MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

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1 . REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but a + b c = a c + b c [Yu can t separate a demminatr, but yu can divide a single denminatr int each numeratr term]. y = a + a + b implies y = a + a + b = a + b a(a + b)+ = a + b a + ab + [The first step is by inverting bth sides, being careful t invert each side as a whle; the secnd step is by multiplying numeratr and denminatr by (a + b)] Nte that cmbining different algebraic fractins can usually nly be dne by multiplying numeratr and denminatr f each fractin in such a way as t create a cmmn denminatr: a + a + b + c = (a + b)c a (a + b)c + a + b ac ac + a(a + b) (a + b)c + ac + a(a + b) = = ac + bc + a + ab c a(a + b) a(a + b)c a c + abc Similarly, a fractin in the denminatr can be simplified by multiplying entire numeratr and entire denminatr by cmmn denminatr f the fractins in the denminatr: a + c a + a+b = a + c a + a+b a(a + b) a(a + b) = a(a + b) = a a(a + b)+ a(a + b) a a+b + ab (a + b)+a = a + ab a + b () Quadratic equatins Knw rts f quadratic equatin ax + bx + c =. (x a) +(y b) = d is circle in dimensins f radius d with center (a, b), cylinder in 3 dimensins. (x a) +(y b) +(z c) = d is sphere in 3 dimensins with center (a, b, c), radius d. Cmpleting the square:. x + ax + y + by + z + cz = d implies. (x + a ) a 4 +(y + b ) b 4 +(z + c ) c 4 = d implies. (x + a ) +(y + b ) +(z + c ) = d + a 4 + b 4 + c 4 [ a sphere abut the pint ( a, b, c ) f radius d + a 4 + b 4 + c 4 (if real)]

2 Ex: Find the center f and radius f the circle: x 3x + y + y =4 ( ) Ans: x 3 ( ) + y + = 3 s center = ( 3, ), radius = 3 (3) Simplifying Pwers: Pwer rules: x a x b = x a+b (x a ) b = x ab x a = x a x a /x b = x a b Of curse, (xy) a = x a y a, but x a y b can nt be simplified in general. Ex: Simplify 5 x 4 3 x 5 Ans: x 9 r x 9. REVIEW OF SOME BASIC TRIGONOMETRY (a) Quadrants Quad I: Sin +/Cs + Quad II: Sin +/Cs - Quad III: Sin -/Cs - Quad IV: Sin -/Cs + Fr psitivity f standard trig functins: All Students Take Calculus. All are psitive in I quadrant. Sine nly is psitive in II quadrant. Tangent nly is psitive in III quadrant. Csine nly is psitive in IV quadrant NOTE: Changing an angle θ by π radians (r 4π, 6π, etc) des nt change it at all. Therefre θ = π/, θ =π + π/ =5π/, θ =4π + π/ =9π/, θ = π + π/ = 3π/ are all the same angle. EX: In what quadrant is θ =3π/3 and is sec θ psitive r negative? Answer: θ =3π/3 =(7 3 )π (7 3 4)π =( 3 )π s it is in the furth quadrant and sec θ is psitive. Values f sin θ and cs θ at 9 degree intervals :. sin θ cs θ Rtating yur hand in a 36 degree circle, yu need t knw these 9 degree values in yur mind. Once yu are cmfrtable with this, yu need t be able t indicate the values f sin nθ and cs nθ fr n =, 3,... This, in fact, will be very imprtant in the third Mdule. Fr example, cnsider the functin sin 3θ. Asθ increases frm degrees t 3 degrees, say, 3θ ges frm degrees t 9 degrees. In fact, each 3 degree increment f θ will change 3θ by 9 degrees. Therefre, when

