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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 1variable 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 onevariable 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) xaxis: 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 xaxis, 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 ...
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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.
 Spring '08
 Auroux
 Multivariable Calculus

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