AREA OF A SURFACE OF REVOLUTION

Size: px
Start display at page:

Download "AREA OF A SURFACE OF REVOLUTION"

Transcription

1 AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7. nd 7.. We wnt to define the re of surfce of revolution in such w tht it corresponds to our intuition. If the surfce re is A, we cn imgine tht pinting the surfce would require the sme mount of pint s does flt region with re A. Let s strt with some simple surfces. The lterl surfce re of circulr clinder with rdius r nd height h is tken to be A rh becuse we cn imgine cutting the clinder nd unrolling it (s in Figure ) to obtin rectngle with dimensions r nd h. Likewise, we cn tke circulr cone with bse rdius r nd slnt height l, cut it long the dshed line in Figure, nd fltten it to form sector of circle with rdius l nd centrl ngle. We know tht, in generl, the re of sector of circle with rdius l nd ngle is l (see Eercise 67 in Section 6.) nd so in this cse it is rl A l l r l rl FIGURE Therefore, we define the lterl surfce re of cone to be A rl. πr cut l r l FIGURE l Wht bout more complicted surfces of revolution? If we follow the strteg we used with rc length, we cn pproimte the originl curve b polgon. When this polgon is rotted bout n is, it cretes simpler surfce whose surfce re pproimtes the ctul surfce re. B tking limit, we cn determine the ect surfce re. The pproimting surfce, then, consists of number of bnds, ech formed b rotting line segment bout n is. To find the surfce re, ech of these bnds cn be considered portion of circulr cone, s shown in Figure. The re of the bnd (or frustum of cone) with slnt height l nd upper nd lower rdii r nd r is found b subtrcting the res of two cones: r A r l l r l r r l r l r l From similr tringles we hve l l l r r FIGURE which gives r l r l r l or r r l r l Putting this in Eqution, we get or A r l r l Thomson Brooks-Cole copright 7 A rl where r r r is the verge rdius of the bnd.

2 AREA OF A SURFACE OF REVOLUTION =ƒ b () Surfce of revolution Now we ppl this formul to our strteg. Consider the surfce shown in Figure, which is obtined b rotting the curve f, b, bout the -is, where f is positive nd hs continuous derivtive. In order to define its surfce re, we divide the intervl, b into n subintervls with endpoints,,..., n nd equl width, s we did in determining rc length. If i f i, then the point P i i, i lies on the curve. The prt of the surfce between i nd i is pproimted b tking the line segment P i P i nd rotting it bout the -is. The result is bnd with slnt height l P ip i nd verge rdius r i i so, b Formul, its surfce re is P i P i- P i P n i i P ip i b (b) Approimting bnd FIGURE As in the proof of Theorem 7.., we hve P ip i s f i* where i * is some number in i, i. When is smll, we hve i f i f i * nd lso i f i f i *, since f is continuous. Therefore i i P ip i f i * s f i * nd so n pproimtion to wht we think of s the re of the complete surfce of revolution is n i f i * s f i * This pproimtion ppers to become better s n l nd, recognizing () s Riemnn sum for the function t f s f, we hve lim n l n i f i * s f i * b Therefore, in the cse where f is positive nd hs continuous derivtive, we define the surfce re of the surfce obtined b rotting the curve f, b, bout the -is s f s f S b f s f With the Leibniz nottion for derivtives, this formul becomes 5 S b d If the curve is described s t, c d, then the formul for surfce re becomes Thomson Brooks-Cole copright 7 6 S d c d d nd both Formuls 5 nd 6 cn be summrized smbolicll, using the nottion for rc

3 AREA OF A SURFACE OF REVOLUTION length given in Section 7., s 7 S ds For rottion bout the -is, the surfce re formul becomes 8 S ds where, s before, we cn use either ds d or ds d d These formuls cn be remembered b thinking of or s the circumference of circle trced out b the point, on the curve s it is rotted bout the -is or -is, respectivel (see Figure 5). (, ) (, ) FIGURE 5 circumference=π () Rottion bout -is: S=j π ds circumference=π (b) Rottion bout -is: S=j π ds EXAMPLE The curve s,, is n rc of the circle. Find the re of the surfce obtined b rotting this rc bout the -is. (The surfce is portion of sphere of rdius. See Figure 6.) SOLUTION We hve d s nd so, b Formul 5, the surfce re is S d Thomson Brooks-Cole copright 7 FIGURE 6 Figure 6 shows the portion of the sphere whose surfce re is computed in Emple. s s s 8

