Gleichungssystem Loesung: Teil 2 Einsetzungsverfahren

Gleichungssystem Loesung: Teil 1 Additionsverfahren, 2. Version

Finding the nth number of an Arithmetic Sequence

INTEGRATION

➪ Integration Key Formula ;

Definition Integration written F(x), is the reversed process of differentiation f '(x) (also derivatives). Functions and their Nature Integration F(x) Function f(x) Differentiation f '(x) 1 4 · x4 ↢↢↢ x3 ↣↣↣ 3x² Constant of Integration Let's consider three functions, defined as follow: f(x) = x5,          its derivative is: f '(x) = 5x4 f(x) = x5 + 2,    its derivative is: f '(x) = 5x4 f(x) = x5 − 12,  its derivative is: f '(x) = 5x4 As you can notice, the derivatives of the three functions are the same and the constants hanging behing the two last functions have disappeared or equal to Zero (0)!!!     ஃ    we can conclude that, the derivative of any constant number equals to 0    ⇒   f(x) = C (also K) and f '(x) = 0 i.e.     i)    f(x) = 7 and f '(x) = 0     ii)    f(x) = −10 and f '(x) = 0     iii)    f(x) = − 1 2 and f '(x) = 0     iv)    f(x) = 11 3 and f '(x) = 0 As we've seen above, the three functions have the same derivative. Now if we go integrating the three of them, we actually need to precise they all have a constant to be sure to find the original functions. The general formula is: ∫ f '(x) dx = f(x) + C So for our three functions we'll have: ∫ 5x4 dx = x5 + C, with C in this case equals to 0 and the function f(x) = x5 ∫ 5x4 dx = x5 + C, with C in this case equals to 2 and the function f(x) = x5 + 2 ∫ 5x4 dx = x5 + C, with C in this case equals to −12 and the function f(x) = x5 − 12 How to find the constant of integration? Integrating a sum or a difference of functions It's about integrating by term or separetly. i.e.    1. Find ∫ x² − x √x dx    ∫ x² − x √x dx = ∫ ( x² x1/2 − x x1/2 ) dx = ∫ x3/2 dx − ∫ x1/2 dx                      = 2 5 x5/2 − 2 3 x3/2 + C    2. Find the integral of 1 + x7 + 1 x² − √x    ∫ ( 1 + x7 + 1 x² − √x ) dx = ∫ ( 1 + x7 + x−2 −x1/2 )                                            = ∫ 1 dx + ∫ x7 dx + ∫ x− 2 dx + ∫ x1/2 dx                                            = x + 1 8 x8 − 1 x − 2 3 x3/2 + C Integrating terms to the power of n: (ax + b)n i.e. Integrate f(x) = (2x + 3)4 to integrate this function, we'll proceed by substitution Let u = 2x + 3   ⇒   u4 = (2x + 3)4, then, we derivate u prior to x: du = 2 dx 1 2 du = dx then after, we derivate u4: 1 2 ∫ u4 du = 1 2 · 1 5 u5 now we replace u by its substituted value: u = 2x + 3: 1 2·5 (2x + 3)5 So,     ∫ (2x + 3)4 dx = 1 2·5 (2x + 3)5 Finding the constant of integration Assume we have a function f(x) = x3 + C, its derivative is f '(x) = dy/dx = 3x2. To find C, we need to know a point on the curve of the function. Let's say P(1, 3) that point on the curve with an equation dy/dx = 3x2 + 4                dy dx = 3x2 + 4         ⇒     y = ∫ (3x2 + 4) dx         ∴     y = x3 + 4x + C         ∴     as (1, 3) is on the curve, we can input x=1 and y=3 in x3 + 4x + C,                to find C:   3 = 1 + 4 + C     ⇒         C = − 2         ∴     y = x3 + 4x − 2 Key Formula

