Problem 14 - 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). $$ f(x, y)=\frac{x^{2}}{y}, \mathbf{p}=(2,-1) $$

Step-by-Step Solution

Answer

The gradient vector at point \( (2,-1) \) is \((-4, -4)\), and the tangent plane equation is \(z = -4x - 4y + 4\).

Step 1: Calculate Partial Derivatives

First, we find the partial derivatives of the function \( f(x, y) = \frac{x^2}{y} \). The partial derivative with respect to \( x \) is calculated using the quotient rule:\[\frac{\partial f}{\partial x} = \frac{2x}{y}\]The partial derivative with respect to \( y \) is:\[\frac{\partial f}{\partial y} = -\frac{x^2}{y^2}\]

Step 2: Evaluate the Gradient at Point \(\mathbf{p}\)

The gradient vector \( abla f \) at point \( \mathbf{p} = (2, -1) \) is evaluated by substituting \( x = 2 \) and \( y = -1 \) into the partial derivatives. For \( \frac{\partial f}{\partial x} \):\[\frac{2 \cdot 2}{-1} = -4\]For \( \frac{\partial f}{\partial y} \):\[-\frac{2^2}{(-1)^2} = -4\]Thus, \( abla f = (-4, -4) \).

Step 3: Find Equation of Tangent Plane

The formula for the tangent plane at point \( \mathbf{p} = (a, b) \) on the surface \( z = f(x, y) \) is given by:\[z - f(a, b) = abla f \cdot \begin{bmatrix} x - a \ y - b \end{bmatrix}\]First, we calculate \( f(2, -1) \):\[f(2, -1) = \frac{2^2}{-1} = -4\]Substitute \( abla f = (-4, -4) \) and \( \mathbf{p} = (2, -1) \):\[z + 4 = -4(x - 2) - 4(y + 1)\]Simplifying:\[z = -4x - 4y + 4\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables. When you have a function like \( f(x, y) = \frac{x^2}{y} \), where \( x \) and \( y \) are the variables, you can think of partial derivatives as a way to measure how the function changes as each variable changes, while keeping the other variable constant.
For the function \( f(x, y) \):

• The partial derivative with respect to \( x \) is found by treating \( y \) as a constant and differentiating with respect to \( x \). In our example, this results in \( \frac{\partial f}{\partial x} = \frac{2x}{y} \).
• Similarly, to find the partial derivative with respect to \( y \), treat \( x \) as constant. Here, this yields \( \frac{\partial f}{\partial y} = -\frac{x^2}{y^2} \).

Thus, partial derivatives provide insights into the local behavior of functions.

Tangent Plane

When dealing with surfaces in three dimensions, a tangent plane is an important concept. Visualize it as the plane that just "touches" a surface at a given point, providing a linear approximation of the surface around that point.
For a function \( z = f(x, y) \), the equation of the tangent plane at a point \( (a, b) \) is formulated as:
\[ z - f(a, b) = abla f \cdot \begin{bmatrix} x - a \ y - b \end{bmatrix} \]
This linear equation represents how changes in \( x \) and \( y \) affect \( z \). In the exercise, by substituting the gradient vector and the point \( (2, -1) \), we arrived at \( z = -4x - 4y + 4 \). This gives the tangent plane's equation that approximates the surface near \( \mathbf{p} = (2, -1) \).
The tangent plane simplifies complex surface interactions into manageable linear functions.

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. It combines all the partial derivatives of a function, offering a powerful insight into its behavior.
For the function \( f(x, y) \) given in the exercise, the gradient is denoted \( abla f \), and calculated by:

• Evaluating \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
• Joining these derivative results into a vector: \( abla f = \begin{bmatrix} \frac{\partial f}{\partial x} \ \frac{\partial f}{\partial y} \end{bmatrix} \).

At the point \( (2, -1) \), this process resulted in \( abla f = (-4, -4) \).
This vector highlights how, at the point \( \mathbf{p} \), the function \( f(x, y) \) changes most steeply. Understanding the gradient is key in optimization problems and in finding tangent planes.

Calculus

Calculus is the branch of mathematics that studies change. In multivariable calculus, which is essential for understanding real-world applications beyond simple functions, both differentiation and integration are extended to functions of several variables.
The problem tackled in the exercise is a typical example of using calculus concepts:

• **Differentiation:** Used to find partial derivatives, allowing us to explore how functions behave locally.
• **Linear Approximation:** Through concepts like the tangent plane, complex surfaces become approachable via linear models.

These principles allow us to tackle problems ranging from physics to engineering and economics. Recognizing how calculus applies in scenarios involving gradients, partial derivatives, and tangent planes enhances our deeper understanding of multidimensional phenomena.