Problem 207 - Calculus 3

Question

Find the linear approximation of each function at the indicated point. $$ f(x, y)=x \sqrt{y}, \quad P(1,4) $$

Step-by-Step Solution

Answer

The linear approximation is \( L(x, y) = 2x + \frac{1}{4}y - 1 \).

Step 1: Understand the Concept

The linear approximation of a function at a specific point is the tangent plane to the function at that point. It approximates the function near the point by a linear function.

Step 2: Identify the Function and Point

The function given is \( f(x, y) = x \sqrt{y} \) and we're asked to find the linear approximation at the point \( P(1, 4) \).

Step 3: Calculate Partial Derivatives

Find the partial derivatives of \( f(x, y) = x \sqrt{y} \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( f_x(x, y) = \sqrt{y} \), and with respect to \( y \) is \( f_y(x, y) = \frac{x}{2\sqrt{y}} \).

Step 4: Evaluate Partial Derivatives at P(1, 4)

Calculate \( f_x(1, 4) \) and \( f_y(1, 4) \). Substituting into the derivatives, we get \( f_x(1, 4) = \sqrt{4} = 2 \) and \( f_y(1, 4) = \frac{1}{2\sqrt{4}} = \frac{1}{4} \).

Step 5: Write the Linear Approximation Formula

The formula for the linear approximation (tangent plane) is \( L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \), where \( (a, b) \) is the point of approximation.

Step 6: Calculate f(1, 4) and Assemble the Equation

Compute \( f(1, 4) = 1 \cdot \sqrt{4} = 2 \). So the linear approximation becomes \( L(x, y) = 2 + 2(x - 1) + \frac{1}{4}(y - 4) \).

Step 7: Simplify the Linear Approximation

First, simplify \( 2(x-1) = 2x - 2 \), and \( \frac{1}{4}(y - 4) = \frac{1}{4}y - 1 \). Combine terms to get \( L(x, y) = 2 + 2x - 2 + \frac{1}{4}y - 1 = 2x + \frac{1}{4}y - 1 \).

Key Concepts

Partial derivatives

Partial derivatives are a crucial concept in multivariable calculus. They measure how a function changes as only one of its variables changes, while the other variables are kept constant.
For example, when you're working with the function \( f(x, y) = x \sqrt{y} \), partial derivatives help determine how much \( f \) changes with a small change in \( x \) or \( y \) separately.

• The partial derivative of \( f \) with respect to \( x \), denoted as \( f_x(x, y) \), tells us how \( f \) changes when \( x \) changes by a small amount while \( y \) remains fixed.
• Similarly, \( f_y(x, y) \), the partial derivative with respect to \( y \), reflects the change in \( f \) as \( y \) changes, keeping \( x \) constant.

In the provided exercise, you calculated these partial derivatives as \( f_x(x, y) = \sqrt{y} \) and \( f_y(x, y) = \frac{x}{2\sqrt{y}} \). Evaluating them at the point P(1,4) helped construct the tangent plane.

Tangent plane

A tangent plane is the linear approximation of a function of two variables at a specific point. Imagine a surface in 3D space—this plane "gently touches" the surface at the point of tangency, closely approximating the surface nearby.
The formula for the tangent plane, or linear approximation, is \( L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \), where \( (a, b) \) is the point of interest.

• \( f(a, b) \) is the value of the function at the point, acting as a constant anchor.
• The terms \( f_x(a, b)(x-a) \) and \( f_y(a, b)(y-b) \) adjust this anchor based on the slope of the function in \( x \) and \( y \) directions.

In our example, substituting the calculated derivatives and the point P(1,4) into this formula provided the tangent plane \( L(x, y) = 2x + \frac{1}{4}y - 1 \). This plane serves as a simple, linear model of the function near the point (1, 4).

Multivariable calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. This branch deals with how functions change in multi-dimensional spaces. Instead of just dealing with simple linear slopes, you deal with much more complex geometries.
Key concepts include:

• Calculating derivatives concerning multiple variables — partial derivatives come into play here.
• Understanding surfaces, curves, and how they can be approximated by planes, like tangent planes.
• Examining how changes along different variables interplay with each other.

In our exercise, understanding multivariable calculus concepts allowed us to work with a 3D surface defined by \( f(x, y) = x \sqrt{y} \) and find its linear approximation at a specific point to simplify complex calculations nearby far easier.

Function of two variables

A function of two variables is a rule that assigns a unique output value based on two input variables, typically represented as \( f(x, y) \). Such functions can describe surfaces in three-dimensional space.
Let's break it down:

• The function \( f(x, y) = x \sqrt{y} \) assigns each pair \((x, y)\) a result by performing a mathematical operation on \( x \) and \( \sqrt{y} \).
• This results in a surface that bends and curves through 3D space.
• Understanding it involves looking at how changes in \( x \) or \( y \) could modify \( f \), leveraging partial derivatives and tangent planes.

Using a function of two variables allows us to model real-world phenomena, ranging from temperature over a region to cost calculations in economics. The exercise demonstrated how to locally linearize such a function near a specific point, aiding in understanding its behavior and interactions with its surroundings.