New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Maverick Koelpin


Maverick Koelpin
GPA 3.77

Charles Egedy

Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Charles Egedy
Study Guide
50 ?




Popular in Course

Popular in Mathematics (M)

This 11 page Study Guide was uploaded by Maverick Koelpin on Tuesday October 13, 2015. The Study Guide belongs to MATH 1552 at Louisiana State University taught by Charles Egedy in Fall. Since its upload, it has received 45 views. For similar materials see /class/222625/math-1552-louisiana-state-university in Mathematics (M) at Louisiana State University.


Reviews for AN GEOM & CALCULUS


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/13/15
Parameterization Study Guide Charlie Egedy April 3 2009 1 Parametric Equations A parameterization is a vector valued function 00 ltftgt htgti If ht 0 then we can talk about a parameterization of a path in R2 otherwise we have a pat in R3 0 ln R2 we can parameterize a line through the point a 12 having slope m by 00 lta t b mt 0gti o A standard way to parameterize a cirle in R2 having center at ab and radius R is 00 lta Rcos 9 b Rsin i9 0 thereby giving the circle a counterclockwise orientation 0 The cycloid in R2 has parameterization Ct Rt 7 Rsint R 7 Rcostgti 0 Given a parameterization 00 we can facilitate the construct of a graph by computing the fol lowing quantitiesi 7 Set the rst coorinate equal to zero to nd y intercepts set the second coorinate equal to zero to nd zinterceptsi e 2 i d 7 Compute derivatives using the equations and 271 it cit l i cit 7 Locate all horizontal and vertical tangents and inflection points 7 Consider extreme behavior of the graph by considering extreme behaviour of the parameter 7 Use all of this information as you would in rst calculus to assemble a graph 7 Mark the direction of parameterizationi If the parameter can be eliminated by solving any of the coordinate equations for t then we can produce an equation that shows the explicit dependence among the dependent variables 2 Arc Length and Speed 0 If 8 represents arc length then speed is de ned to be M 2 2i 0 We compute s by integrating between speci c values ofthe parameteri Thus 8 12 M 2 2dt We can produce an arc length function relative to some initial value of the parameter to by setting t1 t0 t2 t and then replacing every occurrence oft in the integrand by some dummy variable u 0 We can calculate a surface area of rotation about the zaxis by S 2 f yt 2 2dt 3 The Polar Plane 0 We invoke the parameterization lt7 cos 9 T sin 9 0gt and restrict our attention to the rst two coor dinatesl The quantity T will be some function of 9 To convert between the polar plane 7 Given polar coordinates compute the Cartesian coordinates using the given parameteriza tion 7 Given Cartesian coordinates compute a set of polar coordinates by using T 412 y2 and tan Make certain that your choice of T and 9 are consistent with the quadrant containing the point Having obtained this choice you can add 27m radians to the angle to obtain other choices having the same sign for T Other choices can be obtained by adding 7r radians to the angle and multiplying the T coordinate by 71 0 Given an equation in Cartesian coordinates we can obtain one in polar coordinates by a straight forward substitution according to the parameterizationl Given an equation in polar coordinates we can obtain one in Cartesian coordinates by letting T be replaced by xz2 y2 and then using the basic reference triangle to obtain the trig functions of 9 in terms of z and y 0 Some standard polar graphs are 7 The spiral T 9 7 The cirle T k 7 The line 9 k 7 the circles T acost and T asint The roses T sinnz and T cosnz having n petals if n is odd and 2n petals if n is even The limacons T acos t9 7 b and T asint 7 12 Note that the cartioid is a special case with a 12 o The areas swept in a counterclockwise direction between a polar curve and the origin is given by A g ff Ma 0 Arc length of a curve in polar coordinates is given by s f 7 2 2d9l 4 Conic Sections We give the standard form for the conic sectionsl You should be able to locate intercepts and assymp totes as appropriate You should also be able to graph a conic section that has been shifted out of standard positionl Here are the standard forms in standard position 0 The ellipse 2 l specializing to the cirlce when a 12 You should be able to locate intercepts and foci recalling that the pythagorean relationship equating the square of the larger of a and b to the sum of the squares of the other one and cl 0 The hyperbolas 7 2 l and 2 2 ll You should be able to locate the intercepts foci using the relationship 02 a2 122 and symptotes the lines y o The parabolas y having focus 00 and directrix y 7c and z 7 with focus 50 and directrix z 7o Math 1552 Sequences and Series Summary Sheet Sequences Series We assume that for each sequence annk there is a function fX such that fn an k S n lt 00 The sequence annk converges if and only if the function fX has a limit as X approaches 00 A sequence is called monotonically increasing if an S arm and it is called monotonically decreasing if an 2 arm A sequence is bounded above if there is a number M such that an S M for all n A sequence is bounded below if there is a number L such that L S an for all n A sequence that is bounded both above and below is said to be bounded Theorem If a sequence is monotonic and bounded then it converges Notes on the theorem 1 Being monotonic alone is not sufficient