CHAPTER 1. 1 FORMS. Rule DF1 (Linearity). d(αu + βv) = αdu + βdv where α, β R. Rule DF2 (Leibniz s Rule or Product Rule). d(uv) = udv + vdu.
|
|
- Betty Melina Moody
- 5 years ago
- Views:
Transcription
1 CHAPTER 1. 1 FORMS 1. Differentials: Basic Rules Differentials, usually regarded as fanciful objects, but barely mentioned in elementary calculus textbooks, are now receiving our full attention as they deserve. We will see that these fanciful objects have fantastic powers. Since they are fanciful objects, it is rather difficult to tell exactly what they are. However, from our experiences in calculus, spiced by our vivid imagination, we have some good ideas about how they should behave. Let us start with putting down some rules of thumb governing their behavior: Rule DF1 (Linearity). d(αu + βv) = αdu + βdv where α, β R. Rule DF2 (Leibniz s Rule or Product Rule). d(uv) = udv + vdu. Rule DF3 (Chain Rule). du = du dv if u is a (nice) function of v. dv These rules will soon be applied to concrete examples in the routine manner. Before doing so, we describe an idea leading to such rules swiftly, ignoring mathematical rigor in the process. (This idea is a bit misleading; however it serves our purpose well at present.) The expression du here is interpreted as a kind of polished change (or increment) in u, which differs from the real change u by an amount (designated by o( u) in the following identity) so small that practically it may be ignored: u = du + o( u). Here, both u and du are small amounts of u, but their difference o( u), comparing to both u and du, is much smaller. Suppose that we have two quantities u and v, and due to a slight disturbance or other good reasons, u is changed by an amount u to u + u and v is changed by v to v + v. Then their product w uv is changed into (u + u)(v + v) = uv + u v + v u + ( u)( v). Thus, the increment in w is given by w = (u + u)(v + v) uv = u v + v u + ( u)( v). Now we polish this identity: u and v become du and dv respectively, and w becomes dw = d(uv). What happens to the product ( u)( v)? Because both u and v are small, their product is much smaller comparing to each of them and hence can be polished away. 1
2 The final outcome is d(uv) = udv +vdu, namely the Leibniz rule given above. The identity in Rule DF1 can be obtained in the same fashion, except a lot easier. I leave Rule DF3 for you to derive in this fashion. You would say, this looks like something familiar in calculus: d(uv) = udv + vdu d dv seems to be nothing new but the usual product rule dxuv = u dx + v du dx written in a different way. So what is such a big deal? Well, the usual product rule only applies to functions of a single variable x. The rates of change du dv dx and dx of u and v are measured against the change in x. For differentials, what causes du and dv to emerge is not specified and immaterial. It may due to the change of one variable, say x, as you learned in the first year calculus. It may due to the changes of several variables. When I write down the identity d(uv) = udv + vdu, I do not have to specify what variables are, or even how many variables are involved. The cause for the changes du, dv in u, v may well be a mild earthquake I don t care. So the product rule for differentials here has a much wider appeal. In particular, it applies to the several variable case. Let us derive some simple consequences of these rules of thumb. From Rule DF2 it follows immediately that d(u 2 ) = 2udu. By mathematical induction, we can show that holds for all positive integers n. d(u n ) = nu n 1 du (This is a good exercise for reinforcing your skill in induction. Do this.) On the other hand, we have d1 = d(1 2 ) = 2d1. Hence d1 = 0. So, if c is a constant, by Rule DF1, we have dc = cd1 = 0. This is not surprising: we cannot see any change in a constant c and hence its differential should be zero. By Rule DF3, we have: d(1/x) = dx/x 2, d x = dx/2 x, d(1 + x 2 ) = 2xdx, d sin x = cos x dx, d cos x = sin x dx, d tan x = sec 2 x dx, d sec x = tan x sec x dx, de x = e x dx, d log x = dx x, d arcsin x = dx dx, d arctan x = 1 x x 2, etc., (log x here is the same as ln x). Formulae of derivatives you learned in the first year calculus are very handy. Example 1.1. Find d(e cos t ) and d(log sec x). Solution: Let u = cos t. From d(e u ) = e u du we have d(e cos t ) = d(e u ) = e u du = e cos t d(cos t) = e cos t ( sin t dt) = sin t e cos t dt. 2
