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Chapter 14: Problem 2

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

Short Answer

Expert verified
\(\partial f / \partial x = 2x - y\) and \(\partial f / \partial y = -x + 2y\).

Step by step solution

01

Differentiate with respect to x

First, differentiate the given function \(f(x, y) = x^2 - xy + y^2\) with respect to \(x\), treating \(y\) as a constant. \[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2) - \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial x}(y^2)\] Calculate each term:\[\frac{\partial}{\partial x}(x^2) = 2x, \quad \frac{\partial}{\partial x}(xy) = y, \quad \frac{\partial}{\partial x}(y^2) = 0\] Thus,\[\frac{\partial f}{\partial x} = 2x - y\]
02

Differentiate with respect to y

Next, differentiate the function \(f(x, y) = x^2 - xy + y^2\) with respect to \(y\), treating \(x\) as a constant. \[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2) - \frac{\partial}{\partial y}(xy) + \frac{\partial}{\partial y}(y^2)\]Calculate each term:\[\frac{\partial}{\partial y}(x^2) = 0, \quad \frac{\partial}{\partial y}(xy) = x, \quad \frac{\partial}{\partial y}(y^2) = 2y\]Thus,\[\frac{\partial f}{\partial y} = -x + 2y\]

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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 to functions of several variables. While single-variable calculus deals with functions of one variable, multivariable calculus explores functions depending on two or more variables.
For instance, in our exercise, we have the function \(f(x, y) = x^2 - xy + y^2\). This is a function of two variables, \(x\) and \(y\), which means that any changes in these variables will affect the function's output.
In real-world applications, multivariable calculus is used in various fields such as physics, engineering, and economics, where multiple variables are at play."},{

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Most popular questions from this chapter

In Exercises \(71-76\) , 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)? $$ \begin{array}{l}{f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3}} , {-4 \leq x \leq 3,-2 \leq y \leq 2}\end{array} $$

Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6 .\)

In Exercises \(31-38,\) find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=4 x-8 x y+2 y+1\) on the triangular plate bounded by the lines \(x=0, y=0, x+y=1\) in the first quadrant

If \(z=x+f(u),\) where \(u=x y,\) show that $$ x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=x $$

a. Maximum on a sphere Show that the maximum value of \(a^{2} b^{2} c^{2}\) on a sphere of radius \(r\) centered at the origin of a Cartesian \(a b c-\) coordinate system is \(\left(r^{2} / 3\right)^{3}\) . b. Geometric and arithmetic means Using part (a), show that for non negative numbers \(a, b,\) and \(c\) $$(a b c)^{1 / 3} \leq \frac{a+b+c}{3}$$ that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean.
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