Chapter 13: Problem 11
Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y, z)=\frac{x-y-z}{x+y+z} ; P=(2,1,-1) ; \mathbf{a}=-2 \mathbf{i}-\mathbf{j}-\mathbf{k} $$
Short Answer
Expert verified
The directional derivative is \( \frac{4\sqrt{6}}{3} \).
Step by step solution
01
Compute the Gradient of f
The gradient of a function \( f(x, y, z) \) is denoted as \( abla f \) and is found by taking the partial derivatives of the function with respect to each variable. For \( f(x, y, z) = \frac{x - y - z}{x + y + z} \), compute the partial derivatives:1. \( \frac{\partial f}{\partial x} \): Use quotient rule, \( \frac{u}{v} \), \( u = x-y-z \) and \( v = x+y+z \).2. \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial z} \) similarly apply quotient rule.This results in: \[ abla f = \left(\frac{2(y+z)}{(x+y+z)^2}, \frac{-2x - 2z}{(x+y+z)^2}, \frac{-2x - 2y}{(x+y+z)^2}\right) \].
02
Evaluate the Gradient at Point P
Substitute \( P = (2, 1, -1) \) into the gradient found in Step 1.Calculate each component:- \( \frac{\partial f}{\partial x} \) at \( P \) becomes \( \frac{2(1-1)}{(2+1-1)^2} = 0 \).- \( \frac{\partial f}{\partial y} \) at \( P \) becomes \( \frac{-2(2+(-1))}{(2+1-1)^2} = -2 \).- \( \frac{\partial f}{\partial z} \) at \( P \) becomes \( \frac{-2(2+1)}{(2+1-1)^2} = -6 \).Thus, \( abla f(P) = (0, -2, -6) \).
03
Normalize the Direction Vector a
The directional vector is given as \( \mathbf{a} = -2\mathbf{i} - \mathbf{j} - \mathbf{k} \). We need to find its unit vector.The magnitude of \( \mathbf{a} \) is: \[ \| \mathbf{a} \| = \sqrt{(-2)^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \].Normalize \( \mathbf{a} \) to get the unit vector: \[ \mathbf{u} = \left( -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right) \].
04
Calculate the Directional Derivative
The directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{a} \) is given by:\[ D_{\mathbf{u}}f(P) = abla f(P) \cdot \mathbf{u} \]Calculate the dot product:\[ D_{\mathbf{u}}f(P) = (0, -2, -6) \cdot \left( -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right) = 0 \times -\frac{2}{\sqrt{6}} + (-2) \times -\frac{1}{\sqrt{6}} + (-6) \times -\frac{1}{\sqrt{6}} \]This simplifies to:\[ D_{\mathbf{u}}f(P) = \frac{8}{\sqrt{6}} \].
05
Simplify the Directional Derivative
Simplify \( \frac{8}{\sqrt{6}} \):To rationalize, multiply by \( \frac{\sqrt{6}}{\sqrt{6}} \): \[ D_{\mathbf{u}}f(P) = \frac{8\sqrt{6}}{6} = \frac{4\sqrt{6}}{3} \].This is the directional derivative of \( f \) at point \( P \) in the direction of \( \mathbf{a} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient plays a crucial role in evaluating how a function behaves in space. Think of the gradient as a vector that points in the direction of the greatest rate of increase of a function. For a function of multiple variables, such as 3-dimensional space, the gradient is represented by the symbol \( abla f \) and consists of the partial derivatives of the function with respect to each variable. So, if you have a function \( f(x, y, z) \), the gradient is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). This vector provides comprehensive information about how the function changes in different directions.
Partial Derivatives
Understanding partial derivatives is key to comprehending changes in multidimensional functions. A partial derivative measures how a function changes as one of its variables is allowed to vary, keeping the others constant. For the function \( f(x, y, z) = \frac{x - y - z}{x + y + z} \), you'd work out partial derivatives with respect to \( x \), \( y \), and \( z \). These derivatives reveal how the function changes as you tweak just one variable.
- \( \frac{\partial f}{\partial x} \): Finds how \( f \) changes with just \( x \).
- \( \frac{\partial f}{\partial y} \): Captures the function's change with \( y \).
- \( \frac{\partial f}{\partial z} \): Measures the change with \( z \).
Quotient Rule
The quotient rule is an essential tool when dealing with functions that are ratios of two other functions. Suppose you have a function \( f(x, y, z) = \frac{x-y-z}{x+y+z} \). Here, \( u = x-y-z \) and \( v = x+y+z \). The quotient rule states: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \]Applying it ensures you account for both the numerator's and denominator's changes. By doing so correctly, you avoid common pitfalls like missing a change in the denominator that affects the whole fraction, which is necessary for calculating partial derivatives properly.
Unit Vector
A unit vector is anything but a regular vector. It takes any direction's vector and gives you a shorter version of exactly 1 unit in length. Why does it matter? When finding a directional derivative, such as at point \( P \) in the direction of a vector \( \mathbf{a} \), you use a unit vector to specify the precise direction without exaggerating the magnitude. To find a unit vector \( \mathbf{u} \) from \( \mathbf{a} = -2\mathbf{i} - \mathbf{j} - \mathbf{k} \), first calculate the magnitude \( \| \mathbf{a} \| = \sqrt{6} \), and then\[ \mathbf{u} = \left( -\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right) \].With this unit vector, one can properly compute the directional derivative, ensuring the true directional impact is analyzed.