3 You should feel comfortable with the following way of taking the differential of e cos t without introducing a new variable u: d(e cos t ) = e cos t d(cos t) = sin t e cos t dt. Similarly, ( ) 1 d(log sec x) = d log = d( log cos x) cos x = d(log cos x) = d cos x cos x = sin x dx cos x = tan x dx. (Alternatively, d(log sec x) = (sec x) 1 d sec x = (sec x) 1 sec x tan x dx = tan x dx.) Example 1.2. Find dr and d(1/r) for the radius function r = x 2 + y 2 + z 2. Solution: Applying d on both sides of the identity r 2 = x 2 + y 2 + z 2, we obtain d(r 2 ) = d(x 2 ) + d(y 2 ) + d(z 2 ), or 2rdr = 2xdx + 2ydy + 2zdz. Thus we arrive at dr = xdx + ydy + zdz r = xdx + ydy + zdz x2 + y 2 + z 2. Also, d(1/r) = dr/r 2 = (xdx + ydy + zdz)/r 3 = (xdx + ydy + zdz)/(x 2 + y 2 + z 2 ) 3/2. This example illustrates how differentials work for functions of several variables. ( u ) vdu udv Example 1.3. Derive the quotient rule d = v v 2. Solution: Let w u/v. Then u = vw and hence du = vdw + wdv, which gives a relation between du, dv and dw. Use this relation to write dw in terms of du and dv: dw = du wdv v = du u v dv v = vdu udv v 2, which is the desired identity. Example 1.4. It is well-known that the connection between the polar coordinates (r, θ) and the rectangular coordinates (x, y) is given by Find dθ in rectangular coordinates. x = r cos θ, y = r sin θ. (1.1) Solution: We have dx = d(r cos θ) = cos θ dr + r d cos θ = cos θ dr + r( sin θ) dθ, and similarly dy = sin θ dr + r cos θ dθ. This gives sin θ dx + cos θ dy = r dθ. Replacing sin θ by y/r and cos θ by x/r, we have dθ = 1 ( y r r dx + x ) r dy = ydx + xdy r 2 3 = xdy ydx x 2 + y 2, (1.2)
4 which is the final answer. For an obvious reason, dθ is called an angular form. As we will see at the end of the present section, putting the angular form as dθ is misleading. Identity (1.2) will be needed in the future for defining winding number of a loop around the origin. Example 1.5. Find the equation of the tangent line to the ellipse at a point (x 0, y 0 ) on this ellipse. x 2 a 2 + y2 = 1. (1.3) b2 Solution: Applying d to both sides of (1.3), we obtain 2x 2y dx + dy = 0, (1.4) a2 b2 which gives dy/dx = b 2 x/a 2 y. The slope of the required tangent line is dy/dx at (x 0, y 0 ), that is, b 2 x 0 /a 2 y 0. So the equation for the tangent line is y y 0 = ( b 2 x 0 /a 2 y 0 )(x x 0 ), or (x 0 x x 2 0)/a 2 + (y 0 y y 2 0)/b 2 = 0, which can be rewritten as x 0 x a 2 due to the fact that (x 0, y 0 ) is on the ellipse. + y 0y = 1, (1.5) b2 In the rest of the present section we describe how to compute a line integral ω, where is a path and ω is a differential form. Suppose that the variables in ω are x 1, x 2,..., x n, which can be put together as a vector variable x = (x 1, x 2,..., x n ). We may write ω as ω = n k=1 F kdx k where F k = F k (x) F k (x 1, x 2,..., x n ) are functions in variables x j s. The path can be described in parametric equations as x k = x k (t), 1 k n, or, in vector form, x = x(t), where the parameter t is running in some interval, say I = [a, b]. First consider the special case n = 1 and write ω = F (x)dx. When is the usual path in the real line moving from a to b, that is, x x(t) = t (a t b), the integral ω clearly should be interpreted as the usual definite integral b F (t)dt. In the general a case, we pull back the differential form ω in n variables x j s via to get a differential form in one variable t, denoted by ω, and then define ω as b a ω. The pullback ω of ω here is obtained by the substitutions x j = x j (t) (1 j n) into ω. Thus, when ω = n k=1 F kdx k as before, we have ω = n k=1 F k(x(t))dx k (t) = n k=1 F k(x(t))x k (t)dt. The line integral ω is, by definition, the integral of ω over the interval [a, b] : ω = b a ω, with ω = n k=1 F k(x(t)) dx k dt dt, 4 where ω = n k=1 F k(x)dx k.
5 Using this definition, the actual computation of line integrals is quite straightforward. Example 1.6. Compute the line integral xdy + ydz + zdx for three paths linking the origin (0, 0, 0) to the point (1, 1, 1), the first path being = α : x = t, y = t, z = t (0 t 1), the second being = β : x = t, y = t 2, z = t 3 (0 t 1), and the third being = η : x = t 2, y = t 4, z = t 6 (0 t 1). Solution: Put ω = xdy + ydz + zdx. Do the pull-backs α ω = tdt + tdt + tdt = 3tdt, β ω = td(t 2 ) + t 2 d(t 3 ) + t 3 dt = (2t 2 + 3t 4 + t 3 )dt and η ω = (4t 5 + 6t 9 + 2t 7 )dt. Hence we get α ω = 1 0 3tdt = 3 2, β ω = 1 0 (2t2 + 3t 4 + t 3 )dt = = 91 60, and η ω = 1 0 (4t5 + 6t 9 + 2t 7 )dt = 91 60, the same as ω. Notice that is just a reparametrization of β β. This indicates that the fact that a line integral in general depends on the path but not on its parametrization. Example 1.7. Compute the line integral xdy ydx ω x 2 + y 2, where is the path of going around the unit circle once in the anti-clockwise direction described by the parametric equations x = cos t and y = sin t with 0 t 2π, Solution: The differential form ω here is just the angular form dθ in Example 1.4. So the line integral here is the change of the polar angle θ when a point moves around the circle once. Thus the answer should be 2π. But let us just follow the definition described above to compute this line integral. The pull-back ω is ω = x(t)dy(t) y(t)dx(t) x(t) 2 + y(t) 2 = cos t d sin t sin t d cos t cos 2 t + sin 2 t = dt. So ω = 2π dt = 2π. Notice that if goes around the unit circle n times instead of 0 once, then the resulting line integral is 2πn dt = 2πn. 0 A path in the xy plane described by parametric equations x = x(t) and y = y(t) with a t b is called a closed path or a loop if its end points coincide, that is, (x(a), x(b)) = (y(a), y(b)). Given a loop not passing the origin, the integral (2π) 1 ω (where ω is the angular form) is called the winding number of about the origin and is denoted by W(, 0): W(, 0) = 1 ω 1 2π 2π 5 xdy ydx x 2 + y 2.
