/*! 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 92 Let $$f(x, y)=\left\\{\begin{arr... [FREE SOLUTION] | 91Ó°ÊÓ

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Let $$f(x, y)=\left\\{\begin{array}{ll}{0,} & {x^{2} < y<2 x^{2}} \\ {1,} & {\text { otherwise }}\end{array}\right.$$ Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\)

Short Answer

Expert verified
Partial derivatives exist, but the function isn't differentiable at \((0,0)\).

Step by step solution

01

Understanding the function domain

The function \( f(x, y) \) is defined piecewise. It takes the value 0 when \( x^2 < y < 2x^2 \) and the value 1 outside this region. At point \((0,0)\), \( y \) is not between \( x^2 \) and \( 2x^2 \), so it defaults to 1. Therefore, \( f(0,0) = 1 \).
02

Calculate the partial derivative with respect to x

To find \( f_x(0,0) \), we consider the limit definition of the partial derivative: \[ f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} \]We know \( f(0,0) = 1 \). For \( f(h,0) \), since \(0 otin (h^2, 2h^2)\), \( f(h,0) = 1 \). Thus, \[ f_x(0,0) = \lim_{h \to 0} \frac{1 - 1}{h} = 0 \]which means \( f_x(0,0) = 0 \).
03

Calculate the partial derivative with respect to y

To find \( f_y(0,0) \), we use the limit definition of the partial derivative: \[ f_y(0,0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} \]Again, \( f(0,0) = 1 \). For \( f(0,k) \), since \( k otin (0, 0) \), \( f(0,k) = 1 \). Thus, \[ f_y(0,0) = \lim_{k \to 0} \frac{1 - 1}{k} = 0 \]which means \( f_y(0,0) = 0 \).
04

Check differentiability at (0,0)

A function is differentiable at \((0,0)\) if the limit \[ \lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - f_x(0,0)h - f_y(0,0)k}{\sqrt{h^2+k^2}} = 0 \]exists. Substituting \( f_x(0,0) = 0 \) and \( f_y(0,0) = 0 \), we consider specific paths. For \( y = x^2 \), \[ f(h, h^2) = 0 \] and for \( y = 2x^2 \), \( f(h, 2h^2) = 0 \). Thus \[ \frac{0 - 1}{\sqrt{h^2 + (h^2)^2}} = -\frac{1}{h\sqrt{1+h^2}} \] \[ \frac{0 - 1}{\sqrt{h^2 + (2h^2)^2}} = -\frac{1}{h\sqrt{1+4h^2}} \] which goes to infinity as \( h \to 0 \). Therefore, \( f \) is not differentiable at \((0,0)\).

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Key Concepts

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

Partial Derivatives
Partial derivatives allow us to explore how a multivariable function changes as each independent variable experiences a slight modification, keeping all other variables constant. In simpler terms, it offers a way to understand the behavior of a function in relation to one variable at a time.
To compute a partial derivative, we'll use the limit-based formula designed specifically for it. For the function \( f(x, y) \), the partial derivative with respect to \( x \) at the point \((0, 0)\) is calculated using:
  • \( f_x(0, 0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} \)
  • \( f_y(0, 0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} \)
In the exercise, we found both \( f_x(0, 0) \) and \( f_y(0, 0) \) equaled zero. This means the function does not change as \( x \) or \( y \) slightly varies when the other stays at zero, around point \((0, 0)\). This is typical when dealing with piecewise functions. By evaluating different regions and conditions, we determined how changes in each variable affect the function's value individually.
Limit Definition
In calculus, the concept of limits helps us understand the behavior of functions as the inputs approach a particular point. For partial derivatives, we use limits to precisely define the derivative by observing what happens at infinitely small changes from a point.
The exercise implements the limit definition to derive partial derivatives at \((0, 0)\):
  • For \( f_x(0, 0) \), the limit examines the difference \( (f(h,0) - f(0,0))/h \) as \( h \) approaches zero.
  • For \( f_y(0, 0) \), it checks \( (f(0,k) - f(0,0))/k \) as \( k \) tends towards zero.
Since both limits yield 0, the function appears smooth and constant with respect to each variable at \((0, 0)\), from the perspective of each input individually. While this approach works well for detecting behavior along axes, it doesn't assess collective behavior like differentiability, which needs more comprehensive analysis.
Non-Differentiable Points
Despite having partial derivatives equal to zero, a function might not be differentiable at a point. Differentiability hinges on the behavior of a function in all directions around a point, not just the isolated perspective each partial derivative offers.
For fully verifying differentiability at \((0, 0)\), we need to evaluate:
  • The limit: \( \lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - f_x(0,0)h - f_y(0,0)k}{\sqrt{h^2+k^2}} = 0 \)
In the exercise, testing along different paths such as \( y = x^2 \) and \( y = 2x^2 \), showed that the value diverges, leading to infinity as \( h \to 0 \). These paths result in diverging values, indicating the function isn’t smooth or locally linear at \((0, 0)\). Although we have valid partial derivatives, the function exhibits jumps or wrinkles, making it non-differentiable at this specific point.

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Most popular questions from this chapter

In Exercises \(61-64,\) find an equation for the level surface of the function through the given point. $$ g(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}, \quad(1,-1, \sqrt{2}) $$

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