Question

Show that the tangent plane to the quadric surface at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) can be written in the given form. Hyperboloid: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1\) Plane: \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}-\frac{z_{0} z}{c^{2}}=1\)

Short answer

The equation of the tangent plane at the point \((x_{0}, y_{0}, z_{0})\) to the given hyperboloid is \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}-\frac{z_{0} z}{c^{2}}=1\).

Step-by-step solution

Finding the gradient of the function

As our first step, calculate the gradient of the function which represents our quadric surface. Using \(F(x, y, z) =\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}-1\), we calculate the gradient \(\nabla F = \left( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{-2z}{c^2} \right)\)

Substituting the point into the gradient

Now, substitute the point \((x_{0}, y_{0}, z_{0})\) into the gradient. This will yield a normal vector to the surface at that point. At \((x_{0}, y_{0}, z_{0})\), \(\nabla F = \left( \frac{2x_{0}}{a^2}, \frac{2y_{0}}{b^2}, \frac{-2z_{0}}{c^2} \right) \)

Finding the equation of the tangent plane

Equation of the tangent plane in the point normal form is \(n.(r-r_{0})=0\),where \(n\) is the normal vector and \(r_{0}\) is the position vector of a point on the plane, and \(r\) is the position vector of an arbitrary point on the plane. Here, \(n = \nabla F\) at the point (x_{0}, y_{0}, z_{0}) and \(r_{0} = (x_0, y_0, z_0), r = (x, y, z)\). Plugging these values into the equation gives us the equation of the tangent plane: \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}-\frac{z_{0} z}{c^{2}}=1\)

Key concepts

Gradient of a Function

Understanding the gradient of a function is key to solving problems related to the geometry of surfaces like hyperboloids. The gradient is a vector that points in the direction of the greatest rate of increase of the function. Mathematically, for a function \(F(x, y, z)\), the gradient \(abla F\) is found by taking the partial derivatives of \(F\) with respect to each variable. In our exercise, \(F(x, y, z)\) represents the equation of a hyperboloid, and its gradient at any point \((x, y, z)\) is calculated as \(abla F = \left( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{-2z}{c^2} \right)\). The gradient not only gives us the direction of steepest ascent but also serves as a normal vector to surfaces, which is crucial when finding the equation of a tangent plane.

Normal Vector

A normal vector is perpendicular to a surface at a given point. For a 3-dimensional surface, the gradient vector \(abla F\) at a point is a normal vector. Understanding this concept helps in analyzing the properties of the surface and in finding tangent planes.
In the step-by-step solution, after computing the gradient, we substitute the coordinates of the given point to get the normal vector at that specific location on the surface. For our hyperboloid, the normal vector at the point \((x_0, y_0, z_0)\) is found by evaluating the gradient, resulting in \(abla F = \left( \frac{2x_{0}}{a^2}, \frac{2y_{0}}{b^2}, \frac{-2z_{0}}{c^2} \right)\). This vector is essential for determining the tangent plane equation.

Equation of a Tangent Plane

The equation of a tangent plane to a surface at a given point provides a linear approximation of the surface near that point. The general form of this equation is \(n \cdot (r - r_0) = 0\), where \(n\) is the normal vector to the surface at the point, \(r_0\) is the position vector of the point on the plane, and \(r\) is the position vector of any point on the plane.
In our exercise, the normal vector \(n\) is the gradient at the point \((x_0, y_0, z_0)\), and substituting \(n\) and \(r_0\) into the general equation gives the specific equation of the tangent plane to the hyperboloid surface. It is a valuable tool in differential geometry and multivariable calculus for visualizing and understanding the behavior of surfaces.

Position Vector

A position vector, often denoted as \(r\), is a vector that represents the position of a point in space in relation to an origin point. It is essential in formulating the equation of a tangent plane as shown in the provided solution.
For a point with coordinates \((x, y, z)\), its position vector is \(r = (x, y, z)\). In the context of our exercise, we're considering two position vectors: \(r\), the position vector of an arbitrary point on the tangent plane, and \(r_0\), the position vector of the point \((x_0, y_0, z_0)\) where the plane is tangent to the hyperboloid. The difference between these vectors, when dotted with the normal vector, provides the left-hand side of the tangent plane equation.

Hyperboloid

A hyperboloid is a type of quadric surface that can appear in one of two forms, one-sheeted or two-sheeted, both of which are defined by a specific algebraic equation. Our exercise focuses on a one-sheeted hyperboloid, given by the equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1\). This surface has a saddle shape and is symmetric about the origins of the coordinate axes.
The properties of a hyperboloid are vital in the study of geometry and physics. When finding the tangent plane at any point on a hyperboloid, we apply the principles discussed in the previous sections about gradients, normals, and position vectors to arrive at the desired equation.