6 The above example tells is that the winding number of the loop going around the unit circles n times is exactly n. In 3.1 we will see that W(, 0) is always an integer. Forming pullbacks is a nice and easy operation. For example, we have Rule PB1. (g f) ω = f (g ω). Here, g f is the composite of g and f: if f sends x = (x 1,..., x n ) to y = (y 1,..., y m ) and g sends y to z = (z 1,..., z l ), then g f sends x to z: (g f)(x) = z = g(y) = g(f(x)). For a differential form ω in the z-space, g ω is a differential form in y-space with every occurrence of z j (1 j l) in ω replaced by z j = g j (y 1,..., y n ). Similarly, the pull back f (g ω) of g ω is a differential form in x-space obtained by another substitution. Certainly the result of consecutive substitutions by g followed by f is the same as the single substitution by their composite g f. So Rule PB1 is clear. From this rule we can deduce the identity ω = g ω. Indeed, g g ω = b a (g ) ω = b a (g ω) = g ω. A differential form ω is said to be exact if it is the differential of some function f, that is ω = df. In that case, the pull back of ω = df = n k=1 ( f/ x k)dx k is given by ω = n f k=1 x k dx k (t) = d(f(x(t))) d f(x(t)) dt (1.7) x(t) dt (the subscript x(t) indicates the point at which f/ x k is evaluated) and hence we have ω = b a ω = b d a dtf(x(t)) dt = f(x(b)) f(x(a)). We conclude: df = f(the terminal point of ) f(the initial point of ). (1.8) In particular, if is a loop, that is, x(b) = x(a), then ω = 0. Thus the line integral ω depends only on the end points but not on the path linking them. This may be called the path-independence property of line integrals for exact forms. The differential form xdy + ydz + zdx in Example 1.6 does not have this property, because its integrals along two paths α and β with same end points are different. Notice that (1.7) actually shows that df = d f, where f, as a function of t is f(x(t)). This is not surprising because forming pull backs (or substitution) and taking differentials are independent actions like kicking and punching. In general, we have 6
7 Rule PB2. g df = d(g f) d(f g). Now we explain why putting the angular form as dθ is problematic. In rectangular coordinates, the angular form is ω = ( ydx + xdy)/(x 2 + y 2 ), which is defined everywhere except the origin. The expression dθ wrongfully suggests the exactness of ω. If ω were exact, then ω would be zero for any closed path. But, as shown in Example 1.7, when goes around the unit circle once in anticlockwise direction, the line integral ω turns out to be 2π, not 0. The trouble here is caused by the fact that, although ω is defined on the punctured plane R 2 \{(0, 0)} (the Euclidean plane with the origin removed), θ cannot be properly defined on the punctured plane. We may let the domain D for θ be the plane with the negative part of the x-axis removed: D = {(x, y) : y 0 or x > 0}. Then θ becomes a genuine function defined on D satisfying π < θ < π, called the principal value of arg. Exercises 1. Calculate the following differentials: (a) d(e x2 /2 ), (b) d(log log x), (c) d ( sin x) y, (d) d (x y ), (e) d(log r), where r = x 2 + y 2 + z Find the tangent plane to the sphere x 2 + y 2 + z 2 = 3 2 at the point P 0 = (2, 2, 1) by using the method described in the remark after Example Derive the product rule d(uv) = udv + vdu from d(u 2 ) = 2udu. (Hint: compute d (u + v) 2 in two different ways.) 4. Derive the product rule for three functions: d(uvw) = uvdw + uwdv + vwdu. Write an expression for d(u 1 u 2 u n ), giving the product rule for n functions. 5. The logarithmic differential of u, denoted by lu here (this is not a standard notation), is defined to be du/u. Verify each of the following identities: (a) l(uv) = lu + lv, (b) l(u/v) = lu lv, (c) l(u v ) = v((log u)lv + lu), (d) l(e v ) = v lv and (e) l log v = lv/ log v. Part (a) can be generalized as l(u 1 u 2 u n ) = lu 1 + lu lu n. Does it help you to write down the product rule for n functions, which is asked in the last exercise? 6. A point P = (x, t) in the xt-plane is said to be inside the future cone C + if t > 0 and t 2 x 2 > 0. The hyperbolic distance of such a point to the origin is given by r = t 2 x 2. The hyperbolic coordinates (r, θ) of P is related to (x, t) by x = r sinh θ, 7
8 t = r cosh θ, where sinh θ = (e θ e θ )/2, cosh θ = (e θ + e θ )/2 are the hyperbolic sine and cosine functions. Find dr and dθ in terms of dx and dt. 7. Compute the differentials dr x, dr y, dθ x and dθ y, where r x = x r, r y = y r, θ x = y r 2, θ y = x r 2, ( r = x 2 + y 2 ). (This exercise will be useful for computing the Laplacian in polar coordinates.) 8. Consider the spherical coordinates (r, θ, φ) of a point P, where r is the distance to the origin, θ is the longitude and φ is the latitude of P. Its relation to the rectangular coordinates is given by x = r cos θ cos φ, y = r sin θ cos φ and z = r sin φ. (The convention of spherical coordinates here differs from some books.) Express the differentials dr, dθ and dφ in rectangular coordinates. (Hint: The computation is rather tedious. However, Example 1.2 helps.) 9. Compute the line integral xdy ydx, where is the path given by x = cos2 t, y = sin 2 t (0 t π/2). x + y = 1. Note that moves from (1, 0) to (0, 1) along the line What happens when you change the parametrization of this path to x = 1 t, y = t, or to x = 1 t 2, y = t 2? 