3 we wish t envisin the values f sin 3θ thrugh the full 36 degrees, we first draw axes every 3 degrees, and then we can label these axes with the values f the sin 3θ functin, which, beginning at degrees, will be,,,-,, etc. The results f this labeling fr sin 3θ and cs 3θ are given at the tp f the next page. Standard angles:. sin cs. sin 3 cs 3 3. sin 45 cs 45. sin 6 3 cs 6. sin 9 cs 9 sin3θ cs3θ ** There are 4 values f θ between and π where each trig functin has the same value, mdul a plus r minus sign (ie, the same abslute value); namely at θ, (π θ), (π + θ), and (π θ) ** Essential definitins: (b) Inverse trig functins tan x = sin x cs x sec x = cs x, csc x = sin x sin a, cs a, tan a are als called arcsin a, arccs a, arctan a. The functin arctan a is f special interest t us. It is used t slve the equatin tan θ = a The truble is that fr each a>there are tw different angles θ fr which tan θ = a, ne in the first quadrant and ne in the third quadrant. Fr example, tan θ =frθ = π 4 in the first quadrant, and als fr θ = 5π 4 in the third quadrant. Likewise, fr each a< there are tw angles, ne in the secnd quadrant and ne in the furth quadrant, fr which tan θ = a. If we need t slve tan θ = a fr θ, we wuld like t be able t write θ = arctan a Hwever, virtually every cmputer and calculatr will always return a first quadrant angle (in radians, f curse) if yu enter arctan a with a> and a furth quadrant angle (written as a negative angle) if yu enter arctan a with a<. Therefre, t slve tan θ = a, yu must knw which f the tw pssible quadrants the angle is in and then make a crrectin if the angle is in the secnd r third quadrants. Frtunately, the crrectin is easy t remember: just add π

4 T slve tan θ = a, these are the fur pssibilities: () tan θ = a, a > andθ is knwn t be in the first quadrant = θ = arctan a () tan θ = a, a < andθ is knwn t be in the secnd quadrant = θ = arctan a + π (3) tan θ = a, a > andθ is knwn t be in the third quadrant = θ = arctan a + π (4) tan θ = a, a < andθ is knwn t be in the furth quadrant = θ = arctan a Once again, in tabular frm, t slve: tan θ = a If a is and θ is in then θ is psitive first quadrant arctan a negative secnd quadrant (arctan a)+ π psitive third quadrant (arctan a)+ π negative furth quadrant arctan a Hw shuld ne remember this? Rather simple: if θ is in the secnd r third quadrants, just add π radians t the cmputer s evaluatin f arctan a. While this alne is sufficient t get a crrect answer, there are tw simplificatins smetimes uses. First f all, the functin arctan is an dd functin, i.e., arctan( b) = arctan(b) fr any real number b. This cmes int play especially in the secnd and furth quadrants, where the tangent is negative. S, fr example, if tan θ =.5, then θ can be written as θ = arctan(.5) if it is in the furth quadrant, r θ = π arctan(.5) if is is in the secnd quadrant. Ntice that we have fllwed exactly the rules abve, except we have pulled the minus sign ut f the arctangent functin. The secnd simplificatin refers nly t furth quadrant angles. Althugh the cmputer will give furth quadrant angles as negative angles when evaluating the arctangent functin, in sme applicatins it will be desirable t have θ always between and π radians. In this case, nly fr furth quadrant angles, ne will need t add π t btain the same angle written as a psitive angle. If yu have understd all f this, yu will realize that the slutin f tan θ =., when θ is knwn t be in the furth quadrant, can be written three ways: (i) θ = arctan(.); (ii) θ = arctan(.), which give exactly the same θ as the first way; (iii) θ =π + arctan(.5) = π arctan(.5), which gives θ as a psitive angle. (c) Identities () sin x +cs x = (True fr any x; fr example, sin e t +cs e t =) () sin(a ± B) =sina cs B ± cs A sin B (3) cs(a ± B) = cs A cs B sin A sin B Other identities can be derived easily frm these. Example: Divide () by cs x: tan x +=sec x (a) Angle Angle between x-axis and line segment frm (,) t (5,) measured in cunterclckwise directin. Ans: θ = tan ( 5 ) Angle between x-axis and the line segment frm (,) t (-5,-) measured in cunterclckwise directin. Ans: θ = π + tan ( 5 ), nt θ = tan ( 5 ) since θ is clearly in the third quadrant.

