Problem 30 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The given second equation is the tangent plane to the surface at \((x_0, y_0, z_0)\).

Step 1: Understand the Problem

We need to show that the given second equation is indeed the equation of the tangent plane to the given surface at the point \((x_0, y_0, z_0)\). Here, the surface is represented by the equation \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1\).

Step 2: Calculate Gradient of the Surface

To find the tangent plane at a point, compute the gradient of the surface function \(F(x, y, z) = \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} - 1\). The gradient \(abla F\) is given by its partial derivatives:\[abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)\]Calculate each component:- \(\frac{\partial F}{\partial x} = \frac{2x}{a^2}\)- \(\frac{\partial F}{\partial y} = -\frac{2y}{b^2}\)- \(\frac{\partial F}{\partial z} = \frac{2z}{c^2}\)

Step 3: Evaluate Gradient at Point \((x_0, y_0, z_0)\)

Substitute \((x_0, y_0, z_0)\) into the gradient \(abla F\):\[abla F(x_0, y_0, z_0) = \left( \frac{2x_0}{a^2}, -\frac{2y_0}{b^2}, \frac{2z_0}{c^2} \right)\]

Step 4: Write the Tangent Plane Equation

The equation of the tangent plane to the surface at \((x_0, y_0, z_0)\) can be written as:\[\frac{2x_0}{a^2}(x - x_0) - \frac{2y_0}{b^2}(y - y_0) + \frac{2z_0}{c^2}(z - z_0) = 0\]Upon expanding and simplifying:\[\frac{2x_0 x}{a^2} - \frac{2x_0^2}{a^2} - \frac{2y_0 y}{b^2} + \frac{2y_0^2}{b^2} + \frac{2z_0 z}{c^2} - \frac{2z_0^2}{c^2} = 0\]

Step 5: Match with Provided Equation

Re-arrange and simplify the terms to check if it corresponds to the given second equation:\[\frac{x x_0}{a^{2}} - \frac{y y_0}{b^{2}} + \frac{z z_0}{c^{2}} = 1\]Notice that constants from the negative squared terms cancel out the right-hand side, making the left-hand side equal to 1 - the form of the provided second equation.

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics focused on understanding rates of change and slopes of curves. When dealing with surfaces, like in this exercise, differential calculus helps us find tangent planes. Imagine folding a piece of paper over the curved surface; where the paper just touches is analogous to the tangent plane. The key skill is calculating how functions change infinitesimally, which is captured by derivatives. When you compute the derivative of a function concerning a variable, it tells you how the function value changes as that variable changes.
In more practical terms, differential calculus enables us to approximate complex curves and surfaces using simpler, linear elements. This simplification is crucial in various fields such as engineering and physics, where precise calculations are necessary.

• To tackle problems like our exercise, differential calculus provides the tool of a 'gradient'—which we'll dive into next.
• This helps identify how sharp or flat the contact between the surface and tangent plane is.

Surface Gradient

The gradient of a surface is a powerful concept in understanding how a surface behaves at different points. It provides a vector field, which shows the direction in which the surface rises fastest. Imagine standing on a hill—your steepest climb will be in the direction of the gradient.
In mathematical terms, the gradient of a function \(F(x, y, z)\) is denoted as \(abla F\) and consists of its partial derivatives. For our exercise, the gradient \(abla F\) was calculated as:

• \(\frac{\partial F}{\partial x} = \frac{2x}{a^2}\)
• \(\frac{\partial F}{\partial y} = -\frac{2y}{b^2}\)
• \(\frac{\partial F}{\partial z} = \frac{2z}{c^2}\)

This gradient vector acts perpendicular to the tangent plane at a given point. Evaluating it at a specific point, like \((x_0, y_0, z_0)\), gives the necessary coefficients for the tangent plane equation. Understanding the surface gradient provides insight into both the surface's behavior and the intricacies of the tangent plane.

Partial Derivatives

Partial derivatives are fundamental tools used in calculus to handle functions with multiple variables. They measure how a function changes as one specific variable is varied, while other variables are held constant.
To explain with our exercise, consider the function \(F(x, y, z)\). Its partial derivative with respect to \(x\) is given by \(\frac{2x}{a^2}\). This tells us how much \(F\) would change when only \(x\) changes, providing the rate of change in that specific direction.
Similarly, partial derivatives with respect to \(y\) and \(z\) are derived:

• \(\frac{\partial F}{\partial y} = -\frac{2y}{b^2}\)
• \(\frac{\partial F}{\partial z} = \frac{2z}{c^2}\)

In this problem, these partial derivatives are components of the gradient vector, which is crucial in forming the tangent plane's equation. By understanding and calculating these derivatives, we gain the ability to describe the behavior of multi-variable functions, allowing for powerful applications in science and engineering.