Chapter 13: Problem 21
Find the maximum directional derivative of \(f\) at \(P\) and the direction in which it occurs. $$ f(x, y)=2 x^{2}+3 x y+4 y^{2}: \quad P(1,1) $$
Short Answer
Expert verified
The maximum directional derivative at \(P(1, 1)\) is \(\sqrt{170}\), in the direction of \((7, 11)\).
Step by step solution
01
Compute the Gradient of f
The gradient of a function \(f(x, y)\) is a vector consisting of its partial derivatives. First, compute the partial derivative of \(f\) with respect to \(x\), \(\frac{\partial f}{\partial x}\), which is \(4x + 3y\). Next, compute the partial derivative of \(f\) with respect to \(y\), \(\frac{\partial f}{\partial y}\), which is \(3x + 8y\). Therefore, the gradient is \(abla f(x, y) = (4x + 3y, 3x + 8y)\).
02
Evaluate the Gradient at Point P
Substitute the point \(P(1, 1)\) into the gradient. We find \(abla f(1, 1) = (4(1) + 3(1), 3(1) + 8(1)) = (7, 11)\).
03
Calculate the Magnitude of the Gradient
The maximum directional derivative is equal to the magnitude of the gradient vector. Calculate this using the formula \(\sqrt{a^2 + b^2}\). Substitute \(a = 7\) and \(b = 11\), resulting in \(\sqrt{7^2 + 11^2} = \sqrt{49 + 121} = \sqrt{170}\).
04
Identify the Direction
The direction of the maximum directional derivative is along the gradient vector \(abla f(1, 1) = (7, 11)\). This direction is a unit vector in the same direction as the gradient, which is \(\frac{1}{\sqrt{170}} (7, 11)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Vector
The gradient vector is a fundamental concept in calculus, especially when dealing with multivariable functions. Think of it as a tool that tells us the direction in which a function is increasing most steeply. For a function of two variables, say \( f(x, y) \), the gradient is denoted as \( abla f(x, y) \) and is a vector composed of the partial derivatives of \( f \) with respect to \( x \) and \( y \). Specifically, the gradient vector can be expressed as:
- \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
Partial Derivative
A partial derivative represents the rate at which a function changes as one of the variables changes, while holding the other variable constant. This is a cornerstone concept in multivariable calculus, as it helps decompose changes in multivariable functions into simpler single-variable changes.
- For \( f(x, y) = 2x^2 + 3xy + 4y^2 \), the partial derivative with respect to \( x \) is calculated by differentiating \( f \) treating \( y \) as a constant, resulting in \( \frac{\partial f}{\partial x} = 4x + 3y \).
- Similarly, the partial derivative with respect to \( y \) treats \( x \) as a constant, yielding \( \frac{\partial f}{\partial y} = 3x + 8y \).
Magnitude of a Vector
The magnitude of a vector is akin to the length of the vector, giving a sense of its size but not its direction. To calculate the magnitude, one applies the Pythagorean theorem in multiple dimensions. The formula in a two-dimensional space for a vector \( \mathbf{v} = (a, b) \) is:
- \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \)
Unit Vector
A unit vector is a vector that has a magnitude of exactly 1. It is often used to specify a direction without concern for magnitude. To convert a vector into a unit vector that points in the same direction, you divide the vector by its magnitude. If \( \mathbf{v} = (a, b) \), the unit vector \( \mathbf{u} \) in the same direction can be calculated as:
- \( \mathbf{u} = \frac{1}{\sqrt{a^2 + b^2}} (a, b) \)
- \( \mathbf{u} = \frac{1}{\sqrt{170}} (7, 11) \)