Question

sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, $$ y^{2}-4 x^{2}=20,(2,6) $$

Short answer

Curvature is 0.038, radius of curvature is 26.32.

Step-by-step solution

Identify the Conic Section

The given equation is \( y^2 - 4x^2 = 20 \). This resembles the general form of a hyperbola equation \( y^2/a^2 - x^2/b^2 = 1 \). Dividing throughout by 20, we get \( \frac{y^2}{20} - \frac{x^2}{5} = 1 \). Thus, it's a hyperbola centered at the origin with a vertical transverse axis.

Find the Formula for Curvature of a Curve

The curvature \( \kappa \) of a parameterized curve \( (x(t), y(t)) \) defined via a function can be calculated with \( \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \). However, first, we need to convert the given hyperbola equation into explicit or parametric form.

Express y as a Function of x

From \( y^2 - 4x^2 = 20 \), solve for \( y \): \( y^2 = 4x^2 + 20 \) giving \( y = \pm \sqrt{4x^2 + 20} \). We will work with \( y = \sqrt{4x^2 + 20} \) since the point (2,6) is on the upper branch.

Calculate the First and Second Derivatives

Find \( y' \) using the expression \( y = \sqrt{4x^2 + 20} \). Use the chain rule: \( y' = \frac{d}{dx}(4x^2 + 20)^{1/2} = \frac{4x}{\sqrt{4x^2 + 20}} \). Then, find \( y'' \) as follows: \( y'' = \frac{(\sqrt{4x^2 + 20})^3 - 8x^2}{4x^2 + 20} \).

Evaluate the Curvature at the Given Point

Substitute \( x=2 \) into \( y' \) and \( y'' \) to find the curvature at \((2,6)\). First calculate: \( y'|_{x=2} = \frac{8}{\sqrt{36}} = \frac{4}{3} \). Then, substitute \( x=2\) into \( y'' \) to get \( y''|_{x=2} = \frac{5}{81} \approx 0.062 \). Finally, \( \kappa(2) = \frac{|0.062|}{(1+(4/3)^2)^{3/2}} \approx 0.038 \).

Calculate the Radius of Curvature

The radius of curvature \( R \) is the reciprocal of the curvature \( \kappa \). Thus, \( R = \frac{1}{\kappa} \). Use \( \kappa = 0.038 \) to find \( R \approx \frac{1}{0.038} \approx 26.32. \)

Key concepts

Conic Sections

Understanding conic sections is crucial for grasping a wide array of geometrical and algebraic concepts.
Conic sections are the curves obtained by intersecting a plane with a double-napped cone.
Depending on the angle of the intersection, these sections can be one of the following:

• Ellipse: Formed when the plane intersects the cone at an angle less than the cone's slope.
• Parabola: Occurs when the plane is parallel to the slope of the cone.
• Hyperbola: Formed when the plane intersects both napes of the cone.
• Circle: A special case of the ellipse where the plane is perpendicular to the cone's axis.

Each conic section has its own standard equation format, determining its shape and properties.
In this specific exercise, we encounter a hyperbola, displaying unique characteristics like asymptotes, focal points, and a transverse axis.

Hyperbola

A hyperbola is a set of all points in the plane such that the difference of the distances to two fixed points, known as the foci, is constant.
This particular hyperbola is represented by the equation \[ y^2 - 4x^2 = 20 \] which can be re-written in standard form as \[ \frac{y^2}{20} - \frac{x^2}{5} = 1. \]Here are some features that help define a hyperbola:

• Transverse Axis: The line segment that connects the vertices of the hyperbola.
• Vertices: The points where the hyperbola intersects its transverse axis. For a hyperbola with a vertical transverse axis, these points are at distance \( \sqrt{20} \) from the center along the y-axis.
• Asymptotes: Lines that the hyperbola approaches but never touches. They help sketch the curve more accurately.

With a vertical transverse axis, as this hyperbola possesses, the hyperbola opens vertically along the y-axis.
Working with this specific equation, it's easy to determine its orientation and asymptotes.

Radius of Curvature

The radius of curvature gives insight into how sharply a curve bends at a particular point.
It is mathematically defined as the reciprocal of the curvature. Curvature itself is a measure of change in direction at any given point on the curve.When dealing with hyperbolas or any other curves, it's crucial to understand the concept of radius of curvature:

• Determination of Sharpness: A smaller radius of curvature implies a sharper curve, whereas a larger radius indicates a gentler curve.
• Unit Circle Consideration: A radius of curvature that is 1 suggests the curve mimics a section of a circle with radius 1.
• Practical Applications: Radius of curvature finds its application in road design, lens manufacturing, and more, where bending properties are essential.

In the exercise, curvature calculated at the point \((2,6)\) on the hyperbola is \(0.038\), leading to a calculated radius of curvature of approximately \(26.32\).
This helps to understand that at this point on the hyperbola, the curve changes direction gradually.

First and Second Derivatives

Derivatives play a vital role in understanding the behavior of functions, especially in analyzing the rate of change.
The first derivative \( y' \) of a function helps in identifying the slope of the tangent to a curve at any given point.
The second derivative \( y'' \), on the other hand, provides insights into the concavity and inflection points.

• First Derivative \( y' \): Involves finding the rate of change of \( y \) concerning \( x \). For the equation \( y = \sqrt{4x^2 + 20} \), the first derivative is \( y' = \frac{4x}{\sqrt{4x^2 + 20}} \).
• Second Derivative \( y'' \): Gives the rate at which the slope itself changes. For the same function, \( y'' = \frac{(\sqrt{4x^2 + 20})^3 - 8x^2}{4x^2 + 20} \).
• Curvature Implications: These derivatives are used for computing the curvature \( \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \), giving an understanding of how curve behavior influences curvature.

Applying these calculations to the hyperbola at the point \((2,6)\) facilitates the determination of its curvature, illustrating both theoretical concepts and practical results.