Calculus III. Philippe Rukimbira. Department of Mathematics Florida International University PR (FIU) MAC / 50
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1 Calculus III Philippe Rukimbira Department of Mathematics Florida International University PR (FIU) MAC / 50
2 12.3. Change of parameter; Arc Length If t : R R, u t(u) is one-to-one, then r(t) r(u) is called a re-parametrization. Example: r(t) = (3 cos t, 3 sin t, 0), 0 t 2π t : R R, u t(u) = u 3 r(u) = (3 cos u 3, 3 sin u, 0), 0 u 6π 3 PR (FIU) MAC / 50
3 As a set of points, the curve is the same, a circle; just it takes 3 times longer to cover the circle. PR (FIU) MAC / 50
4 Chain rule r(t), t = t(u) dr du = dr dt dt du = dt dr du dt. PR (FIU) MAC / 50
5 Chain rule r(t), t = t(u) dr du = dr dt dt du = dt dr du dt. Example: r(t) = (3 cos t, 3 sin t, 0), t(u) = u 3 dr du = 1 ( 3 sin t, 3 cos t, 0) = ( sin t, cos t, 0). 3 PR (FIU) MAC / 50
6 Arc length parametrization r(t), a t b From Calculus II, the length L a,b of the arc of the curve from t = a to t = b is given by L a,b = b a r (t) dt. Example: r(t) = (4 + 3t) i + (2 2t) j + (5 + t) k. Find L 3,4. PR (FIU) MAC / 50
7 Arc length parametrization Let r(t) be a parameterized curve. Definition t is called an arc length parameter if r (t) = 1. The arc length parameter will be denoted by s. Proposition If r(s) is an arc length parametrization, then L a,b = b a r (s) ds = b a ds = b a. PR (FIU) MAC / 50
8 Re-parametrization by arc length r (t) = r (s) ds dt r (t) = r (s) ds dt = ds dt We see that ds = r (t) dt determines an arc length parameter. s = r (t) dt PR (FIU) MAC / 50
9 Example r(t) = (sin e t, cos e t, 3e t ), t 0. Re-parameterize by by arc length. PR (FIU) MAC / 50
10 Example r(t) = (sin e t, cos e t, 3e t ), t 0. Re-parameterize by by arc length. s = (e t cos e t ) 2 + (e t sin e t ) 2 + ( 3e t ) 2 dt = 2 e t dt = 2e t + s 0 0 = 2 + s 0 So s = 2(e t 1) PR (FIU) MAC / 50
11 Or t = ln ( s 2 + 1) The re-parameterized curve is r(s) = (sin ( s 2 + 1), cos (s 2 + 1), 3( s 2 + 1)). PR (FIU) MAC / 50
12 12.4. Unit Tangent, Normal and Bi-normal Vectors Let r(t) be a vector valued function. We remember that r (t) is a tangent vector. If r (t) O, we may define a tangent vector of magnitude 1, the unit tangent vector, T (t) = r (t) r (t). PR (FIU) MAC / 50
13 12.4. Unit Tangent, Normal and Bi-normal Vectors Let r(t) be a vector valued function. We remember that r (t) is a tangent vector. If r (t) O, we may define a tangent vector of magnitude 1, the unit tangent vector, T (t) = r (t) r (t). Example: Find the unit tangent vector for when t = π 2. r(t) = 4(cos t) i + 4(sin t) j + t k, PR (FIU) MAC / 50
14 12.4. Unit Tangent, Normal and Bi-normal Vectors Let r(t) be a vector valued function. We remember that r (t) is a tangent vector. If r (t) O, we may define a tangent vector of magnitude 1, the unit tangent vector, T (t) = r (t) r (t). Example: Find the unit tangent vector for r(t) = 4(cos t) i + 4(sin t) j + t k, when t = π 2. r (t) = 4 sin t i + 4 cos t j + k r (t) = 17 PR (FIU) MAC / 50
15 So T (t) = 1 17 ( 4 sin t i + 4 cos t j + k) T ( π 2 ) = 1 17 ( 4 i + k) PR (FIU) MAC / 50
16 We also remember that the derivative of a vector valued function of constant magnitude is perpendicular to the original. Therefore, T (t).t (t) = 0. We define the unit normal vector, N(t) by N(t) = T (t) T (t) PR (FIU) MAC / 50
