/*! 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 57 The linearization of \(f(x, y)\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The linearization of \(f(x, y)\) is a tangent-plane approximation Show that the tangent plane at the point \(P_{0}\left(x_{0}, y_{0}, f\left(x_{0}, y_{0}\right)\right)\) on the surface \(z=f(x, y)\) defined by a differentiable function \(f\) is the plane $$ f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)-\left(z-f\left(x_{0}, y_{0}\right)\right)=0 $$ or $$ z=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right) $$ Thus, the tangent plane at \(P_{0}\) is the graph of the linearization of \(f\) at \(P_{0}\)( see accompanying figure).

Short Answer

Expert verified
The tangent plane equation is the linearization of the function at the point \( P_0 \).

Step by step solution

01

Understand the Problem

We need to prove that the equation of the tangent plane at point \( P_0(x_0, y_0, f(x_0, y_0)) \) on the surface \( z = f(x, y) \) is given by the expression \( z = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
Partial derivatives help us to understand how a function changes as each of its input variables change. Imagine a function of two variables, like a surface in three-dimensional space. Here, partial derivatives show the rate of change of the function with respect to one variable, while keeping the other variable fixed.

For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( f_x(x, y) \) and measures how the function changes as \( x \) changes, while \( y \) is held constant. Similarly, \( f_y(x, y) \) measures the change in \( f \) as \( y \) varies, with \( x \) held fixed.
- **Rate of change:** Partial derivatives indicate how steep or shallow a surface is along specific directions.
- **Calculation of tangents:** They are essential in finding the slope of the tangent plane at a given point on the surface.

Understanding partial derivatives is crucial because they form the building blocks needed to comprehend more complex topics like the tangent plane and linear approximations. They capture how changes in each input variable affect the entire function.
Differentiable Function
A differentiable function is one that has well-defined partial derivatives at all points in its domain. In simpler terms, it means that the surface described by the function is smooth and doesn't have any sharp corners or edges. This smoothness is vital for defining things like tangent planes.

For a function \( f(x, y) \) to be differentiable, it must not only have partial derivatives but also obey certain continuity conditions. This ensures that near any given point, the function can be approximated by a plane, making it predictable in a small neighborhood of that point.

- **Smooth surfaces:** Differentiable functions give rise to surfaces that can be approximated accurately with planes, which makes problem-solving more manageable and intuitive.
- **Reliability:** Knowing a function is differentiable allows us to apply various mathematical tools without worrying about irregularities. This reliability makes it much simpler to move forward with calculations involving the function, such as linear approximations.
Linear Approximation
Linear approximation is the idea of approximating a function using a linear function, often making calculations simpler and more intuitive. Imagine taking a small section of a curve and replacing it with a straight line. This is effectively what linear approximation does.

For a function like \( f(x, y) \), the linear approximation at a specific point \((x_0, y_0)\) is given by the tangent plane equation: \[z = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \]This equation captures the essence of the surface around that point with just a flat plane. This is beneficial because:
- **Simplicity:** Approximating with a plane simplifies calculations without greatly sacrificing accuracy for small changes around the point.
- **Predictive Power:** Even though it's an approximation, the tangent plane captures behavior in a small region, effectively mimicking the original function's behavior.

The tangent plane and linear approximation are highly valuable, especially in scenarios where exact solutions are complex or unnecessary. With a linear approximation, predictions about the function's behavior near a point become much easier.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f(x, y)=\left\\{\begin{array}{ll}{y^{3},} & {y \geq 0} \\ {-y^{2},} & {y<0}\end{array}\right.\) Find \(f_{x}, f_{y}, f_{x y},\) and \(f_{y x},\) and state the domain for each partial derivative.

Tangent curves A smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to \(\nabla f\) there. Show that the curve $$ \mathbf{r}(t)=\sqrt{t \mathbf{i}}+\sqrt{t} \mathbf{j}+(2 t-1) \mathbf{k} $$ is tangent to the surface \(x^{2}+y^{2}-z=1\) when \(t=1\)

In Exercises \(55-60,\) verify that \(w_{x y}=w_{y x}\). $$w=x y^{2}+x^{2} y^{3}+x^{3} y^{4}$$

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

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)? $$ \begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3,-3 / 2 \leq x \leq 3 / 2} , {-3 / 2 \leq y \leq 3 / 2}\end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.