Space Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.


 Theodora Singleton
 2 years ago
 Views:
Transcription
1 Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t) y = y(t) z = z(t) Given its prmetric equtions x = x(t), y = y(t), z = z(t), curve C cn be considered to be vector function, tht is function whose domin is in n intervl of rel numbers nd the rnge is set of vectors: r(t) = x(t), y(t), z(t). In this cse, the curve C is the grph of the vector function r(t). Any vlue t = t from the domin of r(t) corresponds to point (x, y, z ) on the curve C. The derivtive of vector function r = x(t), y(t), z(t) is the vector function r (t) = x (t), y (t), z (t) At point (x, y, z ) which corresponds to the vlue t of prmeter t, the vlue of the derivtive r (t ) = x (t ), y (t ), z (t ) represents the velocity vector of the tngent line t (x, y, z ). Note the nlogy with the twodimensionl scenrio. To find tngent line to the curve To find tngent line to the curve x = x(t), y = y(t) t t = t, use x = x(t), y = y(t), z = z(t) t t = t, use Point: (x(t ), y(t )), Point: (x(t ), y(t ), z(t )), Direction vector: x (t ), y (t ). Direction vector: x (t ), y (t ), z (t ). An eqution of the tngent line: An eqution of the tngent line: x = x(t ) + x (t )t x = x(t ) + x (t )t y = y(t ) + y (t )t y = y(t ) + y (t )t z = z(t ) + z (t )t 1
2 Recll tht the length L of prmetric curve x = x(t), y = y(t) with continuous derivtives on n intervl t b cn be obtined by integrting the length element ds from to b. L = The length element ds on sufficiently smll intervl cn be pproximted by the hypotenuse of tringle with sides dx nd dy nd so ds = dx + dy ds = dx + dy = (x (t)) + (y (t)) dt. Anlogously, the length element of spce curve stisfies ds = dx + dy + dz nd so ds = dx + dy + dz = (x (t)) + (y (t)) + (z (t)) dt. The expression (x (t)) + (y (t)) + (z (t)) represents the length of the derivtive vector r = x, y, z. Thus the length of spce curve on the intervl t b cn be found s L = ds = r (t) dt = (x (t)) + (y (t)) + (z (t)) dt. ds. Compre gin the two nd three dimensionl formuls. The length of the curve r = x(t), y(t) for t b is L = b (x (t)) + (y (t)) dt = r (t) dt The length of the curve r = x(t), y(t), z(t) for t b is L = b (x (t)) + (y (t)) + (z (t)) dt = r (t) dt Prctice Problems. 1. Describe the following curves. For those without prmetric representtion, find equtions of prmetric equtions. () The curve given by x = 1 + t, y = t, z = 1 + t. (b) The line segment from (1,, 4) to (3,, 1). (c) The curve given by x = cos t, y = sin t, z =. (d) The curve given by x = cos t, y = sin t, z = t. (e) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. (f) The tringle in the boundry of the prt of the plne 3x + y + z = 6 in the first octnt. (g) The boundry of the prt of the prboloid z = 4 x y in the first octnt.
