Week 8 Summary

18.02 Lecture 18. – Tue, Oct 23, 2007 Change of variables. Example 1: area of ellipse with semiaxes a and b: setting u = x/a, v = y/b, �� �� �� dx dy = ab du dv = ab du dv = πab. (x/a)2 +(y/b)2 <1 u2 +v 2 <1 u2 +v 2 <1 1 a 1 (substitution works here as in 1-variable calculus: du = dx, dv = 1b dy, so du dv = ab dx dy. In general, must find out the scale factor (ratio between du dv and dx dy)? Example 2: say we set u = 3x−2y, v = x+y to simplify either integrand or bounds of integration. What is the relation between dA = dx dy and dA� = du dv? (area elements in xy- and uv-planes). Answer: consider a small rectangle of area ΔA = ΔxΔy, it becomes in uv-coordinates a paral­ lelogram of area ΔA� . Here the answer is independent of which rectangle we take, so we can take e.g. the unit square in xy-coordinates. � � � �� � u 3 −2 x = , so this becomes a parallelogram with sides given by In the uv-plane, v 1 1 y � � �� � � � 3 1 � � 3 −2 � � � � � . vectors �3, 1� and �−2, 1� (picture drawn), and area = det = � =5 =� −2 1 � 1 1 � �� � = 5ΔA, in the limit dA� = 5dA, i.e. du dv = 5dx dy. So For any rectangle ΔA . . . dx dy = �� 1 . . . 5 du dv. General case: approximation formula Δu ≈ ux Δx + uy Δy, Δv ≈ vx Δx + vy Δy, i.e. � � � � �� Δu ux uy Δx . ≈ Δv vx vy Δy A small xy-rectangle is approx. a parallelogram in uv-coords, but scale factor depends on x and y now. By the same argument as before, the scale factor is the determinant. � � ∂(u, v) �� ux uy �� Definition: the Jacobian is J = = . Then du dv = |J | dx dy. ∂(x, y) � vx vy � (absolute value because area is the absolute value of the determinant). Example 1: polar coordinates x = r cos θ, y = r sin θ: � � � � ∂(x, y) �� xr xθ �� �� cos θ −r sin θ �� =� = = r cos2 θ + r sin2 θ = r. yr yθ � � sin θ r cos θ � ∂(r, θ) So dx dy = r dr dθ, as seen before. �1�1 Example 2: compute 0 0 x2 y dx dy by changing to u = x, v = xy (usually motivation is to simplify either integrand or region; here neither happens, but we just illustrate the general method). � � � � � ux uy � � 1 0 � ∂(u, v) � � � � = x, so du dv = x dx dy, i.e. 1) Area element: Jacobian is = � = � y x � vx vy � ∂(x, y) dx dy = x1 du dv. 2) Express integrand in terms of u, v: x2 y dx dy = x2 y x1 du dv = xy du dv = v du dv. 3) Find bounds (picture drawn): if we integrate du dv, then first we keep v = xy constant, slice looks like portion of hyperbola (picture shown), parametrized by u = x. The bounds are: at the top boundary y = 1, so v/u = 1, i.e. u = v; at the right boundary, x = 1, so u = 1. So the inner 1 2 integral is �1 v . The first slice is v = 0, the last is v = 1; so we get � 1� 1 v du dv. 0 v Besides the picture in xy coordinates (a square sliced by hyperbolas), I also drew a picture in uv coordinates (a triangle), which some students may find is an easier way of getting the bounds for u and v. 18.02 Lecture 19. – Thu, Oct 25, 2007 Handouts: PS7 solutions; PS8. Vector fields. F� = Mı̂ + Nĵ, where M = M (x, y), N = N (x, y): at each point in the plane we have a vector � F which depends on x, y. Examples: velocity fields, e.g. wind flow (shown: chart of winds over Pacific ocean); force fields, e.g. gravitational field. Examples drawn on blackboard: (1) F� = 2ı̂ + ĵ (constant vector field); (2) F� = xı̂; (3) F� = xı̂ + yĵ (radially outwards); (4) F� = −yı̂ + xĵ (explained using that �−y, x� is �x, y� rotated 90◦ counterclockwise). Work and line integrals. W = (force).(distance) = F� · Δ�r for a small motion Δ�r. Total work is obtained by summing these along a trajectory C: get a “line integral” � � � � W = F� · d�r = lim F� · Δ�ri . C Δ� r →0 i To evaluate � the line integral, we observe C is parametrized by time, and give meaning to the notation C F� · d�r by � � t2 � d�r � dt. F� · d�r = F� · dt C t1 Example: F� = −yı̂ + xĵ, C is given by x = t, y = t2 , 0 ≤ t ≤ 1 (portion of parabola y = x2 from (0,0) to (1,1)). Then we substitute expressions in terms of t everywhere: d�r dx dy F� = �−y, x� = �−t2 , t�, = � , � = �1, 2t�, dt dt dt � � 1 � 1 � 1 d�r 1 so F� · d�r = F� · dt = �−t2 , t� · �1, 2t� dt = t2 dt = . (in the end things always reduce dt 3 C 0 0 0 to a one-variable integral.) In fact, the definition of the line integral does not involve the parametrization: so the result is the same no matter which parametrization we choose. For example we could choose to parametrize � � π/2 the parabola by x = sin θ, y = sin2 θ, 0 ≤ θ ≤ π/2. Then we’d get C F� · d�r = 0 . . . dθ, which would be equivalent to the previous one under the substitution t = sin θ and would again be equal to 31 . In practice we always choose the simplest parametrization! New notation for line integral: F� = �M, N �, and d�r = �dx, dy� (this is