Problem 17 - Advanced Engineering Mathematics

Question

Find the directional derivative of the given function at the given point in the indicated direction. $$ F(x, y, z)=x^{2} y^{2}(2 z+1)^{2} ;(1,-1,1),\langle 0,3,3\rangle $$

Step-by-Step Solution

Answer

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

Step 1: Find the Gradient of the Function

The first step is to find the gradient \( abla F(x, y, z) \) of the function \( F(x, y, z) = x^2 y^2 (2z+1)^2 \). The gradient is a vector of partial derivatives: \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).- \( \frac{\partial F}{\partial x} = 2xy^2(2z+1)^2 \)- \( \frac{\partial F}{\partial y} = 2x^2y(2z+1)^2 \)- \( \frac{\partial F}{\partial z} = 4x^2y^2(2z+1) \). Therefore, \( abla F(x, y, z) = \left( 2xy^2(2z+1)^2, 2x^2y(2z+1)^2, 4x^2y^2(2z+1) \right) \).

Step 2: Evaluate the Gradient at the Given Point

Now, evaluate the gradient \( abla F(x, y, z) \) at the point \((1, -1, 1)\).- \( \frac{\partial F}{\partial x} \big|_{(1, -1, 1)} = 2(1)(-1)^2(2\cdot1+1)^2 = 18 \)- \( \frac{\partial F}{\partial y} \big|_{(1, -1, 1)} = 2(1)^2(-1)(2\cdot1+1)^2 = -18 \)- \( \frac{\partial F}{\partial z} \big|_{(1, -1, 1)} = 4(1)^2(-1)^2(2\cdot1+1) = 12 \).Thus, \( abla F(1, -1, 1) = (18, -18, 12) \).

Step 3: Normalize the Direction Vector

Normalize the given direction vector \( \langle 0, 3, 3 \rangle \) by dividing it by its magnitude. The magnitude is calculated as:\[ \| \mathbf{v} \| = \sqrt{0^2 + 3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \].The unit vector is:\[ \mathbf{u} = \frac{1}{3\sqrt{2}} \langle 0, 3, 3 \rangle = \left\langle 0, \frac{3}{3\sqrt{2}}, \frac{3}{3\sqrt{2}} \right\rangle = \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \].

Step 4: Compute the Directional Derivative

The directional derivative \( D_{\mathbf{u}} F \) is given by the dot product of the gradient and the unit direction vector:\[ D_{\mathbf{u}} F = abla F(1, -1, 1) \cdot \mathbf{u} \].Calculate this:\[ D_{\mathbf{u}} F = (18, -18, 12) \cdot \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \]\[ = 18 \times 0 + (-18) \times \frac{1}{\sqrt{2}} + 12 \times \frac{1}{\sqrt{2}} \]\[ = 0 - \frac{18}{\sqrt{2}} + \frac{12}{\sqrt{2}} \]\[ = -\frac{6}{\sqrt{2}} \]\[ = -\frac{6\sqrt{2}}{2} \] (multiplying numerator and denominator by \( \sqrt{2} \))\[ = -3\sqrt{2} \].

Key Concepts

Gradient of a Function

In calculus, the gradient of a function provides a way to find the rate and direction of change of a multi-variable function. This concept is essential in fields like physics and engineering for understanding surfaces or performing optimization tasks. The gradient is represented as \( abla F \) and is a vector consisting of partial derivatives with respect to each variable of the function. For example, for the function \( F(x, y, z) = x^2 y^2 (2z+1)^2 \), the gradient is denoted as:

• \( \frac{\partial F}{\partial x} = 2xy^2(2z+1)^2 \)
• \( \frac{\partial F}{\partial y} = 2x^2y(2z+1)^2 \)
• \( \frac{\partial F}{\partial z} = 4x^2y^2(2z+1) \)

This vector gives you the direction of greatest increase of the function at any given point. By finding the gradient, we can predict how changes in the input variables will affect the function's outcome. This is especially useful when evaluating the function at specific points to understand directional tendencies.

Partial Derivatives

Partial derivatives are a cornerstone in calculus and are used to understand how a function changes with respect to one variable, keeping others constant. This can be thought of as taking a slice through the function in a specific direction. For the function \( F(x, y, z) = x^2 y^2 (2z+1)^2 \), partial derivatives are calculated for each variable:

• \( \frac{\partial F}{\partial x} \), which represents the rate of change of the function with respect to \( x \), given as \( 2xy^2(2z+1)^2 \).
• \( \frac{\partial F}{\partial y} \), the rate of change with respect to \( y \), given as \( 2x^2y(2z+1)^2 \).
• \( \frac{\partial F}{\partial z} \), representing the change with respect to \( z \), calculated as \( 4x^2y^2(2z+1) \).

These derivatives are combined to form the gradient vector. Understanding partial derivatives is crucial for tasks such as creating contour maps or evaluating system behavior with multiple influencing factors. When we calculate these derivatives at a specific point, we get a snapshot of how the function behaves at that exact location.

Unit Vector Norm

In vector calculus, ensuring a direction vector has a unit length is essential for computing directional derivatives accurately. This process involves normalizing a vector, which is accomplished by dividing each component of the vector by its magnitude. To normalize the direction vector \( \langle 0, 3, 3 \rangle \), we first calculate its magnitude: \[ \| \mathbf{v} \| = \sqrt{0^2 + 3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]Then, to find the unit vector \( \mathbf{u} \), each component of the vector is divided by the magnitude:\[ \mathbf{u} = \frac{1}{3\sqrt{2}} \langle 0, 3, 3 \rangle = \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \]This normalization step is vital because it ensures that the direction vector only represents direction, not magnitude. Without this, calculations like the directional derivative could incorrectly measure changes in the function’s output.

Vector Calculus

Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, primarily in 3D space. Key operations include gradient, divergence, and curl, which allow for the analysis of scalar functions and vector fields within physical and natural sciences.A directional derivative in vector calculus measures how a function changes as one moves in the direction of a specific vector. It involves computing the dot product of the gradient and a unit direction vector. For the function \( F(x, y, z) \) evaluated at point \((1, -1, 1)\), with the direction vector \( \langle 0, 3, 3 \rangle \), calculating the directional derivative involves:1. Finding \( abla F(1, -1, 1) = (18, -18, 12) \).2. Using the unit vector \( \mathbf{u} = \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \).3. Computing the dot product:\[ D_{\mathbf{u}} F = (18, -18, 12) \cdot \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = -3\sqrt{2} \]This derivative gives us the slope or rate of change of the function in the specified direction, crucial for assessing function behavior in physics and engineering.