17 Example Find the unit normal N for r(t) = (4 cos t, 4 sin t, t) at t = π 2. PR (FIU) MAC / 50
18 Example Find the unit normal N for r(t) = (4 cos t, 4 sin t, t) at t = π 2. T (t) = 1 17 ( 4 sin t, 4 cos t, 1) T (t) = 1 17 ( 4 cos t, 4 sin t, 0) N(t) = T (t) T (t) = 1 17 ( 4 cos t, 4 sin t, 0) 17 4 N( π ) = (0, 1, 0) 2 PR (FIU) MAC / 50
19 The bi-normal vector, B(t), is defined by Notice that B(t) = 1. (Why?) B(t) = T (t) N(t). PR (FIU) MAC / 50
20 The bi-normal vector, B(t), is defined by Notice that B(t) = 1. (Why?) B(t) = T (t) N(t). Remember that if s is the arc length parameter, then s = t 0 r (u) du hence ds = r (t). dt Since, by definition, T = r r (t), we obtain T = dr dt ds dt = dr ds. T (s) = dr ds PR (FIU) MAC / 50
21 12.5. Curvature We define the curvature K of the curve C which is the graph of r(t) by: K (s) = dt ds = d 2 r ds 2 PR (FIU) MAC / 50
22 Using the chain rule, one can show that, in arbitrary parameter t, the curvature is given by: K (t) = 1 r (t) dt dt = r (t) r (t) r (t) 3. PR (FIU) MAC / 50
23 Example Find the curvature of r(t) = (3 cos t, 4 sin t, t) at t = π 2. PR (FIU) MAC / 50
24 at t = π 2. r (t) = ( 3 sin t, 4 cos t, 1) Example Find the curvature of r(t) = (3 cos t, 4 sin t, t) r (t) = ( 3 cos t, 4 sin t, 0) r ( π ) = ( 3, 0, 1) 2 r ( π ) = (0, 4, 0) 2 K ( π ( 3, 0, 1) (0, 4, 0) ) = = 2 2 (10) PR (FIU) MAC / 50
25 12.6. Motion along a curve The acceleration By V (t), we denote the velocity velocity vector r (t) of a curve C which is the graph of r(t). The acceleration vector a(t) is given by a(t) = d dt V (t) = d 2 dt 2 r(t). Next, we look at a decomposition of the acceleration vector. PR (FIU) MAC / 50
26 12.6. Motion along a curve The acceleration By V (t), we denote the velocity velocity vector r (t) of a curve C which is the graph of r(t). The acceleration vector a(t) is given by a(t) = d dt V (t) = d 2 dt 2 r(t). Next, we look at a decomposition of the acceleration vector. Since V (t) = r (t) = ds dt T, where T is the unit tangent vector, we have: PR (FIU) MAC / 50
27 a = d dt (ds dt T ) = d 2 s dt 2 T + ds dt = d 2 s dt 2 T + ds dt dt dt ds dt dt ds = d 2 s dt 2 T + (ds dt )2 K N PR (FIU) MAC / 50
28 We have obtained a decomposition of a into the tangential and normal components a(t) = a T T + a N N = d 2 s dt 2 T + K r (t) 2 N Practically, we have the following formula: a T = a.v V ; a N = a V. V PR (FIU) MAC / 50
29 Example For the curve r(t) = (3t cos t, 3t sin t, 4t), find the tangential and normal vector components of the acceleration. PR (FIU) MAC / 50
30 Example For the curve r(t) = (3t cos t, 3t sin t, 4t), find the tangential and normal vector components of the acceleration. v(t) = (3 cos t 3t sin t, 3 sin t + 3t cos t, 4) a(t) = ( 6 sin t 3t cos t, 6 cos t 3t sin t, 0) a.v = 9t V = 9t a T = 9t 9t PR (FIU) MAC / 50
31 a T T = 9t 9t V a N N = a a T T PR (FIU) MAC / 50
32 12.7. Kepler s Laws of Planetary Motion 1. Law of Orbits : Orbits are conic sections with Sun at focus. 2. Law of Areas : area swept by ray Sun-planet in equal times are equal 3. Law of Periods: The square of the period is a constant times the cube of major axis (T 2 = Ka 3 ) PR (FIU) MAC / 50
