Chapter 3: Problem 142
Find equations of the osculating circles of the ellipse \(4 y^{2}+9 x^{2}=36\) at the points \((2,0)\) and \((0,3)\)
Short Answer
Expert verified
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 by step solution
01
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 \(b=3\).
02
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 \(\frac{dy}{dx} = -\frac{9x}{2y}\).
03
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, \(R=\frac{8}{9}\).
04
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 \((x-2)^2 + \left(y - \frac{8}{9}\right)^2 = \left(\frac{8}{9}\right)^2\).
05
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 \(R=\frac{12}{9}\).
06
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 \(\left(x - \frac{4}{3}\right)^2 + (y - 3)^2 = \left(\frac{4}{3}\right)^2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( (2,0) \).
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 \( R = \frac{12}{9} \).
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.
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.