SEQUENCES AND SERIES

Comparative Summary

Exercises, Convergence and Divergence of Sequences

Arithmetic Sequences Geometric Sequences A. Sequence represents an ordered list of numbers with a common difference d. Ex.: 1, 5, 9, 13, 17, 21, 25, ... , G. Sequence represents an ordered list of numbers with a common ratio r. Ex.: 1, 4, 16, 64, 256, 1024, ... , A. Series and G. Series are the values obtain by adding up all the terms of a sequence. there also called the "sum". Ex.: 1 + 5 + 9 + 13 + 17 + 21 + 25 + ... + (ASs or APs) Ex.: 1 + 4 + 16 + 64 + 256 + 1024 + ... + (GSs or GPs) Common difference is always the same value from one term to the next by adding (or subtracting) both terms. d = an+1 − an Ex: d = 9 − 5 = 4 (ASs above) Common ratio is always the same value from one term to the next by dividing (or multiplying) both terms. r = an+1 an Ex: r = 64 16 = 4 the nth number                 an = a1 + (n − 1)·d the nth number                 an = a1 · rn − 1 Sum of the nth 1st numbers                 Sn = n(a1 + an) 2 Sum of the nth 1st numbers      Sn = a1(rn − 1) r − 1 = a(1 − rn) 1 − r with a1 = a, and the special circumstance that the r is between –1 and 1, | r | < 1. the infinite sum has the following value: S∞ = a 1 − r , with | r | < 1 Sum of a Finite Geometric Series Let's find the sum of the first 7 terms of sequence S, with a1 = 1, r = 2 and n = 7. We would then plug those numbers into the formula and get: S7 = 1(1 − 27) 1 − 2 = (1 − 128) − 1 = − 127 − 1 = 127 Sum of an infinite Geometric Series An infinite number of terms a sequence S in given by: 16 + (-8) + 4 + (-2) + 1 + ..., with a1 = 16 and r = −1/2. r lies within the acceptable limits (− 1 and 1), therefore a finite sum exists when using the infinite term sum formula: S∞ = a1 1 − r = 16 1 − (−1/2) = 16 3/2 = 16 2 3 = 10.67 Convergence and Divergence of Sequences Examples Showing convergence or divergence of the the following sequence: 1, 2, 3, 4, ... Ans. This sequence starts with and goes on bigger and bigger to the infinity (∞), therefore the sequence Diverges. Does the following sequence converges or diverges? 1, 0.1, 0.01, 0.001, 0.0001, ... Ans. The sequence gets smaller and smaller and closer to 0, therefore it converges. Assuming we have the nth number of a sequence define by: an = 3(−1)n n! Ans. To show weither the sequence converges or diverges we have 2 possibilities: we can first find the 3 or 4 first terms of the sequence - a1 = 3(−1)1 1! = −3 - a2 = 3(−1)2 2! = 3 2·1 = 3 2 - a3 = 3(−1)3 3! = −3 3·2·1 = −1 2 - a4 = 1 8 ... As you can see, the terms are getting smaller and smaller and tending to 0, hence the sequence converges, (not obvious though!!!) That's why we will use limits to make it clear... lim n→∞    [ 3(−1)n n! ] = ∞ ∞    (indeterminate form!) we use L'Hôpital's rule (by derivating several times until we get the form allowing the determination of the limits) the second differentiation gives us: lim n→∞    [ 3n(−1)n−1 1! ] = ∞·0 1! = 0 Now, we can clearly see that the sequence converges, as it equals to 0 an = n(n −1) converges or diverges? lim   n(n − 1) = ∞·∞ = ∞ ⇨ the sequence diverges! n→∞