4 AREA OF A SURFACE OF REVOLUTION Figure 7 shows the surfce of revolution whose re is computed in Emple. EXAMPLE The rc of the prbol from, to, is rotted bout the -is. Find the re of the resulting surfce. SOLUTION Using (, ) = we hve, from Formul 8, nd d FIGURE 7 S ds d s Substituting u, we hve du 8. Remembering to chnge the limits of integrtion, we hve S 7 5 su du [ u ] 5 7 As check on our nswer to Emple, notice from Figure 7 tht the surfce re should be close to tht of circulr clinder with the sme height nd rdius hlfw between the upper nd lower rdius of the surfce:. We computed tht the surfce re ws (7s7 5s5).85 6 which seems resonble. Alterntivel, the surfce re should be slightl lrger thn the re of frustum of cone with the sme top nd bottom edges. From Eqution, this is..5(s) 9.8 SOLUTION we hve Using S ds 6 s (7s7 5s5) nd d d d s s d s d su du (7s7 5s5) (where u ) (s in Solution ) Thomson Brooks-Cole copright 7 Another method: Use Formul 6 with ln. EXAMPLE Find the re of the surfce generted b rotting the curve e,,bout the -is. SOLUTION Using Formul 5 with e nd d e

5 AREA OF A SURFACE OF REVOLUTION 5 we hve S d e s e e s u du sec d (where u e ) tn e (where u tn nd ) Or use Formul in the Tble of Integrls. [sec tn ln sec tn ] (b Emple 8 in Section 6.) [sec tn lnsec tn s ln(s )] Since tn e, we hve sec tn e nd S [es e ln(e s e ) s ln(s )] EXERCISES A Click here for nswers. Set up, but do not evlute, n integrl for the re of the surfce obtined b rotting the curve bout the given is.. ln, ; -is. sin, ; -is. sec, ; -is. e, ; -is S Click here for solutions. 7 Use Simpson s Rule with n to pproimte the re of the surfce obtined b rotting the curve bout the -is. Compre our nswer with the vlue of the integrl produced b our clcultor. 7. ln, 8. s, 9. sec,. s e, 5 Find the re of the surfce obtined b rotting the curve bout the -is. 5., , 7. s, 8. cos, 9. cosh,., ,., CAS CAS Use either CAS or tble of integrls to find the ect re of the surfce obtined b rotting the given curve bout the -is..,. s, Use CAS to find the ect re of the surfce obtined b rotting the curve bout the -is. If our CAS hs trouble evluting the integrl, epress the surfce re s n integrl in the other vrible..,. ln, Thomson Brooks-Cole copright 7 6 The given curve is rotted bout the -is. Find the re of the resulting surfce.. s,., 5. s, 6. cosh, 5. () If, find the re of the surfce generted b rotting the loop of the curve bout the -is. (b) Find the surfce re if the loop is rotted bout the -is. 6. A group of engineers is building prbolic stellite dish whose shpe will be formed b rotting the curve bout the -is. If the dish is to hve -ft dimeter nd mimum depth of ft, find the vlue of nd the surfce re of the dish.

6 6 AREA OF A SURFACE OF REVOLUTION CAS 7. The ellipse b b is rotted bout the -is to form surfce clled n ellipsoid. Find the surfce re of this ellipsoid. 8. Find the surfce re of the torus in Eercise in Section If the curve f, b, is rotted bout the horizontl line c, where f c, find formul for the re of the resulting surfce.. Use the result of Eercise 9 to set up n integrl to find the re of the surfce generted b rotting the curve s,,bout the line. Then use CAS to evlute the integrl.. Find the re of the surfce obtined b rotting the circle r bout the line r.. Show tht the surfce re of zone of sphere tht lies between two prllel plnes is S dh, where d is the dimeter of the sphere nd h is the distnce between the plnes. (Notice tht S depends onl on the distnce between the plnes nd not on their loction, provided tht both plnes intersect the sphere.). Formul is vlid onl when f. Show tht when f is not necessril positive, the formul for surfce re becomes S b. Let L be the length of the curve f, b, where f is positive nd hs continuous derivtive. Let S f be the surfce re generted b rotting the curve bout the -is. If c is positive constnt, define t f c nd let S t be the corresponding surfce re generted b the curve t, b. Epress in terms of nd L. S t f s f S f Thomson Brooks-Cole copright 7

