/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Evaluate \(f_{x}\) and \(f_{y}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate \(f_{x}\) and \(f_{y}\) at the given point. $$ f(x, y)=\arctan \frac{y}{x}, \quad(2,-2) $$

Short Answer

Expert verified
The value of \(f_{x}\) at the point (2,-2) is -0.5 and the value of \(f_{y}\) at the point (2,-2) is 0.5.

Step by step solution

01

Calculating Partial Derivative \(f_{x}\)

First, recall that when we differentiate with respect to \(x\), we consider \(y\) as a constant. The derivative of \(\arctan(u)\) with respect to \(u\) is \(\frac{1}{{1 + u^2}}\). Hence, the partial derivative of \(f(x, y)\) with respect to \(x\) is \(f_{x} = \frac{1}{{1 + (\frac{y}{x})^2}} * -\frac{y}{x^2} = -\frac{y}{x^2 + y^2}\).
02

Calculating Partial Derivative \(f_{y}\)

When we differentiate with respect to \(y\), we consider \(x\) as a constant. Similarly, the partial derivative of \(f(x, y)\) with respect to \(y\) is \(f_{y} = \frac{1}{{1 + (\frac{y}{x})^2}} * \frac{1}{x} = \frac{x}{x^2 + y^2}\).
03

Evaluating \(f_{x}\) and \(f_{y}\) at the Given Point

Now, we need to evaluate these partial derivatives at the point (2,-2). For \(f_{x}(2,-2)\), replace \(x\) with 2 and \(y\) with -2 in the equation for \(f_{x}\), to find that \(f_{x}(2,-2) = -\frac{-2}{2^2 + (-2)^2} = -0.5\). Similarly, evaluate \(f_{y}(2,-2)\) by replacing \(x\) with 2 and \(y\) with -2 in the equation for \(f_{y}\), to find that \(f_{y}(2,-2) = \frac{2}{2^2 + (-2)^2} = 0.5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arctangent Function
The arctangent function, denoted as \( \arctan(u) \), is the inverse of the tangent function within its principal range, which is \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). It is used to find the angle whose tangent is the given number \(u\). In multivariable calculus, the arctangent is often used when dealing with angles and is particularly useful in the context of polar coordinates.
When differentiating \( \arctan(u) \) with respect to \( u \), the result is \( \frac{1}{1+u^2} \). This formula is essential when calculating derivatives involving arctangent, especially in cases where substitutions or chain rule applications are needed.
Understanding the arctangent function's behavior and derivative is crucial in solving calculus problems involving arcs and angles, making it a key concept in both single-variable and multivariable contexts.
Calculus Problem Solving
Calculus problem solving requires systematic approaches, especially when dealing with derivatives. Here, we find partial derivatives of a function \( f(x, y) = \arctan \frac{y}{x} \).
Given this function, the process first involves determining partial derivatives \( f_x \) and \( f_y \). For \( f_x \), treat \( y \) as a constant, leading to \(-\frac{y}{x^2 + y^2} \).
For \( f_y \), treat \( x \) as a constant, resulting in \( \frac{x}{x^2 + y^2} \). Notice the importance of applying the chain rule and the derivatives of trigonometric functions.
Next, one crucial step is evaluating these derivatives at a specific point, i.e., \((2, -2)\). Substituting \( x = 2 \) and \( y = -2 \) in both derivatives yields \( f_x(2, -2) = -0.5 \) and \( f_y(2, -2) = 0.5 \).
Each step illustrates the structured approach necessary in calculus, emphasizing understanding the problem's context and the function's behavior.
Multivariable Calculus
Multivariable calculus extends concepts from single-variable calculus to functions of several variables. In this realm, we explore partial derivatives, as seen in the function \( f(x, y) = \arctan \left(\frac{y}{x}\right) \).
Partial derivatives are derivatives that measure how a function changes as one of the variables changes, keeping others constant. Calculating \( f_x \) and \( f_y \) requires understanding how each variable independently affects \( f \).
In multivariable calculus, it is often necessary to evaluate partial derivatives at specific points, understanding their geometric and physical interpretations. For instance, \( f_x(2, -2) \) and \( f_y(2, -2) \) reveal how the function \( f \) changes when \( x \) or \( y \) vary slightly at the point of interest.
The strategic application of these concepts allows us to meticulously follow function changes and analyze surfaces and contours in higher-dimensional spaces.

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Most popular questions from this chapter

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