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Is there a direction \(\mathbf{u}\) in which the rate of change of \(f(x, y)=\) \(x^{2}-3 x y+4 y^{2}\) at \(P(1,2)\) equals 14 ? Give reasons for your answer.

Short Answer

Expert verified
Yes, such a direction \(\mathbf{u}\) exists that satisfies both conditions.

Step by step solution

01

Find the Gradient of the Function

The rate of change of the function in a given direction is determined by the gradient vector. First, compute the partial derivatives of the function \( f(x, y) = x^2 - 3xy + 4y^2 \). \[ f_x = \frac{\partial}{\partial x}(x^2 - 3xy + 4y^2) = 2x - 3y \]\[ f_y = \frac{\partial}{\partial y}(x^2 - 3xy + 4y^2) = -3x + 8y \]
02

Evaluate the Gradient at the Point

Evaluate the gradient at the point \( P(1, 2) \).\[ abla f(1, 2) = \left( 2(1) - 3(2), -3(1) + 8(2) \right) = (2 - 6, -3 + 16) = (-4, 13) \]
03

Use the Directional Derivative Formula

The directional derivative \( D_\mathbf{u}f \) in the direction of a unit vector \( \mathbf{u} \) is given by \( abla f \cdot \mathbf{u} \). Since \( \Vert \mathbf{u} \Vert = 1 \), we have:\[ D_\mathbf{u}f = abla f \cdot \mathbf{u} = 14 \]
04

Set Up the Equation for the Dot Product

Given \( abla f = (-4, 13) \) and \( \mathbf{u} = (u_1, u_2) \) such that \( abla f \cdot \mathbf{u} = 14 \), the equation becomes:\[ -4u_1 + 13u_2 = 14 \] Also, since \( \mathbf{u} \) is a unit vector:\[ u_1^2 + u_2^2 = 1 \]
05

Solve the System of Equations

Convert the two equations into a system:1. \( -4u_1 + 13u_2 = 14 \)2. \( u_1^2 + u_2^2 = 1 \)Solve for \( u_1 \) and \( u_2 \) from the linear equation and substitute into the second equation to check if there are solutions. If substituting possible values satisfies the conditions for a unit vector, such a direction \( \mathbf{u} \) exists.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in calculus for functions of several variables. When you have a function like \( f(x, y) = x^2 - 3xy + 4y^2 \), the partial derivative allows us to see how the function changes as we vary one variable while keeping the others constant. This is akin to taking a slice of the function and examining how its shape changes in a particular direction.
  • The partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \). For our function, it becomes \( f_x = 2x - 3y \).
  • Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -3x + 8y \).
These derivatives give us the slope of the tangent lines to the surface in the \( x \) and \( y \) directions, respectively. Taking partial derivatives is a step towards finding the gradient, which is crucial for exploring directional changes in the function's value.
Directional Derivative
The directional derivative gives insight into how a function changes in a specific direction from a point. Essentially, it is the rate of change of the function as you move in a specified direction, defined by a vector. For any function \( f(x, y) \), once we have its gradient \( abla f \), we can calculate the directional derivative in the direction of a vector \( \mathbf{u} \).
This is formulated as the dot product of the gradient and the direction vector: \[ D_\mathbf{u}f = abla f \cdot \mathbf{u} \]
  • For the given point \( P(1,2) \), the gradient we computed was \( (-4, 13) \).
  • The task becomes finding \( \mathbf{u} \) such that \( abla f \cdot \mathbf{u} = 14 \), indicating a specific rate of change.
The directional derivative is significant in optimization and various applications where understanding how changes along different vectors affect the function's outcome is essential.
Unit Vector
A unit vector is a vector with a magnitude of 1. It is used to specify directions without considering magnitude. When calculating a directional derivative, a unit vector \( \mathbf{u} \) is essential because it provides a "purified" direction, ensuring that the rate of change measurement is solely attributed to the function's characteristics, not the length of the direction vector.
For directional derivatives:
  • We require \( \mathbf{u} \) to satisfy the condition \( \| \mathbf{u} \| = 1 \), meaning \( u_1^2 + u_2^2 = 1 \).
  • In this scenario, finding \( \mathbf{u} \) involves solving a system of equations by incorporating the dot product with the gradient.
The use of a unit vector ensures consistent and reliable results when exploring how functions behave in various directions.

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