7 AREA OF A SURFACE OF REVOLUTION 7 ANSWERS.. S ln s Click here for solutions. s sec tn 5. (5s5 )7 7. (7s7 7s7)6 9. [ e e ].. (5s5 s) [ ln(s7 ) ln(s ) s7 s] 6[ln(s ) s]. 5. () (b) 56s 5 7. [b b sin (s b )s b ] 9. b f s f. r Thomson Brooks-Cole copright 7

8 8 AREA OF A SURFACE OF REVOLUTION SOLUTIONS. =ln ds = +(d/) = +(/) S = π(ln ) +(/) [b (7)]. =sec ds = +(d/) = +(sec tn ) S = π/ π +(sec tn ) [b (8)] 5. = =.So S = π +( ) =π = π 6 5 udu= π 8 +9 [u =+9, du =6 ] 5 u/ = π = +(d/) =+[/( )] =+/(). So S = 9 =π π + d + / 9 = π 9 = π 8 ( +)/ 9 + =π = π =cosh +(d/) =+sinh =cosh.so S =π cosh cosh =π ( + cosh ) = π + sinh = π + sinh or π + e e. = + / /d = + / () = + +(/d) =+ + = +.So S =π + d =π + =π + = π. = = +(/d) =+9.So S =π +(/d) d =π +9 d = π +9 6 d 6 +9 / 5 5 = π 8 = π 7 5. = /d = ( ) / ( ) = / +(/d) =+ S = / = + = π / d =π d =π / =π = π. Note tht this is Thomson Brooks-Cole copright 7 the surfce re of sphere of rdius, nd the length of the intervl =to = / is the length of the intervl = to =.

9 AREA OF A SURFACE OF REVOLUTION 9 7. =ln d/ =/ +(d/) =+/ S = π ln +/. Let f() =ln +/.Sincen =, = = 5. Then S S =π /5 [f() + f(.) + f(.) + +f(.6) + f(.8) + f()] The vlue of the integrl produced b clcultor is 9.6 (to si deciml plces). 9. =sec d/ =sec tn +(d/) =+sec tn S = π/ π sec +sec tn.letf() =sec +sec tn. Since n =, = π/ = π.then S S =π π/ π f() + f +f π + +f 8π +f The vlue of the integrl produced b clcultor is (to si deciml plces). 9π π + f =/ ds = +(d/) = +( / ) = +/ S = π + =π + u + =π u du [u =, du =] +u = π du +u = π +ln u + +u u u = π 7 +ln ln = π 7 +ln = nd = nd. S = π +( ) =π +u du [u 6 =, du =6] = π +u du π = [or use CAS] u +u + ln u + +u = π + ln + = π 6 + ln + 5. Since >,thecurve = ( ) onl hs points with. ( ( ).) The curve is smmetric bout the -is (since the eqution is unchnged when is replced b ). =when =or, so the curve s loop etends from =to =. Thomson Brooks-Cole copright 7 d ( )= d [( ) ] 6 d = ( )( ) + ( ) d ( )[ + ] = 6 d = ( ) ( ) = ( ) ( ) 6 6 ( ) the lst frction is / = ( )

10 AREA OF A SURFACE OF REVOLUTION + d = = = ( +) = for 6=. () S = = π = π ds =π ( + ) = π ( ) + ( )( +) =π 6 + = π ( + )= π = π. Note tht we hve rotted the top hlf of the loop bout the -is. This genertes the full surfce. (b) We must rotte the full loop bout the -is, so we get double the re obtined b rotting the top hlf of the loop: S = π ds=π = + = π / ( +) = π ( / + / ) = π / / = π 5/ / = π + 6 = π 5 8 = 56π (d/) = = d b b = b + d =+ b = b + = b + b / = b + b b b ( / ) b b = + b = b ( ) The ellipsoid s surfce re is twice the re generted b rotting the first qudrnt portion of the ellipse bout the -is. Thus, S = π + d b ( =π b ) = πb ( b ) = πb b u du b [u = b ] = πb b u u + u b sin πb b = b ( b )+ b sin =π b + b sin b b Thomson Brooks-Cole copright 7 9. The nlogue of f( i ) in the derivtion of () is now c f( i ), so S = lim n n i= π[c f( i )] +[f ( i )] = b π[c f()] +[f ()].