3 . For the following curves, find n eqution of the tngent line t the point where t =. Find the normliztion of direction vector t t =. () The curve given by x = cos t, y = sin t, z = t. (b) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. Use the prmetriztion of this curve from problem 1 (e). 3. For the following curves, find the length for t π. Use the clcultor to evlute the integrl in prt (b). () The curve given by x = cos t, y = sin t, z = t. (b) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. Use the prmetriztion of this curve from problem 1 (e). 4. Consider the curve C which is the intersection of the surfces x + y = 9 nd z = 1 y. () Find the prmetric equtions tht represent the curve C. (b) Find the eqution of the tngent line to the curve C t point (, 3, 8). (c) Find the length of the curve from (3,, 1) to (, 3, 8). You cn use the clcultor to evlute the integrl tht you re going to get. 5. Consider the curve C which is the intersection of the surfces y + z = 16 nd x = 8 y z. () Find the prmetric equtions tht represent the curve C. (b) Find the eqution of the tngent line to the curve C t point ( 8, 4, ). (c) Find the length of the curve from (4,, 4) to ( 8, 4, ). Use the clcultor to evlute the integrl tht you re going to get. 6. Find the length of the boundry of the prt of the prboloid z = 4 x y in the first octnt. Solutions. 1. () This curve is line pssing the point (1,, 1) in the direction of 1,,. (b) Any vector coliner with 3,, 1 1,, 4 =,, 5 cn be used s direction vector of the line pssing two points. You cn lso use ny of (1,,4) nd (3,, 1) for point on the line. For exmple, using (1,,4) we obtin prmetric equtions x = 1 + t, y = tz = 4 + 5t. In this cse, the initil point corresponds to t = nd the end point to t = 1. So, the line segment hs prmetriztion x = 1 + t, y = tz = 4 + 5t with t 1. (c) The xyequtions x = cos t, y = sin t, represent circle of rdius 1 in xyplne. Thus, the curve is on the cylinder determined by this circle. The zeqution z = represents the horizontl plne pssing on the zxis. So, this curve is the intersection of the cylinder with 3
4 the horizontl plne: it is circle of rdius 1 centered on the zxis in the horizontl plne pssing z =. (d) The xyequtions x = cos t, y = sin t, represent circle of rdius 1 in xyplne. Thus, the curve is on the cylinder determined by this circle. The eqution z = t hs n effect tht zvlues increse s tvlues increse. Thus, this curve is helix spirling up the cylinder s t increses. Use Mtlb to get precise grph. (e) The curve is the intersection of cylinder with n inclined plne. So, the curve is n ellipse. The xyequtions represent cylinder bsed t the circle of rdius 1 in xyplne. Thus, x nd y cn be prmetrized s x = cos t, y = sin t. To get the z eqution, solve the plne eqution y + z = for z, get z = y nd substitute tht y = sin t. Thus z = sin t. This gives us n eqution of the ellipse to be x = cos t, y = sin t, z = sin t. You cn use Mtlb to get precise grph. 1 (c) Circle in horizontl plne 1 (d) Helix 1 (e) Ellipse (f) The tringle in the boundry of the prt of the plne 3x+y+z = 6 in the first octnt consists of the three line segments, ech of which will hve different set of prmetric equtions. The intersection in xyplne z = cn be obtined by plugging z = in 3x + y + z = 6 nd using x, for exmple, s prmeter. Thus, we hve prt of the line 3x + y = 6 y = 3 3 x between its two intercepts (,) nd (,3) nd so x x = x, y = 3 3 x, z = or x = t, y = 3 3 t, z = with t. Alterntively, the prmetric equtions of this line cn be obtined s equtions of line pssing x nd y intercepts of the plne 3x+y +z = 6, (,, ) nd (,3, ). Using (,3,) s direction vector nd (,3,) s point on the line, we obtin the equtions x = t, y = 3 + 3t, z = with 1 t. Similrly, you cn find equtions of