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# 63 Class Note for MATH 141 at PSU

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Date Created: 02/06/15

Math 141 Calculus with Analytic Geometry II Fall 2006 Penn State University Section 127 167 187 19 A Brief Review of Algebra and Trigonometry The following is a list of questions from the realms of Algebra and Trigonometry that tend to come up in introductory Calculus courses Several of these questions are in fact taken from the Readiness Quiz77 given on the rst day of class Other questions are a compilation of mistakes seen time and time again on homework and exams All of them7 however7 are questions regarding algebra andor trigonometry that the successful calculus student should be capable of answering 1 A Quick Quiz Determine whether each of the following statements is True or False Be careful Try to think of speci c examples before answering with your rst instinct li 5i 1y212y2 9ilfazltaythenzltyi 13ioo Z 6iITyE 10ilfa1aythenzyi l4igl 3i 1 g 7 3x 3x 11 ag 4 g 8 wsinz 12 314159 2 Simplify 15 Simplify h where 312 4 l6 Simplify w where Fz 13 17 Simplify where 7i 5 142 3 18 Simplify m 7 t264 7 t2 12 by getting a common denominator 19 Simplify 2amp2 for n 2 l where N read N factorial represents the product of the rst N positive integers For examplelll2ll223123674ll23424i 3 Inequalities 20 For what values of z is 11 16 positive negative 21 For what values of z is 12 31 03621771 positive 22 Find all values of I such that 31 7 2 2 5 23 Find all values of I such that lt 1 24 Find all values of t such that m 7 t2 64 7 t2 12 is negative 25 Suppose 7l S I S 1 Find upper and lower bounds for 12 and 13 4 Completing the Square and the Quadratic Formula We can solve any quadratic equation of the form 0412 121 c 0 by completing the square The result is known as the Quadratic Formula One application of the quadratic formula is to factor quadratic polynomials In particular 712 V122 7 4acgt I 7 7127 V122 7 4acgt 2a a12bzcaltz7 20 26 Find values A and B such that 12 7 81 7 l 1 142 B by completing the square 27 Find all values of I such that 12 7 51 7 36 0 28 Factor 213 512 7 31 29 Factor 2 cos3 9 5 cos2 9 7 3cos 9 5 The Pythagorean Theorem and the Equation of a Circle ln pictures7 the Pythagorean Theorem says that the area of the green square below is equal to the sum of the areas of the red and blue squaresi If the points P and Q are thought of as the end points of the hypotenuse of a right triangle7 then the distance between these points can be determined using the Pythagorean Theoremi am A circle can be described as the set of points zy that are a xed distance 7 from a central point abi The equation for a circle can thus be derived using the distance formula from above 30 Determine the value of z in the triangle below 31 Determine the value of z in the triangle below H H 12 12 32 Find the distance between the points 714 and 5 2 33 Determine the radius and the center of the circle described by the equation 12101y2 712y 3 Hint Complete the square twice 6 Degrees vs Radians There are two common units for measuring angles degrees and radiansi Most people are familiar with the fact that there are 360 degrees in a circle But why 360 Why not 100 degrees The answer can be traced back to several ancient civilizations The Babylonians used a base60 number system as opposed to our base10 number systemi While the exact reason for the base60 number system is a mystery it does have its advantages over our seemingly more natural number system of today Any multiple of 60 can be divided by 1 2 3 4 5 6 10 12 15 20 and 30 No number smaller than 60 has as many proper divisorsi In particular a circle divided into 360 equal parts can easily be divided into halves thirds fourths fths sixths eighths ninths