11 AREA OF A SURFACE OF REVOLUTION. For the upper semicircle, f() = r, f () = / r. The surfce re generted is r S = π r r + =π r r =π r r r r For the lower semicircle, f() = r nd f () = Thus, the totl re is S = S + S =8π r r r r r,sos =π r r =8π r sin r r r r r r r + r. =8πr π =π r.. In the derivtion of (), we computed tpicl contribution to the surfce re to be π i + i P i P i, the re of frustum of cone. When f() is not necessril positive, the pproimtions i = f( i ) f( i ) nd i = f( i ) f( i ) must be replced b i = f( i ) f( i ) nd i = f( i ) f( i ). Thus, π i + i P i P i π f( i ) +[f ( i )]. Continuing with the rest of the derivtion s before, we obtin S = b π f() +[f ()]. Thomson Brooks-Cole copright 7

Double Integrals over General Regions

Double Integrals over General Regions Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing

More information

Arc Length. P i 1 P i (1) L = lim. i=1

Arc Length. P i 1 P i (1) L = lim. i=1 Arc Length Suppose tht curve C is defined by the eqution y = f(x), where f is continuous nd x b. We obtin polygonl pproximtion to C by dividing the intervl [, b] into n subintervls with endpoints x, x,...,x

More information

Volumes of solids of revolution

Volumes of solids of revolution Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the x-xis. There is strightforwrd technique which enbles this to be done, using

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

So there are two points of intersection, one being x = 0, y = 0 2 = 0 and the other being x = 2, y = 2 2 = 4. y = x 2 (2,4)

So there are two points of intersection, one being x = 0, y = 0 2 = 0 and the other being x = 2, y = 2 2 = 4. y = x 2 (2,4) Ares The motivtion for our definition of integrl ws the problem of finding the re between some curve nd the is for running between two specified vlues. We pproimted the region b union of thin rectngles

More information

Integration. 148 Chapter 7 Integration

Integration. 148 Chapter 7 Integration 48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but

More information

1+(dy/dx) 2 dx. We get dy dx = 3x1/2 = 3 x, = 9x. Hence 1 +

1+(dy/dx) 2 dx. We get dy dx = 3x1/2 = 3 x, = 9x. Hence 1 + Mth.9 Em Solutions NAME: #.) / #.) / #.) /5 #.) / #5.) / #6.) /5 #7.) / Totl: / Instructions: There re 5 pges nd totl of points on the em. You must show ll necessr work to get credit. You m not use our

More information

to the area of the region bounded by the graph of the function y = f(x), the x-axis y = 0 and two vertical lines x = a and x = b.

to the area of the region bounded by the graph of the function y = f(x), the x-axis y = 0 and two vertical lines x = a and x = b. 5.9 Are in rectngulr coordintes If f() on the intervl [; ], then the definite integrl f()d equls to the re of the region ounded the grph of the function = f(), the -is = nd two verticl lines = nd =. =

More information

Section 2.3. Motion Along a Curve. The Calculus of Functions of Several Variables

Section 2.3. Motion Along a Curve. The Calculus of Functions of Several Variables The Clculus of Functions of Severl Vribles Section 2.3 Motion Along Curve Velocity ccelertion Consider prticle moving in spce so tht its position t time t is given by x(t. We think of x(t s moving long

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

Section 7-4 Translation of Axes

Section 7-4 Translation of Axes 62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the

More information

Introduction to Integration Part 2: The Definite Integral

Introduction to Integration Part 2: The Definite Integral Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the

More information

Applications to Physics and Engineering

Applications to Physics and Engineering Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics

More information

Worksheet 24: Optimization

Worksheet 24: Optimization Worksheet 4: Optimiztion Russell Buehler b.r@berkeley.edu 1. Let P 100I I +I+4. For wht vlues of I is P mximum? P 100I I + I + 4 Tking the derivtive, www.xkcd.com P (I + I + 4)(100) 100I(I + 1) (I + I

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values) www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input

More information

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.

