Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \((5,0)\).

Step-by-Step Solution

Answer

The directional derivative is -8.2.

Step 1: 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 \[ (5-2, 0-4) = (3, -4). \]

Step 2: 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 \[ \left( \frac{3}{5}, \frac{-4}{5} \right). \]

Step 3: 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 \( \mathbf{u} \): \[ D_uf = abla f \cdot \mathbf{u} = (-3, 8) \cdot \left( \frac{3}{5}, \frac{-4}{5} \right). \]

Step 4: Calculate the Dot Product

Compute the dot product: \[ (-3) \times \frac{3}{5} + (8) \times \frac{-4}{5} = -\frac{9}{5} - \frac{32}{5} = \frac{-41}{5}. \]

Step 5: State the Directional Derivative

The directional derivative \(D_uf\) at \( (2,4) \) in the direction toward \( (5,0) \) is \(-\frac{41}{5} = -8.2 \).

Key Concepts

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.

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.

• 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.

In our exercise, the gradient vector at the point (\((2,4)\)) is (\((-3, 8)\)). This vector contains all the necessary information about how the function changes around that point. To determine the directional derivative, the gradient vector must be combined with a unit direction vector corresponding to the direction of interest.

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:

• Calculate the magnitude of the vector: \[ \sqrt{3^2 + (-4)^2} = 5 \]
• Divide each component by the magnitude: \[ \left( \frac{3}{5}, \frac{-4}{5} \right) \]

By using unit vectors, we make calculations involving vectors simpler and consistent, especially when determining directional derivatives.

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.

• 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: \[ (-3) \times \frac{3}{5} + 8 \times \frac{-4}{5} = -\frac{41}{5} \]

The outcome, (-\frac{41}{5}), represents the directional derivative, indicating the rate of change of the function in the given direction. The sign of the result also tells us whether the function is increasing or decreasing in that direction. A negative result suggests a decrease.