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# The maximal deflection on an ellipse

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• Radial direction (r): vector from the origin to any point of the ellipse.Normal direction (n): Vector perpendicular to the tangent line.
• Constraint: a condition that a solution to an optimization problem is required by the problem itself to satisfy.objective function is called an optimal solution
• Constraint: a condition that a solution to an optimization problem is required by the problem itself to satisfy.
• ### The maximal deflection on an ellipse

1. 1. The Maximal Deflectionon an Ellipse { Stephanie Moncada
2. 2. Objectives- Find the maximal deflection between the radial direction and the normal direction.- Define the normal direction using properties of gradients.- Find the objective function to be maximized by using lagrange multipliers.- Formulate the angle between the radial and normal direction by applying the dot product.
3. 3. - Consider:An Ellipse centered at the origin, with semi-major axis a, semi-minor axis b, and withthese axes along the x- and y- axesrespectively.
4. 4. Maximal Deflection:- Located in the first quadrant- Angle between the normal and radial vectors- Will not occur at either the X or Y intercepts
5. 5. The maximal deflection occurs where the ellipse meetsthe line from the origin to (a, b).
6. 6. Lagrange MultipliersTo use Lagrange multipliers we need: - An objective function to be maximized.- A constraint.
7. 7. - CONSTRAINTConstraint: a condition to an optimization problem that isrequired by the problem itself to be satisfied.Since we consider only points of the ellipse, itsequation defines the constraint. Accordingly, wedefine the function. Where the constraint is g(x, y)=1
8. 8. - OBJECTIVE FUNCTIONFor the objective function , we want the angle between thenormal and the radial vectors at a point (x,y) on the ellipse.We take r = (x,y) as the radial vector.For the normal vector, we take n = (x/a, y/b), which is one-half of the gradient of g.Then, δ is determined by the equation.
9. 9. Given that δ is determined by the equation Now we can observe that r · n = g(x, y) = 1 for any point on the ellipse. Accordingly, we simplify matters by inverting and squaring to obtainWe define this to be our objective function. That is,
10. 10. For (x, y) in the first quadrant and on the ellipse, we know that δ isbetween 0 and π/2. On this interval, sec^2 δ is an increasing function.Therefore, δ is maximized where f is.Our problem now is to maximize f subject to the constraint g = 1. Thesolution must occur at a point where ∇ f and ∇g are parallel.Thus, this leads to the single equation:From this equation, it is straightforward to derive: This shows that in the first quadrant, the solution to our optimization problem must lie on the line joining the origin to (a, b).
11. 11. As a first step, we compute the partial derivatives
12. 12. Combining these leads to
13. 13. Theorem (1)
14. 14. Several methods to obtain the maximal deflectionon an ellipse:- Direct Parameterization: Using the standard parameterization of the ellipse- Using Slopes: this method expresses everything in slopes.- Symmetry: there is a symmetry that makes the location of the point of maximal deflection natural.
15. 15. ApplicationThe maximal deflection problem has oneapplication.It concerns the ellipsoidal model of the Earth, andtwo ways to define latitude.On a spherical globe, the latitude at a point is theangle between the equatorial plane and the positionvector from the center of the sphere.
16. 16. Sources- The Mathematical Association of America.- Dan Kalman, Virtual Empirical Investigation: Concept Formation and Theory Justification, Amer: Math. Monthly 112 (2005), 786-798.- William C. Waterhouse, Do Symmetric Problems have Symmetric Solutions?, Amer: Math. Monthly 90 (1983). 378-387.