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The Maximal Deflection on an Ellipse

Stephanie Moncada

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.

- Consider:

An Ellipse centered at the origin, with semi-major axis a, semi-minor axis b, and with these axes along the x- and y- axes respectively.

Maximal Deflection:- Located in the first quadrant- Angle between the normal and radial

vectors- Will not occur at either the X or Y intercepts

The maximal deflection occurs where the ellipse meets the line from the origin to (a, b).

Lagrange Multipliers

To use Lagrange multipliers we need: - An objective function to be maximized.- A constraint.

Since we consider only points of the ellipse, its equation defines the constraint. Accordingly, we define the function.

Where the constraint is g(x, y)=1

- CONSTRAINT

Constraint: a condition to an optimization problem that is required by the problem itself to be satisfied.

For the objective function , we want the angle between the normal 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.

- OBJECTIVE FUNCTION

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 obtain

We define this to be our objective function. That is,

Given that δ is determined by the equation

For (x, y) in the first quadrant and on the ellipse, we know that δ is between 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. The solution must occur at a point where ∇ f and ∇g are parallel.Thus, this leads to the single equation:

This shows that in the first quadrant, the solution to our optimization problem must lie on the line joining the origin to (a, b).

From this equation, it is straightforward to derive:

As a first step, we compute the partial derivatives

Combining these leads to

Theorem

(1)

Several methods to obtain the maximal deflection on 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.

ApplicationThe maximal deflection problem has one application.It concerns the ellipsoidal model of the Earth, and two ways to define latitude.

On a spherical globe, the latitude at a point is the angle between the equatorial plane and the position vector from the center of the sphere.

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.