10. For a loop, why is xdy + ydx always zero, while xdy ydx in general is not? For a loop, we have xdy = 1 2 (xdy ydx). Why? 11. We have seen that if a differential form ω is exact, then any line integral ω depends only on the end points of. Prove that the converse of this statement is also true by following the steps described below. Assume that ω = F 1 dx 1 + F 2 dx F n dx n is a differential form having the property that any line integral depends only on the end points of. Fix any point p in the domain of ω. For any point x, let f(x) = ω, where is any path from p to x. This definition of f(x) makes sense because of the assumption on ω here. Complete the proof by checking f/ x k = F k, which uses f x k = lim x h 0 f(x + he j ) f(x) h where e j is the jth vector of the natural basis of R n, namely e j = (δ j1, δ j2..., δ jn ), where δ jk is Kronecker s delta: it is 0 if j k and is 1 if j = k. ω 8
Rule ST1 (Symmetry). α β = β α for 1-forms α and β. Like the exterior product, the symmetric tensor product is also linear in each slot :
2. Metrics as Symmetric Tensors So far we have studied exterior products of 1-forms, which obey the rule called skew symmetry: α β = β α. There is another operation for forming something called the symmetric
More information3. The Theorems of Green and Stokes
3. The Theorems of Green and tokes Consider a 1-form ω = P (x, y)dx + Q(x, y)dy on the rectangle R = [a, b] [c, d], which consists of points (x, y) satisfying a x b and c y d. The boundary R of R have
More informationParametric Curves. Calculus 2 Lia Vas
Calculus Lia Vas Parametric Curves In the past, we mostly worked with curves in the form y = f(x). However, this format does not encompass all the curves one encounters in applications. For example, consider
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
3 Differentiation Rules 3.1 The Derivative of Polynomial and Exponential Functions In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions.
More informationCOMPLEX DIFFERENTIAL FORMS. 1. Complex 1-forms, the -operator and the Winding Number
CHAPTER 3. COMPLEX DIFFERENTIAL FORMS 1. Complex 1-forms, the -operator and the Winding Number Now we consider differentials for complex functions. Take a complex-valued function defined on an open region
More informationVII. Techniques of Integration
VII. Techniques of Integration Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given
More informationHOMEWORK 8 SOLUTIONS
HOMEWOK 8 OLUTION. Let and φ = xdy dz + ydz dx + zdx dy. let be the disk at height given by: : x + y, z =, let X be the region in 3 bounded by the cone and the disk. We orient X via dx dy dz, then by definition
More information1 The Derivative and Differrentiability
1 The Derivative and Differrentiability 1.1 Derivatives and rate of change Exercise 1 Find the equation of the tangent line to f (x) = x 2 at the point (1, 1). Exercise 2 Suppose that a ball is dropped
More informationj=1 ωj k E j. (3.1) j=1 θj E j, (3.2)
3. Cartan s Structural Equations and the Curvature Form Let E,..., E n be a moving (orthonormal) frame in R n and let ωj k its associated connection forms so that: de k = n ωj k E j. (3.) Recall that ωj
More informationPractice Differentiation Math 120 Calculus I Fall 2015
. x. Hint.. (4x 9) 4x + 9. Hint. Practice Differentiation Math 0 Calculus I Fall 0 The rules of differentiation are straightforward, but knowing when to use them and in what order takes practice. Although
More informationMath F15 Rahman
Math - 9 F5 Rahman Week3 7.3 Hyperbolic Functions Hyperbolic functions are similar to trigonometric functions, and have the following definitions: sinh x = (ex e x ) cosh x = (ex + e x ) tanh x = sinh
More informationMath 162: Calculus IIA
Math 62: Calculus IIA Final Exam ANSWERS December 9, 26 Part A. (5 points) Evaluate the integral x 4 x 2 dx Substitute x 2 cos θ: x 8 cos dx θ ( 2 sin θ) dθ 4 x 2 2 sin θ 8 cos θ dθ 8 cos 2 θ cos θ dθ
More informationNOTES ON DIFFERENTIAL FORMS. PART 1: FORMS ON R n
NOTES ON DIFFERENTIAL FORMS. PART 1: FORMS ON R n 1. What is a form? Since we re not following the development in Guillemin and Pollack, I d better write up an alternate approach. In this approach, we
More informationComplex Differentials and the Stokes, Goursat and Cauchy Theorems
Complex Differentials and the Stokes, Goursat and Cauchy Theorems Benjamin McKay June 21, 2001 1 Stokes theorem Theorem 1 (Stokes) f(x, y) dx + g(x, y) dy = U ( g y f ) dx dy x where U is a region of the
More information2016 FAMAT Convention Mu Integration 1 = 80 0 = 80. dx 1 + x 2 = arctan x] k2
6 FAMAT Convention Mu Integration. A. 3 3 7 6 6 3 ] 3 6 6 3. B. For quadratic functions, Simpson s Rule is eact. Thus, 3. D.. B. lim 5 3 + ) 3 + ] 5 8 8 cot θ) dθ csc θ ) dθ cot θ θ + C n k n + k n lim
More informationCHAPTER 6 VECTOR CALCULUS. We ve spent a lot of time so far just looking at all the different ways you can graph
CHAPTER 6 VECTOR CALCULUS We ve spent a lot of time so far just looking at all the different ways you can graph things and describe things in three dimensions, and it certainly seems like there is a lot
More information1.4 Techniques of Integration
.4 Techniques of Integration Recall the following strategy for evaluating definite integrals, which arose from the Fundamental Theorem of Calculus (see Section.3). To calculate b a f(x) dx. Find a function
More informationCalculus II. Calculus II tends to be a very difficult course for many students. There are many reasons for this.