33 13.1 Functions of two or more variables Examples are area of rectangles, volumes of rectangular boxes,... Graphs of functions of several variables Example: f (x, y) = 1 x 1 2 y Graph z = 1 x 1 2 y The graph is a plane. First plot the coordinates axis traces (x=0,y=0; x=0, z=0; y=0,z=0). Connect each pair with a straight line. PR (FIU) MAC / 50
34 Example 2: f (x, y) = We graph the positive section of 1 x 2 y 2 z 2 = 1 x 2 y 2. PR (FIU) MAC / 50
35 Level curves f (x, y), for a constant k, the graph of f (x, y) = k is a curve, called the level curve of height k. Example: f (x, y) = 4x 2 + y 2. The level curve of height 2 is 4x 2 + y 2 = 2 This is an ellipse in the horizontal plane z = 2. PR (FIU) MAC / 50
36 Level Surfaces Example: f (x, y, z) = z 2 y 2 x 2. The level surface of height 1 is z 2 y 2 x 2 = 1 This is a hyperboloid of two sheets! PR (FIU) MAC / 50
37 13.2 Limits and continuity Open and closed sets If D is a set of points in 2-dimensional Euclidean space, then, a point (x 0, y 0 ) is called an interior point of D if there is some circular disc with positive radius and centered at (x 0, y 0 ) and containing only points in D. PR (FIU) MAC / 50
38 A point (x 0, y 0 ) will be called a boundary point of D if any circular disc centered at (x 0, y 0 ) and with positive radius, contains both points in D and points not in D. A set D is said to be open, if everyone of its points is an interior point of D. A set D is said to be closed if its complement is open. PR (FIU) MAC / 50
39 Limits along curves Let f (x, y) be a function of two variables and r(t) = (x(t), y(t)) be a curve whose graph C contains r 0 = r(0) = (x 0, y 0 ). Definition lim f (x, y) = lim f (r(t)) = lim f (x(t), y(t)). (x,y) (x 0,y 0 ), along C t 0 t 0 PR (FIU) MAC / 50
40 Example C is the graph of y = x 2. What is f (x, y) = xy x 2 + y 2 lim f (x, y)? (x,y) (0,0) along C How about limit along C 1 : y = x? PR (FIU) MAC / 50
41 General limits Definition lim f (x, y) = L (x,y) (x 0,y 0 ) if and only if given ɛ > 0, we can find a number δ > 0 such that whenever 0 < f (x, y) L < ɛ (x x 0 ) 2 + (y y 0 ) 2 = d((x, y), (x 0, y 0 )) < δ PR (FIU) MAC / 50
42 Proposition If then for any curve C through (x 0, y 0 ). lim f (x, y) = L (x,y) (x 0,y 0 ) lim f (x, y) = L (x,y) (x 0,y 0 ) along C PR (FIU) MAC / 50
43 If for some curves C 1 and C 2 lim f (x, y) (x,y) (x 0,y 0 ) along C 1 is different from then doesn t exist. lim f (x, y) (x,y) (x 0,y 0 ) along C 2 lim f (x, y) (x,y) (x 0,y 0 ) PR (FIU) MAC / 50
44 Example doesn t exist because lim xy (x,y) (0,0) x 2 + y 2 lim f (x, y) = 0 (x,y) (0,0): along x=0 and lim = 1 (x,y) (0,0): along y=x PR (FIU) MAC / 50
45 Continuity The definition of continuity is the same as in Calculus I In computing limits, convert to polar coordinates if necessary. Example: sin x 2 + y 2 lim (x,y) (0,0) x 2 + y 2 lim (x,y,z) (2,0,1) sin r 2 = lim r 0 r 2 = 1. xz 2 x 2 + y 2 + z 2 = 2 5 PR (FIU) MAC / 50
46 In polar coordinates: 1 x 2 y 2 lim (x,y) (0,0) x 2 + y 2 1 r 2 lim r 0 r 2 the limit doesn t exist! Or = + (in the extended sense). PR (FIU) MAC / 50
47 lim (x,y) (0,0) e 1 x 2 +y 2 x 2 + y = lim e 1 r 2 r 0 r Let X = 1 r, then by l Hopital Rule! lim X + X e X = 0, PR (FIU) MAC / 50
