Problem 35 - Thomas Calculus

Question

(Continuation of Exercise \(34 . )\) Find a Cartesian equation for the plane tangent to the hyperboloid \(x^{2}+y^{2}-z^{2}=25\) at the point \(\left(x_{0}, y_{0}, 0\right),\) where \(x_{0}^{2}+y_{0}^{2}=25\)

Step-by-Step Solution

Answer

The tangent plane equation is \(x_0 x + y_0 y = 25\).

Step 1: Confirm the Point on the Surface

The point \((x_0, y_0, 0)\) must satisfy the hyperboloid equation \(x^2 + y^2 - z^2 = 25\). Since \(z = 0\) at the point, substitute \(z = 0\) into the equation to confirm: \(x_0^2 + y_0^2 - 0^2 = 25\), which simplifies to \(x_0^2 + y_0^2 = 25\). This condition is already given, confirming that the point lies on the hyperboloid.

Step 2: Find the Gradient Vector of the Surface

The gradient vector of the surface defined by \(f(x, y, z) = x^2 + y^2 - z^2\) is obtained by computing its partial derivatives: \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (2x, 2y, -2z)\). At the point \((x_0, y_0, 0)\), the gradient is \((2x_0, 2y_0, 0)\).

Step 3: Write the Equation of the Tangent Plane

The equation of the tangent plane to a surface \(f(x, y, z) = 0\) at a point \((x_0, y_0, z_0)\) with a gradient vector \((A, B, C)\) is given by \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\). Using the gradient \((2x_0, 2y_0, 0)\) at the point \((x_0, y_0, 0)\), the equation becomes: \[2x_0(x - x_0) + 2y_0(y - y_0) + 0(z - 0) = 0\] This simplifies to: \[2x_0 x + 2y_0 y = 2x_0^2 + 2y_0^2\] Since \(x_0^2 + y_0^2 = 25\), it becomes: \[2x_0 x + 2y_0 y = 50\].

Step 4: Finalize the Equation

Simplify the equation: \(2x_0 x + 2y_0 y = 50\) by dividing throughout by 2:\[x_0 x + y_0 y = 25\]. This is the Cartesian equation of the tangent plane at the given point.

Key Concepts

Understanding Hyperboloids

The hyperboloid is a fascinating and complex 3D surface that can appear in two main forms: one-sheeted and two-sheeted. Imagine a shape similar to a cooling tower seen at power plants, or a shape reminiscent of an hourglass. These structures naturally arise in many physical scenarios. In mathematics, they are represented by a specific quadratic equation in three variables. For the two-sheeted hyperboloid at the focus of this exercise, the equation is

• \(x^2 + y^2 - z^2 = 25\).
• Notice the negative sign in front of the \(z^2\) term. This indicates that the hyperboloid opens along the \(z\)-axis.

Points on the hyperboloid are determined by substituting specific values into this equation, such as

• \((x_0, y_0, 0)\) where \(x_0^2 + y_0^2 = 25\).

This verifies that the point lies on the surface. It's essential to grasp these forms vividly, as the concept relates to numerous applications in engineering and physics.

Exploring the Gradient Vector

The gradient vector is a crucial tool in understanding surfaces and finding tangent planes in multivariable calculus. Think of it as a vector pointing in the direction of the steepest ascent of a function. For the function

• \(f(x, y, z) = x^2 + y^2 - z^2\),
• its gradient vector is given by \(abla f = (2x, 2y, -2z)\).

This vector demonstrates how changes in \(x\), \(y\), and \(z\) affect the function's value. At any point, such as

• \((x_0, y_0, 0)\), the gradient becomes \((2x_0, 2y_0, 0)\).

Understanding the gradient also helps in identifying perpendicular directions to a surface at a given point. Essentially, it provides a normal vector to the surface, thus enabling the construction of the tangent plane. This aspect of the gradient makes it an indispensable tool in calculus.

Cartesian Equations and Tangent Planes

A Cartesian equation represents a crucial link between algebra and geometry. It expresses the relationship between Cartesian coordinates \((x, y, z)\) of points creating a specific geometric surface. In the context of tangent planes, these equations allow us to precisely define a plane that tangentially touches a surface at a selected point. For example, to find the tangent plane to the hyperboloid surface at

• \((x_0, y_0, 0)\),

we begin with the gradient vector

• \((2x_0, 2y_0, 0)\) as a normal to the plane.

The equation of the tangent plane is structured using the gradient, resulting in:

• \[ 2x_0(x-x_0) + 2y_0(y-y_0) + 0(z-0) = 0 \],
• which simplifies to \[ x_0 x + y_0 y = 25 \].

This tangent plane calculation elegantly ties together the values on the hyperboloid and the gradient vector, illustrating how the mathematics of surfaces, vectors, and algebra intertwine to describe spatial relationships.