as n does not converge 2 Being bounded alone is not sufficient as sin n does not converge 3 A sequence can converge without being monotonic as for example lquotn 4 A sequence that converges must be bounded Ifthe sequence I anl converges to zero then so does an General We associate with each series S Znk an an underlying sequence an and function fX such that fn an We associate with each series S Znk an a sequence Sn of partial sums given by Sn Zikquotai When we say that the series S converges we mean that the sequence of partial sums converges The harmonic series S Zn1 ln diverges The geometric series S Zn0 arn converges if lrl lt 1 If it converges then it converges to al 7 r If lrl Z 1 then the series diverges We can write the series S Zkaan in an alternate way by writing S Znkbm39 anb The series S Znk ln an b converges as long as k is chosen so that the denominator is never zero Use pfd to find the limit If Earl and 2b both coverge then we can conclude that l ann c 2an 2 2an b 2an 2b 3 2an b 2an 2b The Integral Test 0 The series S Znk an converges whenever the integral lkwfxdx converges and diverges whenever this integral diverges 0 When S converges we can estimate its limit by using the n3911 partial sum The remainder Rn can be estimated by comparing it to two integrals Thus ln1 fxdx S Rn S Inwfxdx Since S Sn R we can write Sn In1 fxdx S s 3 3n Inwfxdx o The pseries S Znk on39p converges when p gt 1 and diverges otherwise The Comparison Tests 0 Comparison Suppose Earl and 2bn are series with positive terms with an S bn for all n larger than some value M 1 If 2bn converges then so does 2a 2 If 2an diverges then so does 2b 0 Limit Comparison Suppose Earl and 2bn are series with positive terms and lim anb c new where c is a real number not equal to zero Then both series converge or the y both diverge The Alternating Series Test and Series with negative terms 0 The series S Znk lquotbn where all ofthe bn s are positive is called an alternating series 0 If S Znk Olquotbn is an alternating series with these two properties 1 bn1 S bn for all n larger than some number M ie if it eventually decreases 2 lim bn 0 new Then the series converges o If either condition fails that is the limit is not zero or there is no large number M for which the sequence bu eventually decreases then the series diverges o If S is a convergent alternating series and SH is the n3911 partial sum then the remainder Rn is given by I Rn I S bn1 that is just the next term 0 A series S Znk oan is called absolutely convergent if the series S Znk anl converges If a series is absolutely convergent then it is convergent If a series is convergent but not absolutely convergent then we say that it is conditionally convergent In particular alternating series can be conditionally convergent Ratio and Root Tests Iflim I an anI L where L lt 1 then the series S Znk an is new absolutely convergent Iflim I an anI L where L gt 1 or ifthe limit is 00 then the series S new Znk oan is divergent Note that a result of L l is inconclusive Iflim I an I 1quot L lt 1 then the series S Znk oan is absolutely I1 00 convergent Iflim I an I 1quot L gt1 then the series S Znk oan is divergent Note that a result of L l is inconclusive Techniques of Integration Study Guide Charlie Egedy April 3 2009 Make certain that you know the basic integrals those listed as 116 in Rogawskils Table of Integrals back part of the book along with the integral fsec3zdz secz tanz lnl secz tanzli You will nd integral 16 to particularly useful in solving partial fraction decomposition problems with minimal effort 1 Substitution Think of substitution as reversing the chain rule for differentiation Find a substitution that simpli es the rule for a function by replacing a repeated element or a particularly complicated element by a single variable Compute the differential of the new variable in terms of the old viariablei If u 91 then du ltgt dzi If the integral is a de nite integral compute limits of integration in terms of the new variable Make the substitution noting that every occurrence of the old variable must be replaced Either complete the process of integration or employ another technique as necessary Integration by Parts Think of integration by parts as reversing the product rule for differentiation The basic theory states that fudv uv 7 fvdui The entire integrand must be incorporated into u and vi Keep in mind the maxim Do No Harml In general you want the processes of differentiation and integration to remove dif culties without introducing new ones If the integrand is the product of a polynomial and either a trig function or an exponential then set u equal to the polynomial and d1 equal to the remainder of the integrand If the integrand is the product of an exponential and a trig function then in general it does not matter how you make the assignment however if you set u equal to the trig function and another round of integration by parts is required then in the second round u must again be set equal to the trig function We make the analogous requirement if u is initially chosen to be the exponential If the integrand is the product of a polynomial and a logarithm then set u equal to the logarithmi 3 Trig Integrals 0 Consider rst fsinmz cos zdz 7 If either m or n is odd then let the corresponding trig function become part of the differential if m is odd then the substitution is u cosz and if n is odd then the substitution is u sinzl 10052x 2 7 If both are even then use the