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Chapter 13: Problem 4

Find the four second partial derivatives of \(f(x, y)=3 x^{2} y+x y^{3}.\)

Short Answer

Expert verified
Answer: The four second partial derivatives are \(f_{xx} = 6y\), \(f_{xy} = 6x + 3y^2\), \(f_{yx} = 6x + 3y^2\), and \(f_{yy} = 6xy\).

Step by step solution

01

Find the first partial derivatives

We have to find the first partial derivatives of the given function \(f(x, y)=3 x^{2} y+x y^{3}\). To find the partial derivatives, we will differentiate the function with respect to x and y. - The first partial derivative with respect to x: \(f_x = \frac{\partial f(x, y)}{\partial x} = \frac{\partial (3 x^{2} y+x y^{3})}{\partial x} = 6xy + y^3\) - The first partial derivative with respect to y: \(f_y = \frac{\partial f(x, y)}{\partial y} = \frac{\partial (3 x^{2} y+x y^{3})}{\partial y} = 3x^2 + 3xy^2\)
02

Find the second partial derivatives

Now, we will differentiate the first partial derivatives with respect to x and y. 1. The second partial derivative with respect to x (i.e., differentiate \(f_x\) with respect to x): \(f_{xx} = \frac{\partial^2 f(x, y)}{\partial x^2} = \frac{\partial (6xy + y^3)}{\partial x} = 6y\) 2. The second partial derivative with respect to x and then y (i.e., differentiate \(f_x\) with respect to y): \(f_{xy} = \frac{\partial^2 f(x, y)}{\partial x \partial y} = \frac{\partial (6xy + y^3)}{\partial y} = 6x + 3y^2\) 3. The second partial derivative with respect to y and then x (i.e., differentiate \(f_y\) with respect to x): \(f_{yx} = \frac{\partial^2 f(x, y)}{\partial y \partial x} = \frac{\partial (3x^2 + 3xy^2)}{\partial x} = 6x + 3y^2\) 4. The second partial derivative with respect to y (i.e., differentiate \(f_y\) with respect to y): \(f_{yy} = \frac{\partial^2 f(x, y)}{\partial y^2} = \frac{\partial (3x^2 + 3xy^2)}{\partial y} = 6xy\) So, the four second partial derivatives are: \(f_{xx} = 6y\), \(f_{xy} = 6x + 3y^2\), \(f_{yx} = 6x + 3y^2\), and \(f_{yy} = 6xy\).

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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 the branch of mathematics that extends calculus to functions of more than one variable.

While single-variable calculus deals with functions of a single variable such as \(f(x)\), multivariable calculus involves functions of two or more variables like \(f(x, y)\).

This expands the capabilities of calculus from dealing with lines and curves to dealing with surfaces and higher-dimensional analogues.
  • It's crucial for applications involving changes across multiple dimensions, such as in physics to describe the world around us.
  • In the context of optimization or economics, understanding how a function behaves across multiple variables allows us to solve real-world problems.
By studying multivariable calculus, students can analyze and understand how functions change in a multi-dimensional space, opening up a new world beyond the simplicity of one-dimensional analysis.
Second Partial Derivatives
Second partial derivatives give insights into the curvature and concavity of surfaces formed by functions of multiple variables.

Just as a second derivative in single-variable calculus expresses the rate of change of the rate of change, second partial derivatives extend this concept to multiple dimensions.
  • They are found by differentiating the first partial derivatives.
  • For example, in our function \(f(x, y)\), the second partial derivatives include \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), and \(f_{yx}\).
Each derivative has a specific interpretation:
  • \(f_{xx}\) corresponds to concavity in the x direction.
  • \(f_{yy}\) indicates concavity in the y direction.
  • \(f_{xy}\) and \(f_{yx}\), often equal across many functions, describe the mixed rate of change respecting both x and y.
Understanding these helps in analyzing the detailed behavior of multi-variable landscapes, critical for tasks like optimizing functions.
Differentiation
Differentiation is the process of finding the rate at which a function is changing at any given point.

In multivariable calculus, differentiation applies to functions with multiple variables by treating each variable separately while treating others as constants.

This leads to the concept of partial derivatives, expressed as \(\frac{\partial}{\partial x}\) or \(\frac{\partial}{\partial y}\).
  • Partial derivatives, like \(f_x\) or \(f_y\), denote sensitivity of the function \(f(x, y)\) to changes in one specific variable.
  • They are foundational for defining directions of greatest increase or decrease across surfaces.
Their calculations help to create a detailed picture of how small changes in inputs affect outputs, crucial for understanding gradient vectors and directional derivatives. Understanding and using these derivatives is essential for exploring and optimizing complex systems.
Functions of Multiple Variables
Functions of multiple variables are a step beyond the basic functions that depend only on a single input. These functions, like \(f(x, y)\), rely on more than one independent variable to produce a result.

They allow for complex modeling of systems where outcomes depend on several different factors.
  • Commonly applied in fields such as physics, engineering, and economic modeling.
  • Analyzing these functions involves examining different dimensions of change - not just linear.
In multivariable functions, each variable represents a potential axis on a coordinate system, thereby creating a multi-dimensional space through which the function creates a surface (or hypersurface in more than three variables).

Understanding how these variables interact through differentiation, including partial derivatives, is what makes solving real-world problems with multi-variable functions both challenging and broadly applicable.

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Most popular questions from this chapter

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum.

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Suppose \(P\) is a point in the plane \(a x+b y+c z=d .\) Then the least distance from any point \(Q\) to the plane equals the length of the orthogonal projections of \(\overrightarrow{P Q}\) onto the normal vector \(\mathbf{n}=\langle a, b . c\rangle\) a. Use this information to show that the least distance from \(Q\) to the plane is \(\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{|\mathbf{n}|}\) b. Find the least distance from the point (1,2,-4) to the plane \(2 x-y+3 z=1\)

Use the method of your choice to ate the following limits. $$\lim _{(x, y) \rightarrow(1,1)} \frac{x^{2}+x y-2 y^{2}}{2 x^{2}-x y-y^{2}}$$
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