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Chapter 14: Problem 116

True or false? Give reasons for your answer. If \(f(x, y)=e^{x+y},\) then the directional derivative in any direction \(\vec{u}\) (with \(\|\vec{u}\|=1\) ) at the point (0,0) is always less than or equal to \(\sqrt{2}\)

Short Answer

Expert verified
True; the maximum directional derivative is \( \sqrt{2} \).

Step by step solution

01

Understand the Problem

We need to determine whether the statement about the directional derivative of the function \( f(x, y) = e^{x+y} \) at the point (0,0) is correct. Specifically, we need to check if the directional derivative is always \( \leq \sqrt{2} \).
02

Calculate the Gradient

The gradient \( abla f \) of the function \( f(x, y) = e^{x+y} \) is given by the partial derivatives: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (e^{x+y}, e^{x+y}) \].At the point (0,0), this gives: \[ abla f(0,0) = (e^0, e^0) = (1, 1) \].
03

Recall the Directional Derivative Formula

The directional derivative of \( f \) in the direction of a unit vector \( \vec{u} = (u_1, u_2) \) is defined as: \[ D_{\vec{u}}f(x, y) = abla f \cdot \vec{u} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \cdot (u_1, u_2) \].
04

Evaluate the Directional Derivative at (0,0)

Using the point (0,0), the directional derivative becomes: \[ D_{\vec{u}}f(0,0) = (1, 1) \cdot (u_1, u_2) = 1 \cdot u_1 + 1 \cdot u_2 = u_1 + u_2 \]. Since \( \|\vec{u}\| = 1 \), we know: \[ u_1^2 + u_2^2 = 1 \].
05

Determine the Maximum of the Directional Derivative

To find the maximum value of \( u_1 + u_2 \) given \( u_1^2 + u_2^2 = 1 \), we use the Cauchy-Schwarz inequality or recognize that cosine of the angle between \( (1, 1) \) and \( \vec{u} \) is maximized. This yields that the maximum occurs when \( u_1 = u_2 = \frac{\sqrt{2}}{2} \), giving \[ u_1 + u_2 = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \].Thus, the maximum directional derivative at (0,0) is \( \sqrt{2} \).
06

Conclusion

Since the maximum value of the directional derivative at (0,0) is \( \sqrt{2} \), the statement that the directional derivative is always \( \leq \sqrt{2} \) is true.

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Key Concepts

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

Gradient
The gradient of a function provides important information about the function's rate and direction of change. Think of it as a multi-variable generalization of the derivative for single-variable functions. Given a function \( f(x, y) \), the gradient \( abla f \) is a vector consisting of the partial derivatives with respect to each variable. For example, with \( f(x, y) = e^{x+y} \), the gradient is:
  • \( \frac{\partial f}{\partial x} = e^{x+y} \)
  • \( \frac{\partial f}{\partial y} = e^{x+y} \)
At any point \((x, y)\), the gradient vector points in the direction of the steepest increase of the function. At the point \( (0,0) \), the gradient is \((1, 1)\), indicating that the function increases equally in both the x and y directions.
Unit Vector
A unit vector is a vector with a length (or magnitude) of 1. It is often used to specify a direction. In mathematical terms, a vector \( \vec{u} = (u_1, u_2) \) is a unit vector if:
  • \( \| \vec{u} \| = 1 \)
  • Which means \( u_1^2 + u_2^2 = 1 \)
Unit vectors are vital in computing the directional derivative, as they ensure the emphasis is solely on the direction without any magnitude affecting it. When evaluating the directional derivative of a function like \( f(x, y) = e^{x+y} \), the unit vector specifies the direction of the slice through the function's surface that we are examining.
Partial Derivatives
Partial derivatives are like the ordinary derivatives but limited to multi-variable functions. They measure the rate at which a function changes as one of its variables is varied, keeping all other variables constant. For a function \( f(x, y) \):
  • The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} \)
  • The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \)
In our exercise, \( f(x, y) = e^{x+y} \), both partial derivatives turn out to be \( e^{x+y} \). At the point \( (0,0) \), each results in \( 1 \). This information forms the components of the gradient vector, which in turn assists in determining the directional derivative.
Cauchy-Schwarz Inequality
The Cauchy-Schwarz Inequality is a key concept in vector mathematics that places bounds on the dot product of two vectors. It states that for any vectors \( \vec{a} \) and \( \vec{b} \):
  • \( |\vec{a} \cdot \vec{b}| \leq \|\vec{a}\| \|\vec{b}\| \)
In our exercise, it helps us to determine the directions where the directional derivative attains its maximum value. As applied here, with \( \vec{a} = (1, 1) \) and \( \vec{b} = \vec{u} \) being a unit vector, the maximum value of the directional derivative is \( \sqrt{2} \) when both components of \( \vec{u} \) are equal, specifically \( \frac{\sqrt{2}}{2} \). Hence, the inequality ensures that no directional derivative can exceed \( \sqrt{2} \) at \( (0,0) \).

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Most popular questions from this chapter

The tangent plane to \(z=f(x, y)\) at the point (1,2) is \(z=3 x+2 y-5\) (a) Find \(f_{x}(1,2)\) and \(f_{y}(1,2)\) (b) What is \(f(1,2) ?\) (c) Approximate \(f(1.1,1.9)\)

In Problems \(90-93,\) check that the point (2,3) lies on the curve. Then, viewing the curve as a contour of \(f(x, y),\) use grad \(f(2,3)\) to find a vector normal to the curve at (2,3) and an equation for the tangent line to the curve at (2,3) $$x^{2}+y^{2}=13$$

Find the directional derivative \(f_{a}(1,2)\) for the function \(f\) with \(\vec{u}=(3 \vec{i}-4 \vec{j}) / 5\) $$f(x, y)=x y+y^{3}$$

True or false? Give reasons for your answer. If grad \(f(1,2)=\vec{i},\) then \(f\) decreases in the \(-\vec{i}\) direction at (1,2)

Give an example of: Formulas for two different functions \(f(x, y)\) and \(g(x, y)\) with the same quadratic approximation near (0,0)
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