identities cos2 and sin2z 0 When arguments differ we have the following identities 7 f sin ax cos bzdz fsina 7 101 sina bzdz 7 f sin ax sin bzdz fcosa 7 101 7 cosa bzdz 7 fcos ax cos bzdz fcosa 7 101 cosa bzdz 0 Consider ftanmz sec zdz 7 lfm is odd then let u seczl 7 If n is even then let u tanzl 7 If m is even and n is odd then convert all tangents to secants via the identity tan2z 2 sec 7 ll 7 Solve f tanm zdz via reduction converting to f tanm 1 sec2 17lldzl Repeat as needed 7 Solve fsecm zdz using integration by parts setting d1 sec2zdz 4 Trigonometric Substitution o If the form a212 122 appears make the substitution ax btanzl o If the form a212 7 122 appears make the substitution ax bseczl o If the form 122 7 a212 appears make the substitution ax bsinzl 5 Partial Fraction Decomposition This technique applies to the integration of rational polynomial integrands o If the numerator is of equal or higher degree than the denominator begin with long division 0 Factor the denominator into products of linear and irreducible quadratic factorsl o For each factor in the denominator of the form I am postulate a term of the form A for I 212Hlml T a o If there are irreducible quadratic factors put them into completed square form b2 16 For each factor of the form zb2 klm postulate a term of the form W for i l 2 i i ml Generate a polynomial equation that equates the numerator of the integrand to the numerator of the postulated decomposition after recombining terms to form a single fraction 0 Evaluate the postulated coef cients using one or both of the following facts 7 If two polynomials are equal then coe cients of terms having the same degree must be equall 7 If two polynomials are equal then their evaluations must be equal for every real number 1 Vector Calculus Study Guide Charlie Egedy April 7 2009 Vectors in R3 Note that to get de nitions and theorems for R2 simply make the third coordinate zero since this will restrict attention to the Xy coordinate plane in i If abc is a point in R3 then ltabcgt is a position vector that is one that begins at the origin and ends at the point with orientation toward the point Given two points we can de ne a vector whose coordinates are computed as the differences between corresponding coordinates Orientation is away from the point whose coordinates were subtracte We de ne vector addition by adding corresponding coordinate and scalar multiplication by mul tiplying each coordinate by the same scalar Thus lta b c 7 lt0 6 lta rd 1 7 6 5 Vector addition has the geometric interpretation of constructing the diagonal of a parallelogram de ned by the two vectors oriented outward from the common point of origin The difference between two vectors is the other diagonal with orientation so that the convention on vector addition is maintained Vector addition is commutative and associative there is a zero vector and each vector has an additive inversei Additionally scalar multiplication is distributive over vector addition HE lta1a2a3gt then HEM a a a i Given any vector other than the zero vector we can de ne a unit vector having the same orien tation as the given vectori Thus given 17 the corresponding unit vector is 617 v 7 W Sometimes it will be useful to use the notation lt1 0 0 lt0 l0gt and I 001 Triangle inequality always applies H17 13H S We orient R3 according to the right hand ruler An equation for a line can be written as a vector valued function of position vectors that are oriented toward points on the line by writing Ht lt10 yo 20gt t ltv1 v2 v3 The rst vector is a position vector that hits some point on the line and the second vector gives the line its orientation The parametric form of the equation for a line comes from writing the coordinates explicitly Thus zzotv1 yy0tv2 andz20tv3i 710 7 7 v1 v2 If any of the coordinates for the orienting vector is zero this means that the cooresponding coordinate does not change with t so we leave that variable out of the string of equalities writing the constant that that coordinate is equal to apart from the equation 7 yiyo 7 The symmetric form for a line comes by equating t in the above equations Thus Given points P and Q we can parameterize a line by writing Ht l 7 t O tObi The point midway between P and Q then occurs when t We can write the equation of a sphere having radius R centered at abc by I 7 a2 y i b2 2 7 c2 R2 We de ne dot product between two vectors as the sum of corresponding productsi Thus viwi 1121112 Usws 1713 The dot product of a vector with itself is the square of its magnitude 1713 H lllll ll acute 1f negative the angle 1s obtusei lf zero the angle 1s r1ghti In the last case we say that vectors are perpendicular or orthogonali The angle between vectors is given by 3086 If the numerator is positive the angle is We can resolve one vector in terms of another by rst computing the magnitude of the parallel resolution and then multiplying this by a unit vector in the desired direction Thus comply v is the component and the projection itself is projlg17 em as the parallel component of 17in the direction of 13 We compute the perpendicular component of the resolution by subtracting the parallel component from the original vector being resolved The cross product of two vectors is a vector that is perpendicular to the two given vectors We i j de ne the cross product as the determinant of a matrix Thus 13 X 17 ul ug ug v1 v2 v3 The magnitude of the cross product is the product of the magnitudes of the two vectors times the