tenths etc Try to do that with a circle divided into 100 equal parts Another civilization the Sumerians used a 360 day calendar to chart the progress of the Suns circular path across the sky While our base10 number system is prevalent today the base60 system does in fact still live on Positions on the globe are still measured in degrees of longitude and latitude Each hour is divided into 60 minutes each minute is divided into 60 seconds Even though the use of both number systems can be easily justi ed they will always have a sense of arbitrariness to themi While the concept of a radian is less widely known it is in fact a much more useful and more logical choice for measuring anglesi Many identities involving trigonometric functions would lose their luster if they were expressed in degrees or any unit other than radiansi The de nition of a radian is based on the following fundamental fact The ratio of the circumference of a circle to its diameter is constant In other words if you take my circle and calculate the circumference of that circle divided by its diameter you always get the same number no matter what circle you started with This constant is denoted by the greek letter 7r for perimeter and yields the familiar formula for the circumference C of a circle of radius 7 2 C 27ml The use of the symbol Tr to represent this number began in the 18th century however the number itself was known to the Babylonians and Egyptians over four thousand years ago In particular a circle of radius 1 has circumference of length 27f and if we mark off one unit length along the perimeter of this circle the angle formed is de ned to be a radiani This construction of a radian is illustrated belowi Thus by de nition there are 27f radians in a circle and 27T radians 360 degrees 34 Complete the following table converting degrees into radiansi degreesl0 15 30l45 60 90 120 135 150 180l210 225 240 270 300 315l330 360 radiansllllllgllllllllllllZW 35 Determine the value of each angle in radians of the triangles shown in Problems 30 and 31 7 The Unit Circle and Trigonometric Functions Consider the unit circle de ned by the equation 12 y2 1 In other words the unit circle is the set of points that is exactly 1 unit away from the origin Now draw a ray that originates from the origin 0 Let 9 be the angle in radians that the ray makes with the positive zaxis measured in the counterclockwise direction Let P represent the point of intersection of the ray and the unit circle The graphs of cost9 and sint9 for 72 g 9 g 27f are shown belowl 7 7W7 711 W 2w Figure 1 Graph of y cost9 l 039 1 l w s l l M NH 1 w s 1 Figure 2 Graph of y sint9 An alternate de nition for the cosine and sine functions is based on a right triangle For 0 lt 9 lt 7r2 consider the triangle shown below A l a where a is the length of the leg adjacent to 9 0 is the length of the leg opposite t9 and h is the length of the hypotenuse The remaining trigonometric functions secant7 cosecant7 tangent and cotangent are de ned in terms of cosine and sine as follows sec E sint9 0 t9 00816 Z tant9 0086 csc 7 7 cost9 7 a 9 1119 0 COM 51119 3 36 Use your answers from Questions 30 31 and 35 to compute the following values exactly a cos7r3 c cos7r4 e cos7r6 b sin7r3 d sin7r4 f sin7r6 37 Use your answers from the previous problem to label the points on the unit circle below by its 1 and y coordinates Each line is labeled by the angle in radians it makes with the positive zaxisi 071 1 38 Compute cosz given that tanz 2 and Tr lt z lt 37r2i 39 Compute the area of the triangle with two sides of length 4 cm and 9 cm resp and an included angle of 7r6i 40 Compute sec 77r6i 41 Draw a graph of y cosz3i 42 Find all the values of 9 such that 1 7 2 cos2 39 0 8 Trigonometric Identities The following is a list of several trigonometric