in fact a differential: if we divide both sides by dt we get the component formula for the velocity d�r/dt). So the line integral 3 becomes � F� · d�r = C � M dx + N dy. C The notation is dangerous: this is not a sum of integrals w.r.t. x and y, but really a line integral along C. To evaluate one must express everything in terms of the chosen parameter. In the above example, we have x = t, y = t2 , so dx = dt, dy = 2t dt by implicit differentiation; then � � 1 � 1 1 2 −y dx + x dy = −t dt + t (2t) dt = t2 dt = 3 C 0 0 (same calculation as before, using different notation). Geometric approach. d�r ds Recall velocity is = T̂ (where s = arclength, T̂ = unit tangent vector to trajectory). dt dt � � � So d�r = T̂ ds, and F · d�r = F� · T̂ ds. Sometimes the calculation is easier this way! C C Example: C = circle of radius a centered at origin, F� = xı̂ + yĵ, then F� · T̂ = 0 (picture drawn), � � so C F� · T̂ ds = 0 ds = 0. � � Example: same C, F� = −yı̂ + xĵ, then F� · T̂ = |F� | = a, so C F� · T̂ ds = a ds = a (2πa) = 2πa2 ; checked that we get the same answer if we compute using parametrization x = a cos θ, y = a sin θ. 18.02 Lecture 20. – Fri, Oct 26, 2007 Line integrals continued. � � � � � Recall: line integral of F = Mı̂ + Nĵ along a curve C: F · d�r = M dx + N dy = F� · T̂ ds. C C C � Example: F� = yı̂ + xĵ, C F� · d�r for C � = C1 + C2 + C3 enclosing sector of unit disk from 0 to π/4. (picture shown). Need to compute Ci y dx + x dy for each portion: � �1 1) x-axis: x = t, y = 0, dx = dt, dy = 0, 0 ≤ t ≤ 1, so C1 y dx + x dy = 0 0 dt = 0. Equivalently, � geometrically: along x-axis, y = 0 so F� = xĵ while T̂ = ı̂ so F� · T̂ ds = 0. C1 2) C2 : x = cos θ, y = sin θ, dx = − sin θ dθ, dy = cos θ dθ, 0 ≤ θ ≤ π4 . So �π/4 � � � π/4 � π/4 1 1 sin(2θ) = . y dx + x dy = sin θ(− sin θ)dθ + cos θ cos θ dθ = cos(2θ)dθ = 2 2 C2 0 0 0 3) C3 : line segment from ( √12 , √12 ) to (0, 0): could take x = √1 2 − √1 t, 2 y = same, 0 ≤ t ≤ 1, ... √1 . 2 but easier: C3 backwards (“−C3 ”) is y = x = t, 0 ≤ t ≤ Work along −C3 is opposite of work along C3 . � � 0 � 1/√2 √ 1 2 1/ 2 y dx + x dy = t dt + t dt = − 2t dt = −[t ] =− . 0 √ 2 C3 1/ 2 0 If F� is a gradient field, F� = �f = fx ı̂ + fy ĵ (f is called “potential function”), then we can simplify evaluation of line integrals by using the fundamental theorem of calculus. Fundamental theorem of calculus for line integrals: 4 � �f · d�r = f (P1 ) − f (P0 ) when C runs from P0 to P1 . � � Equivalently with differentials: fx dx + fy dy = df = f (P1 ) − f (P0 ). Proof: C C � � t1 � t1 dx dy d �f · d�r = (fx + fy ) dt = (f (x(t), y(t)) dt = [f (x(t), y(t))]tt10 = f (P1 ) − f (P0 ). dt dt dt C t0 t0 � E.g., in the above example, if we set f (x, y) = xy then �f = �y, x� = F� . So can be calculated C Ci just by evaluating f = xy at end points. Picture shown of C, vector field, and level curves. Consequences: for a gradient field, we have: � � • Path independence: if C1 , C2 have same endpoints then C1 �f · d�r = C2 �f · d�r (both equal � to f (P1 ) − f (P0 ) by the theorem). So the line integral C �f · d�r depends only on the end points, not on the actual trajectory. � • Conservativeness: if C is a closed loop then C �f · d�r = 0 (= f (P ) − f (P )). � (e.g. in above example, C = 0 + 12 − 12 = 0.) WARNING: this is only for gradient fields! Example: F� = −yı̂ + xĵ� is not a gradient field: as seen Thursday, along C = circle of radius a F� · d�r = 2πa2 . Hence F� is not conservative, and not a gradient field. counterclockwise (F� //Tˆ ), C Physical interpretation. If the force field F� is the gradient of a potential f , then work of F� = change in value of potential. E.g.: 1) F� = gravitational field, f = gravitational potential; 2) F� = electrical field; f = electrical potential (voltage). (Actually physicists use the opposite sign convention, F� = −�f ). Conservativeness means that energy comes from change in potential f , so no energy can be extracted from motion along a closed trajectory (conservativeness = conservation of energy: the change in kinetic energy equals the work of the force equals the change in potential energy). We have four equivalent properties: � (1) F� is conservative ( C F� · d�r = 0 for any closed curve C) � (2) F · d�r is path independent (same work if same end points) (3) F� is a gradient field: F� = �f = fx ı̂ + fy ĵ. (4) M dx + N dy is an exact differential (= fx dx + fy dy = df .) ((1) is equivalent to (2) by considering C1 , C2 with same endpoints, C = C1 − C2 is a closed loop. (3) ⇒ (2) is the FTC, ⇐ will be key to finding potential function: if we have path independence � (x,y) then we can get f (x, y) by computing (0,0) F� · d�r. (3) and (4) are reformulations of the same property).