Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation \(w=T(x, y, z)\) of the tangent "hyperplane" at \(\mathbf{p}\). $$ f(x, y, z)=3 x^{2}-2 y^{2}+x z^{2}, \mathbf{p}=(1,2,-1) $$

Step-by-Step Solution

Answer

The tangent hyperplane is \( w = 7x - 8y - 2z + 3 \).

Step 1: Understand the Hyperplane

The problem requires finding the equation of the tangent hyperplane to the surface defined by the function \( f(x, y, z) = 3x^2 - 2y^2 + xz^2 \) at the point \( \mathbf{p} = (1, 2, -1) \). The equation of a tangent hyperplane at a point \( (a, b, c) \) is given by: \[ w = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c), \] where \(f_x\), \(f_y\), and \(f_z\) are the partial derivatives of \( f \).

Step 2: Calculate Partial Derivatives

Calculate the partial derivatives of the function \( f(x, y, z) \):1. \( f_x(x, y, z) = \frac{\partial f}{\partial x} = 6x + z^2 \).2. \( f_y(x, y, z) = \frac{\partial f}{\partial y} = -4y \).3. \( f_z(x, y, z) = \frac{\partial f}{\partial z} = 2xz \).

Step 3: Evaluate Partial Derivatives at the Point

Evaluate the partial derivatives at the point \( \mathbf{p} = (1, 2, -1) \):1. \( f_x(1, 2, -1) = 6 \cdot 1 + (-1)^2 = 7 \).2. \( f_y(1, 2, -1) = -4 \cdot 2 = -8 \).3. \( f_z(1, 2, -1) = 2 \cdot 1 \cdot (-1) = -2 \).

Step 4: Evaluate the Function at the Point

Evaluate the function at the point \( \mathbf{p} \):\[ f(1, 2, -1) = 3(1)^2 - 2(2)^2 + 1(-1)^2 = 3 - 8 + 1 = -4. \]

Step 5: Form the Equation of the Tangent Hyperplane

Using the formula for the tangent hyperplane, substitute the values:\[ w = -4 + 7(x - 1) - 8(y - 2) - 2(z + 1). \]Simplify the equation:\[ w = -4 + 7x - 7 - 8y + 16 - 2z - 2. \]Thus, the equation becomes:\[ w = 7x - 8y - 2z + 3. \]

Key Concepts

Tangent Hyperplane

In multivariable calculus, a **tangent hyperplane** is akin to the tangent line we are familiar with from single-variable calculus. However, as we move to multiple dimensions, the concept extends into higher dimensions too.

A tangent hyperplane is simply a flat surface that just "touches" a given surface without intersecting it, at least on an infinitesimally small scale. It provides the best linear approximation of the surface at a specific point.

The tangent hyperplane at a point \(\mathbf{p} = (a, b, c)\) on a surface defined by a function \(f(x, y, z)\) can be expressed as:

• \(w = f(a, b, c) + f_x(a, b, c)(x-a) + f_y(a, b, c)(y-b) + f_z(a, b, c)(z-c)\)

This equation uses partial derivatives to capture how the surface behaves as each input variable changes, which is essential for creating the hyperplane.

Partial Derivatives

**Partial derivatives** are the building blocks for understanding how multivariable functions change. When we compute a partial derivative of a function like \(f(x, y, z)\), we find how the function changes concerning one variable while keeping the other variables constant.

In our example:

• \(f_x(x, y, z) = 6x + z^2\) assesses changes along the \(x\)-axis.
• \(f_y(x, y, z) = -4y\) focuses on changes along the \(y\)-axis.
• \(f_z(x, y, z) = 2xz\) relates to changes along the \(z\)-axis.

When evaluating these at a specific point, such as \((1, 2, -1)\), they reveal how the surface bends or slopes in multiple directions at that pinpoint location, which is crucial for defining the tangent hyperplane.

Surface Equation

The **surface equation** in multivariable calculus describes a 3D object using a function of several variables. In this case, \(f(x, y, z) = 3x^2 - 2y^2 + xz^2\) represents a surface in a 3D space.

Understanding how this surface looks at various points requires using principles like partial derivatives. At the point \(\mathbf{p} = (1, 2, -1)\), we evaluate not only its function value but also its first-order partial derivatives.

The calculation of these values provides information essential for higher concept evaluations, such as forming a tangent hyperplane that approximates our surface locally at point \(\mathbf{p}\). This gives insight into the behavior and shape of the surface.

Function Evaluation

**Function evaluation** involves computing the value of a function at a specific point to understand the characteristics of a surface at that location. For the function \((f(x, y, z) = 3x^2 - 2y^2 + xz^2)\), computing it at \(\mathbf{p} = (1, 2, -1)\) tells us the height or depth of the surface at that exact point.

From our calculations, \(f(1, 2, -1) = 3 - 8 + 1 = -4\). This result is critical because it determines where the tangent hyperplane will be anchored in a vertical sense.

By combining this height with slopes determined by the partial derivatives, we form a tangent hyperplane equation to closely approximate the behavior of the surface around that point.