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Magazine Chapter 12: Problem 14
Find the directional derivative of \(f(x, y)=\tan ^{-1}(3 x y)\). What is its value at the point (4,2) in the direction \(\mathbf{u}=(\sqrt{3} / 2) \mathbf{i}-(1 / 2) \mathbf{j} ?\)
Short Answer
Expert verified
The directional derivative is \( \frac{3\sqrt{3} - 6}{145} \).
Step by step solution
01
Compute the Gradient of the Function
The first step is to compute the gradient \( abla f(x, y) \) of the function \( f(x, y) = \tan^{-1}(3xy) \). This is represented by derivatives with respect to \( x \) and \( y \): \[ \frac{\partial f}{\partial x} = \frac{3y}{1 + (3xy)^2} \] \[ \frac{\partial f}{\partial y} = \frac{3x}{1 + (3xy)^2} \] So, the gradient of \( f \) is: \( abla f(x, y) = \left( \frac{3y}{1 + (3xy)^2}, \frac{3x}{1 + (3xy)^2} \right) \).
02
Evaluate the Gradient at the Given Point
Now, we need to evaluate the gradient of the function at the point (4, 2). Substitute \( x = 4 \) and \( y = 2 \) into the gradient:\[ abla f(4, 2) = \left( \frac{3 \times 2}{1 + (3 \times 4 \times 2)^2}, \frac{3 \times 4}{1 + (3 \times 4 \times 2)^2} \right) \]Calculating the value:\[ (3 \times 4 \times 2)^2 = 144 \] \[ abla f(4, 2) = \left( \frac{6}{1 + 144}, \frac{12}{1 + 144} \right) = \left( \frac{6}{145}, \frac{12}{145} \right) \].
03
Normalize the Direction Vector
The given direction vector \( \mathbf{u} = (\sqrt{3}/2, -1/2) \) needs to be a unit vector for the directional derivative calculation. Compute its magnitude:\[ \|\mathbf{u}\| = \sqrt{\left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{-1}{2} \right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]Since its magnitude is already 1, \( \mathbf{u} \) is a unit vector.
04
Compute the Directional Derivative
The directional derivative of \( f \) at (4,2) in the direction of \( \mathbf{u} \) is given by \( D_{\mathbf{u}}f(4, 2) = abla f(4, 2) \cdot \mathbf{u} \), where \( \cdot \) represents the dot product. Calculate the dot product:\[ D_{\mathbf{u}}f(4, 2) = \left( \frac{6}{145}, \frac{12}{145} \right) \cdot \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) = \frac{6\sqrt{3}}{2 \times 145} - \frac{12}{2 \times 145} \]\[ = \frac{3\sqrt{3}}{145} - \frac{6}{145} = \frac{3\sqrt{3} - 6}{145} \].
05
Simplify the Expression
The final step is to simplify the expression:\[ \frac{3\sqrt{3} - 6}{145} \] is the simplified value of the directional derivative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
In the context of multivariable calculus, the gradient is a vector that consists of partial derivatives of a function. It points in the direction of the greatest rate of increase of the function. For a function of two variables, like our example function, the gradient is denoted as \( abla f(x, y) \).
- Components: It consists of the partial derivatives: \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
- Vector Form: Expressed as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
- Purpose: It helps identify the direction where the function increases most rapidly.
Unit Vector
A unit vector is a vector with a magnitude of 1. It is used to indicate direction without scaling the vector in terms of magnitude.
- Norm Calculation: To ensure a vector is a unit vector, you calculate its norm (or length) and ensure it's 1. The norm is found via \(\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are components of the vector.
- Normalization: If a vector isn't already a unit vector, divide each component by the vector's magnitude to normalize it.
Dot Product
The dot product is a mathematical operation that takes two equal-length sequences of numbers, typically coordinate vectors, and returns a single number.
- Formula: For vectors \(\mathbf{a} = (a_1, a_2)\) and \(\mathbf{b} = (b_1, b_2)\), the dot product is \(a_1b_1 + a_2b_2\).
- Applications: Used to find the angle between two vectors or to project one vector onto another.
Partial Derivative
Partial derivatives are used to understand how a multivariable function changes as only one of the input variables is modified while the others are held constant.
- Computation: Denoted as \( \frac{\partial f}{\partial x} \), it measures the rate of change of the function as variable \(x\) changes.
- Evaluation: Similarly, \( \frac{\partial f}{\partial y} \) measures the rate of change as variable \(y\) changes.
- Role in Gradient: These derivatives are the building blocks of the gradient vector.