5 3. PARAMETRIC EQUATIONS: (Stewart, Sectin.) INTRO: In tw dimensins, a path (trajectry) in the x-y plane might be written in Cartesian (rectangular) crdinates, such as the parablic path{ y = x. Hwever, exactly the same parablic path can be written x = t in terms f a parameter, fr example, <t<. That is t say, rather than giving y as a y = t functin f x, bth x and y can be given in terms f a parameter, say t: x = f(t), y= g(t). We wish t study these parametric equatins. In particular, we need t be able t switch between parametric equatins and Cartesian equatins, t recgnize the graphs f parametric equatins, t determine if different parametric equatins have pints f intersectin, etc. These issues and thers will be discussed in this and the next Mdule. In this Mdule and the next, parametric equatins will always be in tw dimensins. Later in the curse, it will be necessary t study trajectries in three dimensins, ie, parametric equatins f the frm x = f(t), y= g(t), z = h(t). In bth cases, it is smetimes useful t think f the parameter t as representing the time, and the parametric equatins as giving the x and y crdinates (r x, y and z crdinates) f a particle fllwing the trajectry. Thus ne sees already a difference between Cartesian and parametric equatins. A Cartesian equatin such as y = x may give the trajectry f a particle, but nthing mre. A parametric equatin such as x = t, y = t gives the same trajectry. but als specifies exactly when the particle is at each pint alng the trajectry. Thus ne can cmpute its velcity, acceleratin, directin f travel, etc, nne f which wuld be knwn if nly the path y = x were given. Examples: { x =3t () y =6t <t<. (half-line frm rigin t infinity) { x =t () y = t <t<. (parabla) { x = t (3) y = t +3 t +4 t<. (half-parabla) { x =sin 4 t (4) y =sin t π <t< 3π. (sectin f parabla in y) What makes this smewhat cnfusing: the fllwing parametric equatins give the same curve as (): { x =4e t (5) y =8e t <t< Even thugh x is given by an expnential in (5), bth () and (5) are the same half-line. It is nt x as a functin f t which determines the trajectry given by parametric equatins, but the relatin f x and y t each ther, via the functin f t. In this case, y =x in bth curves. Of curse, there is a difference between () and (5), even thugh they give the same Cartesian equatin. If ne thinks f t as the time, then a particle fllwing the parametric equatins () is traveling in a straight

6 line at a cnstant speed, while the particle fllwing the parametric equatins (5) travels in the same straight line, but at an expnentially increasing speed. Issues such as speed and acceleratin will be cvered tward the end f this curse. Nte that parametric equatins have a sense f directin. Just evaluate at t =andt = (r any tw nearby values f t) t see the directin f travel. Hw t Find the Crrespnding Cartesian Equatin: Slve fr t (r slve fr a functin f t) and substitute: { x =t () y = t <t< Slve fr t. y = 4 x with <x<. { x = t (3) y = t +3 t +4 t< Slve fr t. y = 4 x + 3 x + 4 with x { x =sin 4 t (4) y =sin t π <t< 3π Slve fr sin t. x = 4 y with <t< { x =4e t (5) y =8e t <t< Slve fr e t. y =x with x> Hw t Graph Try easy values within range f t (estimates kay) { x =t 3 (6) y =4t 4t +.5 t.5 t x y y x.5

7 By cnsidering t =.5 andt =, it is bvius the curve is traveling in the directin tward the right.. QUIZ PREP. Slve an algebraic equatin.. Simplify a prduct f pwers. 3. Given the equatin f a circle r sphere, find the radius and center by cmpleting the square. 4. Find the angle between a line segment and the x-axis. 5. Given an angle, find its quadrant and which trig functins are psitive r negative. 6. Given parametric equatins, identify which is the crrect graph. 7. Given parametric equatins, cnvert them t Cartesian equatins. 8. Lcate the psitin f a particle with mtin specified by parametric equatins (a think-a-little prblem)