Preface Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my class notes they should be accessible to anyone wanting to learn Calculus
More information1 Exponential Functions Limit Derivative Integral... 5
Contents Eponential Functions 3. Limit................................................. 3. Derivative.............................................. 4.3 Integral................................................
More informationf dr. (6.1) f(x i, y i, z i ) r i. (6.2) N i=1
hapter 6 Integrals In this chapter we will look at integrals in more detail. We will look at integrals along a curve, and multi-dimensional integrals in 2 or more dimensions. In physics we use these integrals
More informationLecture 10. (2) Functions of two variables. Partial derivatives. Dan Nichols February 27, 2018
Lecture 10 Partial derivatives Dan Nichols nichols@math.umass.edu MATH 233, Spring 2018 University of Massachusetts February 27, 2018 Last time: functions of two variables f(x, y) x and y are the independent
More informationMATH141: Calculus II Exam #4 review solutions 7/20/2017 Page 1
MATH4: Calculus II Exam #4 review solutions 7/0/07 Page. The limaçon r = + sin θ came up on Quiz. Find the area inside the loop of it. Solution. The loop is the section of the graph in between its two
More informationworked out from first principles by parameterizing the path, etc. If however C is a A path C is a simple closed path if and only if the starting point
III.c Green s Theorem As mentioned repeatedly, if F is not a gradient field then F dr must be worked out from first principles by parameterizing the path, etc. If however is a simple closed path in the
More informationECON 186 Class Notes: Derivatives and Differentials
ECON 186 Class Notes: Derivatives and Differentials Jijian Fan Jijian Fan ECON 186 1 / 27 Partial Differentiation Consider a function y = f (x 1,x 2,...,x n ) where the x i s are all independent, so each
More informationMSM120 1M1 First year mathematics for civil engineers Revision notes 4
MSM10 1M1 First year mathematics for civil engineers Revision notes 4 Professor Robert A. Wilson Autumn 001 Series A series is just an extended sum, where we may want to add up infinitely many numbers.
More informationCALCULUS PROBLEMS Courtesy of Prof. Julia Yeomans. Michaelmas Term
CALCULUS PROBLEMS Courtesy of Prof. Julia Yeomans Michaelmas Term The problems are in 5 sections. The first 4, A Differentiation, B Integration, C Series and limits, and D Partial differentiation follow
More informationPhysics 202 Laboratory 3. Root-Finding 1. Laboratory 3. Physics 202 Laboratory
Physics 202 Laboratory 3 Root-Finding 1 Laboratory 3 Physics 202 Laboratory The fundamental question answered by this week s lab work will be: Given a function F (x), find some/all of the values {x i }
More informationMath 31CH - Spring Final Exam
Math 3H - Spring 24 - Final Exam Problem. The parabolic cylinder y = x 2 (aligned along the z-axis) is cut by the planes y =, z = and z = y. Find the volume of the solid thus obtained. Solution:We calculate
More informationintegration integration
13 Contents integration integration 1. Basic concepts of integration 2. Definite integrals 3. The area bounded by a curve 4. Integration by parts 5. Integration by substitution and using partial fractions
More informationMATH 18.01, FALL PROBLEM SET # 8
MATH 18.01, FALL 01 - PROBLEM SET # 8 Professor: Jared Speck Due: by 1:45pm on Tuesday 11-7-1 (in the boxes outside of Room -55 during the day; stick it under the door if the room is locked; write your
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More information1 Functions of many variables.
