Problem 12 - Advanced Engineering Mathematics

Question

In Problems, find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=4 x+x y^{2}-5 y ;(3,-1), \theta=\pi / 4 $$

Step-by-Step Solution

Answer

The directional derivative is \(-3\sqrt{2}\).

Step 1: Understand the Directional Derivative Formula

The directional derivative of a function \( f(x, y) \) in the direction of a unit vector \( \mathbf{u} = (a, b) \) is given by the formula: \[ D_\mathbf{u}f(x, y) = abla f(x, y) \cdot \mathbf{u} \] where \( abla f(x, y) \) is the gradient of \( f \), and \( \cdot \) represents the dot product.

Step 2: Find the Gradient of the Function

The gradient \( abla f(x, y) \) is a vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y)=4x+xy^2-5y \), calculate: \[ \frac{\partial f}{\partial x} = 4 + y^2 \] \[ \frac{\partial f}{\partial y} = 2xy - 5 \]. Thus, \( abla f(x, y) = (4+y^2, 2xy-5) \).

Step 3: Evaluate the Gradient at the Given Point

Substitute the given point \((3,-1)\) into the gradient vector: \( abla f(3, -1) = (4 + (-1)^2, 2(3)(-1) - 5) = (5, -11) \).

Step 4: Determine the Unit Direction Vector

Since the direction is given by \( \theta=\pi/4 \), the unit vector \( \mathbf{u} \) in this direction is \( \mathbf{u} = (\cos \theta, \sin \theta) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 5: Calculate the Dot Product

Compute the dot product of the gradient and the direction vector: \( (5, -11) \cdot \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = 5 \cdot \frac{\sqrt{2}}{2} + (-11) \cdot \frac{\sqrt{2}}{2} \).

Step 6: Solve the Dot Product

Simplify the expression: \( \frac{5\sqrt{2}}{2} - \frac{11\sqrt{2}}{2} = \frac{-6\sqrt{2}}{2} = -3\sqrt{2} \).

Step 7: Final Answer for the Directional Derivative

The directional derivative of \( f(x, y) \) at the point \((3, -1)\) in the direction \( \theta = \pi/4 \) is \( -3\sqrt{2} \).

Key Concepts

Gradient

The gradient is a crucial concept in multivariable calculus that helps determine the rate and direction of change of a function. It is represented by the symbol \( abla \) and is a vector that consists of the partial derivatives of a function with respect to its variables.

For a function like \( f(x, y) \), the gradient is expressed as:

• \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

In our specific example, where \( f(x, y) = 4x + xy^2 - 5y \):

• The partial derivative with respect to \(x\): \( \frac{\partial f}{\partial x} = 4 + y^2 \)
• The partial derivative with respect to \(y\): \( \frac{\partial f}{\partial y} = 2xy - 5 \)

Calculate these derivatives to form the gradient vector: \( abla f(x, y) = (4 + y^2, 2xy - 5) \).

By evaluating this at a particular point, such as \( (3, -1) \), we get the gradient vector at that point, \( abla f(3, -1) = (5, -11) \). This tells us the direction in which the function increases most rapidly and the rate of that increase.

Unit Vector

A unit vector is a vector that has a magnitude (length) of exactly one. It is typically used to indicate direction. Working with directions often requires using unit vectors because they maintain the direction but not any scaling that might be involved with longer vectors.

To convert a vector into a unit vector, you divide each component of the vector by its magnitude. In our problem, a direction was provided as an angle \( \theta = \pi/4 \). From trigonometry, we know:

• \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \)
• \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \)

Thus, the direction vector is given by \( \mathbf{u} = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

The vector \( \mathbf{u} \) maintains the intended direction towards \( \pi/4 \), without altering the magnitude. Using unit vectors simplifies many calculations in physics and engineering, ensuring scale is not a factor when direction is the focus.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This product can be useful in determining the angle between two vectors or the projection of one vector onto another.

The dot product between two vectors \( \mathbf{A} = (a_1, a_2) \) and \( \mathbf{B} = (b_1, b_2) \) is calculated as:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \)

For the problem provided, we calculated the dot product between the gradient vector at a specific point, \( (5, -11) \), and the unit directional vector \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \). This is executed as follows:

• Compute: \( 5 \cdot \frac{\sqrt{2}}{2} + (-11) \cdot \frac{\sqrt{2}}{2} \)
• Simplify to: \( \frac{5\sqrt{2}}{2} - \frac{11\sqrt{2}}{2} = \frac{-6\sqrt{2}}{2} = -3\sqrt{2} \)

This result, \( -3\sqrt{2} \), represents the directional derivative, indicating the rate of change of the function in the specified direction.