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# Multivariable Calculus & Matrix Algebra (Herron): Week 6 Notes (2/29/16 - 3/4/16) MATH 2010

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This 6 page Class Notes was uploaded by creask on Saturday March 5, 2016. The Class Notes belongs to MATH 2010 at Rensselaer Polytechnic Institute taught by Isom Herron in Spring 2016. Since its upload, it has received 40 views. For similar materials see Multivariable calculus and matrix algebra in Mathematics (M) at Rensselaer Polytechnic Institute.

## Reviews for Multivariable Calculus & Matrix Algebra (Herron): Week 6 Notes (2/29/16 - 3/4/16)

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Date Created: 03/05/16

MATH 2010 - Multivariable Calculus & Matrix Algebra Professor Herron - Rensselaer Polytechnic Institute Week 6 (2/29/16 - 3/4/16) Important : These notes are in no way intended to replace attendance in lecture. For best results in this course, it is imperative that you attend lecture and take your own detailed notes. Please keep in mind that these notes are written speci▯cally with Professor Herron’ s sections in mind, and no one else’ s. Many of the examples done in class had corresponding graphics. For simplicity, I will not be including said graphics in these notes, but they should be easily found on the course page on LMS or elsewhere online. Gradient Vector Fields The gradient of f is de▯ned as: rf = hf ;f x. y 2 2 ~ Suppose that f = x ▯ y . Then: rf = h2x;▯2yi = F = F. F = 2xi ▯ 2yj. This is called a 2D gradient vector ▯eld. A vector ▯eld F is called a conservative vector ▯eld if it is a gradient of some scalar function|that is, if there exists a function f such that F = rf. { In this instance, f is called the potential function for F. ~ Other Types of Gradient Vector Fields p 1. Earlier, we had the function f(x;y) = 1 ▯ x ▯ y , and ~ r = xi + yj = r~ er. ▯xi▯yj ▯r rf = p 2 2= p 1▯r2 1▯x ▯y This is called a radial gradient vector ▯eld, as it is only de▯ned for r < 1. 2. For the function f = ax + by: F = rf = ai + bj ^ This produces a constant vector ▯eld (all of the arrows are the same size and are pointing in the same direction). Example p Given the function f = ln( x + y ), determine the gradient vector ▯eld. p 2 2 1 2 2 { f = ln( x + y ) = 2 ln(x + y ) by the rules of natural logs. 1 { f x 2x 2, fy= 2y 2 x +y x +y ~ r r { F = rf = krk2 = r { The gradient vector ▯eld points away from, but is not de▯ned at, (0;0). Line Integrals Suppose you have a wire or chain, and you desire to know its mass. First, describe the wire as a curve that has arc length: R { l(C) = b kr (t)k dt a R t 0 2 { s(t) = 0 kr (t)k dt Both of these equations measure distance along the curve. 