MA213 Sathaye Notes on Multivariate Functions. 1 Functions of many variables. 1.1 Plotting. We consider functions like z = f(x, y). Unlike functions of one variable, the graph of such a function has to
More informationSolutions to Homework 11
Solutions to Homework 11 Read the statement of Proposition 5.4 of Chapter 3, Section 5. Write a summary of the proof. Comment on the following details: Does the proof work if g is piecewise C 1? Or did
More informationContents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9
MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9 MATH 32B-2 (8W)
More informationGEORGE ANDROULAKIS THE 7 INDETERMINATE FORMS OF LIMITS : usually we use L Hospital s rule. Two important such limits are lim
MATH 4 (CALCULUS II) IN ORDER TO OBTAIN A PERFECT SCORE IN ANDROULAKIS MATH 4 CLASS YOU NEED TO MEMORIZE THIS HANDOUT AND SOLVE THE ASSIGNED HOMEWORK ON YOUR OWN GEORGE ANDROULAKIS TRIGONOMETRY θ sin(θ)
More informationfor 2 1/3 < t 3 1/3 parametrizes
Solution to Set 4, due Friday April 1) Section 5.1, Problem ) Explain why the path parametrized by ft) = t, t 1 ) is not smooth. Note this is true more specifically if the interval of t contains t = 1
More informationIntegration by Parts. MAT 126, Week 2, Thursday class. Xuntao Hu
MAT 126, Week 2, Thursday class Xuntao Hu Recall that the substitution rule is a combination of the FTC and the chain rule. We can also combine the FTC and the product rule: d dx [f (x)g(x)] = f (x)g (x)
More informationMath 122 Test 3. April 15, 2014
SI: Math 1 Test 3 April 15, 014 EF: 1 3 4 5 6 7 8 Total Name Directions: 1. No books, notes or 6 year olds with ear infections. You may use a calculator to do routine arithmetic computations. You may not
More informationChapter 2: Differentiation
Chapter 2: Differentiation Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 82 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L
More informationMath 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3
Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some
More informationMATH 18.01, FALL PROBLEM SET # 2
MATH 18.01, FALL 2012 - PROBLEM SET # 2 Professor: Jared Speck Due: by Thursday 4:00pm on 9-20-12 (in the boxes outside of Room 2-255 during the day; stick it under the door if the room is locked; write
More information7.1. Calculus of inverse functions. Text Section 7.1 Exercise:
Contents 7. Inverse functions 1 7.1. Calculus of inverse functions 2 7.2. Derivatives of exponential function 4 7.3. Logarithmic function 6 7.4. Derivatives of logarithmic functions 7 7.5. Exponential
More informationIntroduction to Differentials
Introduction to Differentials David G Radcliffe 13 March 2007 1 Increments Let y be a function of x, say y = f(x). The symbol x denotes a change or increment in the value of x. Note that a change in the
More information1 Introduction; Integration by Parts
1 Introduction; Integration by Parts September 11-1 Traditionally Calculus I covers Differential Calculus and Calculus II covers Integral Calculus. You have already seen the Riemann integral and certain
More informationMath 32B Discussion Session Week 10 Notes March 14 and March 16, 2017
Math 3B iscussion ession Week 1 Notes March 14 and March 16, 17 We ll use this week to review for the final exam. For the most part this will be driven by your questions, and I ve included a practice final
More informationChapter 2: Differentiation
Chapter 2: Differentiation Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 75 2.1 Tangent Lines and Their Slopes This section deals with the problem of finding a straight line L
More informationMAT137 Calculus! Lecture 6
MAT137 Calculus! Lecture 6 Today: 3.2 Differentiation Rules; 3.3 Derivatives of higher order. 3.4 Related rates 3.5 Chain Rule 3.6 Derivative of Trig. Functions Next: 3.7 Implicit Differentiation 4.10
More informationArc Length and Riemannian Metric Geometry
Arc Length and Riemannian Metric Geometry References: 1 W F Reynolds, Hyperbolic geometry on a hyperboloid, Amer Math Monthly 100 (1993) 442 455 2 Wikipedia page Metric tensor The most pertinent parts
More informationSummary: Primer on Integral Calculus:
Physics 2460 Electricity and Magnetism I, Fall 2006, Primer on Integration: Part I 1 Summary: Primer on Integral Calculus: Part I 1. Integrating over a single variable: Area under a curve Properties of
More informationDifferential calculus. Background mathematics review
Differential calculus Background mathematics review David Miller Differential calculus First derivative Background mathematics review David Miller First derivative For some function y The (first) derivative
More informationArchive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma
Archive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma This is an archive of past Calculus IV exam questions. You should first attempt the questions without looking
More informationν(u, v) = N(u, v) G(r(u, v)) E r(u,v) 3.
5. The Gauss Curvature Beyond doubt, the notion of Gauss curvature is of paramount importance in differential geometry. Recall two lessons we have learned so far about this notion: first, the presence
More informationCalculus II Practice Test Problems for Chapter 7 Page 1 of 6
Calculus II Practice Test Problems for Chapter 7 Page of 6 This is a set of practice test problems for Chapter 7. This is in no way an inclusive set of problems there can be other types of problems on
More informationCalculus & Analytic Geometry I
TQS 124 Autumn 2008 Quinn Calculus & Analytic Geometry I The Derivative: Analytic Viewpoint Derivative of a Constant Function. For c a constant, the derivative of f(x) = c equals f (x) = Derivative of
More informationVector Calculus, Maths II
Section A Vector Calculus, Maths II REVISION (VECTORS) 1. Position vector of a point P(x, y, z) is given as + y and its magnitude by 2. The scalar components of a vector are its direction ratios, and represent
More informationMathematics Engineering Calculus III Fall 13 Test #1
Mathematics 2153-02 Engineering Calculus III Fall 13 Test #1 Instructor: Dr. Alexandra Shlapentokh (1) Which of the following statements is always true? (a) If x = f(t), y = g(t) and f (1) = 0, then dy/dx(1)
More informationAssignment 13 Assigned Mon Oct 4
Assignment 3 Assigned Mon Oct 4 We refer to the integral table in the back of the book. Section 7.5, Problem 3. I don t see this one in the table in the back of the book! But it s a very easy substitution
More informationDay 3 Review of Basic Calculus
Day 3 Review of Basic Calculus Sivaram sivaambi@stanford.edu Institute of Computational and Mathematical Engineering Stanford University September 21, 2011 Differential Calculus Product rule Quotient rule
More informationMath 122 Test 3. April 17, 2018
SI: Math Test 3 April 7, 08 EF: 3 4 5 6 7 8 9 0 Total Name Directions:. No books, notes or April showers. You may use a calculator to do routine arithmetic computations. You may not use your calculator
More informationINTEGRATING RADICALS
INTEGRATING RADICALS MATH 53, SECTION 55 (VIPUL NAIK) Corresponding material in the book: Section 8.4. What students should already know: The definitions of inverse trigonometric functions. The differentiation
More information6.7 Hyperbolic Functions
6.7 6.7 Hyperbolic Functions Even and Odd Parts of an Exponential Function We recall that a function f is called even if f( x) = f(x). f is called odd if f( x) = f(x). The sine function is odd while the
More information1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is
1. The value of the double integral (a) 15 26 (b) 15 8 (c) 75 (d) 105 26 5 4 0 1 1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is 2. What is the value of the double integral interchange the order
More informationA-Level Mathematics DIFFERENTIATION I. G. David Boswell - Math Camp Typeset 1.1 DIFFERENTIATION OF POLYNOMIALS. d dx axn = nax n 1, n!
