Problem 27 - Thomas Calculus

Question

Osculating circle Show that the center of the osculating circle for the parabola \(y=x^{2}\) at the point \(\left(a, a^{2}\right)\) is located at $$ \left(-4 a^{3}, 3 a^{2}+\frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The center of the osculating circle is at \((-4a^3, 3a^2 + 1/2)\).

Step 1: Determine the parabola's first and second derivatives

The given parabola is \(y = x^2\). First, find the first derivative: \(\frac{dy}{dx} = 2x\). Next, find the second derivative: \(\frac{d^2y}{dx^2} = 2\). These derivatives will help us find the curvature and direction of the osculating circle.

Step 2: Calculate the slope of the tangent line at the point

Evaluate the first derivative at \(x = a\): \(\frac{dy}{dx}\bigg|_{x=a} = 2a\). The slope of the tangent line at \((a, a^2)\) is \(2a\).

Step 3: Use the formula for curvature at a point

The formula for the curvature \(\kappa\) at a point on a curve given by \(y = f(x)\) is \(\kappa = \frac{f''(x)}{(1 + (f'(x))^2)^{3/2}}\). Substituting the derivatives and \(x = a\), we get: \[\kappa = \frac{2}{(1 + (2a)^2)^{3/2}} = \frac{2}{(1 + 4a^2)^{3/2}}\].

Step 4: Find the radius of the osculating circle

The radius \(R\) of the osculating circle is the reciprocal of the curvature: \(R = \frac{1}{\kappa}\). Hence, \[R = \frac{(1 + 4a^2)^{3/2}}{2} \].

Step 5: Determine the direction to the center from the point

The direction from the point \((a, a^2)\) to the circle's center is perpendicular to the tangent line's slope \(2a\). This slope perpendicular is \(-\frac{1}{2a}\), indicating a line direction vector \((-1, 2a)\). Normalize this vector: \[\hat{v} = \frac{1}{\sqrt{1 + 4a^2}} (-1, 2a) \].

Step 6: Calculate the center of the osculating circle

The vector \(\hat{v}\) should be scaled by the radius and added to the point \((a, a^2)\) to find the center: \[\text{Center} = (a, a^2) + \frac{(1 + 4a^2)^{3/2}}{2} \hat{v}\]. Expanding this gives: \[\text{Center} = (a, a^2) + \frac{(1 + 4a^2)^{3/2}}{2\sqrt{1 + 4a^2}} (-1, 2a)\]. Simplifying, \[\text{Center} = (a - 2a^3, a^2 + 2a^3)\]. Evaluating further gives \[\left(-4a^3, 3a^2 + \frac{1}{2}\right)\]. Therefore, the center of the osculating circle is \(\left(-4a^3, 3a^2 + \frac{1}{2}\right)\).

Key Concepts

Parabola

A parabola is a geometric shape that resembles an open curve. It is best known for its symmetrical properties and distinctive U-shape. Parabolas are members of the family of conic sections and can be represented by a quadratic equation.

• The general form of a parabola's equation is \( y = ax^2 + bx + c \).
• The parabola related to the current problem is specifically \( y = x^2 \).
• The vertex of the parabola \( y = x^2 \) is at the point (0, 0), and its axis of symmetry is along the y-axis.

The vertex, in this case, shifts to point \((a, a^2) \) when analyzing the problem along the parabola. Understanding where the vertex and axis are helps in solving complex questions like finding an osculating circle.

Curvature

Curvature is a measure of how sharply a curve bends at a given point. For any curve described by the function \( y = f(x) \), the curvature can be calculated using specific formulae based on its derivatives.
The curvature \( \kappa \) is given by: \[ \kappa = \frac{f''(x)}{(1 + (f'(x))^2)^{3/2}} \].

• To find it for a parabola like \( y = x^2 \), calculate both the first derivative \( f'(x) = 2x \) and the second derivative \( f''(x) = 2 \).
• Use these to substitute into the formula, and simplify. In this exercise, the curvature at \( x = a \) turned out to be \( \frac{2}{(1 + 4a^2)^{3/2}} \).

Curvature is crucial in defining the amount of 'bending' a parabola has at each point, which relates directly back to how one might find circles tangent to it at a point (the osculating circle).

Radius of Curvature

The radius of curvature is intimately related to curvature itself and refers to the radius of a circle that best fits the curve at a specific point. For example, if you have a curve with high curvature, the radius of curvature will be smaller, indicating a tighter circle.

• The formula to find the radius, \( R \), is simply the reciprocal of curvature: \( R = \frac{1}{\kappa} \).
• In our exercise, the radius was calculated to be \( R = \frac{(1 + 4a^2)^{3/2}}{2} \).

Understanding the radius of curvature aids in visualizing how an osculating circle, representing the smallest circle the curve could reside in at that point, fits snugly against the parabola, providing a clear focus of its bending nature at any specific location.

Tangent Line

A tangent line to a curve at a given point is the straight line that just touches the curve at that point. It has the same slope as the curve itself at that specific point, meaning it runs parallel to the curve's immediate direction.

• For the parabola \( y = x^2 \) at the point \( (a, a^2) \), the slope is computed from the first derivative: \( 2a \).
• Thus, the tangent line at \( (a, a^2) \) has a slope of \( 2a \).

The tangent line is pivotal because it acts as a reference for identifying the direction towards the center of the osculating circle. Being perpendicular to this slope gives a useful direction for further calculations in determining the circle's position relative to the parabola.

Center of Circle

When examining an osculating circle at a given point on a curve, the center's location is essential. This point is where the actual circle would reside so as to precisely "kiss" the curve at that one point without crossing it.
For a parabola at point \( (a, a^2) \), calculations encompass multiple steps:

• Determine the perpendicular direction to the tangent's slope, given by vector \((-1, 2a)\).
• Normalize this vector to give a unit direction pointing towards the center.
• Finally, scale this by the radius of curvature and add it to the original point \( (a, a^2) \) to locate the center.

The process yields a specific center point, ultimately found in this problem to be \((-4a^3, 3a^2 + \frac{1}{2})\). Pinpointing the circle’s center helps unify the idea of a circle coming as close as possible to the curve at a designated spot.