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In Exercises \(1-22,\) find \(\partial f / \partial x\) and \(\partial f / \partial y\) $$f(x, y)=\sqrt{x^{2}+y^{2}}$$

Short Answer

Expert verified
\( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \) and \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).

Step by step solution

01

Identify the function

The given function is \( f(x, y) = \sqrt{x^2 + y^2} \). This is a two-variable function where we are required to find partial derivatives.
02

Set up the partial derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and find the derivative of \( f(x, y) \) with respect to \( x \).
03

Differentiate with respect to x

Using the chain rule, the derivative of \( \sqrt{x^2 + y^2} \) with respect to \( x \) is \( \frac{x}{\sqrt{x^2 + y^2}} \). Therefore, \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \).
04

Set up the partial derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant and find the derivative of \( f(x, y) \) with respect to \( y \).
05

Differentiate with respect to y

Again, using the chain rule, the derivative of \( \sqrt{x^2 + y^2} \) with respect to \( y \) is \( \frac{y}{\sqrt{x^2 + y^2}} \). Therefore, \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of calculus that deals with functions of several variables. It is crucial when working with real-world scenarios involving three-dimensional space or situations where multiple varying quantities exist. In multivariable calculus, a function can depend on two or more independent variables, such as the function \( f(x, y) \) from the original exercise.

Understanding functions of several variables involves comprehending surfaces in three-dimensional space. The concept is applied in diverse fields, including physics, engineering, and economics. Unlike single-variable calculus, where we deal with curves, in multivariable calculus, we examine surfaces and gradients, which provide directional rates of change.
  • Mapping: Multivariable calculus relates to mapping coordinates from one space to another, often from \( (x, y) \) in 2D to \( (x, y, z) \) in 3D.
  • Level Curves: These are curves where the function \( f(x, y) \) holds a constant value, aiding visual understanding of heights and depressions on a surface.
Chain Rule in Calculus
The chain rule is a fundamental technique in calculus used to differentiate composite functions. In simpler terms, it helps find how a change in one variable affects a function through another variable linked by a nested or composite relationship. This concept extends into multivariable calculus, which was evident in the step-by-step solution.

When dealing with functions that involve transcendental operations, like square roots or exponentials, the chain rule allows us to simplify the differentiation process. For example, with our function \( f(x, y) = \sqrt{x^2 + y^2} \), we used the chain rule to differentiate twice: once for \( x \) and once for \( y \).
  • The outer function is \( \sqrt{u} \) with \( u = x^2 + y^2 \).
  • The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).
  • So, \( \frac{d}{dx} (\sqrt{x^2 + y^2}) = \frac{1}{2\sqrt{x^2 + y^2}} \times (2x) = \frac{x}{\sqrt{x^2 + y^2}} \).
Gradient Calculation
The gradient is vital in multivariable calculus, describing the slope of a function extending into multiple dimensions. It is a vector composed of the partial derivatives of a function, indicating the direction of the steepest ascent at any point on a surface.

For our function \( f(x, y) = \sqrt{x^2 + y^2} \), the gradient is composed of its partial derivatives:
  • \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \)
  • \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \)
The resulting gradient vector is \( abla f = \left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}} \right) \). This vector points in the direction where the function increases most rapidly and its magnitude reflects the slope steepness.

Understanding gradients is crucial for optimization problems, as they help identify maximum or minimum values by locating points where the gradient equals zero (critical points). Consider these points in the context of finding local maxima or minima, which are fundamental in fields like economics for maximizing profit or efficiency.

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Most popular questions from this chapter

Cobb-Douglas production function During the 1920 s, Charles Cobb and Paul Douglas modeled total production output \(P\) (of a firm, industry, or entire economy) as a function of labor hours involved \(x\) and capital invested \(y\) (which includes the monetary worth of all buildings and equipment). The Cobb- Douglas production function is given by $$P(x, y)=k x^{\alpha} y^{1-\alpha},$$ where \(k\) and \(\alpha\) are constants representative of a particular firm or economy. a. Show that a doubling of both labor and capital results in a doubling of production \(P .\) b. Suppose a particular firm has the production function for \(k=\) 120 and \(\alpha=3 / 4 .\) Assume that each unit of labor costs \(\$ 250\) and each unit of capital costs \(\$ 400,\) and that the total expenses for all costs cannot exceed \(\$ 100,000 .\) Find the maximum production level for the firm.

The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity \(Q\) of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula \(Q=\sqrt{2 K M / h}\) , where \(K\) is the cost of placing the order, \(M\) is the number of items sold per week, and \(h\) is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables \(K, M,\) and \(h\) is \(Q\) most sensitive near the point \(\left(K_{0}, M_{0}, h_{0}\right)=(2,20,0.05) ?\) Give reasons for your answer.

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{array}{l}{f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16,-3 \leq x \leq 3} \\\ {-6 \leq y \leq 6}\end{array}$$

Display the values of the functions in Exercises \(37-48\) in two ways: (a) by sketching the surface \(z=f(x, y)\) and \((b)\) by drawing an assortment of level curves in the function's domain. Label each level curve with its function value. $$ f(x, y)=1-|y| $$

The fifth-order partial derivative \(\partial^{5} f / \partial x^{2} \partial y^{3}\) is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: \(x\) or \(y ?\) Try to answer without writing anything down. $$ \begin{array}{l}{\text { a. } f(x, y)=y^{2} x^{4} e^{x}+2} \\ {\text { b. } f(x, y)=y^{2}+y\left(\sin x-x^{4}\right)} \\ {\text { c. } f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { d. } f(x, y)=x e^{y^{2} / 2}}\end{array} $$

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