A-Level Mathematics DIFFERENTIATION I G. David Boswell - Math Camp Typeset 1.1 SET C Review ~ If a and n are real constants, then! DIFFERENTIATION OF POLYNOMIALS Problems ~ Find the first derivative of
More informationP1 Calculus II. Partial Differentiation & Multiple Integration. Prof David Murray. dwm/courses/1pd
P1 2017 1 / 39 P1 Calculus II Partial Differentiation & Multiple Integration Prof David Murray david.murray@eng.ox.ac.uk www.robots.ox.ac.uk/ dwm/courses/1pd 4 lectures, MT 2017 P1 2017 2 / 39 Motivation
More information1.1. BASIC ANTI-DIFFERENTIATION 21 + C.
.. BASIC ANTI-DIFFERENTIATION and so e x cos xdx = ex sin x + e x cos x + C. We end this section with a possibly surprising complication that exists for anti-di erentiation; a type of complication which
More informationQ You mentioned that in complex analysis we study analytic functions, or, in another name, holomorphic functions. Pray tell me, what are they?
COMPLEX ANALYSIS PART 2: ANALYTIC FUNCTIONS Q You mentioned that in complex analysis we study analytic functions, or, in another name, holomorphic functions. Pray tell me, what are they? A There are many
More informationThe Derivative of a Function Measuring Rates of Change of a function. Secant line. f(x) f(x 0 ) Average rate of change of with respect to over,
The Derivative of a Function Measuring Rates of Change of a function y f(x) f(x 0 ) P Q Secant line x 0 x x Average rate of change of with respect to over, " " " " - Slope of secant line through, and,
More informationMA 242 Review Exponential and Log Functions Notes for today s class can be found at
MA 242 Review Exponential and Log Functions Notes for today s class can be found at www.xecu.net/jacobs/index242.htm Example: If y = x n If y = x 2 then then dy dx = nxn 1 dy dx = 2x1 = 2x Power Function
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals 8. Basic Integration Rules In this section we will review various integration strategies. Strategies: I. Separate
More information1 Review of di erential calculus
Review of di erential calculus This chapter presents the main elements of di erential calculus needed in probability theory. Often, students taking a course on probability theory have problems with concepts
More informationChapter 7: Techniques of Integration
Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration
More informationExact Solutions of the Einstein Equations
Notes from phz 6607, Special and General Relativity University of Florida, Fall 2004, Detweiler Exact Solutions of the Einstein Equations These notes are not a substitute in any manner for class lectures.
More information(a) The points (3, 1, 2) and ( 1, 3, 4) are the endpoints of a diameter of a sphere.
MATH 4 FINAL EXAM REVIEW QUESTIONS Problem. a) The points,, ) and,, 4) are the endpoints of a diameter of a sphere. i) Determine the center and radius of the sphere. ii) Find an equation for the sphere.
More information= π + sin π = π + 0 = π, so the object is moving at a speed of π feet per second after π seconds. (c) How far does it go in π seconds?
Mathematics 115 Professor Alan H. Stein April 18, 005 SOLUTIONS 1. Define what is meant by an antiderivative or indefinite integral of a function f(x). Solution: An antiderivative or indefinite integral
More informationIntegration by Parts
Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u
More information9.4 CALCULUS AND PARAMETRIC EQUATIONS
9.4 Calculus with Parametric Equations Contemporary Calculus 1 9.4 CALCULUS AND PARAMETRIC EQUATIONS The previous section discussed parametric equations, their graphs, and some of their uses for visualizing
More informationCHAPTER 7: TECHNIQUES OF INTEGRATION
CHAPTER 7: TECHNIQUES OF INTEGRATION DAVID GLICKENSTEIN. Introduction This semester we will be looking deep into the recesses of calculus. Some of the main topics will be: Integration: we will learn how
More informationM273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3
M7Q Multivariable alculus Spring 7 Review Problems for Exam Exam covers material from Sections 5.-5.4 and 6.-6. and 7.. As you prepare, note well that the Fall 6 Exam posted online did not cover exactly
More informationMAS153/MAS159. MAS153/MAS159 1 Turn Over SCHOOL OF MATHEMATICS AND STATISTICS hours. Mathematics (Materials) Mathematics For Chemists
Data provided: Formula sheet MAS53/MAS59 SCHOOL OF MATHEMATICS AND STATISTICS Mathematics (Materials Mathematics For Chemists Spring Semester 203 204 3 hours All questions are compulsory. The marks awarded
More informationLimit. Chapter Introduction
Chapter 9 Limit Limit is the foundation of calculus that it is so useful to understand more complicating chapters of calculus. Besides, Mathematics has black hole scenarios (dividing by zero, going to
More informationTHE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK. Summer Examination 2009.
OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK Summer Examination 2009 First Engineering MA008 Calculus and Linear Algebra
More informationJUST THE MATHS UNIT NUMBER DIFFERENTIATION 4 (Products and quotients) & (Logarithmic differentiation) A.J.Hobson
JUST THE MATHS UNIT NUMBER 104 DIFFERENTIATION 4 (Products and quotients) & (Logarithmic differentiation) by AJHobson 1041 Products 1042 Quotients 1043 Logarithmic differentiation 1044 Exercises 1045 Answers
More information3. On the grid below, sketch and label graphs of the following functions: y = sin x, y = cos x, and y = sin(x π/2). π/2 π 3π/2 2π 5π/2
AP Physics C Calculus C.1 Name Trigonometric Functions 1. Consider the right triangle to the right. In terms of a, b, and c, write the expressions for the following: c a sin θ = cos θ = tan θ =. Using
More information4.1 Analysis of functions I: Increase, decrease and concavity
4.1 Analysis of functions I: Increase, decrease and concavity Definition Let f be defined on an interval and let x 1 and x 2 denote points in that interval. a) f is said to be increasing on the interval
More informationChapter 5: Integrals
Chapter 5: Integrals Section 5.3 The Fundamental Theorem of Calculus Sec. 5.3: The Fundamental Theorem of Calculus Fundamental Theorem of Calculus: Sec. 5.3: The Fundamental Theorem of Calculus Fundamental
More informationHOMEWORK 3 MA1132: ADVANCED CALCULUS, HILARY 2017
HOMEWORK MA112: ADVANCED CALCULUS, HILARY 2017 (1) A particle moves along a curve in R with position function given by r(t) = (e t, t 2 + 1, t). Find the velocity v(t), the acceleration a(t), the speed
More informationMath 121: Calculus 1 - Fall 2013/2014 Review of Precalculus Concepts
Introduction Math 121: Calculus 1 - Fall 201/2014 Review of Precalculus Concepts Welcome to Math 121 - Calculus 1, Fall 201/2014! This problems in this packet are designed to help you review the topics
More information1. If the line l has symmetric equations. = y 3 = z+2 find a vector equation for the line l that contains the point (2, 1, 3) and is parallel to l.
. If the line l has symmetric equations MA 6 PRACTICE PROBLEMS x = y = z+ 7, find a vector equation for the line l that contains the point (,, ) and is parallel to l. r = ( + t) i t j + ( + 7t) k B. r
More information2.2 The derivative as a Function
2.2 The derivative as a Function Recall: The derivative of a function f at a fixed number a: f a f a+h f(a) = lim h 0 h Definition (Derivative of f) For any number x, the derivative of f is f x f x+h f(x)
More informationGeometry and Motion, MA 134 Week 1
Geometry and Motion, MA 134 Week 1 Mario J. Micallef Spring, 2007 Warning. These handouts are not intended to be complete lecture notes. They should be supplemented by your own notes and, importantly,
More informationHow to Use Calculus Like a Physicist
How to Use Calculus Like a Physicist Physics A300 Fall 2004 The purpose of these notes is to make contact between the abstract descriptions you may have seen in your calculus classes and the applications
More informationd F = (df E 3 ) E 3. (4.1)
4. The Second Fundamental Form In the last section we developed the theory of intrinsic geometry of surfaces by considering the covariant differential d F, that is, the tangential component of df for a
More informationMath 121: Calculus 1 - Winter 2012/2013 Review of Precalculus Concepts
Introduction Math 11: Calculus 1 - Winter 01/01 Review of Precalculus Concepts Welcome to Math 11 - Calculus 1, Winter 01/01! This problems in this packet are designed to help you review the topics from
More informationSolution. This is a routine application of the chain rule.
EXAM 2 SOLUTIONS 1. If z = e r cos θ, r = st, θ = s 2 + t 2, find the partial derivatives dz ds chain rule. Write your answers entirely in terms of s and t. dz and dt using the Solution. This is a routine
More informatione x3 dx dy. 0 y x 2, 0 x 1.
Problem 1. Evaluate by changing the order of integration y e x3 dx dy. Solution:We change the order of integration over the region y x 1. We find and x e x3 dy dx = y x, x 1. x e x3 dx = 1 x=1 3 ex3 x=
More information1. Compute the derivatives of the following functions, by any means necessary. f (x) = (1 x3 )(1/2)(x 2 1) 1/2 (2x) x 2 1( 3x 2 ) (1 x 3 ) 2
Math 51 Exam Nov. 4, 009 SOLUTIONS Directions 1. SHOW YOUR WORK and be thorough in your solutions. Partial credit will only be given for work shown.. Any numerical answers should be left in exact form,
More informationJim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt
Jim Lambers MAT 28 ummer emester 212-1 Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain
More informationDerivatives of Trig and Inverse Trig Functions
Derivatives of Trig and Inverse Trig Functions Math 102 Section 102 Mingfeng Qiu Nov. 28, 2018 Office hours I m planning to have additional office hours next week. Next Monday (Dec 3), which time works
More informationNote: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. = 3 2
Math Prelim II Solutions Spring Note: Each problem is worth points except numbers 5 and 6 which are 5 points. x. Compute x da where is the region in the second quadrant between the + y circles x + y and
More informationInstructions: No books. No notes. Non-graphing calculators only. You are encouraged, although not required, to show your work.
Exam 3 Math 850-007 Fall 04 Odenthal Name: Instructions: No books. No notes. Non-graphing calculators only. You are encouraged, although not required, to show your work.. Evaluate the iterated integral
More information