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clase6-handouts - We say that two curves intersecting on a...

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We say that two curves intersecting on a point ( t , y ) are orthogonal at that point if the tangent lines to both curves at the intersection point are orthogonal. M.I. Bueno Differential Equations and Linear Algebra
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Consider two families F 1 and F 2 . We say that F 1 and F 2 are orthogonal whenever any curve from F 1 intersects any curve from F 2 , the two curves are orthogonal at the point of intersection. M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y = mt M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y 2 + t 2 = C M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y = mt and y 2 + t 2 = C M.I. Bueno Differential Equations and Linear Algebra
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Given a family of curves F , is it possible to find a family of curves which is orthogonal to F ? M.I. Bueno Differential Equations and Linear Algebra
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Given a family F , in order to find the orthogonal family, 1 Find the associated differential equation. 2 Rewrite this differential equation in the explicit form y 0 = f ( t , y ) . 3 Write down the differential equation associated to the orthogonal family y 0 = - 1 f ( t , y ) . 4 Solve the new equation. The solutions are exactly the family of orthogonal curves. M.I. Bueno Differential Equations and Linear Algebra
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M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y 2 + t 2 = Ct M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y 2 + t 2 = Cy M.I. Bueno Differential Equations and Linear Algebra
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Family of curves y 2 + t 2 = Ct and y 2 + t 2 = Cy M.I. Bueno Differential Equations and Linear Algebra
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Another way to compute orthogonal trajectories Consider the family of curves in implicit form f ( t , y , C ) = 0 . Differentiate implicitly with respect to t and get f t + f y y 0 = 0 , f t + f y y 0 = 0 . Then, we get y 0 = - f t ( t , y , C ) f y ( t , y , C ) . Now we solve for C = C ( t , y ) from the equation f ( t , y , C ) = 0. We substitute C ( t , y ) in the above formula for y 0 and get the equation y 0 = g ( x , y ) = - f t ( t , y , C ( t , y )) f y ( t , y , C ( t , y )) . M.I. Bueno Differential Equations and Linear Algebra
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Note that y 0 yields the slope of the tangent line at the point ( t , y ) of a curve of the given family passing through ( t , y ) . The slope of the orthogonal trajectory at the point ( t , y ) must be y 0 ( x ) = - 1 g ( t , y ) . Therefore, the ODE governing the orthogonal trajectories is derived as y 0 = f y ( t , y , C ( t , y )) f x ( t , y , C ( t , y )) . M.I. Bueno Differential Equations and Linear Algebra
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M.I. Bueno Differential Equations and Linear Algebra
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Numerical methods for ordinary differential equations Consider an IVP, y 0 = f ( t , y ) , y ( t 0 ) = y 0 .
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This note was uploaded on 03/14/2010 for the course MATH 3C taught by Professor Jacobs during the Fall '08 term at UCSB.

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clase6-handouts - We say that two curves intersecting on a...

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