More information

Let us recall some facts you have learnt in previous grades under the topic Area.

Let us recall some facts you have learnt in previous grades under the topic Area. 6 Are By studying this lesson you will be ble to find the res of sectors of circles, solve problems relted to the res of compound plne figures contining sectors of circles. Ares of plne figures Let us

More information

Sect 8.3 Triangles and Hexagons

Sect 8.3 Triangles and Hexagons 13 Objective 1: Sect 8.3 Tringles nd Hexgons Understnding nd Clssifying Different Types of Polygons. A Polygon is closed two-dimensionl geometric figure consisting of t lest three line segments for its

More information

CONIC SECTIONS. Chapter 11

CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter 11 11.1 Overview 11.1.1 Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig. 11.1). Fig. 11.1 Suppose we

More information

Lesson 10. Parametric Curves

Lesson 10. Parametric Curves Return to List of Lessons Lesson 10. Prmetric Curves (A) Prmetric Curves If curve fils the Verticl Line Test, it cn t be expressed by function. In this cse you will encounter problem if you try to find

More information

Pure C4. Revision Notes

Pure C4. Revision Notes Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: x n+ x n n + + C, dx = ln x + C, if n if n = In prticulr, this mens tht dx = ln x + C x nd x 0 dx = dx = dx = x + C Integrl of Constnt:

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

Review Problems for the Final of Math 121, Fall 2014

Review Problems for the Final of Math 121, Fall 2014 Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since

More information

m, where m = m 1 + m m n.

m, where m = m 1 + m m n. Lecture 7 : Moments nd Centers of Mss If we hve msses m, m 2,..., m n t points x, x 2,..., x n long the x-xis, the moment of the system round the origin is M 0 = m x + m 2 x 2 + + m n x n. The center of

More information

Area Between Curves: We know that a definite integral

Area Between Curves: We know that a definite integral Are Between Curves: We know tht definite integrl fx) dx cn be used to find the signed re of the region bounded by the function f nd the x xis between nd b. Often we wnt to find the bsolute re of region

More information

Approximate Integration: Trapezoid Rule and Simpson s Rule. y 1. e x 2 dx. y b. f sxd dx < o n

Approximate Integration: Trapezoid Rule and Simpson s Rule. y 1. e x 2 dx. y b. f sxd dx < o n Approimte Integrtion: Trpezoid Rule nd Simpson s Rule There re two situtions in which it is impossible to find the ect vlue of definite integrl. The first sitution rises from the fct tht in order to evlute

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

For a solid S for which the cross sections vary, we can approximate the volume using a Riemann sum. A(x i ) x. i=1.

For a solid S for which the cross sections vary, we can approximate the volume using a Riemann sum. A(x i ) x. i=1. Volumes by Disks nd Wshers Volume of cylinder A cylinder is solid where ll cross sections re the sme. The volume of cylinder is A h where A is the re of cross section nd h is the height of the cylinder.

More information

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006 dius of the Erth - dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.

More information

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Uniform convergence and its consequences

Uniform convergence and its consequences Uniform convergence nd its consequences The following issue is centrl in mthemtics: On some domin D, we hve sequence of functions {f n }. This mens tht we relly hve n uncountble set of ordinry sequences,

More information

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix. APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The

More information

4 Geometry: Shapes. 4.1 Circumference and area of a circle. FM Functional Maths AU (AO2) Assessing Understanding PS (AO3) Problem Solving HOMEWORK 4A

4 Geometry: Shapes. 4.1 Circumference and area of a circle. FM Functional Maths AU (AO2) Assessing Understanding PS (AO3) Problem Solving HOMEWORK 4A Geometry: Shpes. Circumference nd re of circle HOMEWORK D C 3 5 6 7 8 9 0 3 U Find the circumference of ech of the following circles, round off your nswers to dp. Dimeter 3 cm Rdius c Rdius 8 m d Dimeter

More information

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Answer, Key Homework 8 David McIntyre 1

Answer, Key Homework 8 David McIntyre 1 Answer, Key Homework 8 Dvid McIntyre 1 This print-out should hve 17 questions, check tht it is complete. Multiple-choice questions my continue on the net column or pge: find ll choices before mking your

More information

Brief review of prerequisites for ECON4140/4145

Brief review of prerequisites for ECON4140/4145 1 ECON4140/4145, August 2010 K.S., A.S. Brief review of prerequisites for ECON4140/4145 References: EMEA: K. Sdsæter nd P. Hmmond: Essentil Mthemtics for Economic Anlsis, 3rd ed., FT Prentice Hll, 2008.