SEQUENCES AND SERIES - EXERCISES

Exercises Write down the first four terms in ascending power of x of the binomials expansion of (a)  (1 + 3x)12      (b)  (1 − 2x)9 (c)  (2 − x)10      (d)  (1 − x 3 )20      Answers (e)  (2 − 3 2x )7      (f)  ( 3 2 + 2x)9 The sum of the nth terms of a progression is given by Sn where Sn = n(3n − 4). Show that the progression is an AP.     Answer The sum of the nth terms of an AP is Sn where Sn = n² − 3n. Write down the fourth term and the nth term.     Answer Answers to i Notice: To solve these problems, we can use Pascal's Triangle, but that needs a lot of time! An alternative way is to use the Binomial Theorem, which offers a quicker method for expanding any complex binomial series. Only and only when the power n is a positive integer. see (a + b)n (a) (1 + 3x)12      = 12c0·112·(3x)0 + 12c1·111·(3x)1 + 12c2·110·(3x)2 + 12c3·19·(3x)3 + ... + (3x)12      = 1 + 36x + 594x² + 5940x3 + ... + 531441x12 (b) (1 − 2x)9      = 9c0·19·(−2x)0 + 9c1·18·(−2x)1 + 9c2·17·(−2x)2 + 9c3·16·(−2x)3 + ... + (−2x)9      = 1 − 18x + 144x² − 672x3 + ... − 512x9 (c) (2 − x)10      = 10c0·110·(−x)0 + 10c1·19·(−x)1 + 10c2·18·(−x)2 + 10c3·17·(−x)3 + ... + (−x)10      = 1024 − 5120x + 11520x² − 15360x3 + ... + x10 (d) (1 − x 3 )20      = 20c0·120·(−x/3)0 + 20c1·119·(−x/3)1 + 20c2·118·(−x/3)2 + 20c3·117·(−x/3)3 + ... +      = 1 − 20x/3 + 190x²/9 − 1140x3/27 + ... + x20/320 (e) (2 − 3 2x )7      = 7c0·27·(−3/2x)0 + 7c1·26·(−3/2x)1 + 7c2·25·(−3/2x)2 + 7c3·24·(−3/2x)3 + ... + ...      = 128 − 672/x + 1512/x² − 1890/x3 + ... + ... (f) ( 3 2 + 2x)9      = 9c0·(3/2)9·(2x)0 + 9c1·(3/2)8·(2x)1 + 9c2·(3/2)7·(2x)2 + 9c3·(3/2)6·(2x)3 + ... +      = (3/2)9 + (310/27)x + (39/8)x² + (7·(37)/2)x3 + ... + ... Answer to ii the first nth terms are given by an = Sn − Sn−1 ⇒ an = n(3n − 4) − (n − 1) [3(n − 1) − 4] = 3n² − 4n − (n − 1) (3n − 7) = 3n² − 4n − 3n² + 10n − 7) ⇒ an = 6n − 7, therefore, we can find out:       a1 = 6·1 − 7 = − 1 ⤸+6       a2 = 6·2 − 7 = 5 ⤸+6       a3 = 6·3 − 7 = 11 ⤸+6       a4 = 6·4 − 7 = 17 ⤸+6       ...      ...      ...       ...      ...      ...       an = 6·n − 7       and the sequence is: − 1, 5, 11, 17, 23,..., As you can see, the common difference d = 6. By replacing a1 and an in the AP's general formula, we should obtain Sn = n(3n − 4), which is the proof that Sn is an AP progression. By replacing a1 and an in the AP's general formula, we should obtain Sn = n(3n − 4), which is the proof that Sn is an AP progression. Sn = n(a1 + an) 2 = n(−1 + 6n − 7) 2 = 2n(3n − 4) 2 ⇒ Sn = n(3n − 4), the progression is an AP. Answer to ii * The nth term is given by: an = Sn − Sn−1 ⇒ an = n² − 3n − [(n−1)² − 3(n−1)] = n² − 3n − (n² − 5n + 4) = n² − 3n − n² + 5n − 4 ⇒ an = 2n − 4 * The 4th term a4 = 4 Find the first three terms in the expansion of (1 + b)5. Use the substitution x = 1 + b, to find the coefficient of b² in the expansion of (1 + b − 2b²)5. Answer ⤵ * To expand (1 + b)5, we use the Binomial Theorem: ∴ (1 + b)5 = 5C0(1)5(6)0 + 5C1(1)4(6)1 + 5C2(1)3(6)2 + ... + 5C5(1)0(6)5 = 1 + 5 + 10 ⇒ (1 + b)5 = 1 + 5 + 10, which are the 1st three terms of the expansion. * Now x = 1 + b is plugged in the expression (1 + b − 2b²)5 by substitution ⇒ (x − 2b²)5. We can still use the Binomial Theorem, but let's go for the Pascal's Triangle. 0 1 1      1       1 2 1        2 1 3 1    3      3        1 4     1    4 6    4     1 5       1    5   10       10           5   1 As you can see, we're concerned by the 5th line because our power is 5 ∴ we can write: 1(x)5(−2b²)0 + 5(x)4(−2b²)1 + 10(x)3(−2b²)2 + 10(x)2(−2b²)3 + 5(x)1(−2b²)4 + 1(x)0(−2b²)5 and after working out the parenthesis we obtain: x5 − 10x4b² + 40x3b4 − 80x²b6 + 80xb8 − 32xb10. Then, the coefficient of b² is − 10 Close ⤴ The coefficient of b² and b³ in the expansion of (a + b)6 are equal. Find the value of a.

