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Magazine Chapter 11: Problem 21
Find all points at which the direction of fastest change of the function \(f(x, y)=x^{2}+y^{2}-2 x-4 y\) is \(\mathbf{i}+\mathbf{j}\)
Short Answer
Expert verified
The point is \( \left(\frac{3}{2}, \frac{5}{2}\right) \).
Step by step solution
01
Find the Gradient of f
The direction of fastest change of a function at a point is given by its gradient vector. The function is given as \( f(x, y) = x^2 + y^2 - 2x - 4y \). We need to compute the gradient, \( abla f(x, y) \), which involves finding the partial derivatives of \( f \) with respect to \( x \) and \( y \).
02
Compute Partial Derivatives
First, compute \( \frac{\partial f}{\partial x} \):\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2 - 2x - 4y) = 2x - 2. \]Next, compute \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2 - 2x - 4y) = 2y - 4. \]
03
Construct the Gradient Vector
The gradient vector \( abla f(x, y) \) is the vector formed by the partial derivatives:\[ abla f(x, y) = \left(2x - 2\right)\mathbf{i} + \left(2y - 4\right)\mathbf{j}. \]
04
Set the Gradient Equal to the Given Direction
We are given the direction of fastest change as \( \mathbf{i} + \mathbf{j} \), which can be written as the vector \( (1, 1) \). Thus, we set:\[ abla f(x, y) = (1, 1). \]That means we have the system:\[ 2x - 2 = 1 \]\[ 2y - 4 = 1. \]
05
Solve the System of Equations for x and y
Solve the first equation:\[ 2x - 2 = 1 \]\[ 2x = 3 \]\[ x = \frac{3}{2}. \]Solve the second equation:\[ 2y - 4 = 1 \]\[ 2y = 5 \]\[ y = \frac{5}{2}. \]
06
Identify the Points of Fastest Change
The point at which the direction of fastest change of the function is \( \mathbf{i} + \mathbf{j} \) is given by the coordinates \( (x, y) = \left(\frac{3}{2}, \frac{5}{2}\right). \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In calculus, partial derivatives are used to understand how a function changes as individual variables change while keeping the others constant. For the function \( f(x, y) = x^2 + y^2 - 2x - 4y \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), shows how \( f \) changes as \( x \) changes, while \( y \) stays constant.
This is computed as follows:
This is computed as follows:
- Differentiate each term of the function that contains \( x \).
- Ignore \( y \) terms as if they were constants.
- Apply the power rule to get \( \frac{\partial f}{\partial x} = 2x - 2 \).
- Focus on terms with \( y \).
- Treat \( x \) as constant, and differentiate to obtain \( \frac{\partial f}{\partial y} = 2y - 4 \).
Direction of Fastest Change
The gradient vector, indicated as \( abla f \), is crucial in identifying the direction of fastest change of a function. It's constructed by combining the partial derivatives with respect to each variable. For our function, we have the gradient as:
- \( abla f(x, y) = (2x - 2)\mathbf{i} + (2y - 4)\mathbf{j} \).
- The larger the gradient, the steeper the slope.
- It points in the direction where the function increases most rapidly.
System of Equations
A system of equations is a set of equations with multiple variables that can be solved together. When finding points where a function's gradient matches a specific direction, you often encounter such systems. Our task was to solve:
- \( 2x - 2 = 1 \)
- \( 2y - 4 = 1 \)
- To solve \( 2x - 2 = 1 \), isolate \( x \) to get \( x = \frac{3}{2} \).
- Similarly, solve \( 2y - 4 = 1 \) to find \( y = \frac{5}{2} \).