More information

9 CONTINUOUS DISTRIBUTIONS

9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

Surface Area and Volume

Surface Area and Volume Surfce Are nd Volume Student Book - Series J- Mthletics Instnt Workooks Copyright Surfce re nd volume Student Book - Series J Contents Topics Topic - Surfce re of right prism Topic 2 - Surfce re of right

More information

Pythagoras theorem and trigonometry (2)

Pythagoras theorem and trigonometry (2) HPTR 10 Pythgors theorem nd trigonometry (2) 31 HPTR Liner equtions In hpter 19, Pythgors theorem nd trigonometry were used to find the lengths of sides nd the sizes of ngles in right-ngled tringles. These

More information

14.2. The Mean Value and the Root-Mean-Square Value. Introduction. Prerequisites. Learning Outcomes

14.2. The Mean Value and the Root-Mean-Square Value. Introduction. Prerequisites. Learning Outcomes he Men Vlue nd the Root-Men-Squre Vlue 4. Introduction Currents nd voltges often vry with time nd engineers my wish to know the men vlue of such current or voltge over some prticulr time intervl. he men

More information

1 Numerical Solution to Quadratic Equations

1 Numerical Solution to Quadratic Equations cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll

More information

Ae2 Mathematics : Fourier Series

Ae2 Mathematics : Fourier Series Ae Mthemtics : Fourier Series J. D. Gibbon (Professor J. D Gibbon, Dept of Mthemtics j.d.gibbon@ic.c.uk http://www.imperil.c.uk/ jdg These notes re not identicl word-for-word with my lectures which will

More information

The Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx

The Chain Rule. rf dx. t t lim  (x) dt  (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the single-vrible chin rule extends to n inner

More information

Net Change and Displacement

Net Change and Displacement mth 11, pplictions motion: velocity nd net chnge 1 Net Chnge nd Displcement We hve seen tht the definite integrl f (x) dx mesures the net re under the curve y f (x) on the intervl [, b] Any prt of the

More information

MATLAB: M-files; Numerical Integration Last revised : March, 2003

MATLAB: M-files; Numerical Integration Last revised : March, 2003 MATLAB: M-files; Numericl Integrtion Lst revised : Mrch, 00 Introduction to M-files In this tutoril we lern the bsics of working with M-files in MATLAB, so clled becuse they must use.m for their filenme

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

Exponential and Logarithmic Functions

Exponential and Logarithmic Functions Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define

More information

Chapter G - Problems

Chapter G - Problems Chpter G - Problems Blinn College - Physics 2426 - Terry Honn Problem G.1 A plne flies horizonlly t speed of 280 mês in position where the erth's mgnetic field hs mgnitude 6.0µ10-5 T nd is directed t n

More information

r 2 F ds W = r 1 qe ds = q

r 2 F ds W = r 1 qe ds = q Chpter 4 The Electric Potentil 4.1 The Importnt Stuff 4.1.1 Electricl Potentil Energy A chrge q moving in constnt electric field E experiences force F = qe from tht field. Also, s we know from our study

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time

More information

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function. Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while

More information

Solutions to Section 1

Solutions to Section 1 Solutions to Section Exercise. Show tht nd. This follows from the fct tht mx{, } nd mx{, } Exercise. Show tht = { if 0 if < 0 Tht is, the bsolute vlue function is piecewise defined function. Grph this

More information

Mathematics Higher Level

Mathematics Higher Level Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:

More information

Numerical Methods of Approximating Definite Integrals

Numerical Methods of Approximating Definite Integrals 6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils,

More information

SUBSTITUTION I.. f(ax + b)

SUBSTITUTION I.. f(ax + b) Integrtion SUBSTITUTION I.. f(x + b) Grhm S McDonld nd Silvi C Dll A Tutoril Module for prctising the integrtion of expressions of the form f(x + b) Tble of contents Begin Tutoril c 004 g.s.mcdonld@slford.c.uk

