Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-14, find the gradient vector of the given function at the given point \(\mathbf{p}\). Then find the equation of the tangent plane at \(\mathbf{p}\) (see Example 1). $$ f(x, y)=x^{3} y+3 x y^{2}, \mathbf{p}=(2,-2) $$

Step-by-Step Solution

Answer

The tangential equation is \( z = -12x - 16y + 40 \).

Step 1: Find the gradient vector

The gradient vector of a function \( f(x, y) \) is given by \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = x^3 y + 3xy^2 \), we compute the partial derivatives: 1. \( \frac{\partial f}{\partial x} = 3x^2 y + 3y^2 \) 2. \( \frac{\partial f}{\partial y} = x^3 + 6xy \).Evaluating these at the point \( \mathbf{p} = (2, -2) \), we get:- \( \frac{\partial f}{\partial x} \bigg|_{(2, -2)} = 3(2)^2(-2) + 3(-2)^2 = -24 + 12 = -12 \)- \( \frac{\partial f}{\partial y} \bigg|_{(2, -2)} = (2)^3 + 6(2)(-2) = 8 - 24 = -16 \).Thus, the gradient vector at \( \mathbf{p} \) is \( abla f(2, -2) = (-12, -16) \).

Step 2: Calculate the value of the function at the point

Calculate the function \( f(x, y) \) at the given point \( \mathbf{p} = (2, -2) \):\[ f(2, -2) = (2)^3(-2) + 3(2)(-2)^2 = -16 + 24 = 8. \]

Step 3: Write the equation of the tangent plane

The equation of a tangent plane to a surface \( z = f(x, y) \) at a point \( (x_0, y_0, z_0) \) is given by:\[ z - z_0 = \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0), \]where \( z_0 = f(x_0, y_0) \).Substituting \( x_0 = 2 \), \( y_0 = -2 \), \( z_0 = 8 \), and the gradient components, the equation becomes:\[ z - 8 = -12(x - 2) - 16(y + 2). \]Simplifying gives:\[ z = -12x - 16y + 40. \]

Key Concepts

Gradient vector

In calculus, the gradient vector is a vital concept when dealing with multivariable functions. It acts like a compass, pointing in the direction of the steepest ascent on the function's surface. The gradient is represented by the symbol \( abla f(x, y) \) for a function \( f(x, y) \). It is a vector consisting of partial derivatives of the function with respect to each variable, usually \( x \) and \( y \).

To find the gradient vector at a specific point, you first need to compute the partial derivatives:

• \( \frac{\partial f}{\partial x} \) — the rate of change of the function with respect to the variable \( x \) while keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} \) — the rate of change of the function with respect to the variable \( y \) while keeping \( x \) constant.

After calculating these derivatives, you evaluate them at the given point of interest. In our problem, for \( f(x, y) = x^3 y + 3xy^2 \), we find:

• \( \frac{\partial f}{\partial x} = 3x^2 y + 3y^2 \)
• \( \frac{\partial f}{\partial y} = x^3 + 6xy \)

Evaluating these at \( \mathbf{p} = (2, -2) \) gives the gradient vector \( abla f(2, -2) = (-12, -16) \), indicating the path of steepest ascent from that point.

Partial derivatives

When working with functions of multiple variables, such as \( f(x, y) \), partial derivatives help us understand how the function changes with respect to one variable at a time. This is similar to regular derivatives but considers the effects of each variable separately while treating other variables as constants.

To compute a partial derivative, follow these steps:

• Identify the variable to differentiate with respect to, say \( x \), and treat the other variables as constants.
• Apply the rules of differentiation to find \( \frac{\partial f}{\partial x} \).
• Repeat the process for the other variable, e.g., \( y \), to find \( \frac{\partial f}{\partial y} \).

In our original exercise, we have:

• \( \frac{\partial f}{\partial x} = 3x^2 y + 3y^2 \) gives us insight into how changes in \( x \) affect \( f \).
• \( \frac{\partial f}{\partial y} = x^3 + 6xy \) shows the sensitivity of \( f \) to changes in \( y \).

Understanding partial derivatives is crucial as they form the components of the gradient vector and play a key role in formulating equations such as those for tangent planes.

Tangent plane

The concept of a tangent plane is important when analyzing surfaces in three dimensions. It is a plane that "just touches" the surface at a single point, called the tangent point, and has the same slope as the surface at that specific location.

For a function \( z = f(x, y) \), the tangent plane at a point \( (x_0, y_0, z_0) \) is described by the equation:

• \[ z - z_0 = \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0) \]

Here, \( z_0 = f(x_0, y_0) \) is the function's value at the point, and \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are the partial derivatives previously calculated.

In our problem, substituting \( x_0 = 2 \), \( y_0 = -2 \), and \( z_0 = 8 \) (from \( f(2, -2) = 8 \)), we find the tangent plane's equation:

• \[ z = -12x - 16y + 40 \]

This equation gives a linear approximation of the surface around \( \mathbf{p} \), useful for estimating values of \( f \) near this point.

Multivariable functions

Multivariable functions are those that depend on two or more variables, such as \( f(x, y) \). They describe a surface in three dimensions and provide a wealth of information about how different variables interact to influence the outcome.

Key features of multivariable functions include:

• Domains: The set of all possible input pairs \((x, y)\) for which the function is defined.
• Ranges: The set of possible output values \( f(x, y) \).
• Contours: Level curves where the function value remains constant.

When dealing with such functions, we often explore how the function reacts to changes in each variable using partial derivatives. They help us understand the shape of the surface, identify critical points, and analyze behavior.

Moreover, tools like gradient vectors and tangent planes allow us to investigate the function's local behavior, providing insights into optimization problems and stability analysis. Understanding these aspects in our exercise helps grasp the bigger picture of why calculus is vital in modeling real-world phenomena.