Chapter 14: Problem 49
Show that the equation of the tangent plane to the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) can be written as $$\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}+\frac{z z_{0}}{c^{2}}=1$$
Short Answer
Expert verified
The tangent plane is \( \frac{x x_0}{a^2} + \frac{y y_0}{b^2} + \frac{z z_0}{c^2} = 1 \).
Step by step solution
01
Differentiate the Ellipsoid Equation
Start by differentiating the given ellipsoid equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \) with respect to \( x \), \( y \), and \( z \). This gives us the following partial derivatives:\( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2} \).
02
Find the Normal Vector
Using the partial derivatives, we can determine that the normal vector to the ellipsoid surface at any given point \((x_0, y_0, z_0)\) is given by:\( \vec{n} = \left(\frac{2x_0}{a^2}, \frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right) \).
03
Formulate the Tangent Plane Equation
The equation of the tangent plane to a surface at a given point can be described using the normal vector \( (A, B, C) \) and the point \( (x_0, y_0, z_0) \):\[ A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \].
04
Substitute Normal Vector into the Tangent Plane Equation
Plug the components of \( \vec{n} \) into the tangent plane formula:\[ \frac{2x_0}{a^2}(x-x_0) + \frac{2y_0}{b^2}(y-y_0) + \frac{2z_0}{c^2}(z-z_0) = 0 \].
05
Simplify the Equation
Simplify by expanding the terms:\[ \frac{2x_0 x}{a^2} - \frac{2x_0^2}{a^2} + \frac{2y_0 y}{b^2} - \frac{2y_0^2}{b^2} + \frac{2z_0 z}{c^2} - \frac{2z_0^2}{c^2} = 0 \].
06
Reorder and Set Equal to One
Since we know that the point \( (x_0, y_0, z_0) \) lies on the ellipsoid:\( \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + \frac{z_0^2}{c^2} = 1 \), we can subtract this equation from the simplified tangent equation:\[ \frac{x x_0}{a^2} + \frac{y y_0}{b^2} + \frac{z z_0}{c^2} = 1 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ellipsoid
An ellipsoid is a three-dimensional geometric shape that is a generalization of an ellipse. Imagine it as a stretched or squashed sphere. A standard equation of an ellipsoid is given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]where the constants \(a\), \(b\), and \(c\) are the semi-axes lengths along the \(x\), \(y\), and \(z\) axes, respectively. This form shows how far the ellipsoid stretches in each direction, with larger values meaning more stretching.
Unlike a sphere, which is perfectly symmetric in all directions, an ellipsoid may have three different axes lengths, making it more flexible in modeling real-world objects.
Unlike a sphere, which is perfectly symmetric in all directions, an ellipsoid may have three different axes lengths, making it more flexible in modeling real-world objects.
- If \(a = b = c\), then the ellipsoid is a sphere.
- If \(a = b eq c\), it becomes an oblate or prolate spheroid depending on whether \(c\) is smaller or larger than \(a\) and \(b\).
Partial Derivatives
Partial derivatives play a crucial role in calculus and arise when you want to find the derivative of a function of multiple variables with respect to one variable at a time. For the ellipsoid, we have a function in terms of three variables \(x, y, z\).
The process of finding a partial derivative involves treating all other variables as constants and differentiating with respect to the variable of interest. For the ellipsoid given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]the partial derivatives are:
The process of finding a partial derivative involves treating all other variables as constants and differentiating with respect to the variable of interest. For the ellipsoid given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]the partial derivatives are:
- With respect to \(x\), the derivative is \(\frac{2x}{a^2}\).
- With respect to \(y\), the derivative is \(\frac{2y}{b^2}\).
- With respect to \(z\), the derivative is \(\frac{2z}{c^2}\).
Normal Vector
The concept of a normal vector is fundamental when dealing with surfaces in three-dimensional space, like our ellipsoid. A normal vector is a vector that is perpendicular to the surface at a given point.
To determine the normal vector of the ellipsoid at a point \((x_0, y_0, z_0)\), we use the partial derivatives found earlier as the components of the normal vector:\[\vec{n} = \left(\frac{2x_0}{a^2}, \frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right)\]This vector \(\vec{n}\) provides the direction in which the surface is not tangent.
To determine the normal vector of the ellipsoid at a point \((x_0, y_0, z_0)\), we use the partial derivatives found earlier as the components of the normal vector:\[\vec{n} = \left(\frac{2x_0}{a^2}, \frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right)\]This vector \(\vec{n}\) provides the direction in which the surface is not tangent.
- The importance of the normal vector lies in its use for formulating the tangent plane equation, which is the flat plane that "touches" the ellipsoid precisely at the point \((x_0, y_0, z_0)\).
Surface Differentiation
Surface differentiation is a method used to understand how a given surface behaves and changes. It allows us to examine the variations in surfaces like an ellipsoid, particularly how they slope and bend at each point.
For any point \((x_0, y_0, z_0)\) on the ellipsoid, applying surface differentiation yields the tangent plane. The equation for the tangent plane at this point uses the normal vector \((A, B, C)\), formulated as:\[A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\]Substituting the normal vector components derived earlier:\[\frac{2x_0}{a^2}(x-x_0) + \frac{2y_0}{b^2}(y-y_0) + \frac{2z_0}{c^2}(z-z_0) = 0\]The process of surface differentiation helps us establish not just the slope at a point, but also it assists in approximating the surface locally as a plane. This approximation is very useful in various applications, including physics and engineering, where understanding surface interactions are crucial.
For any point \((x_0, y_0, z_0)\) on the ellipsoid, applying surface differentiation yields the tangent plane. The equation for the tangent plane at this point uses the normal vector \((A, B, C)\), formulated as:\[A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\]Substituting the normal vector components derived earlier:\[\frac{2x_0}{a^2}(x-x_0) + \frac{2y_0}{b^2}(y-y_0) + \frac{2z_0}{c^2}(z-z_0) = 0\]The process of surface differentiation helps us establish not just the slope at a point, but also it assists in approximating the surface locally as a plane. This approximation is very useful in various applications, including physics and engineering, where understanding surface interactions are crucial.