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Magazine Chapter 13: Problem 13
Find the directional derivative of \(f\) at \(P\) in the direction of a. $$ f(x, y)=\tan ^{-1}(y / x) ; P(-2,2) ; \mathbf{a}=-\mathbf{i}-\mathbf{j} $$
Short Answer
Expert verified
The directional derivative is \( \frac{\sqrt{2}}{4} \).
Step by step solution
01
Calculate the Gradient of f
First, find the gradient of the function \( f(x, y) = \tan^{-1}(y / x) \). The gradient is a vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( \frac{\partial f}{\partial x} \), apply the chain rule:\[\frac{\partial}{\partial x}(\tan^{-1}(y/x)) = \frac{-y}{x^2+y^2}.\]For \( \frac{\partial f}{\partial y} \), similarly:\[\frac{\partial}{\partial y}(\tan^{-1}(y/x)) = \frac{x}{x^2+y^2}.\]Thus, the gradient is:\[abla f(x, y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right).\]
02
Evaluate the Gradient at P
Substitute \(P(-2, 2)\) into the gradient obtained in Step 1. \[ abla f(-2, 2) = \left( \frac{-2}{(-2)^2+2^2}, \frac{-2}{(-2)^2+2^2} \right). \]This simplifies to:\[ abla f(-2, 2) = \left( \frac{-2}{8}, \frac{-2}{8} \right) = \left( -\frac{1}{4}, -\frac{1}{4} \right). \]
03
Normalize the Direction Vector a
The direction vector is given as \(\mathbf{a} = -\mathbf{i} - \mathbf{j} \), which can be written as \( \mathbf{a} = (-1, -1) \).Calculate its magnitude: \[ \|\mathbf{a}\| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}. \]Therefore, the unit vector in the direction of \(\mathbf{a}\) is \[ \mathbf{u} = \left( \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right). \]
04
Compute the Directional Derivative
The directional derivative of \( f \) at \( P \) in the direction \( \mathbf{a} \) is given by the dot product of the gradient at \( P \) and the unit vector in the direction of \( \mathbf{a} \). \[ D_\mathbf{a} f(P) = abla f(P) \cdot \mathbf{u} = \left( -\frac{1}{4}, -\frac{1}{4} \right) \cdot \left( \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right). \]This becomes:\[D_\mathbf{a} f(P) = -\frac{1}{4}\times\frac{-1}{\sqrt{2}} + -\frac{1}{4}\times\frac{-1}{\sqrt{2}} = \frac{1}{4\sqrt{2}} + \frac{1}{4\sqrt{2}} = \frac{1}{2\sqrt{2}}. \]Finally, rationalize the denominator:\[ D_\mathbf{a} f(P) = \frac{\sqrt{2}}{4}. \]
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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 like \( f(x, y) = \tan^{-1}(y / x) \) is a vector that shows the rate and direction of change of the function. It is composed of partial derivatives with respect to each variable, forming a vector: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). To find these partial derivatives, we apply the chain rule.
- The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), is found using the chain rule. Here, it results in \( \frac{-y}{x^2 + y^2} \).
- Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), results in \( \frac{x}{x^2 + y^2} \).
Partial Derivative
Partial derivatives involve taking the derivative of a function with more than one variable, with respect to one variable at a time while keeping the others constant. In the given exercise:
- The function \( f(x, y) = \tan^{-1}(y / x) \) is dependent on both \( x \) and \( y \).
- The partial derivative \( \frac{\partial f}{\partial x} \) considers changes only in \( x \), calculated as \( \frac{-y}{x^2 + y^2} \).
- The partial derivative \( \frac{\partial f}{\partial y} \) focuses on changes only in \( y \), calculated as \( \frac{x}{x^2 + y^2} \).
Unit Vector
A unit vector is a vector that has a magnitude of 1 and is used to denote direction. It is a normalized version of any vector.
- In the exercise, the direction vector \( \mathbf{a} = (-1, -1) \) was given.
- Its unit vector \( \mathbf{u} \) was found by dividing each component of \( \mathbf{a} \) by its magnitude. The magnitude \( \|\mathbf{a}\| \) is \( \sqrt{2} \), leading to \( \mathbf{u} = \left( \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \).
Dot Product
The dot product, or scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It is calculated by multiplying corresponding entries and summing those products.
- In this solution, the dot product is used to find the directional derivative.
- The gradient at point \( P \) and the unit vector \( \mathbf{u} \) were dotted: \[ D_\mathbf{a} f(P) = abla f(P) \cdot \mathbf{u} = \left( -\frac{1}{4}, -\frac{1}{4} \right) \cdot \left( \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \].
- This operation yields a scalar which represents the rate of change of \( f \) in the direction of \( \mathbf{a} \).
