lec_week8

lec_week8 - MIT OpenCourseWare http://ocw.mit.edu 18.02...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 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 = π ab. (x/a)2 +(y/b)2 <1 u2 +v 2 <1 1 a u2 +v 2 <1 1 (substitution works here as in 1-variable calculus: du = dx, dv = 1 dy , so du dv = ab dx dy . b 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 11 y � � �� � � � 3 1 � � � � � = 5 = � 3 −2 � . vectors �3, 1� and �−2, 1� (picture drawn), and area = det = � � 1 1 � −2 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 � � � . Then du dv = |J | dx dy. Definition: the Jacobian is J = = ∂ (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 � 2 � � = � � = r cos θ + r sin2 θ = r. = ∂ (r, θ) � yr yθ � � sin θ r cos 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 � � � ∂ (u, v ) � � = � 1 0 � = x, so du dv = x dx dy , i.e. 1) Area element: Jacobian is = � � y x � vx vy � ∂ (x, y ) 1 dx dy = x du dv . 1 2) Express integrand in terms of u, v : x2 y dx dy = x2 y x 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 = M ˆ + N ˆ, where M = M (x, y ), N = N (x, y ): at each point in the plane we have a vector ı j � which depends on x, y . F 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 = 2ˆ + ˆ (constant vector field); (2) F = xˆ; (3) F = ıj ı � = −yˆ + xˆ (explained using that �−y, x� is �x, y � rotated 90◦ ı j xˆ + yˆ (radially outwards); (4) F ı j counterclockwise). Work and line integrals. �r W = (force).(distance) = F · Δ� for a small motion Δ�. Total work is obtained by summing these r along a trajectory C : get a “line integral” � � � � �r � W= F · d� = lim F · Δ�i . r C Δ�→0 r i To evaluate the line integral, we observe C is parametrized by time, and give meaning to the � �r notation C F · d� by � � t2 � r� �r � d� dt. F · d� = F· dt C t1 � Example: F = −yˆ + xˆ, C is given by x = t, y = t2 , 0 ≤ t ≤ 1 (portion of parabola y = x2 from ı j (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 r 1 �r � d� dt = so F · d� = F· �−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 �r the parabola by x = sin θ, y = sin2 θ, 0 ≤ θ ≤ π /2. Then we’d get C F · d� = 0 . . . dθ, which would be equivalent to the previous one under the substitution t = sin θ and would again be equal 1 to 3 . In practice we always choose the simplest parametrization! � New notation for line integral: F = �M, N �, and d� = �dx, dy � (this is in fact a differential: if we r divide both sides by dt we get the component formula for the velocity d�/dt). So the line integral r 3 becomes � C �r F · d� = � 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 ˆ ˆ Recall velocity is = T (where s = arclength, T = unit tangent vector to trajectory). d� t dt � ˆ �r �ˆ So d� = T ds, and r F · d� = F · T ds. Sometimes the calculation is easier this way! C C �ˆ � Example: C = circle of radius a centered at origin, F = xˆ + yˆ, then F · T = 0 (picture drawn), ı j � � ˆ � · T ds = 0 ds = 0. so C F � � � �ˆ � �ˆ ı j Example: same C , F = −yˆ + xˆ, then F · T = |F | = a, so C F · T 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. � � � � = M ˆ + N ˆ along a curve C : � · d� = �ˆ Recall: line integral of F ı j Fr M dx + N dy = F · T ds. C C C � � �r Example: F = yˆ + xˆ, C F · d� for C = C1 + C2 + C3 enclosing sector of unit disk from 0 to ı j � 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, � ˆı � �ˆ geometrically: along x-axis, y = 0 so F = xˆ while T = ˆ so j F · T ds = 0. C1 2) C2 : x = cos θ, y = sin θ, dx = − sin θ dθ, dy = cos θ dθ, 0 ≤ θ ≤ π . So 4 �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 1 1 3) C3 : line segment from ( √2 , √2 ) to (0, 0): could take x = 1 √. 2 1 √ 2 − 1 √ t, 2 y = same, 0 ≤ t ≤ 1, ... but easier: C3 backwards (“−C3 ”) is y = x = t, 0 ≤ t ≤ Work along −C3 is opposite of work along C3 . � �0 � 1/√2 √ 1 1/ 2 y dx + x dy = t dt + t dt = − 2t dt = −[t2 ]0 =− . √ 2 C3 1/ 2 0 � � If F is a gradient field, F = �f = fxˆ + fy ˆ (f is called “potential function”), then we can ı j simplify evaluation of line integrals by using the fundamental theorem of calculus. Fundamental theorem of calculus for line integrals: 4 � �f · d� = f (P1 ) − f (P0 ) when C runs from P0 to P1 . r � � 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))]t1 = f (P1 ) − f (P0 ). t0 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 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� = C2 �f · d� (both equal r r � to f (P1 ) − f (P0 ) by the theorem). So the line integral C �f · d� depends only on the end points, r not on the actual trajectory. � • Conservativeness: if C is a closed loop then C �f · d� = 0 (= f (P ) − f (P )). r � 1 1 (e.g. in above example, C = 0 + 2 − 2 = 0.) WARNING: this is only for gradient fields! � ˆ Example: F = −yˆ + x� is not a gradient field: as seen Thursday, along C = circle of radius a ı j 2 ˆ � //T ), �r � counterclockwise (F C F · d� = 2πa . Hence F is not conservative, and not a gradient field. 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: � � �r (1) F is conservative ( C F · d� = 0 for any closed curve C ) � (2) F · d� is path independent (same work if same end points) r � � (3) F is a gradient field: F = �f = fxˆ + fy ˆ. ı j (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�. (3) and (4) are reformulations of the same r property). Ci ...
View Full Document

This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

Ask a homework question - tutors are online