the remining two sides of the tringle. The intersection of xzplne y = cn be obtined by plugging y = in 3x + y + z = 6 nd using x, for exmple, s prmeter. Thus, we hve 3x + z = 6 z = 6 3x nd so x = x, y =, z = 6 3x or x = t, y =, z = 6 3t with t. The intersection of yzplne x = cn be obtined by plugging x = in 3x + y + z = 6 nd using y, for exmple, s prmeter. Thus, we hve y + z = 6 z = 6 y nd so x =, y = y, z = 6 y or x =, y = t, z = 6 t with t 3. (g) The boundry of the prt of the prboloid z = 4 x y in the first octnt consists of three curves, ech of which will hve different set of prmetric equtions. The prmetriztions 4
5 cn be obtined by considering intersections with three coordinte plnes x =, y =, nd z = respectively. The intersection in xyplne z = is circle = 4 x y x +y = 4 which hs prmetric equtions x = cos t, y = sin t. Since we re considering just the prt with x nd y, we hve tht t π. Thus, this curve hs prmetric equtions x = cos t, y = sin t, z = with t π. The intersection in xzplne y = is prbol z = 4 x with x. Using x s prmeter produces prmetric equtions x = x, y =, z = 4 x or x = t, y =, z = 4 t with t. The intersection in yzplne x = is prbol z = 4 y with y. Using y s prmeter produces prmetric equtions x =, y = y, z = 4 y or x =, y = t, z = 4 t with t.. () To find point on the tngent, plug t = in the prmetric equtions of the curve. Get x = 1, y =, z =. To get direction vector, plug t = in the derivtive x = sin t, y = cos t, z = 1. Get, 1, 1. So, the eqution of the tngent is x = 1 + t, y = + 1t, z = + 1t x = 1, y = t, z = t. The direction vector, 1, 1 hs length so its normliztion is, 1, 1. (b) Use the prmetric equtions from problem 1 (d) x = cos t, y = sin t, z = sin t. To find point on the tngent, plug t = in the prmetric equtions. Get x = 1, y =, z =. To get direction vector, plug t = in the derivtive x = sin t, y = cos t, z = cos t. Get, 1, 1. So, the eqution of the tngent is x = 1 + t, y = + 1t, z = 1t x = 1, y = t, z = t. The direction vector, 1, 1 hs length so its normliztion is, 1, () L = π/ (x (t)) + (y (t)) + (z (t)) dt = π/ π/ π/ 1 + 1dt = dt = π. (b) L = π/ the clcultor, L (x (t)) + (y (t)) + (z (t)) dt = π/ ( sin t) + (cos t) + (1) dt = ( sin t) + (cos t) + ( cos t) dt. Using 4. () You cn prmetrize the cylinder x + y = 9 by x = 3 cos t nd y = 3 sin t. From the eqution z = 1 y, you obtin tht z = 1 (3 sin t) = 1 9 sin t. (b) To find direction vector, we need to plug tvlue tht corresponds to the point (, 3, 8) into the derivtive x = 3 sin t, y = 3 cos t, z = 18 sin t cos t. To find this tvlue, set x =, y = 3 nd z = 8 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the first eqution x = 3 cos t = t = ± π. From the second y = 3 sin t = 3 t = π. The vlue t = π stisfies the third eqution z = 1 9 sin π = 1 9 = 8. Thus, t = π. Plugging this vlue in the derivtives produces the direction vector 3,,. So, the tngent line is x = 3t, y = 3 + t, z = 8 + t x = 3t, y = 3, z = 8. 5
6 (c) From prt (b), we hve tht t = π corresponds to the point (, 3, 8). Thus, π is the upper bound. To find the lower bound, determine the tvlue tht corresponds to (3,, 1). Set x = 3, y = nd z = 1 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the first eqution x = 3 cos t = 3 cos t = 1 t =. From the second y = 3 sin t = sin t = t =. The vlue t = stisfies the third eqution z = 1 9 sin = 1 = 1. So, the bounds of integrtion re to π. The length is L = π/ ( 3 sin t) + (3 cos t) + ( 18 sin t cos t) dt = () You cn prmetrize the cylinder y + z = 16 by y = 4 cos t nd z = 4 sin t. From the eqution x = 8 y z, you obtin tht x = 8 (4 cos t) 4 sin t = 8 16 cos t 4 sin t. (b) To find direction vector, we need to plug tvlue tht corresponds to the point ( 8, 4, ) into the derivtive x = 3 cos t sin t 4 cos t, y = 4 sin t, z = 4 cos t. To find this tvlue, set x = 8, y = 4 nd z = nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the second eqution y = 4 cos t = 4 cos t = 1 t = π. From the third z = 4 sin t = sin t = t = nd t = π. The vlue t = π grees with the tvlue we obtined using the yeqution. Plugging this vlue in the xeqution gives you x = 8 16 cos π 4 sin π = 8 16 = 8 which grees with the xcoordinte of ( 8, 4, ). Thus, t = π. Plugging this vlue in the derivtives produces the direction vector 4,, 4. So, the tngent line is x = 8 + 4t, y = 4 + t, z = 4t x = 8 + 4t, y = 4, z = 4t. (c) From prt (b), we hve tht t = π corresponds to the point ( 8, 4, ). Thus, π is the upper bound. To find the lower bound, determine the tvlue tht corresponds to (4,, 4). Set x = 4, y = nd z = 4 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the second eqution y = 4 cos t = cos t = t = ± π. From the third z = 4 sin t = 4 sin t = 1 t = π. So the vlue π obtined from the yeqution cn be discrded nd we obtin tht t = π. Plugging this vlue in the xeqution gives you x = 8 16 cos π 4 sin π = 8 4 = 4 which grees with the xcoordinte of (4,, 4). Thus, the lower bound is t = π. The length is L = π π/ (3 cos t sin t 4 cos t) + ( 4 sin t) + (4 cos t) dt = Recll tht we found prmetric equtions of the three curves in the intersection to be x = cos t, y = sin t, z = with t π, x = t, y =, z = 4 t with t, nd x =, y = t, z = 4 t with t. The three derivtive vectors nd length elements re x = sin t, y = cos t, z = x = 1, y =, z = t x =, y = 1, z = t ds = 4 sin t + 4 cos tdt = 4dt = dt ds = 1 + 4t dt, nd ds = 1 + 4t dt. The totl length cn be clculted s sum of the three integrls below. Using the clcultor for the second two produces π/ dt t dt t dt = π
Final Exam Solutions, MAC 3474 Calculus 3 Honors, Fall 2018
Finl xm olutions, MA 3474 lculus 3 Honors, Fll 28. Find the re of the prt of the sddle surfce z xy/ tht lies inside the cylinder x 2 + y 2 2 in the first positive) octnt; is positive constnt. olution:
More informationUniversity of. d Class. 3 st Lecture. 2 nd
University of Technology Electromechnicl Deprtment Energy Brnch Advnced Mthemtics Line Integrl nd d lss st Lecture nd Advnce Mthemtic Line Integrl lss Electromechnicl Engineer y Dr.Eng.Muhmmd.A.R.Yss Dr.Eng
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationJim Lambers MAT 280 Spring Semester Lecture 17 Notes. These notes correspond to Section 13.2 in Stewart and Section 7.2 in Marsden and Tromba.
Jim Lmbers MAT 28 Spring Semester 29 Lecture 7 Notes These notes correspond to Section 3.2 in Stewrt nd Section 7.2 in Mrsden nd Tromb. Line Integrls Recll from singlevrible clclus tht if constnt force
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 9 Prmetric Equtions nd Polr Coordintes Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition Section 91 Prmetric Curves
More informationCurves. Differential Geometry Lia Vas
Differentil Geometry Li Vs Curves Differentil Geometry Introduction. Differentil geometry is mthemticl discipline tht uses methods of multivrible clculus nd liner lgebr to study problems in geometry. In
More informationCHAPTER 10 PARAMETRIC, VECTOR, AND POLAR FUNCTIONS. dy dx
CHAPTER 0 PARAMETRIC, VECTOR, AND POLAR FUNCTIONS 0.. PARAMETRIC FUNCTIONS A) Recll tht for prmetric equtions,. B) If the equtions x f(t), nd y g(t) define y s twicedifferentile function of x, then t
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationk ) and directrix x = h p is A focal chord is a line segment which passes through the focus of a parabola and has endpoints on the parabola.
Stndrd Eqution of Prol with vertex ( h, k ) nd directrix y = k p is ( x h) p ( y k ) = 4. Verticl xis of symmetry Stndrd Eqution of Prol with vertex ( h, k ) nd directrix x = h p is ( y k ) p( x h) = 4.