identities that calculus students should be at least familiar with if they have not already committed them to memory This list is by no means complete7 however7 all other trigonometric identities can be derived from this collection For a complete list of identities7 see Appendix D page A28 in Stewart7s Calculus 43 Show cost9 7r2 7 sint9 45 Show sina 7 sina cos3 7 cosa sin 44 Show sint9 7r4 sint9 cos t92 46 Show cos 29 1 7 2sin2 t9 47 Compute sin7rl2i 48 9 49 Show sin 29 2 sin9 cos9 Find all angles 9 where cos 29 sin9 50 Find all angles 9 where sin 29 cos9 Solutions Each statement presented in Questions 1 14 is false 1 An easy way to see that this is false is to convert each fraction into it s decimal formi The left hand side becomes 05 03 08g while the right hand side becomes 02 Clearly these two values are not the same What Is True 4 le OlH 4 OHIO OJlC OJlCAD This is simply a generalization of the previous question While there are speci c cases where this identity is true eg I 0 this identity in general is false A more simple example of why this is false is obtained by setting I y 2 ll What Is True Remember in order to add two fractions you must rst nd a common denominator and then and only then may you add numerators That is I 1712 my 2 2 E zzzy 7 71zy 7 i This mistake can be explained by confusing multiplication with addition In particular any common factors that appear in the numerator and denominator of a fraction can be cancelled out Nothing can done with common terms that appear in numerator and denominatori What Is True 1392 2 i We can easily verify that this is false by considering the case when I ll Note that the I l in the numerator is a factor of only one of the terms in the numerator Therefore it cannot be canceled out by the I l in the denominatori What Is True zlzi6 7 12zi6 1lzi2 7 1lzi2 7 I 3z 7 2 7 I lz 7 2 7 13 7 11 5 This common mistake can be chalked up to wishful thinking Exponentiation does not have a distributive property like multiplicationi It certainly would be nice if it did but we can easily see that it does not Again try a simple example like I y l to see that this is not the case 10 What Is True Using the distributive property of multiplication we have Iy2IyIy IIyyry 12zyyzy2 1221yy2 Another example of wishful thinking If we write the statement as I y12 112 y12 we see that this is exactly the same type of error as in the previous question Again exponentiation does not have a distributive property like multiplication A simple example like I y 1 should convince you that the above statement is false What Is True Like in the previous problem there is a way to express xz y as a sumi However in this case it would require an in nite number of terms to do so Exactly how this is done is one of the main topics covered in Math 141 Remember anything in parentheses must be evaluated rst In other words the quantity 31 is being raised to the nth power on the left hand side Again letting n 0 and z 1 should convince you that the statement is false What Is True 31 31 31 31 x ntimes 333zzz WW ntimes 371171 n times Rememeber sin21 denotes the sine of 21 The sine function must be applied to the quantity 21 before dividing by 2 Consider the case when I 7r2 to see that this statement is indeed false What Is True The only simpli cation that can be done here would be to apply a trigonometric identity For example sin21 7 2sinz cosz 2 7 2 sinz cos i This one is a bit tricky The tendency is to simply divide both sides by a But remember if a lt 0 then you must switch the inequality What Is True The following statements are true ifagt0andazltaythenzltyi ifalt0andazltaythenzgtyi Another tricky one This statement is true a is nonzero but if a 0 then I and y need not be the same Again this common error arises when you try to divide