More information

Chapter 6 Solving equations

Chapter 6 Solving equations Chpter 6 Solving equtions Defining n eqution 6.1 Up to now we hve looked minly t epressions. An epression is n incomplete sttement nd hs no equl sign. Now we wnt to look t equtions. An eqution hs n = sign

More information

Volumes as integrals of cross-sections (Sect. 6.1) Volumes as integrals of cross-sections (Sect. 6.1)

Volumes as integrals of cross-sections (Sect. 6.1) Volumes as integrals of cross-sections (Sect. 6.1) Volumes s integrls of cross-sections (ect. 6.1) Te volume of simple regions in spce Volumes integrting cross-sections: Te generl cse. Certin regions wit oles. Volumes s integrls of cross-sections (ect.

More information

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered: Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you

More information

Harvard College. Math 21a: Multivariable Calculus Formula and Theorem Review

Harvard College. Math 21a: Multivariable Calculus Formula and Theorem Review Hrvrd College Mth 21: Multivrible Clculus Formul nd Theorem Review Tommy McWillim, 13 tmcwillim@college.hrvrd.edu December 15, 2009 1 Contents Tble of Contents 4 9 Vectors nd the Geometry of Spce 5 9.1

More information

Section 5-4 Trigonometric Functions

Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form

More information

Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.

Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a. Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles

More information

Lesson 12.1 Trigonometric Ratios

Lesson 12.1 Trigonometric Ratios Lesson 12.1 rigonometric Rtios Nme eriod Dte In Eercises 1 6, give ech nswer s frction in terms of p, q, nd r. 1. sin 2. cos 3. tn 4. sin Q 5. cos Q 6. tn Q p In Eercises 7 12, give ech nswer s deciml

More information

APPLICATION OF INTEGRALS

APPLICATION OF INTEGRALS Chpter 8 APPLICATION OF INTEGRALS 8.1 Overview This chpter dels with specific ppliction of integrls to find the re under simple curves, re etween lines nd rcs of circles, prols nd ellipses, nd finding

More information

Unit 6: Exponents and Radicals

Unit 6: Exponents and Radicals Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -

More information

Chapter 9: Quadratic Equations

Chapter 9: Quadratic Equations Chpter 9: Qudrtic Equtions QUADRATIC EQUATIONS DEFINITION + + c = 0,, c re constnts (generlly integers) ROOTS Synonyms: Solutions or Zeros Cn hve 0, 1, or rel roots Consider the grph of qudrtic equtions.

More information

The Velocity Factor of an Insulated Two-Wire Transmission Line

The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the

More information

10.6 Applications of Quadratic Equations

10.6 Applications of Quadratic Equations 10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,

More information

An Off-Center Coaxial Cable

An Off-Center Coaxial Cable 1 Problem An Off-Center Coxil Cble Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Nov. 21, 1999 A coxil trnsmission line hs inner conductor of rdius nd outer conductor

More information

Using Definite Integrals

Using Definite Integrals Chpter 6 Using Definite Integrls 6. Using Definite Integrls to Find Are nd Length Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Two special Right-triangles 1. The

Two special Right-triangles 1. The Mth Right Tringle Trigonometry Hndout B (length of ) - c - (length of side ) (Length of side to ) Pythgoren s Theorem: for tringles with right ngle ( side + side = ) + = c Two specil Right-tringles. The

More information

y 20 x 1 Solution: a. The x-coordinates of the intersection points are solutions to the equation f(x) = g(x).

y 20 x 1 Solution: a. The x-coordinates of the intersection points are solutions to the equation f(x) = g(x). Mth 8, Exm, Fll 4 Problem Solution. Consider the functions fx) = x 6x nd gx) = 6x 4.. Find the x-coordinte of the intersection points of these two grphs. b. Compute the re of the region bounded by the

More information

11. Fourier series. sin mx cos nx dx = 0 for any m, n, sin 2 mx dx = π.

11. Fourier series. sin mx cos nx dx = 0 for any m, n, sin 2 mx dx = π. . Fourier series Summry of the bsic ides The following is quick summry of the introductory tretment of Fourier series in MATH. We consider function f with period π, tht is, stisfying f(x + π) = f(x) for

More information

Algorithms Chapter 4 Recurrences

Algorithms Chapter 4 Recurrences Algorithms Chpter 4 Recurrences Outline The substitution method The recursion tree method The mster method Instructor: Ching Chi Lin 林清池助理教授 chingchilin@gmilcom Deprtment of Computer Science nd Engineering

More information

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

Multiplication and Division - Left to Right. Addition and Subtraction - Left to Right.

