Chapter 12: Problem 24
Given that \(f_{x}(2,4)=-3\) and \(f_{y}(2,4)=8\), find the directional derivative of \(f\) at \((2,4)\) in the direction toward \((5,0)\).
Short Answer
Expert verified
The directional derivative is -8.2.
Step by step solution
01
Find the Direction Vector
The direction toward \(5,0\) from \(2,4\) is a vector that can be found by subtracting the coordinates. The vector is \[ (5-2, 0-4) = (3, -4). \]
02
Normalize the Direction Vector
To find the unit direction vector, divide the vector by its magnitude. The magnitude is \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \ The unit direction vector is then \[ \left( \frac{3}{5}, \frac{-4}{5} \right). \]
03
Use the Gradient to Calculate the Directional Derivative
The gradient of \(f\) at \( (2,4) \) is given by the vector \( abla f = (f_x, f_y) = (-3, 8). \) The directional derivative \(D_uf\) in the direction of a unit vector \( \mathbf{u} \) is calculated as the dot product of \( abla f \) and \( \mathbf{u} \): \[ D_uf = abla f \cdot \mathbf{u} = (-3, 8) \cdot \left( \frac{3}{5}, \frac{-4}{5} \right). \]
04
Calculate the Dot Product
Compute the dot product: \[ (-3) \times \frac{3}{5} + (8) \times \frac{-4}{5} = -\frac{9}{5} - \frac{32}{5} = \frac{-41}{5}. \]
05
State the Directional Derivative
The directional derivative \(D_uf\) at \( (2,4) \) in the direction toward \( (5,0) \) is \(-\frac{41}{5} = -8.2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Directional Derivative
The directional derivative represents the rate of change of a function as you move in a specific direction. It tells us how quickly the function is increasing or decreasing if you were to take a step in the given direction. To find the directional derivative of a function at a specific point, you must know the direction in which you are interested, typically given by a vector.
Consider a function described by its two partial derivatives, like in our exercise, where we have the partial derivatives at a point (\((2, 4)\)). The important part is that when you want to calculate the directional derivative in a specific direction, this direction needs to be converted into a unit direction vector.
This unit direction vector is then used to find the directional derivative by taking the dot product with the gradient vector of the function.
The directional derivative is thus a powerful tool in calculus that extends our understanding of how functions behave in multidimensional spaces. It finds common applications in physics and engineering to analyze how a variable changes in specified directions.
Consider a function described by its two partial derivatives, like in our exercise, where we have the partial derivatives at a point (\((2, 4)\)). The important part is that when you want to calculate the directional derivative in a specific direction, this direction needs to be converted into a unit direction vector.
This unit direction vector is then used to find the directional derivative by taking the dot product with the gradient vector of the function.
The directional derivative is thus a powerful tool in calculus that extends our understanding of how functions behave in multidimensional spaces. It finds common applications in physics and engineering to analyze how a variable changes in specified directions.
Gradient Vector
The gradient vector is an essential concept in calculus that provides us with critical information about a function's behavior. The gradient of a function consists of its partial derivatives with respect to each variable, effectively forming a vector field.
For a function of two variables, the gradient vector is typically represented as (\( abla f = (f_x, f_y) \)), which points in the direction of the greatest increase of the function.
For a function of two variables, the gradient vector is typically represented as (\( abla f = (f_x, f_y) \)), which points in the direction of the greatest increase of the function.
- Magnitude: The size, or magnitude, of the gradient indicates how steep this increase is.
- Direction: The direction of the gradient vector tells you which direction to move in for the maximum increase.
Unit Direction Vector
A unit direction vector is a normalized vector that points in a specific direction, with a magnitude of (1). This ensures that when it is used in calculations, it doesn't alter the magnitude of other vectors or results.
To find a unit direction vector, take any vector that points in the desired direction and divide it by its magnitude. This transforms any vector into a unit vector by essentially shrinking or stretching it to a length of one.
In this particular exercise, the direction vector from point (\((2, 4)\)) to point (\((5, 0)\)) is (\((3, -4)\)). To convert this direction into a unit vector:
To find a unit direction vector, take any vector that points in the desired direction and divide it by its magnitude. This transforms any vector into a unit vector by essentially shrinking or stretching it to a length of one.
In this particular exercise, the direction vector from point (\((2, 4)\)) to point (\((5, 0)\)) is (\((3, -4)\)). To convert this direction into a unit vector:
- Calculate the magnitude of the vector: \[ \sqrt{3^2 + (-4)^2} = 5 \]
- Divide each component by the magnitude: \[ \left( \frac{3}{5}, \frac{-4}{5} \right) \]
Dot Product
The dot product is a mathematical operation that takes two equal-length sequences of numbers, usually vectors, and returns a single number. This result is a measure of how much one vector "projects" onto another, or how aligned they are.
In the context of calculus, the dot product is particularly useful in calculating directional derivatives because it allows us to conveniently combine the gradient vector and a unit direction vector.
In the context of calculus, the dot product is particularly useful in calculating directional derivatives because it allows us to conveniently combine the gradient vector and a unit direction vector.
- The formula for the dot product of two vectors (\((a, b)\)) and (\((c, d)\)) is: \[ a \times c + b \times d \]
- For example, in our exercise, we have: \[ (-3) \times \frac{3}{5} + 8 \times \frac{-4}{5} = -\frac{41}{5} \]