sine of the angle between themi The orientation of the cross product is given by the right hand ruler Cross product is anticommutative that is swapping the order induces a minus sign The cross product of a vector with itself is the zero vector and cross product distributes over addition The area of a parallelogram is the magnitude of the cross product of the two vectors that de ne it The volume of a parallelpiped is the triple product of the three vectors that de nes it The equation az by CZ d is a plane with normal lta 120 containing the point 10 y020 where are bye 020 di Angles between planes are de ned to be the angles between their normalsi Parallel planes have parallel normals To nd the point of intersection between a line and a plane plug the parametric equations of the line into the line into the equation of the plane and solve for t To nd the point of intersection of two lines give them different parameter names and then equate the corresponding coordinates solving for the two parameters If no solution exists then the lines are parallel with parallel orienting vectors or they are skewi The distance between two planes is the projection of a vector connecting points in each plane with a common normal for the two planes The distance between skew lines is the projection of a vector connecting points in each line with a normal de ned to be the cross product of the vectors orienting the two lines To nd the equation of a line of intersection between two planes rst compute the orienting vector as the cross product of the normals for the two planes Then nd a point common to the two planes by solving the equations of the plane simultaneously choosing any point of convenience Given three points the equation of the plane containing them can be found by rst de ning two vectors oriented in the plane as differences between point coordinates computing the cross product of these two vectors to get the planes normal then computing d using one of the three points 2 There are many variations on the theme contained in Rogawskils problem set 125 Work them alli Quadric surfaces for equations with no cross terms 7 If two variables are missing then the surface is either a plane or two planes 7 If one variable is missing then the surface is a cylinder parabolic elliptic circular or hyperbolic depending upon the conic section depicted by the equation if it were in i 7 If no variables are missing 96 If none are squared then the surface is plane 96 If one is squared then the surface is a parabolic cylinder with rulings that are not parallel to any coordinate axis 96 If two are squared then the surface is either a paraboloid with either circular or elliptical cross sections or it is a hyperbolic parabo oi i 96 If three are squared and the constant on the RHS is positive then the surface is An ellipsoid if all signs are positive It is a sphere if all coef cients are equal a prolate spheriod if two are equal representing minor axes or an oblate spheroid if two are equal representing major axes A onesheet paraboloid if only one sign is negative A two sheet paraboloid if two signs are negative 96 If the RHS is zero then it is a point if all three signs are positive and a cone if one or two of the signs are negative Vector Calculus A vector valued function is any vector that depends upon a parameter A line is a vector valued function as is the helix Ht ltcost sinttgti We take limits or derivatives or integrals of vector valued functions by taking llimits or derivatives or integrals on a coordinate basis We compute speed of a vector valued function as the magnitude of the derivative of the vector valued function The arc length is the integral of the speed In certain cases it is convenient to reparameterize based upon arc length This will happen when the arc length is an invertible function of the original parameter The unit tangent vector is the unit vector in the direction of the derivative of the vector valued function Curvature is the derivative of the unit tangent vector with respect to arc length It can also be calculated by computing the magnitude of the derivative of the unit tangent vector and dividing by the speed function The radius of curvature is the multiplicative inverse of the curvature We can also compute curvature by knowing both the rst and second derivatives of the vector valued function We divide the magnitude of the cross product of these two vectors by the cube of the magnitude of the derivative of the vector valued function The unit normal is the derivative of the unit tangent vector normalized The binormal is the cross product of the unit tangent vector and the unit normal Given a vector valued fucntion 7 The velocity vector is the derivative of the vector valued function lts magnitude is speed 7 The acceleration vector is the derivative of the velocity vector 7 The acceleration vector can be resolved along the unit tangent and unit normal vectorsi 3 Partial Derivatives o A partial derivative is a derivative of a function having more than one independent variable7 taken by treating all but one of the variables as constants and then taking the derivative With respect to the nonconstant variable 0 If derivatives are continuous7 then mixed partials can be computed by taking derivatives in any order


Buy Material

Are you sure you want to buy this material for

50 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Anthony Lee UC Santa Barbara

"I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.