0 ds 0 2 { s (t) = dt = k~r (t)k { ~r(t) = x(t)i + y(t)j p { k~r (t)k = (x (t)) + (y (t)) = ds dt Next, de▯ne the line integral with respect to arc length as: R R p 0 2 0 2 { C f(x;y)ds = c f(x;y) (x ) + (y ) dt R t=b p { = t=af(x(t);y(t)) (x ) + (y ) dt If f is de▯ned on a smooth curve C, the line integral of f along C is: Z X n f(x;y)ds = lim f(x ;y )▯s ,iif this limit exists. C n!1 i j i=1 R R R If f = k (constant), then f ds = k ds = k ds = k l(C): C C C Example: Line Integral of a Constant Function Suppose a thin wire is bent into a semicircle C : x +y = 4;x > 0. The linear density R is ▯ = 3. Find the mass of the wire given by C ▯ds. R R { From the above rule: ▯ds = ▯ ds = ▯l(C): C C { By the equation for C, the radius is equal to 2, so the circumference of the semicircle is equal to 2▯. { Since mass is equal to the linear density ▯ times the circumference l(C): ▯ ▯ l(C) = 3 ▯ 2▯ = 6▯. Types of Line Integrals 2 We’ve discussed the ▯rst type of line integral in the previous example, called line integrals of scalar functions. There are two additional types of line integrals: those over vector ▯elds and those with the generic/original form, also known as the "full de▯nition of the line integral." Line Integrals over Vector Fields R R F ▯ dr = (F ▯ T)dr, where T is the unit tangent vector. C C f = F ▯ T~ Z b r (t) Z b F(r(t)) ▯ kr (t)kdt = F(~r(t)) r (t)dt a kr (t)k a Example: Total Work Find the total work done on an object that traverses the parabolic arc C: ^ 2^ ~ ^ 2^ { ~r(t) = ti + t j;1 ▯ t ▯ 2, subject to force F(x;y) = xyi + y j. R R 2 0 { W = C F ▯ dr = 1 F(~r(t)) r (t)dt { Here, x(t) = t, y(t) = t , and ~ r = i + 2tj.^ { F(~ r(t)) = F(x(t);y(t)) = x(t)y(t)i + (y(t)) j = t i + t j.^ R R ▯ ▯ { = 2ht ;t i ▯ h1;2tidt = 2(t + 2t ) = 1t + t1 6 2 = 99 1 1 4 3 1 4 0 If the curve is smooth and ~ r 6= 0, we can change parameterization without changing the line integral. 1=2 1=2 Instead, use R(u) = u i + uj;1 ▯ u ▯ 4 (then u = f). R4 ~ ~ ~ 0 R4 ▯1 2▯ ~ ~ 3=2 3 { 1 F(R(u)) ▯ R (u)du = 1 2u + u du, where F(R(u)) = hu ;u i. Other Helpful Tips For a given arc such that C = C [ C1[ C :2 3 R R R R { C = C1 + C2 + C 3 ede 3 ^ 3 ^ r(t) = acos ti + asin tj; 0 ▯ t ▯ 2▯ { This creates an astroid: x 2=3+ y 2=3 = a 2=3 Generic/Original Line Integral Form Full de▯nition of a line integral: R { F (x;y)dx + F (x;y)dy, C : y = y(x), a ▯ x ▯ b, dy = y (x) 0 C 1 2 3 { Before vectors, one would follow straight lines from point to point, hence line integrals. Example: Total Work Revisited R 2 2 Evaluate C xy dx + y dy, where C is the parabolic arc y = x from (1;1) to (2;4). { Along the curve, dy = 2xdx. R R { xy dx + y dy = 2x(x )dx + (x ) (2xdx) C 1 R 2 3 5 ▯1 4 1 6▯2 99 { = 1 [x + 2x ] dx = 4t + t3 1 = 4 , as determined before. R ~ ^ ^ Now consider a di▯erent path: C E ▯ dr, where ~r(t) = (1 + t)i + (1 + 3t)j;0 ▯ t ▯ 1. R { 1h(1 + t)(1 + 3t);(1 + 3t) i ▯ h1;3idt 0 R 1 2 { = 0 ((1 + t)(1 + 3t) + 3(1 + 3t) )dt R 1 1 { = (4 + 22t + 30t )dt = [4t + 11t + 10t ] = 0 + 11 + 10 = 25 > 99. 