48 Prove that lim e x 2 +y 2 +z 2 = 1 (x,y,z) (0,0,0) Given ɛ > 0, find δ > 0 such that e x 2 +y 2 +z 2 1 < ɛ whenever x 2 + y 2 + z 2 < δ e x 2 +y 2 +z 2 1 < ɛ iff e x 2 +y 2 +z 2 1 < ɛ e x 2 +y 2 +z 2 < ɛ + 1 x 2 + y 2 + z 2 < ln (1 + ɛ) One can take δ = ln (1 + ɛ) PR (FIU) MAC / 50
49 In computing limits: Simplify if applicable (Factorization) Use continuity of factors if helpful. Polar coordinates as last resort. PR (FIU) MAC / 50
50 Exercises on Continuity 1. Where is f (x, y, z) = 3x 2 e yz cos (xyz) continuous? PR (FIU) MAC / 50
51 Exercises on Continuity 1. Where is f (x, y, z) = 3x 2 e yz cos (xyz) continuous? Answer: Everywhere in R Where is f (x, y, z) = ln (4 x 2 y 2 z 2 ) continuous? PR (FIU) MAC / 50
52 Exercises on Continuity 1. Where is f (x, y, z) = 3x 2 e yz cos (xyz) continuous? Answer: Everywhere in R Where is f (x, y, z) = ln (4 x 2 y 2 z 2 ) continuous? 4 x 2 y 2 z 2 > 0 Equivalent to x 2 + y 2 + z 2 < 4 on the open ball of radius 2. PR (FIU) MAC / 50
53 3. What is the domain of continuity of f (x, y, z) = y + 1 x 2 + z 2 1? PR (FIU) MAC / 50
54 3. What is the domain of continuity of f (x, y, z) = y + 1 x 2 + z 2 1? x 2 + z The complement of an infinite cylinder of radius Where is f (x, y, z) = sin x 2 + y 2 + z 2 continuous? PR (FIU) MAC / 50
55 3. What is the domain of continuity of f (x, y, z) = y + 1 x 2 + z 2 1? x 2 + z The complement of an infinite cylinder of radius Where is f (x, y, z) = sin x 2 + y 2 + z 2 continuous? A composition of two functions: (x, y, z) x 2 + y 2 + z 2 and w sin w which are continuous everywhere, so is the composition! PR (FIU) MAC / 50
56 13.3 Partial Derivatives PR (FIU) MAC / 50
57 16.7 The divergence Theorem Definition of divergence as flux density The divergence of a vector field is the measure of "flux out" of a closed surface. The divergence or flux density of a vector field F is defined by div F(x, y, z) = In rectangular coordinates: Observation: lim volume(σ) 0 div F = F 1 x + F 2 y + F 3 z div curl F = 0 This can be shown directly (do the calculation!) σ F. ds (volume enclosed by σ) PR (FIU) MAC / 50
58 On a doubly connected region, one has the converse: If div G = 0, then G = curl F for some F. Examples of divergence free vector fields, or solenoidal vector fields, are the magnetic fields B. Maxwell s equation states exactly that: div B = 0. Another example: For a magnetic dipole (or a current loop), with constant dipole moment µ, Show that div B = 0. B = µ r 3 + 3( µ. r) r r 5, r 0 PR (FIU) MAC / 50
59 Alternate notation:with =< x, y, z > div F =. F PR (FIU) MAC / 50
60 The divergence Theorem expresses the total flux as the integral of the flux density. S F. n ds = W div F dv where S is the boundary of the region W, oriented by outer unit normals. PR (FIU) MAC / 50
61 Example: Use the divergence Theorem to find the flux of F(x, y, z) = (x 2 y, xy 2, z + 2) across the surface σ of the solid bounded above by the plane z = 2x and below by the paraboloid z = x 2 + y 2. PR (FIU) MAC / 50
62 16.8: Stokes Theorem Stokes Theorem expresses the total circulation of F as the integral of the circulation density: σ curl F. n ds = σ F. dr A special case of Stokes Theorem is Green s Theorem: F =< f, g, 0 > PR (FIU) MAC / 50
63 Example: Verify Stokes Theorem for F(x, y, z) =< x, y, z > where σ the upper hemisphere z = a 2 x 2 y 2 with upward orientation. PR (FIU) MAC / 50
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