Multiplication and Division - Left to Right. Addition and Subtraction - Left to Right. Order of Opertions r of Opertions Alger P lese Prenthesis - Do ll grouped opertions first. E cuse Eponents - Second M D er Multipliction nd Division - Left to Right. A unt S hniqu Addition nd Sutrction

More information

PHY 140A: Solid State Physics. Solution to Homework #2

PHY 140A: Solid State Physics. Solution to Homework #2 PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.

More information

Square Roots Teacher Notes

Square Roots Teacher Notes Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this

More information

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Or more simply put, when adding or subtracting quantities, their uncertainties add. Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re

More information

Lecture 3 Gaussian Probability Distribution

Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike

More information

Section A-4 Rational Expressions: Basic Operations

Section A-4 Rational Expressions: Basic Operations A- Appendi A A BASIC ALGEBRA REVIEW 7. Construction. A rectngulr open-topped bo is to be constructed out of 9- by 6-inch sheets of thin crdbord by cutting -inch squres out of ech corner nd bending the

More information

Sequences and Series

Sequences and Series Centre for Eduction in Mthemtics nd Computing Euclid eworkshop # 5 Sequences nd Series c 014 UNIVERSITY OF WATERLOO While the vst mjority of Euclid questions in this topic re use formule for rithmetic

More information

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems Euler Euler Everywhere Using the Euler-Lgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch

More information

Unit 4. Applications of integration

Unit 4. Applications of integration SOLUTIONS TO 8. EXERCISES Unit 4. Applictions of integrtion 4A. Ares between curves. / 4A- ) (3x x )dx = (3/)x x (/3)x 3 / = /4 b) x 3 = x = x = ± or x =. There re two enclosed pieces ( < x < nd < x

More information

CHAPTER 11 Numerical Differentiation and Integration

CHAPTER 11 Numerical Differentiation and Integration CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods

More information

Basic Math Review. Numbers. Important Properties. Absolute Value PROPERTIES OF ADDITION NATURAL NUMBERS {1, 2, 3, 4, 5, }

Basic Math Review. Numbers. Important Properties. Absolute Value PROPERTIES OF ADDITION NATURAL NUMBERS {1, 2, 3, 4, 5, } ƒ Bsic Mth Review Numers NATURAL NUMBERS {1,, 3, 4, 5, } WHOLE NUMBERS {0, 1,, 3, 4, } INTEGERS {, 3,, 1, 0, 1,, } The Numer Line 5 4 3 1 0 1 3 4 5 Negtive integers Positive integers RATIONAL NUMBERS All

More information

Math 22B Solutions Homework 1 Spring 2008

Math 22B Solutions Homework 1 Spring 2008 Mth 22B Solutions Homework 1 Spring 2008 Section 1.1 22. A sphericl rindrop evportes t rte proportionl to its surfce re. Write differentil eqution for the volume of the rindrop s function of time. Solution

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

Notes for Thurs 8 Sept Calculus II Fall 2005 New York University Instructor: Tyler Neylon Scribe: Kelsey Williams

Notes for Thurs 8 Sept Calculus II Fall 2005 New York University Instructor: Tyler Neylon Scribe: Kelsey Williams Notes for Thurs 8 Sept Clculus II Fll 00 New York University Instructor: Tyler Neylon Scribe: Kelsey Willims 8. Integrtion by Prts This section is primrily bout the formul u dv = uv v ( ) which is essentilly

More information

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk Te Funmentl Teorem of Clculus EXAMPLE: If f is function wose grp is sown below n g() = f(t)t, fin te vlues of g(), g(), g(), g(3), g(4), n g(5).

More information

Diffraction and Interference of Light

Diffraction and Interference of Light rev 12/2016 Diffrction nd Interference of Light Equipment Qty Items Prt Number 1 Light Sensor CI-6504 1 Rotry Motion Sensor CI-6538 1 Single Slit Set OS-8523 1 Multiple Slit Set OS-8523 1 Liner Trnsltor

More information