004967,
-0.050089... |
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 onc... | [
-0.03164226934313774,
0.07194360345602036,
0.031164199113845825,
-0.04241379722952843,
0.04123389720916748,
0.03804660588502884,
0.042182281613349915,
0.03721296787261963,
-0.03331374004483223,
0.056092843413352966,
0.004433308262377977,
-0.1001049056649208,
0.0014004029799252748,
-0.07207... |
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.
... | [
0.059739850461483,
0.04299519211053848,
0.019319044426083565,
0.0019086882239207625,
-0.010413730517029762,
0.0030788020230829716,
-0.08590899407863617,
-0.004896543920040131,
-0.0497145801782608,
0.006664295680820942,
0.07000648975372314,
-0.006107461638748646,
-0.0395393930375576,
0.0868... |
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 val... | [
0.02936205454170704,
0.0525096170604229,
-0.01994895190000534,
-0.02517310529947281,
0.005505033768713474,
0.05583016201853752,
-0.05214155465364456,
0.016795581206679344,
-0.06501811742782593,
0.03795088455080986,
0.07669127732515335,
-0.006813224870711565,
-0.021983817219734192,
-0.02152... |
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... | [
0.0794287696480751,
0.034189485013484955,
0.025797462090849876,
-0.061484191566705704,
-0.02883606031537056,
-0.02434409223496914,
0.019275017082691193,
0.05651194602251053,
-0.09551556408405304,
0.01584555394947529,
0.058406829833984375,
-0.09017116576433182,
-0.016205845400691032,
-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 + 2... | [
0.036003902554512024,
0.06004474684596062,
-0.031926464289426804,
-0.02965027280151844,
-0.04144991934299469,
-0.03659671172499657,
-0.035623736679553986,
0.08752728253602982,
-0.0783684104681015,
0.006719214841723442,
0.048183560371398926,
-0.07993266731500626,
-0.0326470322906971,
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)
Ma... | [
0.08862057328224182,
0.10639328509569168,
0.0490429513156414,
-0.06737717241048813,
-0.0668855682015419,
-0.04618651047348976,
-0.024728238582611084,
0.007077357731759548,
-0.046958811581134796,
0.04410206526517868,
0.03740862384438515,
0.036588914692401886,
0.014011599123477936,
0.0994652... |
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... | [
0.014954634010791779,
0.1344785839319229,
-0.03078564442694187,
-0.014110967516899109,
-0.028600450605154037,
0.008411802351474762,
0.016842108219861984,
0.05713711306452751,
-0.06247341260313988,
0.04259997606277466,
0.09287931770086288,
-0.0483633428812027,
-0.013691396452486515,
0.03636... |
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
... | [
-0.02408967912197113,
0.03779415041208267,
0.08786099404096603,
0.015574335120618343,
0.022342968732118607,
-0.07558796554803848,
0.017391717061400414,
0.021221699193120003,
0.04492919519543648,
0.03772313520312309,
0.03336535766720772,
0.015153896063566208,
0.03667379543185234,
0.00681007... |
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 ... | [
0.06508141756057739,
0.07368186116218567,
-0.02543535642325878,
-0.06432030349969864,
0.00707190902903676,
-0.04174409806728363,
0.04941564053297043,
0.04178319871425629,
-0.09458102285861969,
0.04696584865450859,
0.04823565483093262,
-0.012079882435500622,
-0.014132533222436905,
-0.000061... |
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 ... | [
0.050645094364881516,
0.03229406848549843,
0.04470488056540489,
0.028329741209745407,
0.005772694479674101,
0.0005871674511581659,
-0.011861509643495083,
0.06354962289333344,
-0.03883187100291252,
-0.0003472976968623698,
0.1045459508895874,
-0.059703174978494644,
-0.06329809874296188,
-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 ... | [
-0.040123093873262405,
0.09335164725780487,
-0.020845510065555573,
-0.02968275174498558,
0.017393875867128372,
0.05064249783754349,
0.0007762798923067749,
0.04899068549275398,
-0.0417327806353569,
-0.016094882041215897,
0.08153828978538513,
-0.07153479754924774,
-0.05388902500271797,
-0.02... |
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 you... | [
-0.01246769167482853,
0.07484755665063858,
0.09605453163385391,
-0.06295762956142426,
-0.00037655848427675664,
0.07412883639335632,
0.05846231430768967,
0.002388579538092017,
-0.07022880017757416,
0.05344906821846962,
0.06591147184371948,
-0.09511943906545639,
-0.0041664233431220055,
-0.03... |
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... | [
0.050706151872873306,
0.05772428959608078,
0.04644424840807915,
-0.08210951834917068,
-0.04953503981232643,
0.03515613079071045,
0.040536846965551376,
0.03375213220715523,
-0.07476657629013062,
0.055369164794683456,
0.049393005669116974,
-0.08906559646129608,
-0.0038944557309150696,
0.0424... |
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 ... | [
0.03797698765993118,
0.0666431412100792,
0.0018668669508770108,
0.049065206199884415,
0.009178305976092815,
0.05368335545063019,
0.08316769450902939,
0.12136946618556976,
-0.11155333369970322,
-0.012612785212695599,
0.10970685631036758,
-0.03624430298805237,
-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 func... | [
0.029395515099167824,
0.10340970754623413,
-0.00029126094887033105,
-0.03447156772017479,
-0.023289673030376434,
0.03772219270467758,
-0.006097830832004547,
-0.016995709389448166,
-0.0024088697973638773,
0.04763897508382797,
0.08149439841508865,
-0.015973541885614395,
-0.003570951521396637,
... |
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... | [
-0.04636438935995102,
-0.00045694579603150487,
-0.030932234600186348,
0.02383129671216011,
-0.02062799781560898,
0.05328574404120445,
-0.08649519085884094,
0.024361221119761467,
-0.08895821124315262,
-0.03254534304141998,
0.034827686846256256,
-0.09021524339914322,
-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 dif... | [
-0.03290439024567604,
0.03981836140155792,
0.055290307849645615,
0.016547245904803276,
0.005876890383660793,
0.06557264924049377,
0.002742363838478923,
0.024559400975704193,
0.10044407844543457,
-0.027774862945079803,
0.002301124157384038,
0.042778462171554565,
-0.03802714869379997,
-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... | [
-0.06205613538622856,
-0.005403011571615934,
0.07324870675802231,
-0.01982802152633667,
-0.054619695991277695,
0.05514797568321228,
0.011388523504137993,
0.019227400422096252,
0.006840534042567015,
-0.009105226024985313,
-0.010903405025601387,
-0.08774581551551819,
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 ≠ −... | [
-0.023413505405187607,
0.03669223561882973,
0.05473974719643593,
-0.04938507080078125,
0.0017573012737557292,
0.05240318179130554,
0.04980437457561493,
-0.06957007199525833,
-0.05314742028713226,
-0.03083023801445961,
0.008726219646632671,
-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
... | [
-0.054267436265945435,
0.0539020337164402,
0.09592440724372864,
-0.04780878871679306,
-0.024122437462210655,
0.10988973081111908,
-0.007939374074339867,
0.011648737825453281,
-0.08575966209173203,
0.037758488208055496,
-0.0006808872567489743,
-0.06220179423689842,
0.009136314503848553,
-0.... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.