both sides of am ay by a Remember division by zero is not de ned Whenever you divide both sides of an equations by something you must make sure that you are not dividing by zero What Is True The following statement is true ifaa Oandazaythenzyi 11 12 13 14 15 16 One more tricky one This statement is certainly true if z is nonnegative that is greater than or equal to zero However it is not true if z is negative Try 1 71 What Is True For all real numbers I VIT Il where denotes the absolute value of I We also have that z with the understanding that the left hand side is only de ned when I 2 0 since you cannot take the square root of a negative number This is simply an approximation of 7r which is an irrational number This means that 7r cannot be written as a ratio of integers What Is True 7T7T This may seem like a trivial statement but the point is that 7r is a number Leaving an answer with 7r in it is perfectly acceptable In fact its preferred For example if the answer to a problem is 7r37 then leave your answer as 7r37 since this is the exact answer Do not replace this answer with the approximation 00849079095565 or worse yet 3 1415937 0 084907Wi The ratio 10 is not de ned since division by zero is not allowed The 00 symbol is just that a symboli In nity is not a number it is a concept What Is True The following limits are true 1 lim 7 foo xHO 1 lim 7 00 xao I If the above line looks foreign to you then you are in the right class The concept of a limit is fundamental to the study of Calculus and will be covered in this course Again the ratio 00 is not de ned since division by zero is not allowed regardless of the value of the numerator What Is True There are certain limits that we will eventually refer to as indeterminate There are several types of indeterminate forms 00 being one of themi There are certainly examples of indeterminate forms of type 00 that are equal to 1 But not all of them are Some are equal to zero Some are equal to 00 In fact any number you choose is the value of some indeterminate form of type 00 This topic will be investigated in Math 141 The most common mistake here is to assume that h h or h fh both of which are very rarely the case Remember that the z in can be replaced by anything In this case we want to replace it by z it That means we must also replace 1 by z h in 312 4 That is fzh3zh24 31221hh24 3z26zh3h24 Fzh7Fz 7 1h3713 h 7 h 13312h31h2h3713 h 7 312h31h2h3 7 h 31231hh2 17 First7 note that 735n171 f 71 1 W 735n4 427171 Therefore fn1 7 735n4 421173 n 7 42n71 35n71 735n475n7142n7 s72n71 lt73gt544 35 7 7E 18 t2 M647 t2 7 t2 647 t2 12 7 M647 t2 7 7 M64 7 t2 7 M64 7 464 7 t2 t2 7 M647 t2 M64 7 t2 7 647 t2 7 mim 7 6472 7m 19 To get used to the notation7 it may help to try a few examples For n 17 we have 51712345 21 72345 2 7345 For n 2 we have 7 1234567 4x 4x 41567 4 567 For n 37 we have 91 123456789 a 6 61789 6 789 Now for the general case7 2n3 2n 2n12n22n3 2n 2n 2n 12n 22n 3 10 20 21 22 23 24 25 26 The quantity 11 16 is positive when I gt 0 and z 16 gt 0 ie I gt 0 or when I lt 0 and z 16 lt 0 ie I lt 716 The quantity 11 16 is negative when I gt 0 and z 16 lt 0 notice that there are no values of I such thatzgt0andz16lt0 orwhenzlt0andz16gt0 iiei 716ltzlt0i Notice that the quantity 1231 13x32z 7 1 is positive when 12 f 0 ie I f 0 and a 31 13 gt 0 and 321 71gt 0 or b 31 13 lt 0 and 321 71 lt 0 31 13 gt 0 when 311gt 0 which occurs for z gt 7131 321 71gt 0 when 21 71gt 0 which occurs for z gt 121 Therefore both of these conditions are met when I gt 12 31 13 lt 0 when 311lt 0 which occurs for z lt 713 321 71 lt 0 when 21 71 lt 0 which occurs for z lt 12 Therefore both of these conditions are met when I lt 713 Thus 61 13x32z 71 is positive when I gt 12 or z lt 713 31 7 2 2 5 when 31 7 2 2 5 or 31 7 2 S 751 The rst