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Consider a quadratic surface \(z=a x^{2}+b y^{2}+2 c x y\) with a unique extremum at \(x=y=0 .\) For a suitable choice of coordinates, a smooth function may always be approximated with such a function at any given point. (a) Determine the radius of curvature of the surface along a line in the \(x y\) -plane forming an angle \(\phi\) with the \(x\) -axis. (b) Determine the extrema of the radius of curvature as a function of \(\phi\) and show that they correspond to orthogonal directions.

Short Answer

Expert verified
The radius of curvature is given by \( R(\phi) = \frac{1}{|2a \cos^2 \phi + 2b \sin^2 \phi + 4c \cos \phi \sin \phi|} \). Extrema occur at angles \( \phi \) such that \( \tan 2\phi = \frac{2c}{a-b} \), corresponding to orthogonal directions.

Step by step solution

01

Find the Second Derivatives

To calculate the radius of curvature along a specific line, we first need the second derivatives of \( z \) with respect to \( x \) and \( y \). We have the function \( z = ax^2 + by^2 + 2cxy \). Compute the following second derivatives:\[ \frac{\partial^2 z}{\partial x^2} = 2a, \quad \frac{\partial^2 z}{\partial y^2} = 2b, \quad \text{and} \quad \frac{\partial^2 z}{\partial x \partial y} = 2c. \]
02

Directional Second Derivative

The curvature along a line at angle \( \phi \) is determined by the second directional derivative \( D^2 \) of \( z \) in the direction of the unit vector \( \mathbf{u} = (\cos \phi, \sin \phi) \). This is given by:\[ D^2 = \mathbf{u}^T H \mathbf{u}, \]where \( H \) is the Hessian matrix:\[ H = \begin{pmatrix} 2a & 2c \ 2c & 2b \end{pmatrix}. \]
03

Compute Directional Second Derivative

Substitute \( \mathbf{u} = (\cos \phi, \sin \phi) \) into the directional second derivative formula:\[ D^2 = (\cos \phi, \sin \phi) \begin{pmatrix} 2a & 2c \ 2c & 2b \end{pmatrix} \begin{pmatrix} \cos \phi \ \sin \phi \end{pmatrix}. \]Simplify to find:\[ D^2 = 2a \cos^2 \phi + 2b \sin^2 \phi + 4c \cos \phi \sin \phi. \]
04

Radius of Curvature Formula

The radius of curvature \( R \) along the line forming angle \( \phi \) is given by:\[ R(\phi) = \frac{1}{|D^2|}. \]Therefore, substitute the computed \( D^2 \):\[ R(\phi) = \frac{1}{|2a \cos^2 \phi + 2b \sin^2 \phi + 4c \cos \phi \sin \phi|}. \]
05

Determine Extrema of Radius of Curvature

To find the extrema of \( R(\phi) \), we need to differentiate \( D^2 \) with respect to \( \phi \) and solve\[ \frac{d(D^2)}{d\phi} = 0 \] for \( \phi \). Using trigonometric identities and differentiating gives:\[ -2a \sin 2\phi + 2b \sin 2\phi + 4c \cos 2\phi = 0. \]This simplifies to:\[ (b-a) \sin 2\phi + 2c \cos 2\phi = 0. \]
06

Solve for Extreme Directions

From \((b-a) \sin 2\phi + 2c \cos 2\phi = 0 \), rearrange to find:\[ \tan 2\phi = \frac{2c}{a-b}. \]The solutions of this equation will give the angles \( \phi \) for which \( D^2 \) and therefore \( R \) are extremized.
07

Analyze Extremal Directions for Orthogonality

Once \( \tan 2\phi \) is given, it follows that the solutions for \( \phi \) (let's call them \( \phi_1 \) and \( \phi_2 \)) are separated by \( \phi_1 - \phi_2 = \frac{\pi}{2} \). Thus, the extrema directions are orthogonal, as separated by \( 90^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius of Curvature
In differential geometry, the radius of curvature provides a measure of how sharply a curve bends at a given point. For the quadratic surface described by the equation \( z = ax^2 + by^2 + 2cxy \), the radius of curvature along a line in the xy-plane forming an angle \( \phi \) with the x-axis is a fundamental characteristic that gives insights into the surface's geometry.

The formula to calculate the radius of curvature \( R(\phi) \) along a line at an angle \( \phi \) is:
\[ R(\phi) = \frac{1}{|D^2|} \]
where \( D^2 \) represents the second directional derivative of \( z \). This derivative is computed using the Hessian matrix and the unit direction vector \( \mathbf{u} = (\cos \phi, \sin \phi) \).

The calculation result shows that when substituting \( D^2 \), it becomes:
\[ D^2 = 2a \cos^2 \phi + 2b \sin^2 \phi + 4c \cos \phi \sin \phi \]
Thus, the radius of curvature at any given angle \( \phi \) provides a clear measure of the local geometry, critical in understanding the shape of a quadratic surface.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It plays a crucial role in analyzing quadratic surfaces like \( z = ax^2 + by^2 + 2cxy \).

For the given surface, the Hessian matrix is:
\[ H = \begin{pmatrix} 2a & 2c \ 2c & 2b \end{pmatrix} \]
This matrix encapsulates information about the surface's curvature in the xy-plane.

The role of the Hessian matrix in this context is to help compute the second directional derivative \( D^2 \), which is pivotal for calculating the radius of curvature. The formula for \( D^2 \) using the Hessian \( H \) and a unit vector \( \mathbf{u} \) is:
\[ D^2 = \mathbf{u}^T H \mathbf{u} \]
Thus, it directly relates variations in the surface's z-values to changes in the xy-coordinates, helping discern the surface's extremities and guiding us on how curvature evolves with angle \( \phi \).
Directional Derivative
The directional derivative measures the rate at which a function changes at a point in a specified direction. For the quadratic surface \( z = ax^2 + by^2 + 2cxy \), the second directional derivative helps capture the surface's curvature characteristics.

Choosing a direction represented by \( \mathbf{u} = (\cos \phi, \sin \phi) \), the second directional derivative \( D^2 \) relies on the Hessian matrix, calculated as:
\[ D^2 = \mathbf{u}^T H \mathbf{u} \]
Substituting \( \mathbf{u} \) and \( H \), the result is:
\[ D^2 = 2a \cos^2 \phi + 2b \sin^2 \phi + 4c \cos \phi \sin \phi \]
The derivative \( D^2 \) determines how sharply the surface bends in the direction \( \phi \).

The critical role of \( D^2 \) is in determining the curvature's extrema, and solving for extrema allows finding directions where the surface's curvature is steepest or flattest. This is done by setting the derivative of \( D^2 \) with respect to \( \phi \) to zero and solving, revealing orthogonal directions indicative of maximum and minimum curvatures.

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