Irregular verbs / Liste der unregelmäßigen Verben

infinitive I. Form simple past II. Form past participle III. Form Übersetzung be was / were been sein bear bore borneUK / bornAE etwas tragen (lit.) beat beat beaten schlagen become became become werden begin began begun beginnen, anfangen bend bent bent biegen bet* bet bet wetten bid bid, bade bid, bidden jmd. bitten bind bound bound binden bite bit bitten beißen, stechen bleed bled bled bluten blow blew blown blasen break broke broken zerbrechen breed bred bred züchten bring brought brought herbringen broadcast broadcast broadcast senden, übertragen, ausstrahlen build built built bauen burn burnt, burned burnt, burned brennen, verbrennen burst burst burst (zer)platzen bust* bust bust kaputtgehen buy bought bought kaufen =========== ============== =========== =========== cast cast cast werfen catch caught caught fangen choose chose chosen wählen cling clung clung (sich) klammern, festhalten come came come kommen cost cost cost kosten creep crept crept kriechen schleichen cut cut cut schneiden =========== ============== =========== =========== deal dealt dealt verhandeln, verteilen dig dug dug graben dive divedUK / doveAE dived tauchen do did done tun, machen draw drew drawn zeichnen dream dreamt, dreamed dreamt, dreamed träumen drink drank drunk trinken drive drove driven ein Fahrzeug führen dwell dwelt, dwelled dwelt, dwelled wohnen, leben (formell) =========== ============== =========== =========== eat ate eaten essen =========== ============== =========== =========== fall fell fallen fallen feed fed fed füttern feel felt felt fühlen fight fought fought kämpfen find found found finden flee fled fled fliehen, flüchten fling flung flung schleudern fly flew flown fliegen forbid forbade, forbad forbidden verbieten forecast* forecast forecast vorhersagen, prognostizieren forget forgot forgotten vergessen forgive forgave forgiven vergeben, verzeihen forsake forsook forsaken (jmd.) verlassen, (etw.) aufgeben freeze froze frozen frieren =========== ============== =========== =========== get got gotUK, gotten AE bekommen give gave given geben grind ground ground zerkleinern, mahlen go went gone gehen grow grew grown wachsen =========== ============== =========== =========== hang hung, hanged hung, hanged hängen, aufhängen have had had haben hear heard heard hören hide hid hidden verstecken hit hit hit schlagen, treffen hold held held halten hurt hurt hurt verletzen =========== ============== =========== =========== keep kept kept behalten kneel knelt knelt knien, sich knien know knew known wissen, kennen =========== ============== =========== =========== lay laid laid legen lead led led führen, leiten lean leant, leaned leant, leaned sich beugen, lehnen learn learnt, learned learnt, learned lernen leave left left verlassen lend lent lent verleihen let let let lassen lie lay lain liegen lie lied lied lügen light* lit lit anzünden lose lost lost verlieren =========== ============== =========== =========== make