Problem 53 - Calculus with Concepts in Calculus

Question

Show that an equation of the plane tangent to the ellipsoid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $$ at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}+\frac{z z_{0}}{c^{2}}=1 $$

Step-by-Step Solution

Answer

Differentiate the ellipsoid equation; find and use the gradient as the normal vector at the point to form the tangent plane equation, resulting in \( \frac{x x_0}{a^2} + \frac{y y_0}{b^2} + \frac{z z_0}{c^2} = 1 \).

Step 1: Differentiate the Equation of the Ellipsoid

To find the equation of the tangent plane, 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 partial derivatives: \( \frac{2x}{a^2} \), \( \frac{2y}{b^2} \), and \( \frac{2z}{c^2} \).

Step 2: Gradient Vector of the Ellipsoid

The gradient vector, \( abla f\), at any point \((x, y, z)\) on the ellipsoid is given by \( \left( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2} \right) \). This vector is normal to the ellipsoid at any point \((x, y, z)\) on its surface.

Step 3: Evaluate the Gradient at the Given Point

Substitute the point \((x_0, y_0, z_0)\) into the gradient vector to obtain the normal vector at this specific point: \( \left( \frac{2x_0}{a^2}, \frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right) \).

Step 4: Determine the Tangent Plane Equation

The equation of the tangent plane to a surface at a given point \((x, y, z)\) with a normal vector \( (A, B, C) \) is given by: \( A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \). Inserting our values: \( \frac{2x_0}{a^2}(x - x_0) + \frac{2y_0}{b^2}(y - y_0) + \frac{2z_0}{c^2}(z - z_0) = 0 \).

Step 5: Simplify the Tangent Plane Equation

Expanding and simplifying the expression from Step 4, we have: \( \frac{2x_0}{a^2}x - \frac{2x_0^2}{a^2} + \frac{2y_0}{b^2}y - \frac{2y_0^2}{b^2} + \frac{2z_0}{c^2}z - \frac{2z_0^2}{c^2} = 0 \). Dividing the entire equation by 2, we simplify to: \( \frac{x x_0}{a^2} + \frac{y y_0}{b^2} + \frac{z z_0}{c^2} = 1 \).

Step 6: Verify the Simplified Equation

Ensure the simplified tangent plane equation satisfies the condition for any point on the ellipsoid reaching the given point. The original derivation properly constructs the plane using partial derivatives, aligning with known geometric behaviors of ellipsoids.

Key Concepts

Ellipsoid

An ellipsoid can be thought of as a stretched or squished sphere. It is a three-dimensional surface, and it looks like an egg or a rugby ball. The standard equation for a typical ellipsoid is given by: \[\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1\]In this equation, the coefficients \(a\), \(b\), and \(c\) determine how much the ellipsoid is stretched in each of the coordinate directions:

• \(a\) is the semi-axis along the x-direction,
• \(b\) is the semi-axis along the y-direction,
• \(c\) is the semi-axis along the z-direction.

The values \(a\), \(b\), and \(c\) shape the ellipsoid, affecting its symmetry and size in each direction.
The ellipsoid is a basic shape in geometry and commonly appears in fields like physics, astronomy, and engineering.

Partial Derivatives

Partial derivatives are an essential tool in calculus when dealing with functions of multiple variables. They represent the rate of change of a function with respect to one variable, while keeping the other variables constant. For example, given the ellipsoid's equation:\[\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1\]We compute the partial derivatives with respect to \(x\), \(y\), and \(z\):

• \(\frac{\partial}{\partial x}\left(\frac{x^{2}}{a^{2}}\right) = \frac{2x}{a^2}\).
• \(\frac{\partial}{\partial y}\left(\frac{y^{2}}{b^{2}}\right) = \frac{2y}{b^2}\).
• \(\frac{\partial}{\partial z}\left(\frac{z^{2}}{c^{2}}\right) = \frac{2z}{c^2}\).

Partial derivatives help in understanding how the surface of an ellipsoid tilts or bends at any given point, making them crucial for deriving the tangent plane.

Gradient Vector

The gradient vector is a key concept when analyzing functions of several variables, such as the surface of an ellipsoid. It combines the partial derivatives to represent the direction of the steepest increase of a function. For the given ellipsoid, the gradient vector, \(abla f\), is derived as:\[ \left( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2} \right)\]Here is what it means:

• The gradient points in the direction of the greatest increase of the function.
• Its magnitude tells us how steep the increase is.
• It is perpendicular to the surface of the ellipsoid at any point \((x, y, z)\).

Thus, the gradient vector not only indicates the slope but also defines the normal vector, which is critical for finding the tangent plane to the ellipsoid.

Normal Vector

The normal vector is perpendicular to a surface at a given point. For a surface defined by a function, this vector can be derived from the gradient vector. When dealing with the tangent plane to an ellipsoid, the normal vector is:\[\left( \frac{2x_0}{a^2}, \frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right)\]This is calculated by plugging the specific point \((x_0, y_0, z_0)\) into the gradient vector. The importance of the normal vector includes:

• It provides the perpendicular direction to the tangent plane at a chosen point on the ellipsoid.
• It helps to construct the equation of the tangent plane by incorporating its components \((A, B, C)\) in the plane's equation:

The well-known tangent plane equation is\[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\]Inserting values from the gradient provides a tailored equation specific to the point of interest on the ellipsoid.