0 4 * This is because F is not conservative. In general, F = hf ;f i = hF ;F i. x y 1 2 * @F 2= @F1 if conservative. @x @y Example: A Di▯erent Generic Form Line Integral Problem R Solve the line integral (x ▯ 3xy + y )dx, where C is arc y = 2x ; 0 ▯ x ▯ 1. C R1 R 1 ▯ ▯1 { (x ▯ 3x(2x ) + (2x ) )dx = (x ▯ 6x + 4x )dx = 1x ▯ x + x 4 5 0 0 2 2 5 0 { = 1 ▯ 3 + 4 = ▯ 1 2 2 5 5 This formula can be used to evaluate the line integral: Z Z b r dx 2 dy 2 f(x;y)ds(ds = arc length) = f(x(t);y(t)) ( ) + ( ) dt C a dt dt Orientation of a Curve R R Given a curve C : ~ r(t) = hx(t);y(t)i, where a ▯ t ▯ b : CF ▯ dr = ▯ ▯C F ▯ dr { F is the vector ▯eld. R R F ▯ dr = t F(~r(t)) r (t)dt = 99 , where F = hxy;y i 2 C 1 4 R F ▯ dr = ▯ 99 (opposite sign) ▯C 4 Fundamental Theorem for Conservative Vector Fields Assume that F = rf on a domain D. 4 1. Ifr(t) is a path along C from point P to point Q, then: R ~ { C F ▯ r = f(Q) ▯ f(P) { In particular, the line integral is path independent H 2. The circulation around a closed curve C is zero ( F ▯ dr = 0) C { Because d f(r(t)) = rf(~r(t)) r (t) dt { If F(r(t)) = rf(~r(t)): R R * C F ▯ r = Crf ▯ d~r R t=b Rb d * = t=a rf ▯ r (t)dt = adtf(r(t))dt = f(~r(t))a = f(Q) ▯ f(P) { If P = Q, then the curve is closed. Example: Test for a Conservative Function ~ ^ y^ Suppose F = y i + (2xy ▯ e )j Test: F is conservative if@F2= @F1. @x @y 9 @F 2 @ y > = (2xy ▯ e ) = 2y > @x @x = conservative > @F 1 @ 2 ; @y = @y(y ) = 2y Now ▯nd the potential function f(x;y): { rf = F = y i + (2xy ▯ e )jy ^ { f x y ;f = yxy ▯ e ; take the partial integral of f :x R { fx= y dx ! f = xy + g(y) + c; now take partial derivative with respect to y: { f y 2xy + g (y) { Hence: 2xy + g (y) = 2xy ▯ e ; g (y) = ▯e ! g(y) = ▯e + c y { f(x;y) = xy ▯ e + c Theorem Suppose F is a vector ▯eld on an open, connected region D. R { If F ▯ r is independent of path in D, then F is a constant vector ▯eld in D. C ~ { That is, there exists a function f such that rf = F. Example: Integral over Any Path (Conservative Function Revisited) R(1;1~ ~ 2 y (0;0) ▯ dt, where F = y i + (2xy ▯ e )j over any path. 5 2 y R(1;1) 2 y(1;1) Since F = r(xy ▯ e + c); (0;0) ▯ dr = [xy ▯ e ](0;0)= 1 ▯ e + 1 = 2 ▯ e. R(1;1) R 1 R 1 { y dx + (2xy ▯ e )dy = 0dx + (2y ▯ e )dy (0;0) 0 0 { = 1j + (y ▯ e )j = 1 + (1 ▯ e) = 2 ▯ e, as before. 0 0 This technique may be carried out in general for a gradient F = f x + f y. Gradient ▯eld = path independent integral = integral around a closed loop is zero. Vortex Field ~ ^ ^ 2D: F = F 1 + F 2 3D: F = F 1 + F 2 + F k3 F(x;y;z) { F = rf(x;y;z), with the criterion that rxF = 0 (see Section 16.5). { F = F i1+ F j2+ F k 3 f i x f j y f k z^ { @F1 = @F2; @F2= @F3;@F1 = @F3 @y @x @z @y @z @x 2 3 i j k ▯ ▯ ▯ ▯ ▯ ▯ ~ 4 @ @ @ 5 @F3 @F 2 ^ @F 3 @F1 ^ @F2 @F1 ^ rxF = @x @y @z = ▯ i ▯ ▯ j + ▯ k F F F @y @z @x @z @x @y 1 2 3 A good example is Hill’s Spherical Vortex. Practice Question Let F = rf;f(x;y) = cos(x ▯ 8y). What is the integral of line segment C from (0;0) to (0;▯)? R { F ▯ r = f(0;▯) ▯ f(0;0) C { = cos(▯8▯) ▯ cos(0) = 1 ▯ 1 = 0. End of Document 6

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