More informationUS01CMTH02 UNIT Curvature
Stu mteril of BSc(Semester  I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT 1 Curvture Let f : I R be sufficiently
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More informationLecture XVII. Vector functions, vector and scalar fields Definition 1 A vectorvalued function is a map associating vectors to real numbers, that is
Lecture XVII Abstrct We introduce the concepts of vector functions, sclr nd vector fields nd stress their relevnce in pplied sciences. We study curves in threedimensionl Eucliden spce nd introduce the
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationR(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of
Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationdt. However, we might also be curious about dy
Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For
More information(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence
Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll
More informationReview: Velocity: v( t) r '( t) speed = v( t) Initial speed v, initial height h, launching angle : 1 Projectile motion: r( ) j v r
13.3 Arc Length Review: curve in spce: r t f t i g t j h t k Tngent vector: r '( t ) f ' t i g ' t j h' t k Tngent line t t t : s r( t ) sr '( t ) Velocity: v( t) r '( t) speed = v( t) Accelertion ( t)
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationMath 32B Discussion Session Session 7 Notes August 28, 2018
Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls
More informationMATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The halfangle formula cos 2 θ = 1 2
MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xyplne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More information(6.5) Length and area in polar coordinates
86 Chpter 6 SLICING TECHNIQUES FURTHER APPLICATIONS Totl mss 6 x ρ(x)dx + x 6 x dx + 9 kg dx + 6 x dx oment bout origin 6 xρ(x)dx x x dx + x + x + ln x ( ) + ln 6 kg m x dx + 6 6 x x dx Centre of mss +
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos(  1 2 ) = rcsin( 1 2 ) = rcsin(  1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationcos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves
Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. USubstitution for Definite Integrls A. Th m 6Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationdf dt f () b f () a dt
Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationTotal Score Maximum
Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question 4 5 6 7 Totl Score Mimum 6 4 8 9 4 No clcultors or formul sheets. Check tht you hve 6 pges.. Find
More informationA. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
More information14.4. Lengths of curves and surfaces of revolution. Introduction. Prerequisites. Learning Outcomes
Lengths of curves nd surfces of revolution 4.4 Introduction Integrtion cn be used to find the length of curve nd the re of the surfce generted when curve is rotted round n xis. In this section we stte
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationMathematics of Motion II Projectiles
Chmp+ Fll 2001 Dn Stump 1 Mthemtics of Motion II Projectiles Tble of vribles t time v velocity, v 0 initil velocity ccelertion D distnce x position coordinte, x 0 initil position x horizontl coordinte
More informationMath 6455 Oct 10, Differential Geometry I Fall 2006, Georgia Tech
Mth 6455 Oct 10, 2006 1 Differentil Geometry I Fll 2006, Georgi Tech Lecture Notes 12 Riemnnin Metrics 0.1 Definition If M is smooth mnifold then by Riemnnin metric g on M we men smooth ssignment of n
More informationLecture 3: Curves in Calculus. Table of contents
Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More informationAPPM 1360 Exam 2 Spring 2016
APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationA LEVEL TOPIC REVIEW. factor and remainder theorems
A LEVEL TOPIC REVIEW unit C fctor nd reminder theorems. Use the Fctor Theorem to show tht: ) ( ) is fctor of +. ( mrks) ( + ) is fctor of ( ) is fctor of + 7+. ( mrks) +. ( mrks). Use lgebric division
More informationAnonymous Math 361: Homework 5. x i = 1 (1 u i )
Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rellife exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationMath 211/213 Calculus IIIIV. Directions. Kenneth Massey. September 17, 2018
Mth 211/213 Clculus V Kenneth Mssey CrsonNewmn University September 17, 2018 CN Mth 211  Mssey, 1 / 1 Directions You re t the origin nd giving directions to the point (4, 3). 1. n Mnhttn: go est 4
More informationExploring parametric representation with the TI84 Plus CE graphing calculator
Exploring prmetric representtion with the TI84 Plus CE grphing clcultor Richrd Prr Executive Director Rice University School Mthemtics Project rprr@rice.edu Alice Fisher Director of Director of Technology
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More informationContents. 4.1 Line Integrals. Calculus III (part 4): Vector Calculus (by Evan Dummit, 2018, v. 3.00) 4 Vector Calculus
lculus III prt 4): Vector lculus by Evn Dummit, 8, v. 3.) ontents 4 Vector lculus 4. Line Integrls................................................. 4. Surfces nd Surfce Integrls........................................
