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Let \(f(x, y)=\frac{(x-y)}{(x+y)} .\) Find the directions u and the values of \(D_{\mathbf{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right)\) for which \(\begin{array}{ll}{\text { a. } D_{\mathrm{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right) \text { is largest } \quad \text { b. } D_{\mathrm{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right) \text { is smallest }} \\ {\text { c. } D_{\mathrm{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right)=0} \quad \quad \quad \quad {\text { d. } D_{\mathrm{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right)=-2} \\\ {\text { e. } D_{\mathrm{u}} f\left(-\frac{1}{2}, \frac{3}{2}\right)} {=1}\end{array}\)

Short Answer

Expert verified
The largest directional derivative is \(\sqrt{10}\), smallest is \(-\sqrt{10}\), zero occurs when \(u=(\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}})\), other values involve solving equations.

Step by step solution

01

Find the Partial Derivatives

To find the directional derivatives, we need the gradient of the function. For function \(f(x, y)=\frac{(x-y)}{(x+y)}\), find the partial derivatives with respect to \(x\) and \(y\).The partial derivative with respect to \(x\) is given by \[\frac{\partial f}{\partial x} = \frac{(x+y)-(x-y)}{(x+y)^2} = \frac{2y}{(x+y)^2}.\]The partial derivative with respect to \(y\) is given by \[\frac{\partial f}{\partial y} = \frac{-(x-y)-(x+y)}{(x+y)^2} = \frac{-2x}{(x+y)^2}.\]
02

Evaluate the Gradient at Specific Point

Substitute \((-\frac{1}{2}, \frac{3}{2})\) into the gradient components to find the gradient vector at this point:\[abla f\left(-\frac{1}{2}, \frac{3}{2}\right) = \left(\frac{2\times\frac{3}{2}}{(-\frac{1}{2} + \frac{3}{2})^2}, \frac{-2\times(-\frac{1}{2})}{(-\frac{1}{2} + \frac{3}{2})^2}\right) = (3, 1).\]
03

General Form of Directional Derivative

The directional derivative \(D_{\mathbf{u}}f(x,y)\) in the direction of a unit vector \(\mathbf{u} = (a, b)\) is given by:\[D_{\mathbf{u}}f(x,y) = abla f \cdot \mathbf{u} = (3, 1) \cdot (a, b) = 3a + b.\]
04

Largest and Smallest Directional Derivative

The directional derivative is largest in the direction of the gradient vector and smallest in the opposite direction. Therefore:- Largest: \(u = \frac{(3, 1)}{\sqrt{3^2 + 1^2}} = \left(\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}}\right)\), value is \(\sqrt{3^2 + 1^2} = \sqrt{10}\).- Smallest: \(u = \frac{(-3, -1)}{\sqrt{10}} = \left(-\frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}}\right)\), value is \(-\sqrt{10}\).
05

Zero Directional Derivative

To have a zero directional derivative:\[3a + b = 0\Rightarrow b = -3a.\]The direction vector can be: \(u = \left(\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \text{ or any scalar multiple}.\)
06

Specific Directional Derivative Values

For \(D_{\mathbf{u}} f = -2\):\[3a + b = -2 \Rightarrow \frac{b}{a} = \frac{-2-3a}{a} = -\frac{2}{a}-3.\]Solve for \(a, b\) such that \((a, b)\) is a unit vector.For \(D_{\mathbf{u}} f = 1\):\[3a + b = 1 \Rightarrow \frac{b}{a} = \frac{1-3a}{a} = \frac{1}{a}-3.\]Similarly, solve for \(a, b\) such that \((a, b)\) is a unit vector.

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Key Concepts

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

Partial Derivatives
Understanding partial derivatives is a crucial part of working with functions of multiple variables. When we talk about partial derivatives, we are finding how a function changes as one variable changes while keeping the others constant.

For example, consider the function \(f(x, y)=\frac{(x-y)}{(x+y)}\), which depends on both \(x\) and \(y\). To find its behavior in each direction, we calculate its partial derivatives:
  • The partial derivative with respect to \(x\) shows how \(f\) changes as \(x\) changes and \(y\) remains fixed.
  • Likewise, the partial derivative with respect to \(y\) indicates how \(f\) shifts with adjustments in \(y\) while \(x\) stays the same.
For this function, the partial derivatives are:
  • \(\frac{\partial f}{\partial x} = \frac{2y}{(x+y)^2}\)
  • \(\frac{\partial f}{\partial y} = \frac{-2x}{(x+y)^2}\)
These calculations help pinpoint the rate and direction of change at any point \((x, y)\).
Gradient Vector
The gradient vector has a special place in calculus, especially when dealing with directional derivatives. After computing partial derivatives, they are combined into the gradient vector.
  • It is essentially a vector that points in the direction of steepest ascent on a surface represented by a function.
  • Mathematically, the gradient vector of a function \(f(x, y)\) is denoted as \(abla f\). For our function, \(abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})\).
At the specific point \((-\frac{1}{2}, \frac{3}{2})\), the earlier calculated partial derivatives give us the gradient vector: \[abla f\left(-\frac{1}{2}, \frac{3}{2}\right) = (3, 1).\] This vector, \((3, 1)\), indicates the direction where the function increases the most at that point, while the opposite direction indicates the steepest descent. The length of the gradient vector also tells us how quickly the function increases in its steepest direction.
Unit Vector
A unit vector is a vector that has a length of one. It is crucial in various vector operations, especially when dealing with directional derivatives. The idea of a unit vector comes into play when you want to specify a direction without altering magnitude when multiplying by another vector.

For finding directional derivatives, ensure that the direction is specified by a unit vector. If you have any vector \(\mathbf{v} = (a, b)\), the corresponding unit vector \(\mathbf{u}\) is calculated as:\[ \mathbf{u} = \frac{(a, b)}{\sqrt{a^2 + b^2}}. \]
  • This process normalizes the vector, adjusting it to have a length of one.
  • Using the gradient \((3, 1)\), its unit vector is \(\left(\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}}\right)\), showing that this unit vector keeps the balance of the original vector's direction. It succinctly points in the same way as the original but on a unit scale.
Unit vectors provide a standard, consistent way to describe direction, making calculations like the directional derivative cleaner and more interpretable.

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Most popular questions from this chapter

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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)? $$f(x, y)=x^{3}-3 x y^{2}+y^{2},-2 \leq x \leq 2,-2 \leq y \leq 2$$

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