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Find the linearization \(L(x, y)\) of the function at each point. $$ f(x, y)=x^{3} y^{4} \text { at } \quad \text { a. }(1,1), \quad \text { b. }(0,0) $$

Short Answer

Expert verified
Linearization is \( L(x, y) = 3x + 4y - 6 \) at \((1,1)\) and \( L(x, y) = 0 \) at \((0,0)\).

Step by step solution

01

Understanding Linearization

The linearization of a function at a given point is the equation of the tangent plane to the function at that point. For a function \( f(x, y) \), the linearization \( L(x, y) \) is given by: \[ L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \] where \( f_x \) and \( f_y \) are partial derivatives of \( f \).
02

Compute Partial Derivatives

For the function \( f(x, y) = x^3 y^4 \), compute the partial derivatives.- Partial derivative with respect to \( x \): \( f_x(x, y) = 3x^2 y^4 \). - Partial derivative with respect to \( y \): \( f_y(x, y) = 4x^3 y^3 \).
03

Evaluate at Point (1, 1)

Calculate the values of \( f_x \), \( f_y \), and \( f \) at the point \((1,1)\).- \( f(1, 1) = 1^3 \cdot 1^4 = 1 \).- \( f_x(1, 1) = 3 \cdot 1^2 \cdot 1^4 = 3 \).- \( f_y(1, 1) = 4 \cdot 1^3 \cdot 1^3 = 4 \).
04

Determine Linearization at (1, 1)

Substitute the values into the linearization formula:\[ L(x, y) = 1 + 3(x - 1) + 4(y - 1) \]Simplify the expression:\[ L(x, y) = 3x + 4y - 6 \]
05

Evaluate at Point (0, 0)

Calculate the values of \( f_x \), \( f_y \), and \( f \) at the point \((0,0)\).- \( f(0, 0) = 0^3 \cdot 0^4 = 0 \).- \( f_x(0, 0) = 3 \cdot 0^2 \cdot 0^4 = 0 \).- \( f_y(0, 0) = 4 \cdot 0^3 \cdot 0^3 = 0 \).
06

Determine Linearization at (0, 0)

Substitute the values into the linearization formula:\[ L(x, y) = 0 + 0(x - 0) + 0(y - 0) \]Simplify the expression:\[ L(x, y) = 0 \]

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Key Concepts

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

Linearization
Linearization is a powerful concept in multivariable calculus that simplifies the study of complex functions by approximating them with linear functions. In essence, linearization gives us the equation of the tangent plane to a surface at a given point.
It's an approximation method, often helpful when dealing with functions that are difficult to analyze directly.
To find the linearization of a function at a particular point, we use the formula:
  • \[ L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \]
Here, \(f(a, b)\) is the value of the function at the point, and \(f_x(a, b)\), \(f_y(a, b)\) are the partial derivatives of the function at the same point.
This linear approximation becomes more accurate the closer \((x, y)\) is to \((a, b)\).
Understanding and utilizing linearization effectively allows us to approximate multi-variable functions around a point in a simple, linear form which is computationally efficient, especially in engineering and science applications.
Partial Derivatives
Partial derivatives play a crucial role in multivariable calculus, particularly in the study of functions of several variables. They help us understand how changes in one variable affect the function, while keeping other variables constant.
This is akin to taking the derivative in single-variable calculus, but extended to multiple dimensions.
For a function \(f(x, y)\), the partial derivatives are defined as:
  • The partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), captures how \(f\) changes as \(x\) varies, while \(y\) is fixed.
  • The partial derivative with respect to \(y\), denoted as \(f_y(x, y)\), captures how \(f\) changes as \(y\) varies, while \(x\) is fixed.
When analyzing a function like \(f(x, y) = x^3 y^4\), we differentiate separately with respect to each variable:
  • \( f_x(x, y) = 3x^2 y^4 \)
  • \( f_y(x, y) = 4x^3 y^3 \)
These partial derivatives provide the slopes of the tangent plane along the respective axes.
By computing these derivatives, one can understand the local behavior of the function.
Tangent Plane
A tangent plane is the flat surface that best approximates a curved surface at a given point. Think of it like placing a flat piece of paper against a bumpy hill; the paper represents the tangent plane while it touches the hill at one single point.
In multivariable calculus, this concept helps visualize how a function behaves close to that specific location.
The equation of the tangent plane to a surface \( z = f(x, y) \) at a point \((a, b)\) is derived from the function's linearization. It is represented by:
  • \[ z = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \]
This gives us a linear surface that mirrors the behavior of the original function near the point \((a, b)\).
The coefficients \( f_x(a, b) \) and \( f_y(a, b) \) from partial derivatives act as the slopes in the directions of the \(x\) and \(y\) axes, respectively.
Understanding the concept of a tangent plane is crucial for analyzing real-world surfaces and can be used in fields like physics and engineering to predict how surfaces interact with each other at certain points.

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Most popular questions from this chapter

In Exercises \(57-60,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$ f(x, y)=\sqrt{2 x+3 y-1}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,3) $$

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. $$ f(x, y)=1-|y| $$

a. Around the point \((1,0),\) is \(f(x, y)=x^{2}(y+1)\) more sensitive to changes in \(x\) or to changes in \(y\) ? Give reasons for your answer. b. What ratio of \(d x\) to \(d y\) will make \(d f\) equal zero at \((1,0) ?\)

Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by \(M(x, y, z)=6 x-\) \(y^{2}+x z+60 .\) Where should you locate the radio telescope?

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) $$ \begin{array}{l}{f(x, y, z)=\sqrt{2} \cos x \sin (y+z) \text { at } P_{0}(0,0, \pi / 4)} \\ {R :|x| \leq 0.01, \quad|y| \leq 0.01, \quad|z-\pi / 4| \leq 0.01}\end{array} $$

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