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MIT 18 02 - Change of variables

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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 ab 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 nd 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 A x y it becomes in uv coordinates a paral lelogram of area A 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 5 A in the limit dA 5dA i e du dv 5dx dy So For any rectangle A dx dy 1 5 du dv General case approximation formula u ux x uy y v vx x vy y i e u ux uy x v vx vy y 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 De nition 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 y r sin x y xr x cos r sin r cos2 r sin2 r yr y sin r cos r So dx dy r dr d 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 rst 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 rst 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 nd 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 elds F M Nj 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 elds e g wind ow shown chart of winds over Paci c ocean force elds e g gravitational eld Examples drawn on blackboard 1 F 2 j constant vector eld 2 F x 3 F x yj radially outwards 4 F y xj explained using that y x is x y rotated 90 counterclockwise Work and line integrals W force distance F r for a small motion r Total work is obtained by summing these along a trajectory C get a line integral W F d r lim F ri C 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 y xj 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 de nition 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 2 the parabola by x sin y sin2 0 2 Then we d get C F d r 0 d which would be equivalent to the previous one under the substitution t sin 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 di erential 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 di erentiation 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 di erent notation Geometric approach d r ds Recall velocity is T where s arclength T unit tangent vector to trajectory dt dt So d r T ds and F d r F T ds Sometimes the calculation is easier this way C C Example C circle of radius a centered at origin F x yj then F T 0 picture drawn so C F T ds 0 ds 0 Example same C F y xj then F T F a so C F T ds a ds a 2 a 2 a2 checked that we get the same answer if we compute using parametrization x a cos y a sin 18 02 Lecture 20 Fri Oct 26 2007 Line integrals continued Recall line integral of …


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MIT 18 02 - Change of variables

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