Problem 3 - Calculus Early Transcendentals

Question

I-6 Find an equation of the tangent plane to the given surface at the specified point. $$ z=\sqrt{x y}, \quad(1,1,1) $$

Step-by-Step Solution

Answer

The tangent plane equation is \( z = \frac{1}{2}x + \frac{1}{2}y \).

Step 1: Understand the Formula for the Tangent Plane

The equation of the tangent plane to a surface given by a function \( z = f(x, y) \) at the point \((x_0, y_0, z_0)\) is \( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \), where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.

Step 2: Compute Partial Derivative of z with respect to x

Given \( z = \sqrt{xy} \), let's compute the partial derivative with respect to \( x \). The formula is \( f_x = \frac{\partial}{\partial x}\left( \sqrt{xy} \right) = \frac{1}{2\sqrt{xy}} \cdot y \). Evaluating at \((x_0, y_0) = (1, 1)\), we find \( f_x(1, 1) = \frac{1}{2} \).

Step 3: Compute Partial Derivative of z with respect to y

Next, calculate the partial derivative of \( z = \sqrt{xy} \) with respect to \( y \). The formula is \( f_y = \frac{\partial}{\partial y}\left( \sqrt{xy} \right) = \frac{1}{2\sqrt{xy}} \cdot x \). Evaluating at \((x_0, y_0) = (1, 1)\), we find \( f_y(1, 1) = \frac{1}{2} \).

Step 4: Substitute values into the Tangent Plane Equation

With the derivatives calculated, substitute \( f_x(1,1) = \frac{1}{2} \), \( f_y(1,1) = \frac{1}{2} \), and the point \((1, 1, 1)\) into the tangent plane equation. This gives: \[ z - 1 = \frac{1}{2}(x - 1) + \frac{1}{2}(y - 1) \].

Step 5: Simplify the Tangent Plane Equation

Expanding and simplifying the equation from step 4, we have: \[ z - 1 = \frac{1}{2}x - \frac{1}{2} + \frac{1}{2}y - \frac{1}{2} \].Combining like terms, the equation becomes:\[ z = \frac{1}{2}x + \frac{1}{2}y \].

Key Concepts

Partial Derivatives

Understanding partial derivatives is key to tackling many calculus problems, especially those involving surfaces. In cases where a function has more than one variable, a partial derivative helps us examine how a function changes with respect to one variable while keeping other variables constant.

Take the function \[z = \sqrt{xy} \]used in the problem. To find the partial derivatives:

• The partial derivative with respect to \(x\), denoted as \(f_x\), represents how \(z\) changes as \(x\) changes while \(y\) remains fixed.
• Similarly, the partial derivative with respect to \(y\), \(f_y\), shows the rate of change of \(z\) as \(y\) changes, holding \(x\) constant.

Partial derivatives function similarly to regular derivatives by focusing on one variable at a time. This simplifies complex functions, allowing us to better understand and analyze surfaces in multidimensional spaces.

Surface Equation

The surface equation in this problem is given as \[z = \sqrt{xy} \].

This equation describes a surface in three-dimensional space. Here, \(z\) is the dependent variable, influenced by two independent variables, \(x\) and \(y\).

• Each point on this surface corresponds to particular values of \(x\) and \(y\), calculating \(z\) using the square root of their product \(xy\).
• The surface visually represents a paraboloid, which opens upwards, with its peak at the origin when plotted in 3D space.

Surface equations are critical in calculus for describing shapes and volumes defined over continuous spaces, and are essential for developing deeper insights into multi-variable functions.

Calculus Problem

This exercise exemplifies a typical calculus problem, connecting fundamental concepts like derivatives and tangent planes. The goal is to find the equation of the tangent plane to a given surface at a specific point.

Solving this problem involves understanding the geometry of the situation:

• Identifying tangent planes is akin to finding the "flat" surface that just touches the given surface at the specified point.
• These planes approximate the surface locally around the point of tangency.

Overall, calculus allows us to precisely manage dynamic problems involving rates of change and approximations, helping translate real-world problems into mathematical models for analysis.

Differentiation

Differentiation is a cornerstone of calculus. It helps us understand how a function changes when there's a tiny change in one or more of its input variables.

In the context of the given exercise:

• Differentiation of the function \(z = \sqrt{xy}\) involves taking partial derivatives, \(f_x\) and \(f_y\).
• These derivatives tell us how fast the surface \(z = \sqrt{xy}\) rises or falls as one of the base variables \(x\) or \(y\) changes.

Differentiation provides the tools to tackle complex multi-variable problems by breaking them into simpler one-variable issues, making it indispensable for deriving tangent planes and comprehending the geometry of surfaces.