made made machen (herstellen) mean meant meant bedeuten meet met met treffen =========== ============== =========== =========== pay paid paid bezahlen prove* proved proven beweisen put put put setzen, stellen, legen =========== ============== =========== =========== quit* quit quit verlassen, aufhören =========== ============== =========== =========== read read read lesen (Aussprache!) rid rid rid (jmd. von etwas) befreien ride rode ridden reiten ring rang rung klingeln, läuten rise rose risen aufgehen, ansteigen run ran run rennen =========== ============== =========== =========== say said said sagen see saw seen sehen seek sought sought suchen sell sold sold verkaufen send sent sent schicken set set set setzen, stellen, legen sew* sewed sewn nähen spell spelt, spelled spelt, spelled buchstabieren shake shook shaken schütteln shear* sheared shorn scheren (Schafe), abbrechen shed shed shed abstoßen, ablegen shine shone shone glänzen, leuchten, scheinen shoot shot shot schießen show* showed shown zeigen shrink shrank shrunk schrumpfen, kleiner werden shut shut shut schließen sing sang sung singen sink sank sunk sinken, untergehen sit sat sat sitzen slay slew slain erlegen, bezwingen (lit.) sleep slept slept schlafen slide slid slid ausrutschen sling slung slung werfen, schleudern slink slunk slunk schleichen slit slit slit aufschlitzen smell smelt, smelled smelt, smelled riechen sow sowed sown säen speak spoke spoken sprechen speed* sped sped beschleunigen, flitzen, rasen spend spent spent Geld ausgeben, Zeit verbringen spin spun spun drehen, rotieren spit spat, spit spat, spit spucken split split split teilen spread spread spread (sich) ausbreiten, verteilen spring sprang sprung springen stand stood stood stehen steal stole stolen stehlen stick stuck stuck steckenbleiben sting stung stung stechen stink stank, stunk stunk stinken stride strode stridden (auf etw.) zugehen strike struck struck schlagen, treffen (Blitz, Kugel) string strung strung auffädeln, aufziehen strive strove striven (sich) bemühen swear swore sworn schwören sweep swept swept kehren swell* swelled swollen anwachsen, anschwellen swim swam swum schwimmen swing swung swung schwingen =========== ============== =========== =========== take took taken nehmen, (weg)bringen; dauern teach taught taught lehren, unterrichten tear tore torn zerreißen tell told told erzählen, berichten think thought thought denken thrive* throve thrived gedeihen throw threw thrown werfen thrust thrust thrust stoßen tread trod trodden, trod treten =========== ============== =========== =========== understand understood understood verstehen =========== ============== =========== =========== wake* woke woken aufwecken wear wore worn anhaben, tragen (Kleidung) weave wove woven weben weep wept wept weinen wet* wet wet befeuchten win won won gewinnen win wound wound schlängeln, spulen wring wrung wrung (aus)wringen write wrote written schreiben    * regular form + -ed also possible

