Problem 142 - Calculus 3

Question

Find equations of the osculating circles of the ellipse \(4 y^{2}+9 x^{2}=36\) at the points \((2,0)\) and \((0,3)\)

Step-by-Step Solution

Answer

Osculating circle at (2,0): \((x-2)^2 + \left(y - \frac{8}{9}\right)^2 = \left(\frac{8}{9}\right)^2\). At (0,3): \((x - \frac{4}{3})^2 + (y - 3)^2 = \left(\frac{4}{3}\right)^2\).

Step 1: Write the standard form of the ellipse

The given ellipse equation is \(4y^2 + 9x^2 = 36\). First, divide every term by 36 to obtain the standard form of the ellipse: \[\frac{x^2}{4} + \frac{y^2}{9} = 1\]. Here, \(a^2=4\) and \(b^2=9\), making \(a=2\) and \(b=3\).

Step 2: Find the derivatives

The standard form allows us to differentiate implicitly with respect to \(x\). Differentiating \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) implicitly, we have: \[\frac{x}{2} + \frac{y}{9} \cdot \frac{dy}{dx} = 0\]. Solving for \(\frac{dy}{dx}\), we get \(\frac{dy}{dx} = -\frac{9x}{2y}\).

Step 3: Find the radius of curvature at (2,0)

At the point \((2,0)\), the slope \(\frac{dy}{dx}\) is undefined (as \(y=0\)), indicating a vertical tangent. Use the formula for radius of curvature: \(R = \left(\frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}\right)\) after differentiating \(\frac{dy}{dx}\) implicitly to find \(\frac{d^2y}{dx^2}\). Here, \(R=\frac{8}{9}\).

Step 4: Determine the osculating circle at (2,0)

For a vertical tangent, the osculating circle has its center at \((2, R)=\left(2, \frac{8}{9}\right)\), with radius \(R = \frac{8}{9}\). Its equation is \((x-2)^2 + \left(y - \frac{8}{9}\right)^2 = \left(\frac{8}{9}\right)^2\).

Step 5: Compute at the point (0,3)

Calculate \(\frac{dy}{dx}\) at \((0,3)\): \(\frac{dy}{dx} = 0\) indicating a horizontal tangent. Using the second derivative and radius of curvature formula, find \(\frac{d^2y}{dx^2}\) and \(R=\frac{12}{9}\).

Step 6: Determine osculating circle at (0,3)

As the tangent is horizontal, the osculating circle is centered at \(\left(R, 3\right) = \left(\frac{4}{3}, 3\right)\) with radius \(R = \frac{4}{3}\). Therefore, the osculating circle's equation is \(\left(x - \frac{4}{3}\right)^2 + (y - 3)^2 = \left(\frac{4}{3}\right)^2\).

Key Concepts

Ellipse

An ellipse is a smooth, closed curve that follows the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively. In our problem, the given ellipse equation is \( 4y^2 + 9x^2 = 36 \). To convert this into standard form, divide each term by 36, yielding \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). From this, we recognize that \( a^2 = 4 \) and \( b^2 = 9 \), meaning the semi-major axis \( a = 2 \) and the semi-minor axis \( b = 3 \).

• Ellipses have two focal points, and they are symmetric about both the x-axis and y-axis.
• The standard ellipse exhibits rotation symmetry and can be stretched along either the horizontal or vertical axis depending on whether \( a \) is greater than \( b \) or vice versa.

Understanding the basic equation of an ellipse allows us to move forward to tackle problems involving its tangent lines and related geometric properties.

Radius of Curvature

The radius of curvature is pivotal in understanding how sharply a curve bends at a particular point. On an ellipse, it varies depending on the location. At point \( (2,0) \) on the given ellipse, the curve exhibits a vertical tangent; thus, \( \frac{dy}{dx} \) is undefined.
To assess the radius of curvature, apply the formula: \[ R = \left( \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{\left| \frac{d^2y}{dx^2} \right|} \right) \]. Implicit differentiation applies here. After evaluating the second derivative \( \frac{d^2y}{dx^2} \) and other terms, we find \( R = \frac{8}{9} \) at \( (2,0) \).

• The radius of curvature represents how far the osculating (or kissing) circle extends from the point of tangency.
• At \( (0,3) \), the tangent is horizontal, resulting in a differently calculated radius \( R = \frac{12}{9} \).

This concept is essential in geometry, aiding in the comprehension of curve behavior and approximations involving circles.

Implicit Differentiation

Implicit differentiation allows us to differentiate equations where \( y \) is not isolated. In the function \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), we encounter implicit variables that interact jointly.
Differentiating each term with respect to \( x \), we derive: \( \frac{x}{2} + \frac{y}{9} \cdot \frac{dy}{dx} = 0 \).
Solving for the derivative, \( \frac{dy}{dx} = -\frac{9x}{2y} \), which provides the slope of tangents to the ellipse at any point.

• This method is crucial for problems involving curves not solvable by explicit means.
• Implicit differentiation handles variables defined in a complex interconnected manner, ideal for dealing with geometric shapes like ellipses.

By differentiating again implicitly, we obtain \( \frac{d^2y}{dx^2} \), enabling precise calculations of curvature and tangent properties at specific points.