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationJim Lambers MAT 280 Spring Semester Lecture 26 and 27 Notes
Jim Lmbers MAT 280 pring emester 200910 Lecture 26 nd 27 Notes These notes correspond to ection 8.6 in Mrsden nd Tromb. ifferentil Forms To dte, we hve lerned the following theorems concerning the evlution
More informationTime : 3 hours 03  Mathematics  March 2007 Marks : 100 Pg  1 S E CT I O N  A
Time : hours 0  Mthemtics  Mrch 007 Mrks : 100 Pg  1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new
More informationSolution to HW 4, Ma 1c Prac 2016
Solution to HW 4 M c Prc 6 Remrk: every function ppering in this homework set is sufficiently nice t lest C following the jrgon from the textbook we cn pply ll kinds of theorems from the textbook without
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More information7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?
7.1 Integrl s Net Chnge Clculus 7.1 INTEGRAL AS NET CHANGE Distnce versus Displcement We hve lredy seen how the position of n oject cn e found y finding the integrl of the velocity function. The chnge
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationEdexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks
Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1
More informationPARABOLA EXERCISE 3(B)
PARABOLA EXERCISE (B). Find eqution of the tngent nd norml to the prbol y = 6x t the positive end of the ltus rectum. Eqution of prbol y = 6x 4 = 6 = / Positive end of the Ltus rectum is(, ) =, Eqution
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
More informationPractice Final. Name: Problem 1. Show all of your work, label your answers clearly, and do not use a calculator.
Nme: MATH 2250 Clculus Eric Perkerson Dte: December 11, 2015 Prctice Finl Show ll of your work, lbel your nswers clerly, nd do not use clcultor. Problem 1 Compute the following limits, showing pproprite
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationZ b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...
Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.
More informationr 0 ( ) cos( ) r( )sin( ). 1. Last time, we calculated that for the cardioid r( ) =1+sin( ),
Wrm up Recll from lst time, given polr curve r = r( ),, dx dy dx = dy d = (r( )sin( )) d (r( ) cos( )) = r0 ( )sin( )+r( ) cos( ) r 0 ( ) cos( ) r( )sin( ).. Lst time, we clculted tht for crdioid r( )
More informationMath Advanced Calculus II
Mth 452  Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationGeometric and Mechanical Applications of Integrals
5 Geometric nd Mechnicl Applictions of Integrls 5.1 Computing Are 5.1.1 Using Crtesin Coordintes Suppose curve is given by n eqution y = f(x), x b, where f : [, b] R is continuous function such tht f(x)
More information4. Calculus of Variations
4. Clculus of Vritions Introduction  Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationMTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008
MTH 22 Fll 28 Essex County College Division of Mthemtics Hndout Version October 4, 28 Arc Length Everyone should be fmilir with the distnce formul tht ws introduced in elementry lgebr. It is bsic formul
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More informationNot for reproduction
AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type
More informationMathematics 19A; Fall 2001; V. Ginzburg Practice Final Solutions
Mthemtics 9A; Fll 200; V Ginzburg Prctice Finl Solutions For ech of the ten questions below, stte whether the ssertion is true or flse ) Let fx) be continuous t x Then x fx) f) Answer: T b) Let f be differentible
More informationKEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a
KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider
More informationChapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1
Chpter 5. Let g ( e. on [, ]. The derivtive of g is g ( e ( Write the slope intercept form of the eqution of the tngent line to the grph of g t. (b Determine the coordinte of ech criticl vlue of g. Show
More informationLoudoun Valley High School Calculus Summertime Fun Packet
Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!
More informationHIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)
HIGHER SCHOOL CERTIFICATE EXAMINATION 999 MATHEMATICS UNIT (ADDITIONAL) Time llowed Three hours (Plus 5 minutes reding time) DIRECTIONS TO CANDIDATES Attempt ALL questions ALL questions re of equl vlue
More informationLine Integrals. Chapter Definition
hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xyplne. It
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More information