CONNECTED RATES OF CHANGE

Example , Exercises , Answers

DEFINITION OF 'RATE OF CHANGE' Rate of change is the speed at which a variable changes over a specific period of time. It's often used when speaking about momentum, and it can generally be expressed as a ratio between a change in one variable relative to a corresponding change in another. Graphically, the rate of change is represented by the slope of a line. To solve problems regarding the rate of change, we use the chain rule. EXAMPLE If a given function y = f(x) has a rate of change with respect to r; we can also find the rate of change of x with respect to r. * The volume, Vcm3, of a sphere of radius r cm is given by V = 4 3 πr3. V is increasing at the rate of 1.5cm3 per second. Find the rate of increase of r with respect to time when r = 2 ANSWER Given: dV dt = 1.5cm3 (rate of change of V) and the volume V = 4 3 πr3. Searched: dr dt Derivating V with respect to r gives us: dV dr = 4πr2 We know that: dV dr = dV dt · dt dr ⇒ 4πr2 = 1.5 · dt dr ⇒ 4πr2 1.5 = dt dr ;   Multiply both fractions by (−1) ⇒ 1.5 4πr2 = dr dt , with r = 2 ⇒ dr dt = 1.5 16π = 0.0298 cm/s Therefore, The radius increases at the rate of 0.0298 cm per second. EXERCISES The equation of a curve is y = x² − 5x. A point P is moving along the curve so that the x-coordinate is increasing at the constant rate of 0.2 units per second. Find the rate at which the y-coordinate is increasing when x = 4 Answer i. The equation of a curve is y = 4 − 1 x A point P is moving along the curve so that the y-coordinate is increasing at the constant rate of 0.01 units per second. Find the rate at which the x-coordinate is increasing when x = 1 Answer ii. The equation of a curve is y = 1 2 − x A point P is moving along the curve so that the y-coordinate is increasing at the constant rate of 0.5 units per second. Find the rate at which the x-coordinate is increasing when x = 1 Answer iii. The volume, Vcm3, of a cube of the edge xcm increasing at the constant rate of 3cm3 per second. Find the rate at which x is increasing when x = 10 The volume, Vcm3, of a sphere of radius xcm is given by V = 4 3 πx3. The volume is increasing at the constant rate of 0.2 cm3 per second. Find the rate of increase of the radius when the radius is 5cm A rectangular water tank (see below) is being filled at the constant rate of 20 l sec The base of the tank has dimensions w = 1m and L = 2m. What is the rate of change of the height of water in the tank? (in cm sec ) Answer iV. ANSWERS Answer i. Given:   y = x² − 5x ⇒   dy dx = 2x − 5 The x-coordinate increases at a constant rate of 0.2 units/s ⇒ dx dt = 0.2 Searched:     dy dt , for x = 4 Using the chain rule:   dy dx = dy dt · dt dx ⇒ 2x − 5 = dy dt · 1 0.2 therefore,      dy dt = (2x − 5)· 0.2 for x = 4,      dy dt = (2·4 − 5)· 0.2 = 3 · 0.2 = 0.6 The y-coordinate increases at a rate of 0.6 units per second. Answer ii. Given: y = 4 − 1 x   ⇒   dy dx = 1 x²   and   dy dt = 0.01 (y-coordinate rate of change) Searched:    dx dt ,  for x = 1 With the chain rule:     dy dx = dy dt · dt dx   ⇒   1 x² = 0.01 · dt dx therefore,   dt dx = 1 0.0x²   or   dx dt = 0.0x² 1 for x = 1,    dx dt = 0.01² 1 = 0.01 The x-coordinate increases at a rate of 0.01 units per second. Answer iii. Given: y = 1 2 − x   ⇒   dy dx = 1 (2 − x)² the y-coordinate is increasing at the constant rate of 0.5 units per second  ⇒  dy dt = 0.5 Searched:    dx dt ,  for x = 1 Assuming the chain rule:     dy dx = dy dt · dt dx   ⇒   1 (2 · x)² = 0.5 · dt dx therefore,   1 (2 − x)² · 0.5 = dt dx   or   dx dt = (2 − x)² · 0.5 1 for x = 1,    dx dt = 0.5 1 = 0.5 The x-coordinate increases at a rate of 0.5 units per second. Answer iV. Given: · The volume V of water in the tank by: V = w·L·H · The rate of change of the volume by: dV dt Searched: · The rate of change of the height H of water: dH dt V and H are functions of time. So let's differentiate: dV dt = W·L· dH dt , (with W and L as constants) therefore, dH dt = 1 W·L · dV dt Convert liters into cm3 and meters into cm as follows: 1 liter = 1 dcm3 = 1000 cm3 and 1 meter = 100 cm and, dH dt = 1 100cm·200cm · 20·1000cm3 = 1 cm sec The rate of change of the height H of water is 1 cm sec . TOP