situation occurs when I 2 73 and the second situation occurs when I S 711 lt 1 when 71 lt 217375 lt 1 Multiplying each side by 3 yields 73 lt 21 7 5 lt 3 Adding 5 to each side yields 2 lt 21 lt 8 And nally dividing by 2 yields 1 lt z lt 4 Recall that in Problem 18 we determined that 64 7 t2 7 t264 7 t2 12 7 64 7 2t2 64 7 t2 Notice that the above function is de ned as long as 64 7 t2 gt 0 ie 78 lt t lt 8 since you cannot take the 521 is negative whenever 64 7 2t2 lt 0 ie t2 gt 32 since the square root of a negative number Also the ratio denominator is always positive for 78 lt t lt 8 Therefore 34 is negative when 78 lt t lt 7S2 and x 32 lt t lt 8 From the graphs below we can easily see that 0 S 12 S 1 and 71 S 13 S 1 if 71 S I S 1 1474 andB717i That is 1278171z742717 11 27 28 29 30 31 32 33 Using the Quadratic Formula we have that 12 7 51 7 36 0 when Six25144 I 2 isix169 2 75i13 2 Therefore 12 7131 36 0 when I 9 and z 74 213 512 7 3x 1212 51 7 3 211 712z 3 121 7 lz 3 If we let 213 512 7 31 then fcos t9 2 cos3 9 5 cos2 9 7 3cos 9 Using our answer to the previous problem we have 2cos3t95cos2t97 30086 cost92cost97 lcost93 Using the Pythagorean Theorem we have 12z2l 2121 35212 z 2 Using the Pythagorean Theorem we have 12212 1 121714 1234 z32 Using the distance formula we have that the distance between 71 4 and 5 2 is 51227423674 By completing the square we have 12101y2712y 152725y762736 Therefore 152725y7627363 or equivalently 152y76232536 64 which is the equation of a circle of radius 8 centered at the point 75 6 12 34 degrees l 0 l 15 l 30 l 45 l 60 l 90 radians l 0 l 7 l 70 l 300 l 315 l 330 l 360 12lll 3 ll lllll5 lllyl 35 A A I A I A 36 a cos7r3 12 c cos7r4 22 e cos7r6 2 b sine3 Z d sine4 m2 f sin7r6 12 37 To help remember these values7 notice that all of these numbers are of the form i 2 fori01 27341 7 7 071 lt 3 1 NH is 5 E W g E 710 7r 0 170 77f 117r 6 6 1 u 54 77 5 4 44 7 re eff 3 W 38 Since 2 tanz sin 1 cos I we must have sinz 2 cos 1 Using the identity cos2 zsin2 z 1 see Trigonometric Identities7 we have 7 071 7 g 1 cos2 I sin2 I cos2 I 4cos2 z 5cos21 AN Therefore COSQI 7 15 which implies that cosz i1 Since 7r lt z lt 37r27 we must conclude that cosz 7 39 The given triangle is drawn below 13 E4sin7r6 I 9 Therefore the area of the triangle is given by Base gtlt Height 7 9 X 4sin7r6 40 2 2 7 9 X 4 gtlt12 2 9 sec77r6 cos77r6 7 2 41 42 76W 7 2 73 J L 37r J 2 71 jr 2v 67f i Notice that 1 7 2cos2 39 0 When cos2 39 12 In turn7 this happens When cos3t9 i12i Therefore 39 2n 17r4 Where n is any integer In other words7 9 must be of the form 2n 17r12 for some integer n 43 0086 7r2 cost9 cos7r2 7 sint9 sin7r2 cost9 0 7 sint9 1 7 sint9 44 sint9 7r4 sint9 cos7r4 sin7r4 cost9 sint9 22 22 cost9 2sint9 cost92 45 sina 7 sina sina cos7 sin7 cosa sina cos 7 sin cosa 46 00826 0086 9 cost9 cost9 7 sint9 sint9 cos2 9 7 sin2 9 1 7 sin2 7 sin2 9 1 7 2 sin26 14 47 Using the identity from Problem 46 with 9 7rl2 we have cos7r6 17 2 sin27rl2 Solving for sin27rl2 yields sin2 Trl2 1 7 HEW6 2 ii 2 f 242 7 4 Taking the square root of both sides produces i 27 sin7rl2 2 But since 7rl2 is in the rst quadrant sin7rl2 must be positive and therefore 2 7 3 sin7rl2 48 Using the identity from Problem 46 we have 17 2sin2t9 sint9 or equivalently 02sin2 9sin 971 2 sint97 lsint9li Therefore either sint9 71 ie 9 37r2 27m where n is any integer or sint9 12 ie 9 7r6 27m or 9 57r6 27m where n is any integer 49 sin 29 sint9 9 sint900st9 sint900st9 2 sin 9 cos 9 50 Using the identity from the previous problem we have 2sint900st9 cost or equivalently 2sint900st9 7 3086 2 sint9 7 l 3086 0 Therefore either cost 0 ie 9 7r2 27m or 9 37r2 27m where n is any integer or sint9 12 ie 9 7r6 27m or 9 57r6 27m where n is any integer 15

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